text:
How do phase transitions occur in the solid phase?
For example, how does graphite transform into diamond as pressure increases at a certain (adequate) temperature? I'm not sure if this is a great example to look at the bigger picture, so another example would be pure iron. At atmospheric pressure, as liquid iron cools down it solidifies to $\delta\text{-}\ce{Fe}$, with a bcc crystal structure. Further cooling down brings it to $\gamma\text{-}\ce{Fe}$, which has a fcc. Then, it keeps cooling down and it solidifies to $\alpha\text{-}\ce{Fe}$, again with a bcc structure. I don't really see how do the atoms physically rearrange to form the different crystal structures, neither do I see why would they jump back and forth between these. To add to this, I believe the difference between $\alpha$ and $\delta$ are their paramagnetic properties.
I guess the general question is, how do the atoms in the solid phase rearrange themselves by changes in pressure or temperature? (in contrast to, for example, a liquid-gas phase transition). And of course, why? What drives this change?

Magnetic properties have nothing to do with all this and are not the cause of difference between $\alpha$ and $\delta$. Indeed, $\alpha$ iron may or may not be ferromagnetic (dependent on temperature), while its crystal structure remains the same.

So they're both bcc, yet different?

In a way, they can be thought of as not different at all. Imagine one phase which is stable until certain temperature, then ceases to be stable, then becomes stable again, then melts.

Then why call them differently? And so, why would this happen?

Because one exists in one range of temperatures, and another exist in a different range, and never the twain shall meet. As to why, that's a very broad question, about as broad as chemistry itself. Solid phase structures are notoriously hard to predict.

Well, hopefully someone will be nice enough to give a broad but brief overview of what needs to be taken into consideration.

@IvanNeretin - well, the twain definitely can meet on binary phase diagrams, where it is quite clear that the two regions of bcc are the same since they are often connected. Now, your answer is correct - taking pure iron you see phase changes at different temperatures and each was given a name.

What do you know about thermodynamics of phase transitions, chemical potential etc?

Only what I learned in PChem as an undergrad. I'm a chemistry grad student, but mostly do OChem, so certainly it wouldn't be my strongest suit, yet I can probably follow some texts.

Phase transitions in solids occur in the same manner as in liquids or gases, through nucleation and growth. Yes, those processes require atomic motion. Yet,atoms diffuse in solids all the time, through various mechanisms. While potentially slower, the thermodynamics of solid phase transitions are easily seen in calorimetry measurements for example.
Nucleation and growth of precipitates is a key part of many engineering alloys. On the other hand, some phase changes (like diamond to graphite) are very hard to make happen at near-human-tolerable temperatures.
The solid phase also sees the martensitic phase transitions, which are second-order diffusionless atomic rearrangements. The name comes from the martensite phase observed in steels. Shape memory alloys are another example of martensitic phases.
A classic materials science text would be 'Phase Transformations in Metals and Alloys", by D.A. Porter and K.E. Easterling. In my opinion, the second edition is vastly preferable to the more recent third edition where the changes mainly seemed to be reintroducing all the errors that were removed in the second edition (sigh).

The white allotrope of phosphorus transforms slowly to red even at STP, especially when exposed to light. See https://en.wikipedia.org/wiki/Allotropes_of_phosphorus


text:
Same set of carboxylate reagents yielding two different products with an alkylide
I recently came across two contrasting products for the same set of reagents - carboxylic acid and alkyllithium, and wish to understand why exactly any difference exists between the two.
Reaction 1

Reaction 2

I understand that in steps A, B, and E, the alkylide (hydride in E) is acting as a base and abstracting the acidic proton. I'm fine with that.
However, compare step C and F. Both involve the same reagents (carboxylate and an alkylide). In step C, nucleophilic addition of the ethylide has taken place on the carbonyl group, which is a reasonable step according to me. However, in step F, an ester has been formed instead. I am unable to grasp how an ester can even form in this condition (considering both ethylide and carboxylate are negatively charged), and even if it can form, why wasn't it formed in C as well?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q64 and 27 in Alcohol, Ethers and Epoxides

There has to be a mistake somewhere. Based on charge conservation, the $\ce{Li}$ ends up associated with the counterion, $\ce{Li^+}$, i.e. a $\ce{Li2}$ molecule gets formed, which I believe has only been observed in the gas phase.

@ralk912 Yes, that's reasonable. So, you're suggesting that reaction 1 is correct, and reaction 2 is incorrect?

Yes, I don't see anyway reaction 2 would be right.

Step F has to be wrong. I suspect it is a typo and should be MeI

@Waylander There isn't any iodine here...

@GaurangTandon  I mean that the reagent should be MeI instead of MeLi as currently written

@Waylander oh, great. That could definitely be the intended case of the author. Thanks! :)

The product of F is cyclohexylmethyl ketone. The same reaction as in Reaction 1. Reviewed by Margret Jorgensen in Organic Reactions.

@user55119 Thanks for getting that data! I updated my answer below accordingly. Thanks!

@Waylander: see my comment above.

Reaction 2 is wrong. Reaction 1 is correct.
Later in the book (q87), another question is asked with the same set of reagents (carboxylic acid and methyllithium). Reaction 1 has been used in that as well.
Reaction 2 is wrong because, as ralk912 noted above, charge conservation has to be followed. If methyl lithium reacts with a negatively charged species, the product has to be negatively charged as well. If the ester is neutral, then lithium must be negatively charged, and would associate with the $\ce{Li+}$ formed in step E to form $\ce{Li2}$, which is only observed in gas phase.
Not to forget that there simply isn't any mechanism wherein a carboxylate would react with an alkylide to form an ester. It does not simply occur. Reaction 2 is wrong.

Waylander has suggested a good reason for a misprint i.e. the reagent $\ce{CH3Li}$ might have been intended to be $\ce{CH3I}$ instead. Then, it would make sense for the carboxylate anion to attack an alkyl halide, and displace an iodide ion (which could easily attach with the $\ce{Li+}$ ion to form lithium iodide) forming the methyl ester.

user55119 has confirmed that the product of the step F is cyclohexylmethyl ketone, and that it is obtained by reaction 1 only. The book containing this is "Margret Jorgensen in Organic Reactions."

Maybe add that cyclohexyl methyl ketone would form after aqueous workup, as in Reaction 1.

@ralk912: The geminal bis-lithium alkoxide is implicit from reaction 1. Its stability is the reason the methyl ketone is not formed, and over reacts with methyllithium, prior to addition of water.

@user55119 I'm afraid I don't follow?

If the product of C were not stable in solution, it would collapse to Li2O and a ketone. The ketone would react with methyllithium to form a tertiary alcohol. The same is true of F. Read "Reaction of Organolithium Reagents with Carboxylic Acids",  Margaret J. Jorgenson ,Vol.18, in Organic Reactions.

But I never said it is not stable. I actually say that to get the ketone back from the geminal dialkoxide, an acidic workup is needed.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Why is quinoline less basic than pyridine?
Quinoline is less basic than pyridine, even though it has more resonating structures. Why? Does the negative inductive effect reduce basicity of quinoline?

This is not due to the inductive effect, but rather, due to solvation. The positive charge of quinoline on the nitrogen is harder to solvate due to steric repulsion of the nearby ring, whereas in pyridine, the small hydrogen atoms in place do not pose a strong steric hindrance for the solvation of the ion.

The neighbour benzene also have a -I effect on Nitrogen how?

@user60369 It can withdraw electrons through resonance (though wouldn't this be the reason for stabilizing the conjugated acid too?). Actually, in the gas phase, quinoline is predicted to be more basic than pyridine, so this is not the reason you're looking for.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Acid-base reaction vs. salt dissolving reaction
I was watching a video in Khan academy about acids and bases. In the video, they asked the $\mathrm{pH}$ of a solution that is $\pu{0.15M}$ of $\ce{NH3}$ ($K_\mathrm{b}=1.8\times10^{-5}$) and $\pu{0.35M}$ of $\ce{NH4NO3}$. So, it was obvious to me that the first reaction is going to be that of a weak base because the $K_\mathrm{b}$ is given. But what wasn't clear to me that the second reaction is going to be that of a salt dissolving reaction that is dissolved completely as is claimed in the video.
Since this problem is presented in a beginners learning video I guess there is a way to determine that $\ce{NH4NO3}$ is a salt that is dissolving completely in water and not an acid or base according to the information given in the video.

tl;dr

What wasn't clear to me that the second reaction is going to be that of a salt dissolving reaction that is dissolved completely as is claimed in the video.

Remember the rule: "All ionic salts and strong acids/bases are 100% dissociated in aqueous solution"

I guess there is a way to determine that $\ce{NH4NO3}$ is a salt that is dissolving completely in water and not an acid or base

Realize that all such types of species of the form $\ce{A+B-}$ are ionic salts. If you try draw out the Lewis structure of $\ce{NH4NO3}$, you'll quickly realize it to be very different from that of, say, water or acetic acid.
In ammonium nitrate, you have two separate ions - ammonium cation and nitrate ion - held together by strong electrostatic forces of attraction (often called an "ionic bond"). Water, having a high dielectric constant, is able to break these electrostatic forces easily and dissolve the salt in 100% concentration.
Hence, we say that "All ionic salts and strong acids/bases are 100% dissociated in aqueous solution" i.e. when dissolved in water, they'll dissociate completely.  It seems you already knew the latter part, while the part for ionic salts is new for you.
As I said before, note that ionic salts are different from weak acids or weak bases like $\ce{CH3COOH}$. The $\ce{O-H}$ bond here is covalent and breaking it is an altogether different job as compared to breaking ionic bonds. Here, the reaction (or "equilibrium") has to favor the dissociation of acetic acid. This is ensured by the high stability of the conjugate base. If you've taken GOC classes, you would quickly realize that the conjugate base $\ce{CH3COO-}$ is much weaker than say $\ce{HSO3-}$ or $\ce{NO3-}$ (guess which acids the latter two are? ;) )
Again, realize that the rule is actually not completely correct. Later down the equilibrium road, you'll come across the equilibrium constant $K_\mathrm{sp}$ that defines exactly how much soluble certain salts are in aqueous solutions. Certainly, many ionic salts would be in that list, and that too with very low solubilities instead! Again, the way these people simplify calculations at an elementary level is to implicitly assume that "All ionic salts whose $K_\mathrm{sp}$ is provided are only partially soluble, while those whose $K_\mathrm{sp}$ is not provided are completely soluble (i.e. 100% dissociated)"

The solution which is $0.15$ M in $\ce{NH_3}$ and $\ce{0.35 M}$ in $\ce{NH_4NO_3}$, acts as nothing but a basic buffer solution. Applying the Henderson-Haselbach equation for basic buffers, $$\ce{pOH = pK_b + log({[salt]/[base]})}$$ Here the salt ( which is actually the salt formed by the base after reacting with strong acid) is $\ce{NH_4NO_3}$ because it is formed by the reaction of weak base $\ce{NH_3}$ with strong acid $\ce{HNO_3}$ and the salt concentration is $0.35$ M.  $\ce{pK_b}$ = $\ce{-log(1.8* 10^{-5}})$ =$4.745$. So, the $\ce{pH}$ will be ,$$\ce{pH= 14 - pOH =14 -4.745 -log(0.35/0.15) = 8.887}$$ So, the pH can be determined simply by considering the buffer solution. For your query, about the ionic property of $\ce{NH_4NO_3}$ , the $\ce{NH_4^+}$ ions and $\ce{NO_3^-}$ ions are simply held by pure electrostatic attraction. So, when you dissolve them in a solution they will  purely dissociate, because there is no other force ( like covalent bonding ) to hold them back or dissociate partially. They have to fully dissociate in the solution in $\ce{NH_4^+}$ and $\ce{NO_3^-}$.


text:
How to separate sodium carbonate from sodium hydroxide?
I got a drain cleaner in dust form, which consists of $\ce{Na2CO3}$ and $\ce{NaOH}$. I wanted to extract the $\ce{NaOH}$ so I tried to heat the mixture in order to just melt the sodium hydroxide, but it didn't really work as expected. Is there any easy way to do it?

