text:
Why does the derivation Gibbs-Helmholtz equation hold when the enthalpy is temperature-dependant
The derivation shown here assumes H and U are independant of temperature, right? But what if $H(T) = H_{0}+C_V \Delta T $?
Then $dG = C_V\, dT - SdT$, right? So
$ \displaystyle \left( \frac{\partial G}{\partial T}\right)_p=C_V -S$ ?
$$\begin{align}
\left(\frac{\partial (G/T)}{\partial T}\right)_p &= \frac{T(\partial G/\partial T)_p - G(\partial T/\partial T)_p}
{T^2} \\[8pt]
&= \frac{T(C_V-S) - G(1)}{T^2} \\[8pt]
&= \frac{TC_V-TS-G}{T^2} \\[8pt]
&= -\frac{H_0}{T^2}
\end{align}$$
as desired (since $G = H_0+C_V\Delta T - TS$).
But this way we see that the H in this equation would only be containing the constant term? Of course we insert H(T) to get thee exact G, but where in my thinking process did I go wrong? How would this derivation be made correctly, assuming that H changes with Temperature?

No, there are no assumptions. You should consider that S is temperature-dependent as well if your plan is to derive the temperature-dependence of G by adding the temperature dependence of H and S. I bet something will cancel out.

Should not it to be Cp for G, resp. Cv for A? As H and G are at constant pressure, while U and A at constant volume.

dG=Cp.dT - S.dT - TdS=Cp.dT - S.dT - dQ= - S.dT // dG/dT= -S

I edited the first term in the derivation to something that seems sensible, starting from a "mauled" form that doesn't make sense.

Your equation $( \frac{\partial G}{\partial T})_p=C_V -S$ is wrong. It should read $( \frac{\partial G}{\partial T})_p=-S$ as in the answer you link to.

@NightWriter Okay, but why if H is dependant on T?

See Poutniks comment a little further above: If $G=H_0  +C_p \Delta T -TS$ then  $dG=C_p dT -SdT-TdS$. For a reversible process at constant p,  $dq = C_p dT=TdS$

@NightWriter A bit of misunderstanding can also arize from the fact that ${H_0}$ and ${H}$ are two different concepts. One is at standard conditions, which include a predefined temperature. The other is not.

@StianYttervik Here I only refer to $H_0$ as the value at some (potentially arbitrary) reference T, no more. Not same as $H^\circ$ at 1 bar.

The derivation of the van't Hoff equation does not require that you specify the T-dependence of H. It arises from computing the derivative (limiting slope) of $G/T$ so you need to consider a polynomial expansion of $G/T$ to linear terms in $T$ around the temperature of interest (that is, an infinitely  small interval dT). At the risk of redundancy:
$$\begin{align}
 (d(G/T))_p &= ((1/T)dG-G(dT/T^2))_p \\
&= ((1/T)(VdP-SdT)-(H-TS)(dT/T^2))_p \\
&= (-(S/T)dT-(H/T^2-S/T)dT)_p \\
&= (-(H/T^2)dT)_p
\end{align}
$$
The answer is therefore found in differential Calculus.
Another comment explains that, if
$G(T)=H(T) -TS(T)=H_0  +C_p \Delta T -TS(T)$
where $H(T)$ is expressed as a linear polynomial in T ($C_p$ and $p$ assumed constant) then
$dG=C_p dT -SdT-TdS$
Since for a process at constant pressure,  $dq = C_p dT=TdS$ (since $C_p = (dH/dT)_p$ and $dq_p= dH$; also for a reversible process $dq= TdS$), it follows that
$\left(\frac{ \partial G}{\partial T} \right)_p=-S$
and the van't Hoff expression provided in the linked answer also follows.

Thank your very much! Could you also explain how we get to $C_pdT=TdS$?

$C_p = (dH/dT)_p$ and at constant p $dQ_p= dH$. Also for a reversible process $dQ= TdS$.


text:
Why mixing water with ethanol produces heat energy?
Sorry if I'm asking something stupid but I'm trying to understand where this energy gets from. As far as I could find, it is not a reaction as both chemicals left unchanged, so it might have something to do with hydrogen bonding but I can't connect these two facts together.
UPD: it is not homework question (I have a degree in a completely unrelated field) and I did try to find a reasonable explanation in search engines but with no luck. Atm my theory is that ethanol creates H-bonds with water and those bonds have less energy than the ones between H2O-H2O and the excess of the energy goes away as heat but I don't have anybody who could confirm or reject that. As for the forums - Stack Exchange branches to me is the most credible source so I went asking here first.

There is not much to connect. Some compounds like to stay together, some don't.

why minus? if you see how to improve the question - let me know, otherwise your downvote is kinda pointless ;)

I don't know why minus. Guess someone doesn't like something about your question. To me, it's OK. Here, have a plus.

@skwisgaar, don't be discouraged by minus or pluses. These upvotes or downvotes don't mean anything in real life (may be in the virtual world). It is a good question.

I didn't downvote, but I think it's a homework question discussed in the majority of introductory chemistry books, and yes, you are right, it's about H-bonding.

@andselisk my field is pretty far from chemistry so I'd really appreciate if you advice some book where this topic is discussed. For now all I could find is that H-bond is responsible for H2O higher melting/boiling point (compared to, e. g. H2S), so, I presume, it keeps molecules together and requires additional energy to break this bond. But I'm not sure how it is connected to mixing with ethanol - maybe it creates bonds with smaller energy or simply breaks some of existing ones in the water, can't find anything definitive

@skwisgaar Note that the liquid also decreases it's volume after mixing and cooling down. It provides a hint about better intermolecular bonding.

source: Stack Exchange

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

sort: votes


text:
Concentration in a solution
If I mix up 46.29 grams of Potassium Nitrate powder (KNO3) in 1000mL of water and then take 1 ml of that solution and put it into 1 gallon of water what will the ppm of "K" and "NO3" be separately.  For the 1000 mL solution and the 1 gallon of water.  A calculator I was using says it should add 7.5 ppm NO3 and 4.73 ppm of K to the 1 gallon.  If you could tell me how you came up with your answer that would be great.  I can come up with the 7.5 ppm but if I use the same logic for the K it does not work so I must not be doing it correctly.  Thank you for any help!

There is no "K" or "NO3" in solution. Potassium nitrate dissociates completely to $\ce{K+}$ and $\ce{NO3-}$ with equal ppm concentration on molar basis. 2. You didn't mention what basis you are calculating ppm for. 3. You cannot selectively pick different amounts of "K" or "NO3" (whatever these are) from one solution and transfer it to another. It seems to me the question needs improvements in terminology, notations and extra details.

Thank you for the quick reply.  I don't know how to answer what you have said.  This is for a liquid fertilizer.  I was using this calculator https://rotalabutterfly.com/nutrient-calculator.php and I input the numbers above.  I am just trying to figure out how they came up with the end results because it didn't seem correct compared to other solutions.

The calculator most likely intend ppm as mg in one kg of solution (at this concentration usually density is taken to be almost one, so most of the time ppm are used as mg/l. This can be fine or can be wrong, it just depends on concentration and purposes). Assuming you now the atomic mass involved, you must end up to the right values without calculator.

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


text:
Derivation of relationship betwee standard Gibbs energy of reaction and equilibrium constant?
How does one derive $\Delta_\mathrm r G^\circ = -RT\ln(K)$? I don't see the connection mathematically or conceptually; how does Gibbs at all relate to the equilibrium constant of a reaction?
This was part one of a three part question which I've split into 3 individual questions as advised. I left this question as the one associated with the original title; the other one is here  and I will edit the third in when my post cooldown allows me to post it, if anyone wishes to visit those as well now.

@KarstenTheis The question has been split, as requested.

Related: https://chemistry.stackexchange.com/questions/104950/how-to-derive-the-relation-between-gibbs-energy-and-equilibrium-constant/104964#104964]


text:
Can I use the Henderson–Hasselbalch for calculating the pH of non-buffer systems?
Can I use the Henderson–Hasselbalch equation on reactions that are not buffers? When I read up on the derivation of the equation, I noticed that there is absolutely no assumption on the relative concentrations of the acid and conjugate base at least for a monoprotic acid.
Here's a quick derivation of the Henderson Hasselbalch equation:
The reaction:
$$\ce{HA(aq) + H2O(l) <=> A-(aq) + H3O+(aq)}$$
$$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
Then I take the logarithm of each side, resulting in
$$\log K_\mathrm{a} = \log\frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]} = \log\frac{[\ce{A-}]}{[\ce{HA}]} + \log [\ce{H3O+}]$$
Consequently,
$$-\mathrm{p}K_\mathrm{a} = -\mathrm{pH} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

Can I use the Henderson-Hasselbalch equation on reactions that are not buffers?

If you know the equilibrium concentrations of weak acid and conjugate base, you can use the equation. You can also use it when you know the $\mathrm{pH}$ and the $\mathrm pK_\mathrm a$ and want to know the ratio of protonated and deprotonated species (for example for relating color and $\mathrm{pH}$ when working with pH indicators). Written as a formula, the following is always correct for a weak acid and conjugate base at equilibrium:
$$ \mathrm{pH} = \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}}$$
or
$$ \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}} = \mathrm{pH} - \mathrm pK_\mathrm a $$
The equation is especially convenient for a buffer because when you mix high concentrations of weak acid and its conjugate base, the concentrations don't change much when the system reaches equilibrium, so you can approximate the $\mathrm{pH}$ by plugging in the initial concentrations. So for buffers,
$$ \mathrm{pH} = \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}} \approx \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{initial}}}{[\text{weak acid}]_{\text{initial}}}$$
For a solution of a weak acid only or weak base only, that does not work (one of the initial concentrations is zero, and you get a nonsense result). The approximation is best for high and similar concentrations of acid and base when the $\mathrm pK_\mathrm a$ is close to $7$.

Can I use the Henderson–Hasselbalch equation on reactions that are not buffers?

Yes, but you should still keep in mind some limitations. The equation you provide
$$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
should more generally be written
$$K_\mathrm{a} = \frac{a_\ce{A-}a_\ce{H3O+}}{a_\ce{HA}a_\ce{H_2O}}$$
It reduces to your equation$^\dagger$ when the concentrations can be used to represent activities $a_i (=[i]/C^\circ$), and the activity of water is 1, which generally holds in the limit of very dilute solutions. Therefore it should not be assumed to be accurate (only approximate) at higher concentrations.

$\dagger$ Ignoring one other "minor" issue, namely that $K_a$ should be unitless, but that is a common fault in the way the equation is written.
As helpfully explained in a comment, this is solved by writing
$$K_a = \frac{[\ce{A^-}][\ce{H_3O^+}]}{[\ce{HA}]C°}$$
where $C°$ is the reference standard concentration (usually $\pu{1 M}$).


text:
What interesting things can be seen in spectra between 400-700nm?
The visible spectrum (400-700nm) tends to give a very blurry spectral picture, as it is not a wide band relative to the typical peak width of features you can find there. Further, the vibrational modes tend to fall below it in IR, and the electron transitions that get physicists excited tend to fall above it in UV.
What are some examples of interesting features that a sightseeing student could find with a visible-spectrum spectrometer? They are quite common in low-budget labs but tend to only get used for measuring concentration with Beer's law.

Absorbance spectrum of colored solutions may be boring for a "sight seeing" student, as you call yourself. Emission spectrum is far more interesting for visual learners. Anything luminous, which you can see with your eyes, should have a spectrum in the visible range. This includes the sky, moon, a red hot stove, fluorescent lamps, computer screen, flames, bulb, LED etc.

The visible spectrum (400-700nm) tends to give a very blurry spectral
  picture

I gather you are talking about absorbance spectrum of solutions but this is not universally true. If you ever had a chance to try holmium oxide "solution" placed in a spectrometer, its absorbance spectrum has very sharp features.
Emission spectrum on the other hand is far more interesting. One can see very sharp emission lines even with a CD spectroscope. All you have to look at sodium street lamps, white fluorescent bulbs or the moon (one can see very weak solar spectrum with dark lines).
If are looking for low budget "instruments" I would suggest that you play with a CD or DVD as a diffraction grating rather than using a lab based absorbance measuring instruments. That is just a black box.
https://hackaday.io/project/28529-portable-dvd-spectroscope


text:
Which quantities change when choosing a different standard state?
Chemistry and biochemistry use a different standard state. For example, the standard hydrogen ion concentration is the one at pH = 0 in one case, and the one at neutral pH in the other case.
Which quantities change when choosing a different standard state, and which quantities have special values at standard state?

Quantities that are dependent on the standard state
When calculating the equilibrium constant K or the reaction quotient Q from a set of observed concentrations, these concentrations are given relative to the standard concentration of the species. For example, in aqueous solutions the solvent water has a different standard concentration (that of pure water) than solute (roughly 1 M). Therefore, the value of K and Q depend on choice of standard state.
Whenever a thermodynamic quantity has a Plimsoll symbol appended, it refers to a system at standard state, e.g. $\Delta_r G^\circ, \Delta_r H^\circ, S_f^\circ, \Delta H_f^\circ$ or to the substance at standard state. The choice of standard state changes the value of these quantities. For the biochemical standard state, the quantity is appended with a prime, e.g.  $\Delta_r G^\circ {'}$ to indicate that difference.
Quantities that have special values at standard state
At standard state, activities are equal to one. Consequently, the reaction quotient Q is also equal to one.
$$Q = 1 \ \ \ \ \text{(at standard state)}$$Also, and this is a bit of a tautology but useful in deriving some relationships, when concentrations are at standard state (or more generally if $Q = 1$), quantities are equal to the standard quantities. Take for example the Gibbs energy of reaction:
$$\Delta_r G =  \Delta_r G^\circ + R T \ln Q$$
When concentrations are at standard state, this reduces to:
$$\Delta_r G =  \Delta_r G^\circ \ \ \ \ \text{(at standard state)}$$
Of course, that is the idea in calling $\Delta_r G^\circ$ the standard Gibbs energy of reaction.


text:
Gibbs Free Energy Notation
What's the difference between $G$, $\Delta G$, $\Delta G^\circ$, and $\Delta_\mathrm r G^\circ$?
I've seen the first two used interchangably, and seen $G$ specifically referred to as change in Gibbs Free energy. Is this common? Are there cases where $G$ refers to absolute Gibbs Free Energy?
For those with the circle ($^\circ$), I've seen that referred to as "standard", and this means that they are in reference to standard pressure or concentration. Does this make their use noticably different?

This is a good staring point: https://chemistry.stackexchange.com/a/41864/72973

@Karsten Theis I did read that, and it clarifies a bit on the use of standard GFE over plain GFE. However, I don't understand what the $\Delta_r$ signifies, and how it is different from a normal $\Delta$ or just nothing at all, as some texts seem to use.

$G$ is the Gibbs free energy of a system. It is a conceptual quantity in the sense that there is no reference point that defines $G = 0$ for a substance (unlike entropy). Whenever you see a plot of $G$, notice that the scale is not labeled with absolute values (no indication of where the origin is).
$\Delta G$ is the change in Gibbs free energy, i.e. $G_\text{final} - G_\text{initial}$. Because it is a difference, there is no issue with the lack of reference point.
$\Delta_\mathrm r G$ is the change in $G$ as a reaction proceeds, i.e. $\Delta_\mathrm r G = \frac{\mathrm dG}{\mathrm d\xi}$ with $\xi$ the extent of reaction variable. Because of this definition, it is an intensive quantity with dimensions energy per amount of substance. It is called the Gibbs energy of reaction.
$\Delta_\mathrm r G^\circ$ is the standard Gibbs energy of reaction, for the special case when all reactants and products are at standard state.
$\Delta G_\mathrm f^\circ$ is the standard Gibbs energy of formation of a substance, the $\Delta_\mathrm r G^\circ$ for the reaction that makes the substance from the elements (where all products and reactants are at standard state).
$\Delta G^\circ$ is for any process at standard state (e.g. ice melting at a pressure of 1 bar). It is also sometimes used instead of $\Delta_\mathrm r G^\circ$, but that is confusing because the meaning and the dimensions are different.

I've seen the first two used interchangably, and seen $G$ specifically referred to as change in Gibbs Free energy. Is this common? Are there cases where $G$ refers to absolute Gibbs Free Energy?

$G$ should always refer to absolute Gibbs energy, and never to a change in Gibbs energy.

For those with the circle ($^\circ$), I've seen that referred to as "standard", and this means that they are in reference to standard pressure or concentration. Does this make their use noticably different?

Yes, because the Gibbs energy is concentration-dependent. With $Q$ as the reaction quotient,
$$\Delta_\mathrm r G = \Delta_\mathrm r G^\circ + R T \ln Q$$
As concentrations change in a reaction, $\Delta_\mathrm r G$ changes but $\Delta_\mathrm r G^\circ$ does not. When a reaction approaches equilibrium, $\Delta_r G$ approaches zero (but $\Delta_\mathrm r G^\circ$ doesn't change, as just mentioned).
The reason that $\Delta_\mathrm r G$ is concentration-dependent is that the Gibbs energy has an entropy component ($G = H - T S$), and there is an entropy of mixing that depends on concentrations (the $R \ln Q$ part).

This makes sense, thank you. I know in a reaction that the gibbs free energy change is the sum of the $\Delta G_f^\circ$ of the products minus the sum of the $\Delta G_f^\circ$ of the reactants - is this because $\Delta G$ is a state function, so you can add up the formation energies in and subtract the ones that come out?

$\Delta G^\circ$ is just the generic symbol for the standard free energy change, for any process such as a phase change. By the way, you don't mention what "standard" refers to (ie 1 bar) but ok.

@Vedvart1 - Yes, Hess' law and other additivity of thermodynamic quantities rely on these quantities being state functions. It also allows us to evaluate irreversible processes (i.e where the entropy of the universe increases) by comparing them to near-equilibrium processes that get the system from the same initial state to the same final state, just on a different path.

@Night_Writer - I added your info on $\Delta G^circ$. Standard state is not trivial, see this question and answer: https://chemistry.stackexchange.com/a/114698/72973


text:
Does changing the concentration of reactants alter the equilibrium constant

$$\ce{S(s) + O2(g) -> SO2(g)}$$
The reaction of sulfur with oxygen is written in equation form above. This equation can be interpreted in all of the following ways except
(A) either $\ce{S(s)}$ or $\ce{O2(g)}$ will be completely used up
(B) $Q$ must be close to $1.0$ since there is  one mole of gas on each side of the equation
(C) this reaction goes to completion
(D) adding $\ce{O2}$ will change the equilibrium constant

I do not know how to extrapolate $Q$ from just an equation with no other data. However, I also learned that the equilibrium constant is, well, constant, and adding O$_2$ should change the equilibrium concentrations, not the equilibrium constant.
The correct answer is (B). How would (D) be a valid interpretation?

I'd say it wouldn't. Your interpretation is correct, and the test is inaccurate.

D is an invalid statement because the ''equilibrium constant'' for a certain reaction is constant as long as the temperature of the system is constant.
When the concentration of a species Is changed according to Le Chatelier's principle the equilibrium system will change it's rates of the forward and backward reactions so as to balance the concentrations thus keeping the equillibrium constant the same.


text:
Volume of 0.5 M HCl needed to make a 10 % 500 ml solution

What volume of $0.5\ \mathrm M$ HCl solution is needed to prepare a $500\ \mathrm{ml}$ $10\ \%$ concentration solution, whose density is $1.05\ \mathrm{g/cm^3}$.

I've started off with finding the mass of the final solution, which would be $500\ \mathrm{cm^3} \times 1.05\ \mathrm{g/cm^3} = 525\ \mathrm g$.
Next I find the mass of the HCl in it, $52.5\ \mathrm g$.
After that, the moles of HCl, approx. $1.4384\ \mathrm{mol}$.
But after that I'm not sure how to proceed. I've tried dividing the molarity by the moles to get the volume, but that didn't work. I think I might have to make a system of equations but I'm bad at math so if that's how it's supposed to be done, I'd like for someone to show me how.

As you have calculated, 500 ml of a 10 % HCl solution contain about 1.44 mol HCl. Thus the concentration is about 2.9 mol/l. Therefore, you cannot prepare this solution by diluting a solution with c = 0.5 mol/l HCl.

Thanks, I guess I've forgotten that sometimes there just aren't any solutions (pun not intended) for such things!

There is a way, using the fact HCl forms high boiling azeotrop with water, near 20% of HCl. So one can by rectification column evaporate excessive water, cool it down and dilute to 500 ml. :-)


text:
Why isn't nylon as strong as kevlar?
Why is Kevlar stronger than nylon? They both contain Hydrogen bonding and should have equal strength but this is not the case instead Kevlar is fire resistant. What makes it stronger. Greater hydrogen bonds?!

Hydrogen bonds are the same in both. But kevlar is conjugated all through.

This suggests aromatic stacking: https://en.wikipedia.org/wiki/Kevlar

"Strong" is an ambiguous term here. nylon is strong in the sense that it is hard to break (but is very easy to stretch and deform). Kevlar is also hard to break but is also much more rigid and harder to deform. Stiffness and "strength" are not the same and Kevlar is useful because it has both.