@Mithoron Well, I managed to melt it. The melting point of NaOH according to Wikipedia is 323°C not a difficult temperature to achieve. The idea was that by heating the mixture to let's say to 400°C would melt the NaOH but not the Na2CO3 which melts at 851°C and separate them this way. What I didn't know is that Na2CO3 dissolves in liquid NaOH so separation in this way is not possible.

There is a forum which discusses the similar problem of separating sodium hydroxide and sodium carbonate from drain cleaner. The solution is taken from there(slightly edited):

This is relatively easy because sodium carbonate is practically
  insoluble in 50% sodium hydroxide solution so a glass frit type filter
  is needed to remover the insoluble sodium carbonate. Most commercial
  sodium hydroxide contains only a little sodium carbonate from
  absorption of atmospheric $\ce{CO2}$. So for carbonate free sodium
  hydroxide solution, make up one litre at a time of say 60% solution.
  To do this, add about 300g of sodium hydroxide pellets of flakes to
  500ml of ice-cold water in a large beaker and stirr until practically
  dissolved, cover and leave to cool . Allow it to cool to about 20-25 C
  and the add another 300g of sodium hydroxide and stirr until most has
  dissolved (much less heat is generated on the second addition), dilute
  to one litre, pour into a 1 L thick-walled polythene bottle and leave
  it to stand for about 1 month in a cool place and then decant the
  clear sodium hydroxide solution.
If the material is say 50% sodium carbonate do the following:
Add 500g of mixed sodium salts to 500ml of ice-cold water, stirr the
  slurry periodically for a couple of hours and then filter through a
  glass frit funnel (a vacuum filtration set-up will be needed) and wash
  the cake with a little water, say 50-100ml, to remove most of the
  remaining caustic soda; some sodium carbonate will dissolve but this
  will precipitate on the next addition. The cake will be almost pure
  sodium carbonate but you can always treat it with saturated sodium
  bicarbonate solution to ensure that it is.
The clear filtrate should be placed in a calibrated vessel and another
  500g of material added and again stirred periodically for a couple of
  hours then filter and wash as before. Hopefully there should be about
  700-800ml of very strong sodium hydroxide solution. Place it in a
  polythene bottle and let it stand of a month so the sodium carbonate
  can crystallize out, decant the clear solution and dilute as required.
NOTE:

60% caustic soda solution is rather viscous and will not
  filter without suction even on a fairly coarse frit, 50% is still
  pretty viscous and difficult to filter.
The sodium carbonate that
  crystallises out sinks very slowly hence the long standing time.
Sodium carbonate dissolves rather slowly in water form some reason so
  with vacuum filtation the residence time of the wash water will
  actually dissolve very little sodium carbonate, too much wash water
  will dilute the solution below 50% and make the carbonate more
  soluble.
Hot 50% caustic soda solution is extremely aggresive
  towards skin! When handling strong sodium hydroxide solutions, keep a
  bowl full of dilute citric around in case of emergency.

There are other method such as treating a more dilute solution with
  slaked lime, filtering (difficult because the precipitate is fine
  grained) and evaporating down.

SAFETY RE:keep a bowl full of dilute citric around in case of emergency. // You need a lot more than a bowl full of citric acid. You should have eye glasses, an apron, gloves, an eye wash, and a safety shower in case of an accident. You should also not be working alone with such chemicals.

Thank you very much for your response. I didn't know that solubility of Na2CO3 decreases when NaOH is presented. That explains the white precipitate in an exerinment I did using the drain cleaner.


text:
Standard state of a substance - definition

In chemistry, the standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. IUPAC recommends using a standard pressure $p^\circ=10^5\ \mathrm{Pa}$. Strictly speaking, temperature is not part of the definition of a standard state. For example, the standard state of a gas is conventionally chosen to be unit pressure (usually in bar) ideal gas, regardless of the temperature.

Source: Wikipedia
Now, as we can see the definition for the standard state of gas is independent of its temperature as well as volume. But the properties of gas depend on those and standard state is a reference point for calculating properties. So shouldn’t the definition of standard state for an ideal gas include temperature or volume?

Let us continue this discussion in chat.

Related: What is the significance of the “Standard Temperature” in Standard Enthalpies of Formation tables?

The IUPAC definition of standard state is:

State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p°, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p°. For a solute in solution it is the (hypothetical) state of solute at the standard molality m°, standard pressure p° or standard concentration c° and exhibiting infinitely dilute solution behaviour. For a pure substance the concept of standard state applies to the substance in a well defined state of aggregation at a well defined but arbitrarily chosen standard pressure.

Specifying that the standard state is a specific temperature and pressure, would mean that the standard state is not even remotely realizable for many gases.  For example, how could you have pure water vapor at standard temperature and pressure?

Well, there is some water vapor in equilibrium with liquid water at standard temperature and pressure.

@ralk912 but you don't have pure gas phase water at STP

Well no, but you implied that there is no water vapor at STP at all.

@ralk912 I'll add the word "pure" to the last sentence to clarify.

So you’re saying all gases can’t exist in gaseous state at a fixed temperature, but that is not the case with pressure? @DavePhD

@MollyCooL That's kinda what I'm saying, but not exactly.  For example, above 100 degrees C, you can have water at 1 bar.  Above the boiling point of any liquid, it can be at 1 bar.  But note also, the IUPAC doesn't commit itself to any particular pressure.  Instead it says "arbitrarily chosen standard pressure".  But hopefully someone else will give you a better answer.  I don't totally like my answer right now.

Oh! But for 1 bar pressure all gases can exist in the gaseous state right? Just asking, I get it that they say “arbitrarily chosen standard pressure”.

Shouldn't the reasoning be reversed: first consider the condition, and then decide whether the standard state is gas, liquid or solid? How to decide the standard state to consider for a given substance...?


text:
Is Argus lab accurate?
I was trying to calculate the resonance energy of benzene by using Argus Lab. These values are weird and very different from the experimental data. I have two questions regarding them.

What is the difference between Final SCF Energy and heat of formation?
When I tried to calculate the energy, It gave me two kinds of energy - Final SCF Energy and Heat of Formation. I know the definition of heat of formation. But what I found weird is that Heat of Formation of $\ce{H2}$ molecule was $\pu{-13.3564 kcal/mol}$. It should be zero instead.
Is there any mistake in following procedure:
a. I calculated the energy of cyclohexane($x$), cyclohexene($y$), and benzene ($z$).
b. Energy change while converting cyclohexane to cyclohexene ($a$) $= y-x$
c. Expected energy for benzene ($b$) $= x+3a$
d. Resonance energy of benzene = $z-b$
(Answer should be negative, implying that resonance energy is stabilizing in nature)
By using the SCF Energy data, I got  resonance energy $\pu{415 Kcal/mol}$ and while using the heat of formation data I got $\pu{-41.8 Kcal/mol}$.

Here is the data (in $\pu{Kcal/mol}$) -:
$$
\begin{array}{|l|c|c|c|} \hline
\text{Energy type} & \mathrm x & \mathrm y & \mathrm z\\
\hline
\text{SCF Energy using PM3} & {-21556.7995} & {-20680.6471} & -18512.6189 \\
\text{Heat Of Formation} & {-37.43} & {-3.0101
} & 23.96 \\ \hline
\end{array}
$$
Is this a reliable software for calculating the resonance energy as there is a difference about $\pu{5 Kcal/mol}$ or I am doing some mistake by overlooking any concept?

You are modelling hydrogenations, but I do not see any hydrogen in your equations. Even in quantum chemistry you have to use stoichiometric equations.

I think "a" is the energy difference between 2 C-H bonds and C=C Pi bond.

A reaction you can model is $C_6H_{10} + H_2 \to C_6H_{12}$. But you have to consider the hydrogen, so $a = E_\text{SCF}(C_6H_{12}) - E_\text{SCF}(C_6H_{10}) - E_\text{SCF}(H_2)$.

Also keep in mind that you have calculated energies. When comparing with (free) enthalpies measured by experiment you have to consider this.

Ok got my mistake in case of SCF energy.I earlier assumed that SCF of H2 is 0. Leave it.
My question is regarding the "heat of formation", at least heat of formation of H2 is 0. So it should give precise results.


text:
Equilibrium, temperature change, and prediction in constant volume
When it's constant volume, change of $T$ (Kelvin temperature) in a reaction equilibrium system won't change the volume. Let's say we change the temperature from $T$ to $2T$ of this reaction at equilibrium
$$\ce{3A + B -> C + D}$$
Because of the change of temperature, $K_\mathrm{p}$ will change, depending on whether the reaction is exo- or endothermic. At the same time, because of constant volume and change of temperature, reaction quotient $Q_\mathrm{p}$ at $2T$ is newly formed (originally $Q_\mathrm{p}(T)=K_\mathrm{p}(T)$ in equilibrium turn to $Q_\mathrm{p}(2T)$)
Here's a question. Say, this reaction is exothermic, and $T$ is changed to $2T$. So, $K_\mathrm{p}$ should be lowered, and with $K_\mathrm{p}$ lowered, reaction also proceeds to reverse. This can also be explained by Le Chatelier's principle.
But when it is at constant volume, $Q_\mathrm{p}$ could newly form simultaneously. And if $Q_\mathrm{p}(2T)<K_\mathrm{p}(2T)$, the reaction could proceed forward even though it's exothermic with the increase in temperature.
I've done some math and it seems it only depends on how big $\Delta H$ is. So can that really happen? Or am I missing something? If what I'm saying is right, is this an expception of the Le Chatelier's principle?
(I've explained everything here, but I can explain with an example if anything is still unclear)

I don't understand what you mean by "$Q_\text{p}$ could newly form simultaneously".

I think the problem with this particular case is that a change in temperature at constant volume will also increase the total pressure, thus according to Le Chatelier's, the equilibrium will shift towards the side with less moles of gas (products in this case). Depending on numbers, this effect might be more predominant than the reverse shift for it being an exothermic reaction.