Kevlar is an A,B co-polymer where the monomers are terephthalic acid and 1,4-diaminobenzene. The amide linkage, together with the aromatic rings in the polymer, makes a very rigid polymeric structure that stacks the chains in an organized way, maximizing the effectiveness of hydrogen bonding between amide linkages, together with pi-stacking of the rings:

http://pubs.acs.org/cen/coverstory/8005/8005olympics.html
The chains form sheets that stack together along the axis of the fiber. In order for projectiles to break these fibers, these strong, inter-strand attractions must be overcome.
Standard Nylon is a similar polymer, where the aromatic moieties are replaced by aliphatic chains. These are much more flexible and give more degrees of freedom to the polymer chains. This results in a more random arrangement of chains that still hydrogen bond, but not in such an organized way, giving a weaker fiber.

https://www.researchgate.net/publication/305313239_Nylon_66_Nonwoven_Fabric_Separates_Oil_Contaminates_from_Oil-in-Water_Emulsions

First, it's important to define what you mean by "nylon", a term which is often used generically to refer to many different aliphatic polyamides. The most common are "nylon 6" and "nylon 6,6". Nylon 6 is the polymer of 6-aminohexanoic acid (aka 6-aminocaproic acid), while nylon 6,6 is from alternating units of 1,6-diaminohexane (aka hexamethylene diamine) and 1,6-hexanedioic acid (aka adipic acid).
Kevlar is one of the trade names of the polymer of alternating p-phenylenediamine and terephthalic acid. This is one example of the general class of aramid polymers, which means polyamides made from aromatic monomers.
As mentioned in the comments, both types of polyamide can form interstrand hydrogen bonds that strengthen the bulk material. However, the fact that kevlar is made entirely from rigid aromatic monomers means that the individual strands are much less flexible than the strands of the aliphatic polyamides. The aromatic groups can further participate in interstrand interactions that are stronger than the interstrand aliphatic interactions. Both of these factors contribute to making very strong and inflexible fibers. For this reason, Kevlar is used for ballistic protection applications.
On a side note, the fire resistance you mentioned is more closely associated with the aramid made from isophthalic acid and meta-phenylenediamine, which is marketed by Dupont under the name Nomex. The meta- orientation of the linkages means that the individual strands, while still rigid, are not linear, so they do not align in the same arrangement as in Kevlar. This leads to a much weaker material that is not effective for ballistics protection, but works well for making flame-resistant fabrics.


text:
Origin of the name of barbituric acid
Recently, I have encountered two different accounts explaining the origins of the name "barbituric acid". Both accounts do mention that the name of this compound was given by the famous organic chemist Adolf von Baeyer. Li mentions that von Baeyer named the compound after his girlfriend at the time, with the name Barbara$^{1}$. Quin & Tyrell, on the other hand, informs that von Baeyer named the compound after the patron saint of artilleryman, St. Barbara$^{2}$. This is a seemingly trivial question but I'm genuinely just curious.
References

Li, J. J. Name Reactions, 3rd ed.; Springer: Heidelberg 2006; p. 15.
Quin L. D.; Tyrell, J. A.; Fundamentals of Heterocyclic Chemistry: Importance in Nature and in the Synthesis of Pharmaceuticals. John Wiley & Sons: Hoboken, New Jersey 2010; p. 119.

Unrelated to barbiturates but of similar interest, the ketone barbaralone was named for the late Professor Barbara M. Ferrier for work she accomplished while she was a postdoctoral student with Professor William v. E. Doering at Yale during the 1960’s. As to the eponymy of the related bullvalene, see ref. 15 in footnote 2. https://de.wikipedia.org/wiki/Barbara_M._Ferrier

Interesting question but the sad answer is we will never know the true story. Here is the original text (thanks to Google Books) dated 1863, right from the horse's mouth Adolf von Baeyer.

Dieser Gedanke hat mich bei der folgenden Arbeit geeitet (spelling?), und ich habe
mich zunächst bemüht, die Kenntnis des Materials zu vervollständigen.
Man wird sehen, wie sich dasselbe in einfachster Weise um die Substanz
$\ce{N2CO3H4}$, die ich Barbitursäure nennen will, gruppieren lässt und wie
also die Frage nach der Konstitution der Harnsäure und ihrer Derivate
auf die Untersuchung dieser Substanz zurückgeführt ist.

[DeepL Translation] This thought led (?) me to the following work, and I first tried to complete the knowledge of the material. One will see how the same can be grouped in the simplest way around the substance $\ce{N2CO3H4}$, which I want to call barbituric acid, and how the question of the constitution of uric acid and its derivatives can be traced back to the investigation of this substance.
He never explained his choice- no footnotes anywhere. However the unabridged Oxford English Dictionary (not open access) attributes it to a woman's name Barbara.
< French barbiturique (Ann. de Chimie et de Physique, 1865), < German barbitur in barbitursäure (Baeyer 1863, in Ann. d. Chemie u. Pharm.), < Barbara, a woman's name.
Cited from Adolf von Baeyer.

German Wikipedia has some referenced explanations for the Barb(ara) part.

The Journal of Chemical Education also has a whole article dedicated to it. The end story is that we will never know. I showed the original quote and this is as factual as it can be. The rest is mostly speculation.


text:
Which is better: Increase in collision frequency or increase collision energy
What would have a greater effect on the rate of a reaction, increasing the frequency of collisions or increasing the collision energy of the particles? (assuming that using a catalyst is not an option)
I got this from a past paper question:
Explain in terms of collision frequency and collision energy how the rate would change if the temperature was increased, and which of these causes the greater effect.
I know the first part of the question.
I assume that the collision energy is more significant, which is why we use catalysts, but not sure how.

Did you learn about the Arrhenius equation including the pre-exponential factor?

You can have as many collisions as you like but if the activation barrier is not exceeded then no reaction can occur. Thus it is energy that matters.

Collision energy is what matters
Both the frequency of collisions and the average energy of those collisions are increased at higher temperatures. But the frequency doesn't matter much for the rate of reaction because collisions are already very very common and are not the limiting factor in causing a reaction to happen.
The distribution of energy in moving particles (and therefore the energy involved in collisions) follows a statistical distribution. Simplifying a bit (by ignoring the orientation of colliding molecules and a bunch of other probabilistic factors) the reason a reaction follows a collision is that the collision has enough energy to overcome some threshold energy barrier for a reaction to take place (simplifying again, collisions without enough energy just cause molecules to bounce off each other perhaps exchanging some kinetic or vibrational energy in the process). Only when the net energy involved is sufficient to, for example, break a bond in one of the molecules, does a reaction result. A lot of collisions are happening all the time and most don't have enough energy to cause a reaction to happen. In fact, if the population of collisions with enough energy is too low there will be no reaction at all no matter how many collisions there are and increasing the number will make no difference at all.
As the temperature increases, more of the molecules are pushed into the "have sufficient energy" part of the distribution (which might be a very small proportion of all the molecules in the mixture). Then reactions start to happen.
But the number of collisions is very large compared to the number leading to a reaction so it is the number of molecules with enough energy that dominates the result.


text:
Change in heat of photosynthesis
Consider a plant that uses the energy it receives from the sun to do work (to undergo photosynthesis) to produce glucose. The minimum amount of work this plant can do to produce 1 mole of glucose is equal to the change in Gibbs free energy that goes into producing one mole of glucose. I calculated the change in Gibbs free energy that goes into producing one mole of glucose to be $\Delta G =2878.94\ \mathrm{kJ}$. Therefore the plant does $2878.94\ \mathrm{kJ}$ of work. Also if we calculate the change in enthalpy for the same process we get, $\Delta H = 2803.04\ \mathrm{kJ}$. Therefore the heat from photosynthesis under these conditions is, $Q=\Delta H - \Delta W = -69.9\ \mathrm{kJ}$.
This confuses me because it implies that photosynthesis under these conditions gives off heat. How can this be?

The reaction is still endothermic. The sun does the work on the reaction, which is endothermic and gives off heat. It only sounds confusing if you are more used to discussing processes that run in the absence of work. The heat release is important to comply with the 2nd law.

Generally speaking (without need to refer to your particular problem), the sign of the free energy change and thus the magnitude of maximum possible work does not depend on $\Delta H$ or $\Delta S$ alone, it depends on the combination:
$$w_{\mathrm{max},\text{non-}pV} = \Delta G = \Delta H - T \Delta S.\tag{1}$$
This equation describes the maximum amount of non-$pV$ work that can be obtained from the system or minimum amount that needs to be done if the process is performed reversibly at constant $T$ and $p.$ In addition, when work other than mechanical $(pV)$ is performed, the value of $\Delta H$ is no longer equal to the heat evolved, in fact then $\Delta H = q_p + w_{\mathrm{max},\text{non-}pV}.$ Finally, it remains that at constant $T$ the entropy change of the surroundings is related to the amount of heat evolved as
$$\Delta S_\mathrm{surr} = \frac{-q_p}{T}.\tag{2}$$
This allows the above to be rewritten in terms of entropies of the system and the surroundings (at constant $T$ and $P):$
$$w_{\mathrm{max},\text{non-}pV} =  w_{\mathrm{max},\text{non-}pV} + q_p - T \Delta S = w_{\mathrm{max},\text{non-}pV} -T(\Delta S_\mathrm{surr} + \Delta S_\mathrm{sys}) $$
Removing the cancelling terms we conclude that
$$\Delta S_\mathrm{surr} + \Delta S_\mathrm{sys} = 0\tag{3}$$
that is, the entropy of the universe remains constant, which is consistent with reversibility.
That clears up the question of how it is possible for a process to be formally endothermic (in the sense that $\Delta H > 0$) and yet heat is released $(q < 0).$ This does not however explain why the net process of glucose generation by photosynthesis, assumed ideal (amongst other things reversible) generates heat.
The answer is that the released heat is responsible for an increase in the entropy of the surroundings which offsets a reduction in the entropy of the system when glucose is generated, which can be shown by writing
$$\Delta S_\mathrm{sys} = -\Delta S_\mathrm{surr} = \frac{q_p}{T} < 0\tag{4}$$
Without that increase in the entropy of the surroundings life would not be possible.


text:
How to measure quantities at standard state when this state is a hypothetical one?
Consider the reaction
$$\ce{A(g) + H2O(l) <=> B(aq) + C(g)}$$
If all gases and solutions were ideal, the standard state would be pure water, the solute B at a concentration of 1 M, the gases A and C each at a pressure of 1 bar, and the system at a pressure of 1 bar.
For non-ideal gases and solutions, it would be the hypothetical state where all the activities are 1 (i.e. equal to the activity of an ideal gas at 1 bar, activity of a solvent with infinitely diluted solutes, and equal to the activity of a solute behaving ideally at a concentration of 1 M).
This hypothetical state can be quite different from a real state. How can the pressure be 1 bar while two gases are at a partial pressure of 1 bar each? What if the solute B is barely soluble, how do you get an activity of 1? How can you have pure water when it is acting as a solvent?
So how to you measure quantities like $\Delta_\mathrm{r} G^\circ$ when the standard state is a hypothetical one?

The standard states are limiting states. In these limits you can write the chemical potential of the component as
$\mu = \mu^\circ+RT\log(x)$
where $x$ is a measure of the concentration of solute (for Henry's law, where the standard state is a hypothetical state related to the behavior as $x \rightarrow  0$) or solvent (for Raoult's law, where the standard state is approached as $x \rightarrow  1$). In other words, in the limiting states the activity goes to 1. The meaning of $\mu^\circ$ depends on the limiting law (hypothetical with properties of an infinitely dilute solution or pure solvent).
A standard state can be hypothetical (as may happen when Henry's law is invoked), but it can also be real (as in Raoult's law, for a pure solvent). It is a state in which the sample behaves ideally as defined by the above equation, in the limit of infinite dilution or of a pure solvent. This limiting behavior provides a convenient and predictable reference, like setting the potential energy to zero when particles are separated by an infinite distance. In the case of Raoult's law, the standard state is the (not particularly mysterious) pure substance. In the case of Henry's law, the reference state is found by extrapolating behaviour at infinite dilution to the standard concentration (1 molal or 1 M, conventionally). This may be a hypothetical state.
Using reference standard states (at 1 bar of course) can simplify combining or comparing data for different substances or conditions. Of course when considering real solutions, activities may diverge from concentrations and activity coefficients deviate from unity.
For a mixture of substances, for each component you can describe individual thermodynamic properties relative to the ideal standard states. By ideal it is also meant that individual solutes do not interact.