For this particular scenario, the system in equilibrium satisfies (assuming ideal gases):
$$K_{\text{p}}^{(T)} = \frac{p_\text{C} p_\text{D}}{p_{\text{A}}^3 p_\text{B}} = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(T)\,2}}}$$
The equilibrium constant at temperature $2T$ can be calculated as:
$$K_\text{p}^{(2T)} = K_\text{p}^{(T)} \exp \left(\frac{\Delta H}{2RT}\right)$$
After increasing the temperature at constant volume, a new total pressure, $p_\text{tot}^{(2T)}$ is attained. Since the molar fractions are not changed at the beginning, the reaction quotient is:
$$Q_{\text{p}}^{(2T)} = \frac{p_\text{C} p_\text{D}}{p_{\text{A}}^3 p_\text{B}} = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(2T)\,2}}}$$
So, for the reaction to proceed forward, we want $Q_{\text{p}}^{(2T)} < K_\text{p}^{(2T)}$, that is:
$$\frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(2T)\,2}}} < K_\text{p}^{(T)} \exp \left(\frac{\Delta H}{2RT}\right) = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(T)\,2}}} \exp \left(\frac{\Delta H}{2RT}\right)$$
Simplifying:
$$\frac{1}{{p_\text{tot}^{(2T)\,2}}} < \frac{1}{{p_\text{tot}^{(T)\,2}}} \exp \left(\frac{\Delta H}{2RT}\right)$$
But $\displaystyle p_\text{tot}^{(T)} = \frac{n_\text{tot}RT}{V}$ and $\displaystyle p_\text{tot}^{(2T)} = \frac{2n_\text{tot}RT}{V}$, so:
$$\frac{1}{4} < \exp \left(\frac{\Delta H}{2RT}\right)$$
$$\Delta H > - 2RT \ln{4}$$
As long as $\Delta H$ satisfies this relationship (which will always imply a range of exothermic reactions since $T$ is always positive), this reaction will proceed when increasing the temperature at constant volume.
Notice that if this was a reaction where $\Delta n = 0$, the $p_\text{tot}$ cancels out and we get:
$$1 < \exp \left(\frac{\Delta H}{2RT}\right)$$
$$\Delta H > 0$$
Which implies that, when change in number of moles is not a factor to consider, the equilibrium will only shift to the right upon temperature increase at constant volume for reactions with $\Delta H > 0$, i.e. endothermic reactions.

That is exactly what I was looking forward to. Thank you very very much for your answers and solutions. I am also amazed by chemistry itself again and your skills of handling equations too.

Could I ask you one more slight thing? In the situation above, let's say we have q, reaction heat. Since it's constant volume, this q is equal to delta E( internal energy) and delta H is equal to q + delta n(mol)RT ? thanks again...

Whether it is at constant volume or not, the relationship $H = U + PV$ is the definition of $H$. So, $\Delta H = \Delta U + \Delta (PV) = \Delta U + \Delta (nRT)$ (for an ideal gas). If the process is at constant temperature, then $\Delta H = \Delta U + RT\Delta n$. If the volume is also constant, then $\Delta U = q$ as you mentioned, so $\Delta H = q + RT\Delta n$. As a side note, notice that the condition of spontaneity at constant temperature and volume is now $\Delta A < 0$, and not $\Delta G < 0$.


text:
What is the volume of the produced SO2 gas at STP?

In a electricity producing plant $3.1\times10^7\ \mathrm{kg}$ coal with $2.4\ \%$ rhombic sulfur is burnt every year. What is the volume of the produced $\ce{SO2}$ gas at STP?

My effort  for solving this answer :
In one day how much sulfur is burnt. Then I wrote the equation
$$\ce{S + O2 -> SO2}$$
With the unitary method I tried but I am confused from where to start, how to start? Please give me a hint how to start the answer.

You have an excess of oxygen, right...?
So the limiting reactant is the sulfur, which is $2.4$% of the coal.
Spoiler:
Step 1:

 So calculate the sulfur content, which is $3.1 \times 10^7$ kilograms multiplied by $2.4 \times 10^{-2}$ which is $7.44 \times 10^5$ kilograms of sulfur.

Step 2:

 The $\ce{SO_2}$ is $50$% sulfur by mass (32 amu of sulfur vs. 64 amu total) so there is $1.488 \times 10^6$ kilograms of $\ce{SO_2}$ produced.

Step 3 (for moles, optional):

 If you want moles, divide by $64$, which is $2.325 \times 10^4$ moles (someone check my math)

Then again, you emphasized STP, so proceed with caution (sulfur doesn't react at $20$ degrees Celsius).

Thanks , I have started to think your hint , soon I will get my answer :)


text:
What is so special about positive hydrogen ions that they are used in pH calculation?
I just learned about pH. If I understand it correctly, all it really tells is the concentration of positive hydrogen ions (protons) there are in a given liquid.
Why is this pH scale entirely dedicated to positive hydrogen ions? What is so special about protons/hydrogen ions? I would expect the scale to would count all positive ions, even if they are of atoms different from hydrogen. I don't understand why is the proton ion a "superstar", being so special that people made a scale dedicated just to it.

There's also the $pOH$ scale as well, which works in the same way, but for $OH^{-}$ ions instead... The 'p' denotes $log_{10}$. They're interchangeable; at $25^{°}C$, $pH + pOH = 14$, just like $K_{w} = [H^{+}][OH^{-}]$. They're just logarithmic scales to make it easier to represent small numbers.

https://chemistry.stackexchange.com/questions/5047/on-the-scale-of-stuff-with-a-ph

What we're really talking about with the acidity of a solution is the proportion of water which is dissociated into hydroxide $OH^-$ or hydronium ions $H_3O^+$ (hydronium is usually a more accurate representation than a free proton).
Other ions do affect the acidity of a solution, but by inducing water to dissociate. Check out the example on the wiki page. Pure water alone has a $pH$, because of spontaneous dissociation of some water molecules into ions. The $pH$ of pure water is 7, because the $pK_a$ of water as a weak acid is 14.
Basically, water is the special thing, and its constituent ions happen to be hydroxide and hydrogen / hydronium. Those ions are what we measure with a $pH$ electrode, so the scales are named after them. Water has a concentration around 55 M, vastly predominating in aqueous solutions.

What is this pKa?

The pKa is the negative log of the acid constant, the equilibrium constant for an Arrhenius acid to give up its proton. The pKa of water is 14, the pKa of vinegar is ~4.8, HCl is -7 (a very strong acid).


text:
Why are acids and bases are defined as both donors and acceptors of species in the acid-base theories?
So, a Brønsted-Lowry acid is a proton donor, and a Brønsted-Lowry base is a proton acceptor. On the other hand, a Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor.
Why is it that in these acid-base theories, acids and bases are defined as both donors and acceptors of species? Or is there a part of acid-base theory I don't quite understand?

I honestly don't understand your question. How else do you want acids or bases to be defined as? Some ions will always be donated into the solution, and some will always be accepted from it. What other way is there for acids or bases to function?

These are primarily old theories that were kind of evidence or observation based. These are not completely correct theories as there are may loop holes in them. For eg. many types of compounds with significant acidic and basic characters cannot be predicted by these theories. Also, to a certain extent the Bronsted and Lewis concepts are complementary.Wikipedia says - ""

A Lewis base is often a Brønsted–Lowry base as it can donate a pair of electrons to H+; the proton is a Lewis acid as it can accept a pair of electrons. The conjugate base of a Brønsted–Lowry acid is also a Lewis base as loss of H+ from the acid leaves those electrons which were used for the A—H bond as a lone pair on the conjugate base.

and

In 1923, Lewis wrote An acid substance is one which can employ an electron lone pair from another molecule in completing the stable group of one of its own atoms. The Brønsted–Lowry acid–base theory was published in the same year. The two theories are distinct but complementary.

Basically, proton is a $H^+$ ion and we all know from our simple understanding that compounds that give $H^+$ or proton are acids and those that accept $H^+$ or proton are bases.

They're distinct alright, but complementary as mentioned. I prefer the Brønsted-Lowry theory though... It's easier to understand for me.

@KianStevens You know, its not absolutely fine I  would see. For eg. if you are reading Friedel Crafts Alkylation you would come across the need of a Lewis acid like AlCl3 or BF3. If you are acquainted with the Lewis theory, you would face there as AlCl3 is not predicted by the Bronsted Lowry concept.


text:
Clarification in a particular application of Le Chatelier principle
There is something I don't understand about a particular application of the Le Chatelier principle. The following system is in equilibrium: $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$ If I add some $\ce{H3O+}$ the reaction will shift left. Now if I keep adding $\ce{H3O+}$, it will keep shifting left until the $\ce{CH3COO-}$ would deplete? And would the $\ce{H3O+}$ will remain at its initial concentration? So I can't understand how a buffer work, because if you have a reaction like above and you are adding acid, the added $\ce{H3O+}$ and the $\ce{CH3COO-}$ will bind together to form $\ce{CH3COOH}$ and water thus keeping the $\ce{H3O+}$ exactly the same. Where am I wrong?

https://chemistry.stackexchange.com/questions/81192/why-doesnt-a-buffer-solution-change-phappreciably?rq=1

This question does not add any value for future visitors unless you clarify what was the need to bring in a buffer here, when the corresponding sodium ethanoate salt is not even present in the solution. Till then, I'm voting to close the question as unclear. You may always [edit] the question later to reflect changes.

@GaurangTandon There is no requirement of "sodium" for something to be a buffer.  Other spectator ions would do.  Why does OP need to include spectator ions to be understood?

@DavePhD Oh, I am sorry. I was about to write "metal ethanoate salt" or simply "ethanoate salt", but then substituted it with "sodium" for some reason. My fault. The reason why I said the question is unclear, instead, is that the OP says "The following system is in equilibrium:" but didn't ever say whether it was only acetic acid or acetic acid and its salt (only latter works). Moreover, the part after "So I can't understand how a buffer work..." simply wasn't clear. That said, I tried my best to answer the OP based on what I understood, and advised the OP to edit their question for clarity.

@GaurangTandon as long as there is some acetic acid and some acetate (some acid and some conjugate base) it is a buffer.  It don't think you should be distracted by the spectator ions, whether they be metal ions or quaternary amine ions or whatever.

@DavePhD Yes, I agree. There was no need to be distracted over the spectator ions.

For starters, I wouldn't classify your equilibrium as a buffer. A buffer requires excess weak acid, and it also requires either a strong base or the salt of the weak acid (which you don't have). The equilibrium you stated is the dissociation of CH3COOH into its conjugate base and a conjugate acid.

Say you had the following buffer:
$$CH_{3}COOH + NaOH ⇌ CH_{3}COONa + H_{2}O$$
The equilibrium of this would be:
$$CH_{3}COOH ⇌ CH_{3}COO^{-} + H^{+}$$
And the ionic equation would be:
$$H^{+} + OH^{-} → H_{2}O$$
As per definition, a buffer is a system that minimises pH change on small additions of acid or alkali. How? All down to equilibrium...
If you were to increase $[H^{+}]$:

In terms of the equilibrium, the added $H^{+}$ would react with the base, $CH_{3}COO^{-}$, to form
$CH_{3}COOH$. This is Le Chatelier's principle in the fact that increasing $[H^{+}]$ shifts the equilibrium to the left.

If you were to increase $[OH^{-}]$:

In terms of the ionic equation, the added $OH^{-}$ would react with the $H^{+}$ to form $H_{2}O$. This is Le Chatelier's principle in the fact that increasing $[OH^{-}]$ shifts the equilibrium to the right.

In both of these, it's clear that the addition of $H^{+}$ or $OH^{-}$ would just be an acid-base reaction. The products formed minimise the $pH$ of the system due to their relative strengths compared with the added acids/alkalis.
The $pH$ does change though, but just not as much as it would if it wasn't in the buffer system. A buffer just keeps the equilibrium central.