When considering the chemical potential of a solute at standard state we must specify the pressure or only the concentration $1$ M? Because in t

@adosar the standard pressure is in general 1 bar (this can be assumed if not mentioned otherwise).

You don't measure $\Delta G^\circ$. You measure $K$ and then compute $\Delta G^\circ = -RT\ln(K)$. Crudely put, the value of $K$ will depend on the units you use for concentration and partical pressures so we have decided to use 1 M and 1 barr, respectively, when computing $\Delta G^\circ$.
Furthermore, we assume that the gases and solutions are ideal (i.e. that the molecules don't perturb each others properties) so if you measure $K$ as high concentrations and partial pressures you have to correct for that by extrapolating to infinite dilution/ideal gas conditions or estimating the activity coefficients.
So you don't have to measure $K$ at standard conditions. In fact it's better to measure $K$ at much lower concentrations and conditions where you have ideal gas behaviour.


text:
Relationship between Ka and Kb

Considering the $K_\text{a}$ of the weak acid and the $K_\text{b}$ of the weak base that form the salt, we conclude that the solution will be acidic if $K_\text{b}$ > $K_\text{a}$ and basic if $K_\text{a}$ > $K_\text{b}$

I was wondering if anyone could clarify what the book is saying.
My question is If we mix a solution with a $K_\text{b}$ of 0.1 with a solution with a $K_\text{a}$ of 0.001 would the final solution be acidic or basic?

If there is a solution of mix of a weak acid/base and it's alkaline/acidic salt in molar ratio 1:1, then

The solution is acidic, if $K_\mathrm{a} \gt K_\mathrm{b}$, like acetic acid + sodium acetate(alkaline base)
The solution is alkalic(ine)/basic, if $K_\mathrm{a} \lt K_\mathrm{b}$, like ammonium chloride ( acidic salt) + ammonia(base)

Note that $$\begin{align}
K_\mathrm{w}&=\ce{[H+][OH-]} \\
K_\mathrm{a}&=\frac{\ce{[H+][A-]}}
{\ce{[HA]}}  \\
K_\mathrm{b}&=\frac{\ce{[HA][OH-]}}
{\ce{[A-]}}  \\
&=\frac{\ce{[HA]  K_\mathrm{w}}}{\ce{[A-][H+]}}  \\
&=\frac{\ce{  K_\mathrm{w}}}{K_\mathrm{a}}  \\
K_\mathrm{w}&=K_\mathrm{a} \cdot K_\mathrm{b} \\
\end{align}$$
Similarly, if there is such a 1:1 (molar amount) solution of an acid and a base, that are not in the mutual relation acid/salt or base/salt, the above is valid as well.

Ka and kb are related as -
Kw = ka × kb
Here, kw is equlibrium constant of self ionisation of water. At a particular temperature kw is constant. E.g value of kw at 25 °C is 10^(-14).
So, ka and kb are inversely propotional . Acidic and basic strength is inversely proportional to ka and kb respectively bescause measure of its strength is negative of log(ka or kb).

Thank you for your answer. If we mix a solution with a Kb of 0.1 with a solution with a Ka of 0.001 would the final solution be acidic or basic

That is s wrong question. Ka is a constant of acidic form of a compound, independently in if it exists as an acid or is neutralised, becoming a basic salt , or can be in any middle step.


text:
why do atoms need to be ionised in a mass spectrometer?
why do atoms need to be ionized in a mass spectrometer? why can't you determine the mass of the atoms when they are not ionised? Also what is the mass-to-charge ratio

The mass spectrometer "separates" on the basis of mass to charge ratio. The x axis of a mass spectrum is actually m/z, where z is the charge.

why do atoms need to be ionized in a mass spectrometer? why can't you determine the mass of the atoms when they are not ionised?

There are various types of mass spectrometer
https://en.wikipedia.org/wiki/Mass_spectrometry#Mass_selection
but all of them require neutral particles to be charged in order to be accelerated and then detected. The spectrometers use electromagnetic fields to accelerate the ions, this effect only occurs if the species are charged as the field will not accelerate a neutral particle.
For example, in the magnetic sector machine, the ions are formed in an ionization chamber, accelerated in a uniform electric field and shot into the magnetic sector where they are sorted by their path through the field. The higher the charge on the particle, the more it is defected, the higher the mass, the less it is deflected:

http://orgchemguide.blogspot.com/2011/04/magnetic-deflection-or-sector-mass.html

Also what is the mass-to-charge ratio

So the ions are sorted not by their charge or mass alone, but by the ratio of the two, the mass to charge ratio. The machine cannot tell the difference between an ion with mass 4 amu and charge 1+ and another with mass 8 amu and charge 2+, both have the same mass to charge ratio and will behave identically in the field.
Most detectors also work only if they are struck by a charged particle.
Here is a great guide to the simple principles of how a MS works:
https://www.chemguide.co.uk/analysis/masspec/howitworks.html


text:
What is wrong in this reaction?

What is wrong in this reaction?

I cannot find anything wrong, please help. Any hints would be appreciated.

The acidic hydrolysis would prolly be interfered by the presence of the basic nitrogen on the ring

More than that, Grignard reagents are also strong bases. The proton from the NH would quench any Grignard that formed. You need to protect the NH.

@YUSUFHASAN Actually it's like only thing that would work. Even the substrate could react with itself!

@Mithoron Won't the H+ be wasted in reacting with the N of the substrate?

@YUSUFHASAN It's quenching of alkoxide - last step, you'd need to use different method to make it in the first place, though.

The problem is with the free $\ce{NH}$. Grignard reagents are strong bases ($\mathrm{p}K_\mathrm{a}\sim 50$), strong enough to deprotonate $\ce{R2NH}$. Any Grignard that formed would be immediately quenched by the free $\ce{NH}$. You need to protect, this is typically done with tBOC, more discussion here.

This seems weird to me..I mean, I assumed that once the Grignard reagent starts forming,would it abstract a proton from another Grignard molecule, considering repulsion b/w two negatively charged species?

the quench is most likely internal. So the net reaction is reduction of the secondary bromide to methylene


text:
Naming ester as a substituent to carboxylic acid
Can carboxylic acid functional group -COOH and ester's functional group R-COO-R' be in one structure? If yes, carboxylic acid will be the parent chain and ester will be the substituent as c.a. is higher than ester in the priority table. Then, how do we name the structure where ester becomes a substituent? Thank you.

As you have correctly stated in the question, the $\ce{-COOH}$ group has higher priority than the $\ce{-COOR}$ group. In fact, the $\ce{-COOH}$ group has the highest priority (excepting cations) in IUPAC nomenclature.
The $\ce{-COOH}$ group should be included as the suffix of the parent chain, while the $\ce{-COOR}$ is indicated using a prefix.
In fact, Section P-65.6.3.2.2 of the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names, 2013 (aka the Blue Book) specifies the rules for this.

When, in an ester with the general structure R-CO-O-R' or R-S(O)XO-R', another group is present that has priority for citation as the principal group or when all ester groups cannot be described by the methods prescribed for naming esters, an ester group is indicated by prefixes as 'acyloxy' for the group R-CO-O—, and 'alkyloxy(alkany1)...oxo' or 'alkyl(allcanyl)oxycarbonyl' for the group —CO-OR'.

I quote the example provided in the text:

And another one (not from the Blue Book though)

3-Ethoxy-3-oxopropanoic acid

References: Favre, H. A., Powell, W. H. Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 R. Soc. Chem 2013 DOI: 10.1039/9781849733069


text:
How to find oxidation state of Co and Cr in [Co(NH₃)₄(NO₂)₂][Cr(NH₃)₃(NO₂)₃]?

Which of the following options represents the oxidation state of $\ce{Co}$ and $\ce{Cr}$ in the given complex?
$$\ce{[Co(NH3)4(NO2)2][Cr(NH3)3(NO2)3]}$$
(A) $2, 3$
(B) $3, 2$
(C) $3, 3$
(D) $2, 2$

I am unable to judge the net charge on complex when it breaks into ions and therefore I am having trouble with this question. I assumed that some unknown charge with opposite signs must be on the complex ions.

Conventionally, first complex ion is positively charged and the latter is negatively charged. Suppose thess charges are $p+$ and $q-$, respectively. Since two ion ratio is 1:1, $|p+|= |q-|$.
Suppose oxidation number of $\ce{Co}$ is $n$ and that of $\ce{Cr}$ is $m$. Hence, for cation complex (Note that the net charges of $\ce{NH3}$ and $\ce{NO2-}$ is $0$ and $-1$, respectively):
  $$n+0\times4+(-1)\times2=+p = n-2$$
and for anion complex:
 $$m+0\times3+(-1)\times3=-q=m-3$$
Since, $|p+|= |q-|$,
$$m-3= +p \; \text{and  }  n-2=-p$$
Or, $m+n=5$.
The only answer agree with this is (A) or (B):
If (A) is correct, then, $\ce{Co^2+}$ and $\ce{Cr^3+}$. Substitute these values in original two equation: $+2+0\times4+(-1)\times2=+0 $ and $+3+0\times4+(-1)\times3=+0 $. This can't be.
If (B) is correct, then, $\ce{Co^3+}$ and $\ce{Cr^2+}$. Substitute these values in original two equation: $+3+0\times4+(-1)\times2=+1 $ and $+2+0\times4+(-1)\times3=-1 $. This is correct.
Therefore, your answer is (B), and the oxidation numbers of $\ce{Co}$ and $\ce{Cr}$ ared $+3$ and $+2$, respectively.


text:
Chemical reaction energy barrier width
Well chemistry before the discovery of quantum world was saying:"In order for a chemical reaction to happen , reactants must have enough energy to be converted into products."I was wondering now that we have discovered quantum worlds and quantum properties and quantum effects and quantum tunneling , maybe if the energy barrier is very narrow then we dont have to add this "Activation energy" and the reactants pass through this energy barrier to create products .

Cant the reactants tunnel through the energy barrier to form products?

Quantum tunneling can have a stunning effect on the rate of reactions, see for instance the example (https://science.sciencemag.org/content/299/5608/867) of work by Zuev et al. or the perspective by R. McMahon (https://science.sciencemag.org/content/299/5608/833).

Can this be done to all reactions?

Please modify your question by clicking on the "edit" button in addition to seeking clarification here. Or, post this as a new question ("how important is tunneling as a reaction mechanism" or similar, although this might get shot down as too broad or requiring further research on your part). For the record tunneling can be important for any reaction where there is a proton or electron transfer involved and the electron/proton donor/acceptor are placed in a "sweet spot" that increases the importance of this over other mechanisms.

Tunneling can be important for any reaction where there is a proton or electron transfer involved and the electron/proton donor/acceptor are placed in a "sweet spot" that increases the importance of this over other mechanisms.
As explained here, the conditions for appreciable contribution of tunneling to a reaction mechanism are that (1) the activation barrier be large enough (otherwise tunneling is not necessary); (2) the width of the barrier (the tunneling distance) should not be too great and (3) the particle must have significant wave (delocalized) character, that is, small mass (such as a proton or electron).