Your question is good-intentioned but there are quite a few things that would need clarifying.
First, the above setup is not a buffer. You need additional metal ethanoate for a buffer to form. The above dissociation is only of a simple weak acid dissociating in water.
Secondly, yes, adding more $\ce{H3O+}$ to the above solution will shift the reaction to the left ("suppress the dissociation of acetic acid") but it would still increase the overall concentration of $\ce{H3O+}$, contrary to "the $\ce{H3O+}$ will remain at its initial concentration" (note that the backward shift is almost always negligible compared to the moles of hydronium we add (corrected thanks to @TanYongBoon!)).
If this doesn't make sense, consider waiting a bit more for your course work to proceed towards completion. Because sometimes, certain problems automatically clear themselves as you proceed later in the coursework :)


text:
Best way to raise pH without alkali metal?
I have a process that requires pH~14-14.5 to work. I’ve been achieving this with concentrated NaOH solutions, but the Na cations are later causing side reactions which are detrimental.  Is there any organic base/non-alkali base that has the solubility in water and pKa to reach these pH values?

Do you mean non-alkali metal, i.e. non Group I metal? All "bases" are alkalis.

There's a whole lot of them, so better edit your question quickly, 'cause right now you could get either simple yes as an answer or you might find a whole book somewhere.

Did you try KOH? LiOH?

K and Li are Alkali metals...

Mithoron - I'd take a book. You got one?

Can you say what the process is? Very high pH requires very high concentration; very high basicity may be obtained with unique bases. Maybe there is a way to avoid the side reactions.

@tungstenbronze What I meant was, you said sodium is a problem, so why blame all alkali metals?  Are they all problems?

Tetramethylammonium hydroxide
This is a strong base like NaOH.
See Behavior of Polyethylene Oxide Based Nonionic Surfactants
in Silicon Processing Using Alkaline Solutions J. Electrochem. Soc. 1995   volume 142, 621-627:

alkaline cleaning solutions based on quarternary ammonium hydroxides such as choline ($pK_b = 0.1$) and tetramethyl ammonium hydroxide ($pK_b = 0$)

and Ceramic Nanomaterials and Nanotechnology II which says:

In this research, tetramethylammonium hydroxide (TMAH) was used, instead of KOH, to adjust pH of the precursor solution to 14.0

Also, TETRAMETHYLAMMONIUM HYDROXIDE Journal of the Chemical Society (1905):

An estimation of the strength of the base by means of the velocity
  of saponification of methyl acetate in N/80 solution showed that it was
  somewhat weaker than sodium hydroxide. The velocity constants obtained at 25 [degrees] were 0.0106 and 0.0115 respectively, so that if the
  strength of sodium hydroxide is represented as 100, that of tetramethylammonium
  hydroxide, under the above conditions, will be represented by 92.

@OscarLanzi this thesis says pKb is zero https://digital.library.unt.edu/ark:/67531/metadc822794/m2/1/high_res_d/dissertation.pdf

Now we're good.

@OscarLanzi I think Wikipedia is wrong.  Another reference says "In this research, tetramethylammonium hydroxide (TMAH) was used, instead of KOH, to adjust pH of the precursor solution to 14.0". https://books.google.com/books?id=Gb9RAAAAMAAJ&q=%22tetramethylammonium+hydroxide%22+complete+dissociation&dq=%22tetramethylammonium+hydroxide%22+complete+dissociation&hl=en&sa=X&ved=0ahUKEwjb9rnMnPLZAhVNyGMKHUjvCO8Q6AEISDAI

@OscarLanzi I added better references to the answer.

Water solutions of amines can get to pH+ 12 or so. Piperidine, $\ce{(CH2)5NH}$, is about pH 11.
To get much higher pH+ requires somewhat exotic fluorinated organic chemicals, such as $\ce{9-COO-Me-fluorene}$, $\ce{(4-Me-C6F4)2CHCN}$ or $\ce{(4-Me-C6F4)(C6F5)CHCN}$. See Paenurk et al, A unified view to Brønsted acidity scales: do we need solvated protons? and download the PDF for an interesting explanation.

You shouldn't make a "sentence link" especially with what looks like acids; you mean their salts or what?


text:
Is the Hartree-Fock procedure the same thing as the self-consistent-field (SCF) theory?
Is it right to think that the Hartree procedure is called the SCF theory, as the notion of self-consistent field comes from approximating multi-electron wavefunction as multiple one-electron wavefunctions. Since we don't know the final charge density, we must iterate until it is self-consistent?

These are two different questions, and really should be divided. (It's site policy to only ask one question at a time, in a given Question.) Please re-ask your second paragraph in a new question. Check out the [help] for more information on site policies.  Thanks for asking here -- welcome to the site!

HF and SCF are used interchangeably quite often (whether you agree with it or not).

To those who think this question is too broad: it has a concrete answer which is given below and it is definitely not too broad. It is a fundamental part of the theory that is not discussed in modern literature.

As an extension of HF and SCF being used interchangeably, I often see (and use myself) SCF to cover both HF and DFT, as the Kohn-Sham equations are solved using an SCF procedure that is identical to the one used for HF.

No.
The Hartree-Fock method encompasses a set of assumptions used to formulate and solve a series of equations embodying a mean-field, time-independent Schrödinger equation.
The self-consistent field method is a computational technique used in myriad applications where the equations to be solved are implicit—that is, where it is impossible to isolate an analytically solvable expression for a variable and its derivatives.  To the best of my knowledge, any solution method involving a "guess-and-check" approach can be considered a 'self-consistent solution method'.
The Hartree-Fock method is an example of a use-case for a self-consistent field procedure, but they are not synonymous.

Excellent answer. I would note that the author's confusion likely stems from old literature using the phrase 'SCF' when describing HF, as back then it was the first widely used SCF method. Nowadays, as you've pointed out, there are numerous SCF methods that are not HF.


text:
SCF procedure performed for some quantum-chemical methods, but not others. Why?
The SCF procedure is used to find solutions to the Schrödinger equation, and obtain energy of the molecular system. It uses a "guess-and-check" approach, where an initial guess is needed to solve the equation, giving the charge density, which can be used again and again until the results become self-consistent.
So, why don't some quantum-chemical methods perform SCF, while others like HF do?

Most computation methods write the wavefunction($\Psi$) in terms of Slater determinants ($\Phi$)
$\Psi = \sum_i^L c_i\Phi_i$
which in turn are made from MOs ($\phi$), which are linear combinations of basis functions ($\chi$)
$\phi_i = \sum_\mu^K C_{\mu i}\chi_\mu$
The molecular energy depends on the MO coefficients ($C_{\mu i}$) in a non-linear fashion, so if they are variationally optimized then you need to iterate (i.e. SCF). This is the case for HF (where $L$ = 1 and $c_1$ = 1) and MCSCF (where both $\{c_i \}$ and $\{C_{\mu i} \}$ are optimized).
For methods like CI, perturbation theory, and coupled cluster, the MO coefficients are not changed, so you don't need an SCF.

It can also be said more generally that there are two main methods for solving the Schroedinger equation: variation and perturbation. When you use variation, you alter some coefficients of basis "things" (basis functions in SCF, Slater determinants in CI or both in MC-SCF) to make the initial guess of WAVE FUNCTION for the solution more and more consistent with the Hamiltonian of the system. When you do perturbation it is the HAMILTONIAN itself which is gradually improved to arrive at more precise wave function (which should be already made consistent with the initial Hamiltonian, though). Perturbation method is very general and hence versatile, used not only for energy calculations; historically, it was the first expansion beyond the HF - DFT came in much later and does not compute the actual wave function, only the electron density (and also the energy therefore).
Answering @Umkay comment on whether DFT uses SCF:
DFT (more precisely, Kohn-Sham version of it) DOES perform SCF. The crucial point is that DFT uses a different approach to the description of the system than the usual SCF/perturbation methods. These methods try to find the most reasonable approximation to the wave function of the system; DFT instead takes notice of many properties being a direct function of election density instead of the wave function itself; and, as the density has only three arguments (x, y and z) instead of 3*N for N-electron wave function, it tries to go round and compute the density not from the "best-approximated" wavefunction.
It appears, though, that quite efficient way for this is to construct a fictional system which is defined to have the same density as the "true" one but has a wave function which is less undertaking to compute. Namely, this system have elections without direct interaction between them (i.e., without corresponding terms in the Hamiltonian) but, like in HF, having some self-consistent term describing their interactions with less computational burden. The difference with the HF is that KS-DFT uses this term to describe interactions between electrons which arguably can be described by such a term: the correlated motion of [distinct pairs of] electrons. In HF it is only the electrostatic field created by all the electrons which is described in this way (KS-DFT uses also that but not just that). The probable impossibility to have such CORRELATIONAL term is supported by the still ongoing search for the "best" density functional. Hence also the debate on whether DFT counts as an ab initio method. More importantly, it means that the orbital-related properties computed with DFT are quite nominally corresponding to those of the real system - DFT was simply not designed to get them, after all. On the other hand, one may say that to get density "good enough", you probably also need the "fictional system" close enough to the real one. For example, it was once shown by Stowasser and Hoffman in 1999 that for ionization properties KS-DFT orbitals look very similar to the "actual" orbitals for this case - the Dyson orbitals - computed with some high-level method.
But KS-DFT still makes use of SCF for computing the wave function of its "fictional" system. And there are the double-hybrid functionals which also employ perturbation approach. But these are not quite consistent with the initial formulation of DFT, because, as A. Becke noted in 2014 review, they use unoccupied orbitals which are merely by-products of employing SCF for DFT as they have no direct connection with the electron density at all.

Welcome to Chem.SE, Igors! Nice to see you made the hop over from CCL.

Since DFT does not compute the actual wave function, it doesn't perform SCF, correct? Because I read somewhere on this site that SCF covers both HF and DFT, as the Kohn-Sham equations are solved using the SCF procedure identical to HF.

@hBy2Py: I just got to read the CC Daily mail and saw that I am capable of answering :-)

@Umkay: I have edited the answer itself, I hope it does rely to your comment.


text:
Most basic nitrogen in Adenine
I need to find the most basic site of Adenine:

The $\ce{NH2}$ group and $\ce{NH}$ cant be strong base as electrons are delocalised. This leaves us with $1,7,3$. I thought that electron density on $1,3$ nitrogens is enhanced due to $+R$ effect of $\ce{NH2}$.
But is this correct, I still cannot decide which is the strongest basic site.
To add: Both $1,3$ nitrogen have two nitrogen in neighbourhood which destablise their conjugate acids by $-I$ effect. So still I can't decide.

I wonder if this could get a proper answer. See this related question about purine. Theoretical comparison was near to impossible in that case.

@Gaurang in that we can argue $1$ is farther away to the two $N$ at $7,3$ but $3$ is closer to $9,1$ that makes it easy to say  conjugate acid of $3$ is more destablised by $-I$ effect. So we can say $1>7,3>9$ but still not sure about $7,3$

Martin here analysed purine  at the DF-B97D3/def2-TZVPP level of theory for such info and still put a follow-up question because he felt his analysis may be not good enough. I can assure you won't get here anything better.

According to Ref.1:

Adenine undergoes two ionization reactions. Density function calculations in the gas phase carried out by Russo et al. (1998) indicate that protonation occurs preferentially on the N1 nitrogen of the 6-member ring, in good agreement with the assignment given by Christensen et al. (1970b). The anionic species with a proton-deficient 5-member ring is the most stable.

Here, "Russo (1998)" is the Ref.2, which says (emphasis added):

HOMO of adenine is formed essentially by the lone pairs of all nitrogens. The orbital of the amine group contributes with the higher coefficient, and that of $\ce{N}$1 with the lower one. On the basis of this information, $\ce{N}$1 protonation might be the less favorable process. On the contrary, the molecular electrostatic potential map (Fig. 5) indicates that $\ce{N}$1, $\ce{N}$3, and $\ce{N}$7 are the better candidates for proton attachment. In particular, the $\ce{N}$1 and $\ce{N}$3 regions appear to be favored over $\ce{N}$7. Experimentally, the preferred protonation sites of adenine are $\ce{N}$1 and $\ce{N}$3 with the former slightly favored.[reference 18] Our computation favors protonation at $\ce{N}$1 over those at $\ce{N}$3 and $\ce{N}$7 by $1.12$ and $\pu{7.38 kcal mol-1}$, respectively. This trend is in agreement with previous theoretical computations.[reference 17] From bond orders analysis (see Fig. 6), and due to the small energy difference it is possible to view the $\ce{N}$1 and $\ce{N}$3 protonated forms in $\ce{N}$1 the double bonds occur in $\ce{C}$4—$\ce{C}$5 and in the $\ce{C}$2—$\ce{N}$3 positions, whereas, in the second, in the $\ce{C}$4—$\ce{C}$5 and $\ce{N}$1—$\ce{C}$2 positions as two resonance hybrids of a unique structure with a consistent charge delocalization. This observation allows better insight into the reason for which the protonation on the five-membered ring is unfavorable. In fact, as shown in Figure 6, the positive charge appears to be distributed only on the $\ce{C}$8 and $\ce{N}$9 atoms, because it is evident that no interactions exist between the two rings upon protonation.

Note: "Christensen" of the Ref.1 is the same as reference 18 of the Ref.2, and is listed here as Ref.3. Reference 18 mentioned in the Ref.2 is listed here as Ref.4.
References:

Erik Balodis, Melerin Madekufamba, Liliana N. Trevani, and Peter R. Tremaine, "Ionization constants and thermal stabilities of uracil and adenine under hydrothermal conditions as measured by in situ UV–visible spectroscopy," Geochimica et Cosmochimica Acta 2012, 93, 182-204 (DOI: https://doi.org/10.1016/j.gca.2012.06.023).
Nino Russo, Marirosa Toscano, André Grand, and Franck Jolibois, "Protonation of thymine, cytosine, adenine, and guanine DNA nucleic acid bases: Theoretical investigation into the framework of density functional theory," Computational Chemistry 1998, 19(9), 989-1000 (DOI: https://doi.org/10.1002/(SICI)1096-987X(19980715)19:9<989::AID-JCC1>3.0.CO;2-F).
James J. Christensen, J. Howard Rytting, and Reed M. Izatt, "Thermodynamic pK, ΔH°, and ΔS° and ΔCp° values for proton dissociation from several purines and their nucleosides in aqueous solution," Biochemistry 1970, 9(25), 4907–4913 (DOI: https://doi.org/10.1021/bi00827a012).
Janet E. Del Bene, "Molecular orbital study of the protonation of DNA bases," J. Phys. Chem. 1983, 87(2), 367–371 (DOI: https://doi.org/10.1021/j100225a040).

Can it be explained on the basis of Hydrogen bonding? We need to compare 1,3, 7. In 3 and 7 hydrogen bonding is possible while its not possible in the case of 1 so 1 is most basic site?

@Abcd since the calculations are in the gas phase, I don't see how hydrogen bonding can be involved.

Hydrogen binding is possible in the gas phase, ain't it?\

@Abcd definitely possible, but what molecule do you propose is hydrogen bonding to Adenine?

I was referring to intramolecular hydrogen bonding.

@Abcd sorry, I'm not understanding.  How is intramolecular hydrogen bonding possible at N-3 and N-7?

Here: https://imgur.com/lGzlNi8 ... Similarly its possible for N7 also

@Abcd personally, I think the angles are too awkward for that to happen.  You could ask another question about whether that happens.  Maybe someone else knows better.

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


text:
Why was rhenium used as an igniter in flash bulbs?

It looks like it typically had a current passed though it to ignite zirconium threads.
What properties of rhenium made it the best metal to use for this?

Well, I think this is a dumb question, but what is a "flashbulb"? Are you asking about the photography flashbulb one? But, then that WP page doesn't mention "rhenium" anywhere...

@GaurangTandon For the youngsters out there: there was an era when photographs were taken using photographic film. For much of this era electronic means of producing bright light (xenon electronic flash devices and white LEDs) were rare. So people used chemical means of creating flashes of bright light (originally magnesium powder, later a variety of enclosed bulbs containing things like zirconium wire the could be ignited electrically or mechanically). Some of these clearly used rhenium igniters. ;-)

Given the price of rhenium, it is quite likely the igniter is an alloy. However, Wikipedia does suggest that rhenium filaments for, e.g. mass spectrometers, are more stable then alternatives in air/oxygen atmospheres. Combined with the refractory nature, one could imagine that getting the filament really hot really fast in air (so the zirconium threads burn rapidly) is pretty stressing on the filament, and rhenium would be a good choice.

A friend suggested that perhaps rhenium was chosen for its spectral output when burned. These bulbs were around at the beginning of color photography, so good color rendering index would have been desirable. I will need to test the spectrum of these compared to, say, burning iron or aluminum!

Rhenium is also an extremely rare and expensive metal. This type of flashbulb was single use, disposable; and professional photographers used them by the dozens. So any rhenium content would have been extremely small. Something like a plating on another element. Now, most lightbulbs are either evacuated, or filled by an inert gas. Flashbulbs are instead filled with oxygen, to get the brightest possible burn from their fuel. Now here I speculate: prolonged storage in an oxygen atmosphere would likely be corrosive, leading to failures. The rhenium may be an anticorrosion plating for reliability.

New tungsten alloys containing rhenium were developed as a filament material in flashbulbs for the purpose of improving their reliability. Weak batteries (anything but rare in the carbon-zinc era) could easily result is a slightly-delayed flash or no flash at all, spoiling the photo in either case.
An important characteristic of batteries is their internal resistance. This impedance is small in a fresh dry-cell battery but increases as the battery wears out, eventually reaching a value high enough to make the battery unusable. The rhenium acts to "increase the cold resistance of tungsten wire filament so as to permit sufficient power to be obtained over large variations in impedance."
As the prior post suggested, anti-oxidation properties of the rhenium also helped greatly  reduce the chance of a filament burn-out and consequent flash lamp ignition failure.
Because photography wasn't cheap back in the 50's and 60's, the vast majority of cameras and flash units sat in closets at least nine-tenths of the time. Even unused, their carbon-zinc batteries could go bad in three years. Considering the many millions of flash  pictures attempted with nearly-dead batteries in that era, this reliability feature looks like an excellent idea.
For details, please see United States Patent No. 3,123,993 granted 3/10/1964 to George W. Cressman, Chagrin Falls, and Louis A. Demchock, Jr., Cleveland, Ohio, assignors to General Electric Company.


text:
Sigma bond behaving as stronger nucleophile despite presence of pi bond?
In my book, the following reaction is given

The solution states that the sigma bond between hydrogen and carbon (3 degree) in isobutane attacks $\ce{H+}$ from $\ce{HF}$ to generate tert-butyl carbocation. This carbocation then adds on to ethene to generate another carbocation followed by some rearrengements until the given product is obtained.
My doubt is, since pi bonds are stronger nucleophiles (because of loosely held electrons) then why in this reaction the carbon-hydrogen sigma bond attacks $\ce{H+}$ instead of $\ce{HF}$  adding to the pi bond?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Ref. Q.26 MS Chouhan - Alkenes


text:
First and last unit of polyethene
Is a hydrogen atom bonded to the first and the last unit of polyethene?
If yes, where does the hydrogen come from?

There are several hydrogens on every unit of polyethene. And they all come from the original monomer ethene. So, sorry but I don't understand your question? What else are you trying to ask? Did you have a look at polyethene's structure?

There may be a hydrogen, or oxygen, or whatever other atom attached to the first and the last unit of pretty much any polymer. It comes from nowhere, and it does not matter. Ditto for covalent solids. Do you know the structure of graphite or diamond, for example?

Yes, in The Characterization of Linear Polyethylene SRM 1475. II. Determination of Total Methyl Content by Infrared Spectrophotometry and references cited therein, it is confirmed that the polymer chains terminate with methyl groups (the repeating unit plus an additional hydrogen as you phrase it).

Oh, but isn't the last carbon atom (before the terminal methyl group) bonded to two hydrogen atoms? (apart from being bonded to the terminal methyl group and the previous carbon of the chain)

@GaurangTandon yes, the second and penultimate carbons have 2 hydrogens each and the first and last carbons have 3 hydrogens each.


text:
Which of the following has maximum nucleophilicity?
Question

The answer should be (D) as the inductive effect of 3 methyl groups increases negative charge on oxygen and thus makes it a good nucleophile.But the answer is (A).

D is the worst nucleophile because of the steric hindrance

According to me, the answer should be (C) in general cases but this might get changed to A) in special cases. The best way to judge their nucleophilicity order  is to check their basicity. The weak bases are generally better nucleophiles in polar protic solvents . The acidity order of the  conjugate acids are $$\ce{CH_3OH < (CH_3)_3COH  < NH_3 < CH_4}$$So, the strongest acid is $\ce{CH_3OH}$ and weakest acid is $\ce{CH_4}$. So, weakest base is $\ce{CH_3O^-}$  and strongest is $\ce{CH_3^-}$. So, the maximum nucleophilic character should be of $\ce{CH_3O^-}$,not generally $\ce{CH_3^-}$ as basicity is inversely proportional to nucleophilicity in general cases. Coming to D), the inductive effect of three methyl groups actually decreases the acidity of tert-butyl alcohol. So, the tert-butoxide ion is more prone to take $\ce{H^+}$ rather than to nucleophilically attack.  It also creates steric crowding, and preferably takes easily ionisable hydrogen ion if possible from the compound.So,  looking only at increased $\ce{+I}$ effect doesn't tell that its nucleophilicity is increased. About A), this $\ce{CH_3^-}$ is usually seen in Grignard reagents, whose basic activity is much much greater than its nucleophilic activity. If acidic hydrogen is present in the compound, it preferably takes that. But, if a compound doesn't possess any acidic hydrogen (special case), then its nucleophilic character is more as being less electronegative than $\ce{O}$, it is very less reluctant to keep yhe lone pair on itself like the case in $\ce{CH_3O^- or (CH_3)_3CO^-}$. Thus in the absence of any acidic hydrogen $\ce{CH_3^-}$ is better nucleophile. So, from the above discussion, it can be concluded that in many of the cases, where the attacked compound doesn't possess any acidic hydrogen, A) is maximum nucleophillic. But in the opposite  cases, C) is maximum nucleophilic.