In complement to this I would like to point out that Science magazine has editted and published a special disussing the tunneling of heavy elements which can be found here: https://science.sciencemag.org/content/299/5608/833. Now, it is hidden behind a paywall but it shows that tunneling in chemical reactions is a thing. Plus, in the public synopsis in that link the editor explains that the tunneling of electrons and hydrogens is important for many reactions, even some enzymatically catalyzed reactions.


text:
Why only ketone forms enolate in Claisen condensation between ester and ketone?
Claisen condensation involves nucleophilic substitution in esters by an enolate (formed by removing an alpha-H from an ester or a ketone) giving a beta-keto ester or a diketone.
My textbook says that for the reaction between ketone and ester, the ketone always forms the enolate by losing it's alpha-H . But why can't the ester form the enolate here? What will happen if the ester has alpha-Hs and the ketone dosen't?
Attempts:
Does it the O on the alcohol side of the enolate ester help in delocalizing the negative charge via an inductive effect making it less effecive as a nucleophile?
(I can't find anything on other sites I've seen:
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch21/ch21-3-3.html
https://en.wikipedia.org/wiki/Claisen_condensation
http://www.chtf.stuba.sk/~szolcsanyi/education/files/Organicka%20chemia%20II/Prednaska%205_Aldehydy%20a%20ketony_Reakcie%20II/Doplnkove%20studijne%20materialy/Carbonyl%20compounds_full_Blackmond.pdf
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Claisen_Condensation )

Wikipedia, at the very least, states that the combination is ketone plus non-enolisable ester. But to be fair, ketones (pKa ~ 20) are more acidic than esters (pKa ~ 25), so I am more curious about why the ketone doesn't just do an aldol with itself instead of a Claisen, unless the ester is used in large excess.

This link may help clarify the issue: https://chemistry.stackexchange.com/questions/93813/which-alpha-h-will-be-extracted-in-intramolecular-condensation-rxn-between-ester/93860#93860

The pKa of a ketone alpha hydrogen is of the order of 20, the pKa of an ester alpha hydrogen is of the order of 25 source. Bear in mind that this is a log scale, meaning the proportion of ketone enolate to ester enolate is 100000:1.
In answer to the comment by @orthocresol: the ketone does self-condense, but this is reversible. The condensation with the ester is irreversible due to formation of the anion of the product B-diketone.


text:
Observations when mixing potassium Triiodide and Sodium thiosulfate
I have a reaction equation, it's all balanced but I don't understand what the physical change is. It is as follows:
$$\ce{2Na2S2O3 + KI3 \to 2Na2S4O6 + 2NaI + KI}$$
It was used to prove if $\ce{S2O3^2-}$ was in the solution or not. I'm not sure but I think it changes color to violet and back to colorless due to $\ce{I-}$ but if someone could definitively say what you are supposed to observe it would be helpful.

$\rm I_3^-$ is colored pretty heavily (though I wouldn't call that color violet); $\rm I^-$ is colorless. What's unclear about that?

The keyword you should search is iodometric titration. Triiodide solutions are dark brown, just like the tincture of iodine. If you were to add thiosulfate ion (say from a buret) to  the triiodide solution would become colorless. The change is not that sharp. Therefore this is not the way end point is detected. A small drop of starch is added. Starch makes a deep bluish purple complex with free iodine. The moment all the iodine (in the form of triiodie $\ce{I2.I-}$ ) is consumed by thiosulfate, there is a very sharp transition from bluish purple to a colorless solution.

Do you mean that triiodide ions make deep blue color with starch indicator?


text:
What is the most common element in the human body?
I was wondering that since H2O was 66% of the human body, hydrogen would be the most. But apparently carbon is. Can someone explain that?

You have to clarify what you mean by most, whether that means number of atoms or by mass percentage. @JacobLee

And you could take most of the water out of a human body and still have a body (a dead one, admittedly).

I don't think a lack of clarity on atomic vs mass % is sufficiently unclear that the question cannot be answered.

No, carbon is not the most common element of the human body.  The human body is mostly water, so the most common element is bound to be hydrogen or oxygen.
Which one wins depends on the rules of your game.  If you are counting atoms in all those $\ce{H2O}$ molecules, you choose hydrogen.  If you go by mass instead, oxygen is the most common element in our bodies.


text:
Rate expression for the final product

Let's say we have:
$$
\begin{align}
\ce{A &-> B}&\quad&\text{(fast equilibrium)}\label{rxn:1}\tag{1}\\
\ce{2 B + C &-> D}&\quad&\text{(slow)}\label{rxn:2}\tag{2}\\
\ce{D + E &-> F}&\quad&\text{(fast)}\label{rxn:3}\tag{3}
\end{align}
$$
Find the rate expression for the formation of $\ce{F}$.

Since I am writing the rate law for $\ce{F}$, should I write it as
$$\text{rate} = k_3[\ce{D}][\ce{E}]$$
because that's the reaction step that actually includes $\ce{F}$ as the product, or do I always have to write is in terms of the slow step i.e.
$$\text{rate} = k_2[\ce{B}]^2[\ce{C}]?$$
What happens with the leftover $[\ce{B}]$ and $[\ce{A}]$ in the steps \eqref{rxn:2} and \eqref{rxn:1}, respectively — do I include those concentrations as well?

I don't know how you call it in English but you can apply this approximation $\frac{d[D]}{dt} = 0$ because (3) is fast and $D$ have a really short lifetime, so you can say that
$$k_2[B]^2[C]=k_3[D][E]$$
This approximation is why the teacher is always saying that only the slow reaction is important
Fast equilibrium means that the equilibrium is quickly governed by thermodynamics laws and so the constant
$$K = \frac{[B]}{[A]}$$
is not variating, so you can express $B$ in function of $A$
$$\frac{d[F]}{dt} = k_2K^2[A]^2[C]$$
I don't know how far you need to go but if you really want to solve for $[F]$ you need to write an equation of conservation
$$[N] = [A] + [B] + [C] + [D] + [E] + [F] = cst$$
$$d[A] + d[B] + d[C] + d[D] + d[E] + d[F] = 0$$
$$d[D] = 0$$
$$d[E] = 0$$
You can also say that $d[C] \approx 0$ because it looks like it's just a partner of collision, (it's the solvent) because the solvent is in much larger quantity you can say the concentration stay pretty the same
$$d[A] + d[B] = - d[F]$$
$$[A] - [A]_0 + [B] = -[F]$$
$$[A](1+K) = [A]_0 - [F]$$
$$[A]^2= \frac{([A]_0 - [F])^2}{(1+K)^2}$$
$$\frac{d[F]}{dt} = K_0\frac{([A]_0-[F])^2}{(1+K)^2}$$
with $K_0 = k_2K^2[C]_0$

Steady state approximation $dD/dt=0$

thank you for the precision, when you solve you obtain the equation of $F(t)$, you can also change the constant to see the influence https://www.desmos.com/calculator/7pq68olwtj

To second @porphyrin, a situation like the advancement of your reaction (3) being determined by the slowest reaction (2) of the three is referred as Bodenstein principle, too (https://en.wikipedia.org/wiki/Max_Bodenstein).


text:
Can pi backdonation occur on non-metal centers?
The common example of back-donation is the interaction of a CO molecule with a metal center (d-orbitals) on a surface.
Can a similar mechanism occur between CO and a non-metal center, like oxygen on a surface (like in an oxide)? Can the p-orbitals of the surface oxygen be involved in such an interaction?

It is a matter of perspective: back donation of pi-electrons is not limited to M-CO interactions. The more important question would be whether you would like to define it that way and look at it that way or not, and whether or not it will have a stabilising effect. (Related negative hyperconjugation) Carbon suboxide can certainly be explained that way: $\ce{O#C\bond{->}C\bond{<-}C#O <-> O=C=C=C=O}$

Pi back-donation can and often does occur in all-nonmetal molecules. See this answer in relation to chalcogen halides, where pi back-donation affects the stoichiimetry we see.

Yes.
This is a hot topic in main group chemistry, and has been used to explain stable CO complexes with main group elements.
Here is a very nice open-access article by Sergeieva et al. where they look at the pi-back donation with some state-of-the-art computational chemistry.
Here's a graphical demonstration of the  Si←CO σ-donation and Si→CO π-back donation.

Figure 2 Plot of deformation densities Δρ1–3 (isovalue=0.003) of the pairwise orbital interaction and shape of the most important occupied and vacant orbitals (isovalue=0.03) in silylene−carbonyl complex 1 with the orbital interaction energies ΔEorb (in kcal/mol) and their eigenvalues ν (in e). For the deformation densities, the direction of the charge flow is red→blue. The eigenvalues ν indicate the amount of donated (negative numbers) and accepted charge (positive numbers). The occupied orbitals are shown in yellow and blue for the different phases, while the unoccupied orbitals are in cyan and orange. Reproduced with permission according to CC BY 4.0 license.

The references contained therein detail the long history of Si--CO backbonding in particular.
There is nothing to stop back-donation to non-metal centers, but it is much rarer with main group elements.
This is likely just because it happens to be that the overlap is more likely to occur with metals due to, e.g., ionic radius, d-orbital shape, etc..

Good discussion, but pi back-donation is not rare at all with main-group elements. It's just more subtle. For example, we can draw a structure for disulfur dichloride without pi back-donation, but that interaction accounts for $\ce{S2Cl2}$ being favored over $\ce{SCl2}$.


text:
Does the word HA always refer to a weak acid?
I was wondering if HA always represents a weak acid or can it also represent strong acids.
Thank you all in advance

There is absolutely no info about how the strong the acid is in this notation. HA may refer to both weak and strong acids. https://en.wikipedia.org/wiki/Acid_strength

In fact, it need not to refer to an acid at all. E.g, in a biochemistry context, A may mean a particular coenzyme or a substrate, and HA may mean it's hydrogenated form. Considering A as the rest of a (weak) Acidic molecule is implied from the given context or it is given explicitly.


text:
Stability of Cycloalkyl Carbocations
What is the order of stability of the cycloalkyl carbocations?
I think the stability should increase as we move from cyclopropyl carbocation to cyclohexyl carbocation, since the angle strain is continuously decreasing. I understand this as, the difference between the stable 120 degree angle of a carbocation and the actual bond angle of the cycloalkyl cation is decreasing on moving from cyclopropyl to cyclohexyl carbocation.
But someone recently told me that the cyclopentyl carbocation is more stable than the cyclohexyl carbocation, citing the faster rate of SN1 reaction observed in 1-chloro-1-methylcyclopentane than in 1-chloro-1-methylcyclohexane.
So what is the actual stability order of these cations?