Nucleophilicity depends upon many other factors other than electron density . A predominating effect is that of steric crowding.
The tertiary butyl carbanion is  overcrowded this restricts the ability of the nucleophile to donate its electron pair.Thus making it a bad nucleophile.

Finding out the best nucleophile from the remaining options is pretty simple.
Over here we will use electronegativity as the primary factor for determining the nucleophillic order.
More electronegative an element is stronger the pull on the electrons.
Thus the methyl carbanion bearing the negative charge on the carbon is the best nucleophile.  (Carbon is the least electronegative element between carbon,nitrogen and oxygen )

Note
MasterOrganicChemistry has a well written post, differentiating between nucleophillicity and basicity

What about (C)? It doesn't have crowding and has CH3 attached to it which further increases its nucleophilicity.

@IshaRoy Oxygen is more electronegative than carbon so the negative charge is more strongly bound to oxygen(option C) than carbon (option A)

Basicity is not a subset of nucleophilicity.

https://chemistry.stackexchange.com/questions/20061/whats-the-difference-between-a-nucleophile-and-a-base

Edited it .  ......


text:
What is the charge of the amino acid lysine at pH 2?

What is the charge of the amino acid lysine at $\mathrm{pH}=2$? $\mathrm{p}K_\mathrm{a}$'s are carboxylic: 2.18; amino: 8.95; side chain: 10.79.

The side chain of lysine is an amine group, so I assume that when fully protonated it has positive charge ($\ce{NH3+}$).
At $\mathrm{pH} = 2$, everything will be protonated because their $\mathrm{p}K_\mathrm{a} > \mathrm{pH}$.
This gives a net 0 (carboxy) + 1 (amino) + 1 (side chain) = +2 charge. Why do sites such as this say that at $\mathrm{pH} = 2$, lysine's charge is only +1, not +2?
There's probably something really simple I don't know. If anyone can help point it out, that would be great.

At $\mathrm{pH}=2$, the charge will be:
(charge near neutral pH, 2 groups positive, one negative) + (the fraction of the pKa = 2.18 group that is protonated)
=1 + (the fraction of the $\mathrm{p}K_\mathrm{a} = 2.18$ group that is protonated)
$=1 + 1/(1 + 10^{\mathrm{pH}-\mathrm{pK}}) =1 + 1/(1 + 10^{2-2.18})=1.60$
So the charge is +1.60, which is closer to 2 than 1, as you are saying.

Well it worth to pointing that's average charge and modal value is +2

I'll mark this answer as correct in a second, but if +2 is the average charge, then what is 1.6?

@LittleDragon 1.60 is the average charge.  1.60 is the charge that you would measure by electrophoresis.

[OP] This gives a net 0 (carboxy) + 1 (amino) + 1 (side chain) = +2 charge.

This is approximately correct. See DavePhD's answer for a more accurate treatment.

[OP] Why do sites such as this say that at $\mathrm{pH} = 2$, lysine's charge is only +1, not +2?

The table you cite is for proteins. When lysine gets incorporated into a protein, it forms peptide bonds with its carboxylate and its main chain amino function groups. Therefor, these groups no longer exist and don't get protonated or deprotonated (unless the lysine is the C-terminus or the N-terminus, in which case the table gives the wrong value). In short, the table is about the charge state of the side chain, not the original free amino acid.

[question cited] What is the charge of the amino acid lysine at $\mathrm{pH}=2$? $\mathrm{p}K_\mathrm{a}$'s are carboxylic: 2.18; amino: 8.95; side chain: 10.79.

The $\mathrm{p}K_\mathrm{a}$ values given are correct for lysine in water. In a protein, they might differ depending on the chemical environment of the specific lysine residue you are looking at. They would also depend on ionic strength and on cosolvents, if any.


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How do quaternary ammonium salts catalyse SN2 reactions?
I am asked to identify the "best catalyst" for the reaction:
$$\ce{CH3(CH2)8CH2Br->[\ce{KCN}][\text{benzene}]CH3(CH2)8CH2CN}$$
The options are (A) benzyl chloride (B) aniline (C) N,N,N-trimethylbenzylamine chloride, and (D) N-phenylethanamide.
Now, at first, I immediately thought that the catalyst chosen may have to prevent a possible ring formation, given there are already ten carbon atoms. However, I am not sure how any of those catalysts help prevent that (I doubt if a ring would even form though, because there isn't any nucleophile, but those ten carbons look suspicious!)
Anyway, I realized that this catalyst may have to be polar aprotic for the given SN2 to occur. But, all of the given catalysts are pretty protic (resonance delocalisation of the benzyl carbanion with the ring) I would rank them on proticity as (D) > (C) > (B) > (A).
Hence, although all are protic, I may make a vague guess that (A) is the least protic and hence the answer. However, the answer given is (C) and the solution instead states that:

Solvation of nucleophile is very less in presence of quaternary ammonium salt.

So, how? Where did the part about acidity go? And why worry about solvation? There doesn't appear to be any factor related to it at all (I read from here). Even so, how exactly would quaternary ammonium salts be poorer "solvators" than the others, say option (A)? Can I even predict this theoretically?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q11 in Amines

Trimethylbenzylammonium chloride is in this reaction a good example of a phase transfer catalyst. It first forms the trimethylbenzylammonium cyanide in aqueous phase (or on the surface of solid phase) which is soluble in the organic phase. Because of the polar nature of the cyanide ion it is not going to be heavily solvated by organic solvent (in this case benzene) molecules so the result is 'naked' cyanide which is highly reactive. This reacts with the substrate forming, in this case, the cyanodecane and trimethylbenzylammonium bromide. The trimethylbenzylammonium bromide is then free to migrate back to the aqueous phase and exchange anion to reform trimethylbenzylammonium cyanide. Most phase transfer reactions benefit from vigorous stirring to accelerate the exchange between the phases.
Further reading in this PDF.

Interesting, thanks for your answer! Can you comment on why the other options are not good examples of phase transfer catalysts? Is it possible to compare this property theoretically? Also, which page was I supposed to read in that PDF?

Benzyl Chloride - not good as it is an alternative substrate for the nucleophile. Aniline and phenylethanamide - not good as it is not soluble in aq. phase or able to draw cyanide ions into solution from solid phase.  Section 1.2.3 of the PDF mentions the concept of the 'naked' ion. I am not aware of any theoretical model to predict PT catalytic property - this does not mean it does not exist


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Decarboxylation of cyclohex-1-ene-1,2-dicarboxylic acid by electrolysis method

What is the major product when cyclohex-1-ene-1,2-dicarboxylic acid is electrolysed?

My try:

I got 2 cases but I am unable to decide which product is major, and need help here.

From Wikipedia,

Cyclooctyne ($\ce{C8H12}$) is the smallest cycloalkyne capable of being isolated and stored as a stable compound

This is because, again quoting the simple language:

Because of the linear nature of the $\ce{C–C≡C–C}$ alkyne unit, cycloalkynes can be highly strained and can only exist when the number of carbon atoms in the ring is great enough to provide the flexibility necessary to accommodate this geometry. (self-emphasis)

Hence, your case I is invalidated.
Regarding case II, the middle four-member ring has $\ce{4\pi}$ electrons in conjugation in a planar system. By Hückel's rule, this ring will be anti-aromatic, and therefore, will not exist.
Therefore, we can say that both the cases are not favorable.

Correction: I initially mentioned that the electrolysis is probably unfavorable, both due to the theoretically unfavorable molecules I and II, and a lack of favorable research data on Google Scholar for "electrolysis of "cyclohexene-1,2-dicarboxylic acid"" or "decarboxylation of "cyclohexene-1,2-dicarboxylic acid"". However, Avnish's answer demonstrates dimerization of cyclohexyne to a more stable form, though not exactly obtained from electrolysis, but which seems very plausible. So, I am compelled to say - at least with this much data - that the electrolysis seems favorable.
With that said, however, case I and II are certainly still not the major products (not even minor).

Although I do not have any reliable reference, I would guess that almost certainly the bond lengths in the central 4 membered ring would distort (the lengths of the two bonds coinciding with the 6 membered ring will be shorter than the other two and will have much higher double bond character than the other two bonds of the four membered ring), hence avoiding conjugation and anti-aromaticity; it will rather be more towards non-aromatic.

Turns out a similar reaction has been performed,   the reaction proceeds via the following mechanism

SOURCE

The cyclohexyne dimerizes to 1,2,3,4,5,6,7,8-octahydrobiphenylene. Then further rearranges. So I would say that electrolysis is feasible with rearrangement as depicted in the picture .
 The second compound(1,2,3,4,5,6,7,8-octahydrobiphenylene) which you've drawn is more stable since very unstable compounds (cyclohexyne) do not form even more unstable compounds. The second compound will not be the final product/major product owing to its high instability due to anti aromaticity.

PS
 I haven't delved into the details regarding the instability of cyclohexyne and 1,2,3,4,5,6,7,8-octahydrobiphenylene since they have been well covered in Gaurang's answer

Without data on the actual products, no serious prediction can be made. Therefore I will make a wild guess:
The diradical, if it formed would be vinylic: very reactive. It would abstract hydrogen atoms from allylic positions of the unreacted cyclic diacid dianion so the first product would be 1,3-cyclohexadiene.
More likely, the electrolysis would produce a monoradical which would pick up a single hydrogen to give cyclohexene 1-carboxylic acid as a first product. If this became the major product and was decarboxylated in a second step, it would also give 1,3-cyclohexadiene.
However, with radicals being formed from a starting material that has allylic hydrogens, I predict that the major product will be tar. Poly-tar.


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Deriving a relation between entropy, volume and particles through the chemical potential
I'm a teaching assistant on a first year course on Introduction to Thermodynamics based on the book "Molecular Driving Forces" by Dill & Bromberg.
There's a question which asks the students to derive the following relation between S, V and N:
$S(V,N) = Nkln(V)$
A simple solution that is hinted is to use the differential form of the fundamental entropy equation and realizing the following identities (assuming an ideal gas):
$\frac{P}{T} = (\frac{\partial S}{\partial V})_{U,N}=\frac{Nk}{V}$
$ds = Nk\int \frac{dV}{V} \rightarrow \Delta S = Nkln(\frac{V_2}{V_1})$
Now, I've meditated about and have been asked by several students the following question: could we arrive to the same (or equivalent) expression through a chemical potential route? After all, we need an expression that relates S both to N and V. Unfortunately, I have not been able to give a satisfactory answer.
Here's what I've tried, arriving to some ugly looking equation that intuitively looks incorrect to me:
Use the definition:
$\frac{\mu_j}{T}=-(\frac{\partial S}{\partial N_j})_{U,V,N_{j\neq i}}$, which also stems from the fundamental differential entropy equation.
Now I make the assumption that the chemical potential is equal to the density of molecules in the ideal gas. This is what I'm most dubious about, although I couldn't think of another way in which to have a relation to it. If that holds, I can then write the following identities:
$\mu = \frac{n}{V}$ then $P=\mu kT$
$\frac{-P}{T^2k} = \frac{\mu}{T} = (\frac{\partial S}{\partial N_j})_{U,V,N_{j\neq i}}$
Now because $P = \frac{nkT}{V}$,
$\frac{-P}{T^2k} = \frac{-nkT}{VT^2k}=\frac{-n}{VT}=(\frac{\partial S}{\partial N_j})_{U,V,N_{j\neq i}}$
Which I think can be solved:
$\int dS = \int \frac{-n}{VT}dn \rightarrow \Delta S = -\frac{\Delta n^2}{2VT}$
This result doesn't really make much sense to me (a negative entropy (?)). I'm a bit confused, because the first 'route' seems so simple but this second attempt looks fairly involved.
So, what I wrong in my logic? Apologies beforehand if I'm doing something egregiously wrong, it's been awhile since I've had to solve these 'pen and paper' problems.

I'm not sure how to help, but I'm sure that $\int n , dn \neq \Delta n$.

Also, I think that using this relationship between chemical potential and entropy won't lead to your result, since this relationship holds $V$ constant, and you want to get to entropy as a function of a variable $V$ and $n$. But, that I might have wrong.

@ralk912 d'oh! Thanks for the heads up on the integral, will update the correct result now.
Also, as far as I'm aware, isn't that what's happening on the first correct solution to the problem? Where $N$ is being treated as a constant, although $S$ depends on both $V$ and $n$.

Well, $N$ is treated constant because it's for that particular case (i.e. this is valid for gas expansions/compressions). If both $N$ and $V$ were to vary, then the first derivation wouldn't be valid. You're trying to get to an expression where $N$ is constant and $V$ varies from one where $V$ is constant and $N$ varies.

Also, from a dimensional analysis, you're assumption says that $\mu$ has $\pu{mol dm-3}$ units, while the definition implies the correct $\pu{J mol-1}$ units for chemical potential. So I think that assumption is not valid.

I see your point, after all the chemical potential is a kind of energy (the Gibbs free energy at constant $T$ and $P$), and not the density. I just used that assumption because I could not think of another way to have the chemical potential appear in the general gas equation. Also, the final expression I've arrived to implies that the entropy becomes more negative as the number of particles increases for a fixed $V$ and $T$, which does not resonate well with my chemical intuition. After all, shouldn't more particles in a system have more dofs to explore and increase the entropy?


text:
Help with a reaction: naphthalic anhydride to diol with LAH
First time asking a question but I've been trying to perform this reaction:

for quite a while now. Most of the times I've only gotten this far:

Here is the procedure I've been following:
Dry THF (20 mL), LiAlH4 (6 mmol) and anhydrous ZnCl2 (3 mmol) were added in a 100 mL round‐bottom flask fitted with a CaCl2 guard tube and mounted over a magnetic stirrer. 1,8‐Naphthalic anhydride (0.99 g, 5 mmol) was added into the flask slowly at room temperature. After 6 hr, the excess LiAlH4 was destroyed by addition of 5 mL water, and the mixture was adjusted with 1 N HCl to weak acidity. The resultant two‐phase solution was extracted with ethyl acetate (3 × 15 mL), washed with brine and then dried over anhydrous MgSO4. After the solvent was evaporated, the crude product was recrystallized with benzene and methanol (1/1) to give white crystals 0.790 g (84.1% yield). M.p.: 161–163°C (lit.2c Boekelheide, B. and Vick, G. K. 1956. A synthesis of pleiadiene and acepleiadylene. J. Am. Chem. Soc., 78: 653
 [Google Scholar]
 157–158°C); ESI‐MS (m/z): 171 (M + H − H2O)+, 189 (M)+; IR (cm−1): 3345, 3241, 1601, 1510,1425, 1366, 1320, 1058, 1035, 1013, 998, 826, 774.724; 1H NMR (600 MHz, CD3COCD3), δ4.54 (s, 2H), 5.27 (s, 4H), 7.44 (t, 2H, J = 7.5 Hz), 7.64 (d, 2H, J = 7.2 Hz), 7.87 (d, 2H, J = 7.8 Hz).
from this paper:
https://www-tandfonline-com.proxy.bc.edu/doi/full/10.1081/SCC-120038510?scroll=top&needAccess=true
I've tried using ZnCl2, but it's hygroscopic so picks up water in the atmosphere and destroys my LAH (this is my hypothesis). I also tried refluxing for 8 hours and again only got the second product before my THF evaporated. Do you guys have any suggestions? Thanks in advance.

ZnCl2 is commercially available in ether soln under sureseal. Worth trying this rather than solid ZnCl2

The above option is a good one. Also, how are you drying your THF? Have you tried adding more equivalents of $\ce{LiAlH4}$?

My THF comes from an SPS. It takes me about 30 seconds to go from SPS to nitrogen-flushed two-neck with a dry needle. I'll certainly try both suggestions though.


text:
Do amides oxidize to double-bonds?
Do amides from primary or secondary amines ever oxidize to a double-bond on the nitrogen? For example, like from an $\mathrm{R \bbox[yellow]{\color{red}{-}}(NH)-(C=O)-R'}$ to an $\mathrm{ R \bbox[yellow]{\color{red}{=}}N-(C=O)-R'}$?
Seems like a favorable product, given the creation of that conjugated pi-bonded system! But, I can't seem to find any references to such a reaction -- would it perhaps require some kind of elimination groups on the amine as well?

The metabolism of acetaminophen to NAPQI looks like that, although it does involve the formation of the quinone as well.


text:
Friedel Crafts reaction of anisole
Does anisole undergo friedel-crafts reaction or does the oxygen atom complex with Lewis acid?
We know that yields of this reaction are reduced in case of phenol but does the same happen in anisole?

It does

It does indeeed.
This reference describes the F-C reaction of anisole under AlCl3 or BF3 catalysis in good yields with the expected o/p direction.
J. Am. Chem. Soc. 106(1984) 5284
There are numerous other references
It should be noted that Anisole + AlCl3 is a method for demethylation to the phenol.


text:
Can the solutions for the Briggs Rauscher experiment be made ahead of time and stored?
I will be performing the Briggs Rauscher Oscillating reaction as a part of an investigation into the effect of heat on the antioxidants.
This will require 3 solutions to be mixed together, as shown below, to obtain a reaction which will oscillate between clear to yellow to a dark blue.
Solution A:
- 9%  $\ce{H2O2}$
Solution B:
- Acidified 0.2 M $\ce{KIO3}$ solution
Solution C:
- Solution of soluble starch/Malonic acid/$\ce{MnSO4}$
However, as I will be able to work on this in 45 minute blocks, I am unable to make up the solutions fresh every single time. Hence, I was wondering if it would be possible for me to make up enough solution at the beginning and store them for subsequent trials.
Does anyone know if these solutions can be made ahead of time and stored, without obviously affecting the reaction. If so, how for how long, and are any special conditions eg. refrigeration required?

Should be no problem, even next day.

Thanks. I'm just concerned will the starch fall out of solution?


text:
Why are there so many DFT Methods?
There are many different DFT method available, is this simply because individuals find a better DFT method to suit specific types of molecules?

The issue is that unlike for the methods based on Hartree-Fock, there isn't a systematic path to improving DFT. Unlike for HF, where we can treat the correlation as a pertubation (MP-n) or include excited Slater determinants (CI and CC) to generally improve our description of a molecule, we still don't have a way of determining the true form of the density functional (specifically the part governing exchange correlation) in DFT, so for now we use educated guesses for the general form (Local Density Approximation, Generalized Gradient Approximation, etc) and optimize them for specific systems.
In principle, if the exact form of the functional were known, DFT would be exact. This is why some authors like to use a separate term DFA (Density Functional Approximation) to describe the instances of DFT in use today.

I think there's an image that's useful to conceptually understand this crucial point - when we apply DFT, we are not actually solving electron systems - we are solving "particles quite-similar-to-but-not-exactly-like-electrons" as defined by interactions (exchange and correlation) that are similar to electrons, but not the same. So if we found the actual functional for electrons, the very same methods we use now would be exact for all cases, but for now we're stuck with solving "almost-electron" systems and using them as approximations for the actual electron systems.

Might be of further interest: Is density functional theory an ab initio method? (DFA always :D)

There are not that many DFT methods, at least not much more than WFT methods. DFT itself is a method, though. And if you really mean DFT as a method, than there are these methods, because people are working on this field and try to solve there scientific problems using DFT methods.
I guess, you probably mean, why there are so many (exchange correlation) functionals around. That is, because we don't know the exact functional and it is neither trivial to find nor is it easy to improve, if you have found a nice approximation.
And yes, you are right, many functionals exist for special purposes. But the goal is to find a functional that is able to treat as many cases as possible as best as possible.


text:
Is dH work dependent?
In the equation $\ce{d}H=\ce{d}U+\ce{d}(PV)$
Under constant pressure is $\ce{d}(PV)$  equal to the work done ($W$) on the system? (where $W= P_{\text{ext}}\mathrm{d}V$ (for irreversible process) and W = $P_{\text{int}}\mathrm{d}V$ (for reversible process)) And is this $W = \Delta{n_g}RT$?

What is your question exactly ??

I've edited it. please do check it out

Enthalpy is a physical property of a substance (including even a mixture of substances undergoing a chemical reaction).  It is thus a function of state, and its change depends only on the initial and final states, and not on the process path between the initial and final states (of the substance or reacting mixture).  Since work is a process-path-dependent entity, enthalpy can generally be regarded as independent of work.

When a system undergoes any thermodynamic process( without undergoing any chemical conversion), it is assumed to be a closed system, i.e mass of the system doesn't change. So, in a closed system $\mathrm{\Delta n_g =0}$.So, in that closed system under constant pressure, $\mathrm{d(PV) =PdV}$. and also differentiating both sides of the ideal gas equation $$\mathrm{d(PV) =d(n_gRT) -> PdV = n_gRdT (as~ dn_g =0~ by~ definition)}$$So, the work in reversible process is $\ce{P_{int}dV = n_gRdT}$.But work in a irreversible process is different which is just $\mathrm{P_{ext}(V_2 -V_1) = P_{ext}n_gR(T_2/P_2 - T_1/P_1)}$. But, if you consider a process where a chemical conversion occurs at a constant Temperature,there no of gaseous molecules in the reactants and the products are different. There you can write,$\mathrm{dH = dU + \Delta n_gRT}$, as , $\mathrm{PV_{final} -PV_{initial} = \Delta n_gRT.}$ But I would not prefer to call the last term on the R.H.S of the equation as work .


text:
What is the reason behind the phenomenon of Joule-Thomson effect?
For an ideal gas there is no heating or cooling during an adiabatic expansion or contraction, but for real gases, an adiabatic expansion or contraction is generally accompanied by a heating or cooling effect. What is the reason behind such a phenomenon? Is it related to the property of real gases or is it something else?

In a reversible adiabatic expansion or compression, the temperature of an ideal gas does change.
In a Joule-Thompson type of irreversible adiabatic expansion (e.g., in a closed container), the internal energy of the gas does not change.  For an ideal gas, its internal energy depends only on its temperature.  So, for an irreversible adiabatic expansion of an ideal gas in a closed container, its temperature does not change.  But, the internal energy of a real gas depends not only on its temperature but also on its specific volume (which increases in an expansion).  So, for a real gas, its temperature changes.  The Joule-Thompson effect is one measure of the deviation of a gas from ideal gas behavior.
ADDENDUM
This addresses a comment from the OP regarding the effect of specific volume on the internal energy of a real gas.
Irrespective of the Joule-Thompson effect, one can show (using a combination of the first and second laws of thermodynamics) that, for a pure real gas, liquid, or solid (or one of constant chemical composition), the variation of specific internal energy with respect to temperature and specific volume is given by:
$$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$The first term describes the variation with respect to temperature and the second term describes the variation with respect to specific volume.  For an ideal gas, the second term is equal to zero.  However, for a real gas, the second term is not equal to zero, and that means that, at constant internal energy (as in the Joule-Thompson effect), the temperature will change when the specific volume changes.  This is a direct result of the deviation from ideal gas behavior.