Stability and reactivity are relative concepts and they are both rely on reference frames. The both are not linked and one cannot infer from one to the other. Sometimes these arguments seem reasonable, and it has been done before, but usually this is coincidental. Your friend might be right, but that doesn't mean you're wrong. The relative stability of carbocations is no easy feat to determine, and there might not be a suitable textbook concept yet. In any case, you need to find a common reference frame for both ions.


text:
Energy balance in battery thermodynamics
I am trying to understand the derivation of a general energy balance in battery thermodynamics. The following relation is frequently found to determine the heat generation of a battery:
$\dot{Q} = \dot{Q}_\text{rev} + \dot{Q}_\text{irrev} = IT\,\frac{\mathrm dE_0}{\mathrm dT} + I(E-E_0)$
where $\dot{Q}$ is the heat generation, $I$ current, $T$ temperature and $E$ cell voltage. The index $0$ denotes the open circuit voltage. Furthermore, $\Delta S = zF\,\frac{\mathrm dE_0}{\mathrm dT}$, with $F$ as Faraday constant and $z$ as number of exchanged electrons.
The derivation is as follows:
First law of thermodynamics: $\mathrm dU = \mathrm dQ - \mathrm dW \tag{1}$ with $\mathrm dW = p\,\mathrm dV + \mathrm dW_\mathrm{el}$
$\mathrm dH = \mathrm dU + p\,\mathrm dV + V\,\mathrm dp \tag{2}$
Substituting (2) in (1):
$\mathrm dH - p\,\mathrm dV -V\,\mathrm dp = \mathrm dQ - p\,\mathrm dV - \mathrm dW_\mathrm{el} \tag{3}$
with $\mathrm dp=0$ follows:
$\mathrm dH = \mathrm dQ -\mathrm dW_\mathrm{el}$, differentiation with respect to time:
$$\frac{\mathrm dH}{\mathrm dt} = \frac{\mathrm dQ}{\mathrm dt} - \frac{\mathrm dW_\mathrm{el}}{\mathrm dt} = \dot{Q} - EI$$
So far, everything is fine. But now:
$$H = G + TS \rightarrow \frac{\mathrm dH}{\mathrm dt} = \frac{\mathrm dG}{\mathrm dt} + T\,\frac{\mathrm dS}{\mathrm dt} + \frac{\mathrm dT}{\mathrm dt}S$$
Here my first problem in understanding arises: In general $\frac{\mathrm dG}{\mathrm dt}$ and $\frac{\mathrm dT}{\mathrm dt}S$ cancels each other, since, $S=-\frac{\mathrm dG}{\mathrm dT}$ and $\frac{\mathrm dG}{\mathrm dt}$ can be expanded with $\mathrm dT$: $\frac{\mathrm dT}{\mathrm dt}S = -\frac{\mathrm dT}{\mathrm dt}\frac{\mathrm dG}{\mathrm dT}$. This means, that the influence of the open circuit voltage in the given energy balance would vanish.
I assume, that this is prevented by simply stating isothermal conditions: $\frac{\mathrm dT}{\mathrm dt}S = 0$. But this makes no sense to me, since the whole purpose of this calculation is the temperature increase with time during cycling a cell. Can somebody explain this to me?
Now if $\mathrm dT=0$ the derivation would proceed like this:
$$\frac{\mathrm dG}{\mathrm dt} + T\,\frac{\mathrm dS}{\mathrm dt} = \dot{Q} - EI$$
$$\frac{\mathrm d(-znFE_0)}{\mathrm dt} + T\,\frac{\mathrm d\left(znF\,\frac{\mathrm dE_0}{\mathrm dT}\right)}{\mathrm dt} = \dot{Q} - EI$$
where $znF=C$ is the charge and $\frac{\mathrm dC}{\mathrm dt} = I$
And here my second problem arises: to get to the equation at the very beginning of this post, it is necessary to set $E_0$ and $\frac{\mathrm dE_0}{\mathrm dT}$ constant, so that the chain rule does not apply. And I do not understand why this should be valid?
Can anybody help me with this?

The answer to this problem explains how to arrive at
$$ dQ_p = - dw_{ele} +  dG + TdS $$
This expression is generally true for electrical work at constant T and p. From it the differential equation in the OP can be derived by taking the time derivative on both sides and making appropriate substitutions.

$\frac{\mathrm dG}{\mathrm dt}$ can be expanded with dT:  $\frac{\mathrm dT}{\mathrm dt}S = -\frac{\mathrm dT}{\mathrm dt}\frac{\mathrm dG}{\mathrm dT}$

But remember that you defined
$G=H-TS$
and you should use that when computing $\frac{\mathrm dG}{\mathrm dt}$. But introducing that just results in a circular argument (a lot of terms cancelling to no use).

to get to the equation at the very beginning of this post, it is necessary to set $E_0$ and $dE_0/dT$ constant

That's ok. $E_0$ is the voltage when no current is running (that is, battery is not being depleted, as there is no reaction, and therefore composition is constant over time), and is thus constant with respect to time.

But this makes no sense to me, since the whole purpose of this calculation is the temperature increase with time during cycling a cell.

That's tricky and I do not have a complete answer. The equation derived is for a specific T. To solve for the time-dependence of the temperature evidently requires additional steps, presumably incorporating expressions for the heat capacity.

I am still thinking about the first part of your answer but considering the second part: How is $E_0$ defined to be t-independent? If you consider a discharge curve of a battery: Discharging with interruptions to get a course of $E_0$ values shows that $E_0 = f(x)$ and since $x$ (composition) and $t$ are connected, $E_0$ is time dependent, isn't it?

Here's an alternative attempt. There's a lot of cancellation, which I leave to the end.
Besides, definitions are used in the second line only. When terms vanish, I tried to show why with an arrow.
$$\require{cancel}
\dot{Q} = \dot{U} + \dot{W} = \dot{H} - \frac{d}{dt}(PV - W)\\
= \dot{G} - \frac{d}{dt}(PV - W - TS) = \frac{d}{dt}(G - PV + W + TS)\\
= -\frac{d}{dt}(znFE_0) \cancelto{0}{- V\dot{p}} \cancel{- p\dot{V} + p\dot{V}} + \cancelto{IE}{\dot{W}_\text{el}} + \frac{d}{dt}(znFT\frac{d}{dT}E_0)\\
= -\cancelto{I}{\frac{d}{dt}(znF)}E_0 - znF\cancelto{0}{\frac{d}{dt}E_0} + IE + \cancelto{I}{\frac{d}{dt}(znF)}T\frac{d}{dT}E_0 + znF\cancelto{\frac{dE_0}{dt} = 0}{\frac{dT}{dt}\frac{dE_0}{dT}} + znFT\cancelto{\frac{d}{dT}\left(\frac{d}{dt}E_0\right) = 0}{\frac{d}{dt}\frac{d}{dT}E_0}\\
= I(E - E_0) + IT\frac{E_0}{dT}
$$
The assumption that $E_0$ is (almost) independent of $t$ is used. If this doesn't hold, you'll end up with a new term of course: $znF\frac{d}{dt} \left( E_0 + T \frac{d}{dT} E_0\right)$.

Ok, I still don't understand why $E_0$ should be time-independent... anyway... should your derivation and the one I postet in my question not lead to the same result. Even if I take $\frac{dE_0}{dt}=0$, the problem with $dT=0$ persists.

No, the only assumption with my proof is that $\frac{d}{dt}E_0 = 0$, there's no assumption on $dT$ (do you mean $\frac{d}{dt}\left( \frac{d}{dT} E_0 \right) = \frac{d}{dT}\left( \frac{d}{dt} E_0 \right) = 0$?).

No, I made that assumption ($dT=0$) in order to get to the desired result in my original question. The part referenced with "Here my first problem in understanding arises". You didn't need that assumption and therefore, I'm wondering, because the resulting equation should be identical, independent of the ansatz.

Now I managed to get my original derivation to match yours, without stating $dT=0$. But to do this, I have to ignore, that G would cancel and perform the derivation until I got: $znF\frac{dT}{dt}\frac{dE_0}{dT}$ to get rid of it, assuming $E_0$ is constant. But this feels like mathematical cheating^^

My derivation uses the chain rule to state that $\frac{dT}{dt}\frac{E_0}{dT} = \frac{E_0}{dt}$ (last equality).

Yes I know, thats exactly what I did (now), but anyway, in general it is possible to cancel G even before this term is showing up. Therefore, it feels like a kind of a trick. Maybe this is the case... I don't know.


text:
Current and reversible heat in battery reaction
In the book Battery Technology Handbook by Kiehne [1] (Google Books) on page 21, equation (32):

The reversible heat effect per time unit can be related to current flow, because each multiple of the cell reaction requires the current amount $n\cdot F$:"
$$\frac{\mathrm dQ_\mathrm{rev}}{\mathrm dt} = \frac{Q_\mathrm{rev}}{nF}\cdot i \label{eqn:32}\tag{32}$$
with $n$: number of exchanged electrons; $F$: Faraday constant $(\pu{96485 A s/equivalent})$; $i$: current in $\pu{A}$.

Two questions:

How is the current related to $nF$? current and $nF$ have different dimensions?

How can one derive equation \eqref{eqn:32}?

References

Battery Technology Handbook, 2nd ed.; Kiehne, H. A., Ed.; Electrical engineering and electronics; Marcel Dekker: New York, 2003. ISBN 978-0-8247-4249-2.

Update in order to respond to comments
I know that $\Delta G$ is connected with a reversible process and in principle, your derivation makes sense. My problem lies in the formula:
$\Delta S = \frac{Q_{rev}}{T}$ is a basic thermodynamic relation and the Gibbs Energy is defined as $\Delta G = \Delta H - T\Delta S$. So, following your derivation $\Delta H$ would be zero. Since this problem is connected with my other question, which is linked in the comments, I would expect two different heat terms, a reversible one and a irreversible one. And from the formulas there, I expect $Q_{rev}$ to be connected only with entropy.
Therefore, $Q_{rev} = -T\frac{\text{d}G}{\text{d}T}$ rather than $Q_{rev} = -\Delta G$ as the combination of your equations (2) and (4) implies.
I is quite hard to explain my problem...
There is another way to get to the equation in question, but this seems to be too simple:
Since, $\frac{\Delta G}{nF} = \text{Voltage}$, $\frac{T\Delta S}{nF} = \text{formal Voltage}$. And $\textit{Voltage}\cdot I = P = \dot{Q} = \frac{\text{d}Q}{\text{d}t}$, equation (32) seems perfectly valid, but this is a kind of a weak derivation.

How is the current related to nF? current and nF have different dimensions?

$nF$ is the charge $C$ transferred during the reaction, while current $i$ is the rate of charge transfer ($dC/dt$).
The following is a derivation, now edited to be more rigorous. The equation you provide is an expression for electrical Joule heating, which follows in the steady-state from the expression for the power (rate of work) generated by the battery. From Ohm's law
$P=\left(\frac{dw_{rev}}{dt}\right)=-i^2R=-i E_0 \tag{1}$
where $E_0$ is the (open-circuit) electric potential, $i$ the current, $w_{rev}$ is the reversible electrical work. Note that $E_0$ is the potential when reversible work is done.
Now assume a steady state, with constant P and T constraints, with work done and heat generated cancelling (so that the internal energy $U = constant$), that is
$Q_{rev} = -w_{rev} \tag{2}$
and therefore also power and rate of heat dissipation equal:
$\left(\frac{dQ_{rev}}{dt}\right)=-\left(\frac{dw_{rev}}{dt}\right)=i E_0 \tag{3}$
Now according to the Nernst equation,
$\Delta G=w_{rev} =-nFE_0 \tag{4}$
In the steady-state the dissipated heat is equal to the electrical work, so that, combining (2) and (4), we have that
$\frac{Q_{rev}}{nF}=-\frac{w_{rev}}{nF}=E_0 \tag{5}$
which leads, combined with (3), to the equation in the book:
$\left(\frac{dQ_{rev}}{dt}\right)=i \frac{Q_{rev}}{nF} \tag{6}$

Aside
Note I apply the opposite sign convention for work from that in the OP (apologies) - in the convention I use work is positive when performed on the system (charging a battery is positive work, discharging negative work). This does not affect the result of the derivation, since I apply the same sign convention for heat (for an exothermal process heat is negative).

On $\Delta H$ and an alternative (longer) derivation
For a process at constant p, it is common to encounter the expression
$$\Delta H = Q_p$$
However if there is non-PV work, the more general form of this equation is
$$\Delta H = Q_p + w_{non-pV}$$
This expression is general and applies when the process is carried out either reversibly or irreversibly. In differential form,
$$d\Delta H = dQ_p + dw_{non-pV} \tag{a1}$$
Note also that in the preceding derivation non-pV work is electrical.
When the process is carried out reversibly a maximum amount of work is done and
$$w_{non-pV,rev} =\Delta G = \Delta H - T \Delta S = \Delta H - Q_{rev}$$
This leads to the following expression at constant T, equal to that in the linked problem (but note the different work sign convention here):
$$ d\Delta H = dQ_{rev} + dw_{rev} = d\Delta G + Td\Delta S \tag{a2}$$
where I dropped the "non-pV" subscript since the work is assumed to be electrical.
Since H is a state function it must be equal for reversible and irreversible processes, and we can equate (a1) and (a2), leading to the following general expression:
$$ d\Delta H = dQ_p + dw_{ele}=  d\Delta G + Td\Delta S $$
which leads finally to
$$ dQ_p = - dw_{ele} +  d\Delta G + Td\Delta S  \tag{a3}$$
Taking the time derivative of this equation results in the final expression in the linked problem (making sure to account for differences in the work sign convention):
$$\dot{Q} = \dot{Q}_\text{rev} + \dot{Q}_\text{irrev} = IT\,\frac{\mathrm dE_0}{\mathrm dT} + I(E-E_0) \tag{a4}$$
To obtain the equation in this problem it is only necessary to apply the reversibility condition starting from either (a3) or (a4). Starting from (a3),
$$ dQ_{rev} = - dw_{ele,rev} +  d\Delta G + Td\Delta S  $$
But $ dw_{ele,rev} =  d\Delta G  $ which leads to (somewhat trivially)
$$ dQ_{rev} = Td\Delta S  $$
Taking the time derivative and inserting the Nernst expression for $\Delta S$ (twice!)  gives
$$\dot{Q}_\text{rev} = \left(\frac{dQ_{rev}}{dt}\right) = IT\,\frac{\mathrm dE_0}{\mathrm dT} = I\frac{Q_{rev}}{nF}$$
which is the desired expression. The same result can be arrived at by applying the reversibility condition to equation (a4).

Thank you. I have the feeling that in this particular field there is a distinct sloppiness with mathematical formulations and nomenclature, or is it just me? What I still don't understand: Why do they claim $I=nF$?

I can't say who's being sloppy, but I also don't follow, where is it claimed that I=nF? In the other question you posted it is claimed that C = znF (=charge) which seems consistent.

But the book states: $Q_{rev} = T\Delta S$, what means, that your ansatz in equation (2) would be wrong, wouldn't it? Furthermore, $\Delta G$ is connected with the OCP $E_0$ (not the standard state!) and the electrical power is connected with a voltage under load. Compare to my other question: link

$n$ is the number of exchanged electrons not the amount of substance... so $nF$ is not C, $znF$, where n is the amount of substance and z the valence would be the charge... or is this, because the whole thing is related to one formula conversion, i.e. per mol?

@user230821 Here I use n as the amount of electrons exchanged in moles, but not as the stoichiometric coefficient (usually written as $\nu$), so that $\Delta G$ is not a molar quantity as written. I am not sure if that's consistent with the definition of "number of exchanged electrons" as you mean it but the derivation appears consistent. Also, zn in the other problem and n here seem to be identical.

"But the book states: Qrev=TΔS, what means, that your ansatz in equation (2) would be wrong". I don't see why but maybe you can elaborate. To make things clearer I added "rev" labels to Q. Plus the Nernst equation as written can be applied to a reversible process, making it consistent with reference to reversible heat.

Ok, that might be the source of my misunderstanding between nF and current. This problem seems solved. For the other part please see the updated question above.

@user230821 I added an explanation to my answer. That is an understandable point of confusion, because of the way $\Delta H$ is introduced, but the problem is resolved when you consider how the term for non-pV work makes its way into the expression.

Sorry, I don't understand your your extension about $\Delta H$. If I use this, $\Delta G$ cancels and I end up with $T\Delta S = Q$, again.

Let us continue this discussion in chat.

I added what I hope is a clear derivation of the expression in the linked problem, and hopefully it's clear how it's related to this problem. I understand now why you brought up H repeatedly - sorry  I had missed your point, and hope this clarifies things.

I have to think about that. But one question: Isn't this an explanation for the linked answer? At the moment I don't get the connection to the problem in this question. But I have to think about it... so therefore it might be only hidden to me ;)

I think if you understand the meaning of the final equation (which basically comes from the fact that dH(rev process) = dH(irrev process) you will get pretty far understanding the full derivation.

@user230821 The derivation I added is relevant mostly to the other question, but since you added comments to your question here, and since the questions are closely related, I added the derivation here. Note it is self-contained and general, except for the assumption of constant T and p.


text:
Protonation of pyridine
Recently, I have been reading the Nobel Lecture given by Kenichi Fukui when he won the Nobel Prize in 1981, for his contributions to the development of Frontier Molecular Orbital Theory. One of the most well-known ideas in this theory is that principally, the reactivity of molecules can be understood by looking at their HOMOs and LUMOs. On p. 15 of the lecture notes, it is written:

If HOMO or LUMO happens to be inadequate owing to its extension, the
  symmetry, or the nodal property, the next orbital should be sought for. One of
  the simplest examples of such an instance is the protonation of pyridine. In this case, the nitrogen lone-pair orbital is not HOMO, but the addition of proton to the nitrogen lone-pair so as not to disturb the π conjugation will evidently be more advantageous than the addition to higher occupied π orbitals which may intercept the π conjugation. Thus, the reason why proton dare not add to the positions of large amplitude of π HOMO in this case will easily be understood. It is not completely satisfactory to dispose of a disagreement between the HOMO-LUMO argument and the experimental fact formally as an exception to the theory. A so-called exception does possess its own reason. To investigate what the reason is will possibly yield a novel finding.

In the above paragraph, Fukui is essentially telling us that consideration of the HOMO of the pyridine molecule would not allow us to determine the site of protonation. As Fukui writes, the HOMO of this molecule is one of $\pi$ MOs of the aromatic system while in this protonation reaction, the lone pair orbital on the nitrogen atom reacts. He also ends off by saying that further investigation of this example may "possibly yield a novel finding".
Firstly, I would like to ask if any further investigation has been done into this and whether any novel finding has been derived to date? Secondly, I would like to put forth my own rationalisation of this phenomenon. Personally, I do not think that this example poses as an exception to the theory because this acid-base phenomenon is controlled by thermodynamics. That is to say, whether the protonation occurs, it should result in the resultant conjugate acid being of the greatest stability (i.e. Gibbs free energies need to be considered). Thus, kinetic factors, like orbital interactions between frontier orbitals, become much less relevant. Would you agree with my rationalisation?


text:
Convert an image of a chemical structure to a ChemDraw file?
Is there a way to convert a bitmap (JPG, PNG) image of a chemical structure to a ChemDraw file so that you don't have to draw a structure you find in a paper yourself?

In short, yes. But I don't know of a program that does this. Also, it would not be simple, since it would require features to identify perspective and depth (to disentangle overlaping features), all far from trivial. But for simple molecules you may be able to write your own script.

You question immediately brought two terms in my mind: "chemical" and "OCR" (Optical Character Recognition). You should Google "Optical OCR" and you will find some interesting resources.

You would probably need to train a machine learning algorithm. Also SMILES would likely be an easier output, which you can convert into a ChemDraw file trivially anyway. I think it would be a fun side project, although the benefits are perhaps not commensurate with the required effort...

I asked the same question (https://chemistry.stackexchange.com/questions/103229/reverse-image-searching-a-chemdraw-figure) a few months back with little in the way of answers, unfortunately. Hopefully there is more luck here.

Mathpix (https://mathpix.com/blog/convert-image-to-chemdraw) claims to do it. Costs $ (not much, free educational usage), have never used it, just showed up in a Google query, and any other disclaimers I've forgotten about...

There is a new tool for this: pic2mol.com. It's a cloud-based optical structure recognition service and an add-in for ChemDraw. Like Mathpix it is also paid, but there is a free trial available. If you have ChemOffice, you can give it a try. Disclaimer: I am affiliated with pic2mol.


text:
Citric acid and ammonium nitrate
Soil based question. I have soil with to much Ammonium Nitrate in it. I am wondering if I use citric acid and water mixed will that flush out the extra Ammonium Nitrate that has built up in the soil.
Will citric acid bond to Ammonium Nitrate?
Any incite on these to compounds will be greatly appreciated.

Ammonium nitrate is very soluble in water, so watering the soil will remove it, though, depending on the amount and dilution, it could then pollute streams, lakes and other bodies of water, possibly causing an algae bloom.
Your best choice might be to experiment with nitrate-tolerant plants, such as lettuce, ornamental cabbage and rhubarb, or perhaps bell peppers.
Citric acid will not help in any way, and would make the soil less likely to support life.
In nature, denitrifying bacteria will slowly reduce the $\ce{NH4NO3}$. In an anaerobic environment, it might be converted to $\ce{NH4+}$.

I genuinely don't think citric acid would do that for large amount of soil. Yet, there are  some microbial processes to remove nitrates from soil though (Ref.1):

Abstract: The microbial processes of denitrification and dissimilatory nitrate reduction to ammonium (DNRA) are two important nitrate reducing mechanisms in soil, which are responsible for the loss of nitrate ($\ce{NO3-}$) and production of the potent greenhouse gas, nitrous oxide ($\ce{N2O}$). A number of factors are known to control these processes, including $\ce{O2}$ concentrations and moisture content, $\ce{N, C,}$ pH, and the size and community structure of nitrate reducing organisms responsible for the processes. There is an increasing understanding associated with many of these controls on flux through the nitrogen cycle in soil systems. However, there remains uncertainty about how the nitrate reducing communities are linked to environmental variables and the flux of products from these processes. The high spatial variability of environmental controls and microbial communities across small sub centimeter areas of soil may prove to be critical in determining why an understanding of the links between biotic and abiotic controls has proved elusive. This spatial effect is often overlooked as a driver of nitrate reducing processes. An increased knowledge of the effects of spatial heterogeneity in soil on nitrate reduction processes will be fundamental in understanding the drivers, location, and potential for $\ce{N2O}$ production from soils.