Could you elaborate on the internal energy dependency of real gases in the Joule-Thompson effect?

See my Addendum.

It's my understanding that, in a Joule-Thompson expansion, the internal energy can change, and what stays constant is the enthalpy, i.e., U + PV.

@theorist There are actually two versions of JT.  One is the version you referred to involving steady flow through a porous plug or valve.  The other version is a closed system containing two chambers separated by a partition.  The initial pressures in the two chambers are unequal, and the partition is either totally removed or punctured.  In this case, the total internal energy is constant.

I believe what you were initially describing is typically referred to as a Joule expansion, as distinguished from a Joule-Thomson expansion.  At least that's how I've always seen the two distinguished (e.g., https://www-thphys.physics.ox.ac.uk/people/AlexanderSchekochihin/A1/2011/handout11.pdf)  (though that author really shouldn't be putting deltas in front of W or Q).


text:
Coordination compounds

Which of the following complex is diamagnetic?

$\ce{[Cr(Cl3)(NMe3)]}$
$\ce{[Co(EDTA)]-}$
$\ce{[Cu(CN)(NO2)(NH3)(Py)]}$
$\ce{[Fe(NH3)6]^2+}$

For (4) I found that since $\ce{NH3}$ is a weak field ligand it cannot pair the electrons of $\ce{Fe^2+}$. Thus, (4) will remain paramagnetic.
But, for the rest of the options, I'm unable to clearly understand if they are paramagnetic or diamagnetic. I need help here.

1) Cr3+ has 3 electrons in 3d orbital. Obviously, it is a tetrahedral splitting, so configuration will be eg2 t2g1 which shows it is paramagnetic.
2) Here you need to know that for Co3+ all complexes are low spin except [CoF6]3- and [Co(H2O)3F3]. So, the complex given will be low spin with all 6 electrons paired & thus it will be diamagnetic.
3) Cu2+ has 3d9 configuration. So, it can never be diamagnetic as one electron will always be unpaired.
4) What you thought is right. But I would like to add that NH3 doesn't always behave like a weak field or strong field ligand. It depends on many factors which can't be explained using CFT.
Hope it helps!

If  complexes [Co(F6)3- and [Co(H20)3F3] are diamagentic exceptionally why is [Co(EDTA)]- diamagnetic ?It should also be paramagnetic

Why do you think the first one has an octahedral splitting? There are only four ligands.

@kishansharma the two complexes I mentioned are the only high spin complexes of Co3+, so other complexes including the one you asked will be low spin (with all 6 electrons paired and therefore diamagnetic)

@Justanotherchemist Thanks...i have corrected it


text:
Which mechanism will this reaction follow?
 What could be a possible mechanism? Please explain.

I think in the first step, all the ethers are hydrolysed because they are easily cleaved in acidic medium which is provided here by $\ce{H_3O^+}$ and after hydrolysis, the gem-diols will be formed which will be immediately converted into ketone  by loss of $\ce{H_2O}$ molecules . Then, the rearrangement will be a keto-enol tautomerism which will yield Resorcinol. After  that aqueous bromine solution will exhibit activated electrophilic aromatic substitution to yield C). The mechanism is as follows,


text:
What is the annelation effect?
My textbook had this question:

Hydrogenation of napthalene with excess hydrogen gas stops cleanly at 1,2,3,4-tetrahydronapthalene. What conclusion can be drawn from this experiment?

I observed that only one of the two rings was reduced. I did expect both the rings to reduce though, because the hydrogen gas was in excess.
In the solution manual, it is given that this is the "annelation effect", which isn't given in my textbook. I checked the internet, and could only find two research papers, which only detailed some typical aspect of annelation, not annelation itself. Wikipedia instead has no article on the "annelation effect" (but annulation instead, they definitely don't seem to be related though)
So, what is the annelation effect? I am not asking for the mechanism (unless it is as simple as catalytic reduction of alkenes, which I guess it isn't) nor the internal details of its procedure, but a simple overview of what it is and where it is applicable, with an example or two so I can easily identify it. And, how does the "annelation effect" help the fact that the above hydrogenation "stops cleanly" at one ring, even with excess hydrogen?

Annulation/annelation is  just adding another ring and can stabilise or destabilise a ring in effect, that's it. Also no idea why are you holding to overly laconic wording from your (not particularly helpful) book.


text:
How to pass a Plane of symmetry?
While passing a plane of symmetry through any compound inorganic or organic should it contain a minimum number of atoms or can it be passed without containing any atoms ?

Yes , the plane of symmetry need not cross/pass through any atoms.
The compound that you've drawn can show cis-trans isomerism. I believe you're talking about  the cis form.

The cis form will have a plane of symmetry

The trans form will not have a plane of symmetry

While the principle information (plane of symmetry does not have to pass through atoms) is correct, it is wrong that cis-1,2-dichlorocyclohexan will have such a plane. The ring is in a chair conformation with one chlorine atom equatorial, the other axial (in the ground state). See also this related question: Is cis-1,2-dimethylcyclohexane a meso compound?

@Martin-マーチン While what you say is correct, it is also incomplete. The interconversion rate of the conformers might matter. In other words, the molecule might have an effective plane of symmetry on average if the conformers interconvert fast enough.

@matt_black I seriously doubt that, as far as I remember, there is no symmetrical transition state that would interconvert the two. Also, symmetry is not something you can measure, or is it?

@Martin-マーチン I don't know whether this compound interconverts or is locked in a single configuration at normal conditions. But, if it interconverts, what I said is correct. The key question is whether the chlorines are distinguishable under normal conditions.

@matt_black As with the linked question, the molecule will most definitely interconvert, even at low temperatures. I have trouble understanding what average symmetry is, or how it could have an effective plane of symmetry. Even if it interconverts ultrafast, you would be able to observe both enantiomers simultaneously. If you mean with effective plane of symmetry that the molecule appears achiral, then that is true, but wouldn't then every racemic mixture have this property? Anyway, this is not the place to discuss that (maybe [chat]), I just pointed out a quite common mistake in the answer.


text:
A basic question about tetramethylammonium hydroxide --strong or weak?
This is a follow-up to the question here.  There is an answer claiming that tetramethylammonium hydroxide (TMAH) is a suitable strong base that can be used instead of the usual alkalies to reach pH 14.  However, Wikipedia says that TMAH is only a weak base.
But, when I state this I get comments that other sources document TMAH as a strong base.  The other sources, however, just mention it in passing and do not offer direct proof.  Can anyone provide a reference that actually does settle this question?  Thanks!
Note: At the time the question was posed Wikipedia did make the weak base claim. It has since been edited, and it now appears that the edit is correct.  See the answers.

Remark: Wikipedia backs its claim with this German thesis "Changes in high-molecular weight compounds during
beech litter decomposition" as a reference. On printed page 12, the $\mathrm{p}K_\mathrm{b}$ is given to be 4.2. This citation was added to that WP page in Dec 2014 in this revision by user "Czeror". Might help in organizing clues to resolve the dilemna.

Wikipedia clearly says it's a very strong base, I have no idea why you'd have doubts about that. Also title needs to be rewritten.

@Mithoron the title is a great pun, I love it.  Several journal articles also say 4.2, maybe the oldest being https://www.sciencedirect.com/science/article/pii/S0165237010000811 (2010) and then newer articles citing to that.  Overall, I think it is a good question.

@DavePhD Well, I feel title should be informative and clear in the first place, and then it may be also entertaining.

TETRAMETHYLAMMONIUM HYDROXIDE Journal of the Chemical Society 87 (1905): 955-961:

Practically the preparation of tetramethylammonium hydroxide
  from its salts resolves itself into a question of solubility as follows.
  In general, the equation
$$\ce{NMe4X + MOH = NMe4OH + MX}$$
will represent a real action proceeding nearly to completion if M, X,
  and the solvent are so chosen that all the substances except MX shall
  be soluble, or at least that MX shall be much less soluble than either
  of the original reacting substances. This principle was applied by
  Hofmann in his preparations with water as solvent. For M and X
  he chose either the pair Ag,I or Ba,SO4, in both of which cases MX is
  practically insoluble in water.
It is clear that if the general application of the above principle is
  justifiable, tetramethylammonium hydroxide may be prepared from a
  tetramethylammonium salt by means of potassium hydroxide if we so choose X and the solvent that of the substances represented in the equation
$$\ce{NMe4X + KOH = NMe4OH + KX}$$
all shall be soluble except KX.

...

An estimation of the strength of the base by means of the velocity
  of saponification of methyl acetate in N/80 solution showed that it was
  somewhat weaker than sodium hydroxide. The velocity constants obtained at 25 [degrees] were 0.0106 and 0.0115 respectively, so that if the
  strength of sodium hydroxide is represented as 100, that of tetramethylammonium
  hydroxide, under the above conditions, will be represented by 92.

So by two independent methods it is shown that tetramethylammonium hydroxide  is much stronger than pKb=4.2.  Firstly, the halide cannot not be converted to the hydroxide by adding metal hydroxide, unless an insoluble halide is produced to force the equilibrium forward.  Seconding, it is shown by velocity of saponification to be only slightly weaker then NaOH.
See also Essential Principles of Organic Chemistry (1956)

It follows that a quaternary ammonium hydroxide is completely dissociated and should be a strong base, as is sodium hydroxide. The experimental facts are entirely consistent with this view. Although trimethylamine in water solution gives only a small concentration of hydroxide ion, tetramethylammonium hydroxide gives a full molar equivalent of hydroxide ion in solution.

DavePhD is right, and I am accepting and up-voting his answer.  But first ... the real key to the conundrum is a little outside the box.  It is trimethylamine that has a $\mathrm{p}K_\mathrm{b}$ of 4.2.  Apparently some chemists mistook the basicity of TMAH for that of trimethylamine.

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


text:
Is Co[Hg(SCN)4] diamagnetic?
I was reading a book and it was mentioned that $\ce{Co[Hg(SCN)4]}$ is diamagnetic. I know that it is made up of $\ce{Hg^2+}$ and $\ce{Co^2+}$. $\ce{Co^2+}$ has 7 electrons which means that it should always have an unpaired electron but then why is this compound diamagnetic? Am I missing something?

There was definitely a typo and the book meant to write $\ce{Ca}$ instead of $\ce{Co}$. In the magnetic properties of a coordination complex, both the counter-ions and the coordination sphere metal ion are contributors. Since $\ce{Hg^2+}$ is diamagnetic and the overall compoud is given to be diamagnetic, the counter-ion has to be diamagnetic as well. Hence, it cannot be $\ce{Co^2+}$, which happens to have unpaired electrons.
What the book might have meant to say is that calcium is the counter-ion here. When checking diamagnetism or paramagnetism, you have to instead check the metal ion inside the coordination sphere, which in this case is $\ce{Hg^2+}$. And note that $\ce{Hg^2+}$ will have a $\ce{5d^10}$ electronic configuration.