Denitrification and dissimilatory nitrate reduction to ammonium (DNRA) are two of several important processes that are responsible for nitrogen cycling in soil. Other key nitrogen transforming processes in soil include anaerobic ammonium oxidation (Anammox), which oxidizes ammonium to nitrogen gas ($\ce{N2}$) gas using nitrite ($\ce{NO2-}$) as the electron acceptor. Nitrification, denitrification, and DNRA are all capable of producing $\ce{N2O}$ (See Figure). Both denitrification and DNRA are processes that reduce nitrate ($\ce{NO3-}$) through various intermediate steps to N2 and ammonium ($\ce{NH4+}$), respectively, and are the only soil microbial processes capable of both removing soil $\ce{NO3-}$ and producing $\ce{N2O}$. Denitrification produces $\ce{N2O}$ when abiotic conditions or gene complement prevent its reduction to $\ce{N2}$ and DNRA releases $\ce{N2O}$ as a by-product of the reduction process.

References:

M. Giles, N. Morley, E. M. Baggs, T. J. Daniell, “Soil nitrate reducing processes – drivers, mechanisms for spatial variation, and significance for nitrous oxide production,” Frontiers in Microbiology 2012, 3, 407 (Rewiew with 16 pp.) (doi: 10.3389/fmicb.2012.00407).

This is an interesting answer but you don't really address the specifics of the original question!

@Night Writer: I appreciate your feedback. I don't think citric acid would do that. Also, OP didn't indicate how much soil he/she is talking about. Thus, I gave a alternative solution. I edited my answer accordingly. :-)

Citric acid will not help flush Ammonium Nitrate from soil, as the other poster says, water will do that fine. For a fertiliser spill on soil, after removing whatever fertiliser you can, add sugar (sucrose) first then water. Microbial action will clean it up.


text:
Can a reactor be operated at both reflux and above atmospheric pressure?
I'm not a chemist by trade so my experience in the lab has probably not been as in-depth as most of you. I've been tasked with investigating a proposed reaction in which an organic mixture (mostly formic acid) is to be heated at reflux for several hours.
The complication being that a pressure of approx. 4 bar is required during the reaction. How would one go about achieving this? The reflux condensors I've browsed through are all open to atmosphere on one end.
Is there some sort of pressure regulator I can attach on the output of the condensor to maintain the pressure at 4 bar?
I'm told that the reaction may produce $\ce{CO2}$ and $\ce{NH3}$ by-products which means pressure is expected to increase and fluctuate as the reaction proceeds. Ideally I'd like to occasionally vent these to drive the reaction equilibrium forward.
The formic acid is quite corrosive to anything not plastic or glass.

In a mostly formic acid solution I can't imagine that much $\ce{NH3}$ is going into the gas phase. Essentially all the  $\ce{NH3}$ will be converted into  $\ce{NH^+4}$ by the formic acid. // Would venting the $\ce{CO2}$ drive the reaction to completion?  $\ce{CO2}$ isn't very soluble in an acidic solution.

Another question. Do you need just a limited quantity, or is this for industrial production?

Conceptually 4 bar isn't a problem. I have no idea what kind of setup is available commercially, but maybe some sort of teflon lined stainless steel with teflon gaskets. A chemist can work out the experimental conditions, but to build the reactor for industrial scale production you'll need the expertise of a chemical engineer.

@MaxW you are most likely correct, most NH3 would most likely end up as ammonium formate, the low solubility of CO2 would need to be vented occasionally to drive the reaction forward
Im required to only make a small quantity in the lab to prove the reaction and determine what kind of yields are possible, the industrial scale would then be handled by a chemical engineering team

You could theoretically boil at the higher temperature but why would that be a requirement? Why would you require reflux specifically if you can actually heat the formic acid to higher temperatures under pressure? Heating should work on its own and degassing to drive the equilibrium should work with a back pressure regulator. However,  most lab glassware will not take that pressure. You normally need some stainless steel pressure equipment.
One way might be to use plug flow reactors rather than bombs.

I cooked a number of things under reflux since the reaction rate was slow. The 4 bar limit might be some limit imposed by the equipment. In other words to get the reaction to go fast you'd need maybe 20 bar, but the equipment could only handle 4 safely. // I'd agree that ordinary glassware won't take 4 bar safely.

Reflux is just a condition that basically means, "heat to the boiling point and ensure the liquid can condense back into the flask/reactor etc." Heating to reflux is not that informative either, especially if you have azeotropes. Operating under atmospheric pressure limits you to the boiling point of the mixture. Working under pressure does not (depending on the temperature you need of course).

Well that is no doubt why the OP wants 4 bar, to get a higher boiling point.

But the reflux instruction or requirement under pressure serves no purpose, and certainly not in most glassware, especially no need for condensers

My point is simply that formic acid would boil at a higher temperature at 4 bar than it does at 1 bar. Not sure what that temperature would be.

The OP's task, given as written, serves no purpose. They just need to heat so that nothing boils off at 4 bar or above and use a gas-liquid separator.  It's not a standard procedure for regular lab glassware. Specialist kit is needed

Hi, apologies if this does not make sense. The instructions I received from the engineer stated that they plan to run the process industrially at 4 bar for a variety of operational/pumping reasons. He stated that the solute biurea has very low solubility in formic acid and to get any kind of reasonable reaction rate it would need to be done under high temperature for several hours leaving reflux as a requirement.
If glassware is not capable of these pressures is there an alternative to stainless steel? Formic acid does not seem to be compatible with SS

Hastelloy,  Monel,  Incelloy ; nickel SS alloys. Expensive though

@SimpleJack I would ask your engineer what temperature and pressure they propose to run the process at. I would surmise that the temperature used at the pressure, does not cause "reflux".

@Beerhunter - Given that CO2 would be created and must have pressure relief, just releasing gas would release CO2 saturated with formic acid. Having a pressure relief above the condenser would reduce the amount of formic acid in the gas significantly.

@MaxW Under pressure there's no reason to boil a liquid. I take your point though.

First of all I would consider the necessity, price and complexity of the pressure and try to carry out the reaction in standard laboratory glassware for longer time (checking the content of the desired product every duration period pointed in your instruction, e.g. 3 hours, 6 hours, etc.). If you get a good yield at ambient pressure this may save your company some money and dealing with pressure equipment at the price of little money and a working day in a simple experiment.
If you have to stick to the elevated pressure at laboratory scale you could use a vertical 1 - 1 1/2 inch quartz tube with the lower end welded or PTFE tube in a steel sleeve. I would recommend you to use a 1-1.25 m long tube but not less to have enough space for the vapor. I would not suggest to fill the tube more than 1/8 of its total height(=volume).
Do not forget to cover the bottom of the reactor vessel with broken glass/ceramic balls to avoid sprinkling.
Please make sure to take precaution measures in case the reactor ruptures, which could not be excluded. Carrying out the experiment in a fume chamber and having some non-burning mesh/cloth around it is a must/standard laboratory technique.
Another standard safety measure is to have weak top cover so in case something goes wrong the risk of injuring something/somebody around will be diminished.

Trying to make some kind of homemade rig is a huge safety problem in my mind. There must surely be some company that sells such reactor vessels for industrial research. // The OP made it quite clear that the vessel needed to be pressurized to test the feasibility of the full size industrial setup.


text:
Relation between ionic resonance energy and polarity

The ionic resonance of the bonds are given as follows:
$$
\begin{array}{cr}
\ce{A-B} & 24.3 \\
\ce{A-C} & 50.6 \\
\ce{A-D} & 102.3 \\
\ce{A-E} & 105.9
\end{array}
$$
Which of the bonds is the most polar?
a) $\ce{A-B}$
b) $\ce{A-C}$
c) $\ce{A-D}$
d) $\ce{A-E}$

I thought that more the resonance energy, more is the electron delocalisation and less polar is the bond, but this is not the case with the answer given.
I was getting a), but the answer is d).

Ionic resonance means including contributions to the structure from species such as $A^+B^- $  or $A^-B^+$ in a molecule of type AB. In this case (d) would seem to be correct.

I'm not too familiar with resonance theory, but I'll try:
According to modern VBT, the general idea is that the state of the molecule is not described by a single state wave function, but rather a linear combination ("superposition" if you will) of multiple different state wave functions. In your particular case, the resonance is between a "dominant" covalent structure (involving a completely equal sharing of electron density) and a "minor contributor" ionic structure (with unequal electron density).
Since resonance energy is defined as the amount of energy needed to convert the true delocalized structure into that of the most stable contributing structure (which can be interpreted as the amount of stabilisation provided by the linear combination of the state wave functions), a higher resonance energy would mean a greater contribution (and hence weighing coefficient) from the ionic structure. This means that there is a greater ionic character to the bond (and it is more polar).
Hope this helped.


text:
I think elements can be broken down, but their definition says otherwise, why?
Elements are made up of quarks, right?
and physicists have smashed atoms into one another to get quarks, right?
Then if the first two statements are true then elements can be broken down into simpler substances.
If I am right, then can someone give me a better definition?

Elements can't be broken down chemically. Yes in a nuclear reaction they can. The breaking down of elements is what gives a nuclear bomb its BOOM!

If quarks existed independently then what you say might be true. But they don't. We can infer that nuclei are made of quarks by smashing things together and observing the mess that creates, but we don't ever get an isolated, simpler substance that we can separate into something simpler than elemental nuclei. Under normal lab conditions the normal chemical definition of elements holds.

The atoms of chemical elements cannot be broken down by chemical reaction. You have simply shortened your definition of an "element* to the point where it breaks, if taken out of the chemical context.

What definition, where? The short answer is given in the IUPAC Gold Book:

chemical element

A species of atoms; all atoms with the same number of protons in the atomic nucleus.
A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under 1, but mostly the term chemical element is used for both concepts.

A longer answer is: Yes, physicists have managed to split atoms further down than to nuclei and electrons. It is also possible to "transmute" an element into another, such as creating Plutonium from Uranium. This involves a lot of machinery and energy, whereas splitting a substance into its constituent elements can be as simple as heating or dissolving in an acid and could be done in an afternoon with a few glass beakers.
Looking at it from a more theoretical perspective, one can (snarkily) define Chemistry as "the Physics concerned with the outer electron shell". Once you work on other areas of the atom (that is, the nucleus) or an electron shell of an atom can no longer be reasonably defined (plasma as in the sun), you tend to leave the area of chemistry (and the applicability of the term "element").

I really like your definition of chemistry: the physics concerned with the outer electron shell! Is it your own invention or someone else said it (just for quoting purposes)?

@M.Farooq A guy I went to high school with and whose name I can't remember said that about my choice of university studies. Like I said, snarky.

All science is either physics or stamp-collecting. ;-)

While there's a lot of truth in defining chemistry that way it does miss quite a few things - chemical thermodynamics, NMR and Mossbauer, and various core level involving spectroscopic techniques (e.g. EXAFS) are the ones that come immediately to mind


text:
Why different Asbestos types are hydrophobic and hydrophilic
I was doing research into the structure of asbestos to see that white asbestos chrysotile is hydrophilic where as both brown and blue asbestos is hydrophobic.
Could you please explain why this is?

.... homework. ;-) What have you found out? 99% of the people here likely to answer also have to look it up on wikipedia or someplace before ...