text:
Why is the second acid dissociation constant for sulphuric acid much lower than the first in water?
In aqueous solution, $\ce{H2SO4}$ ionises in two steps :
$$\ce{H2SO4(aq) -> H+(aq) + HSO4-(aq)}; \quad K_\mathrm{a_1}=1\times10^3$$
$$\ce{HSO4-(aq) -> H+(aq) + SO4^2-(aq)}; \quad K_\mathrm{a_2}=1.2\times10^{-2}$$
The value of $K_{\mathrm{a}_2}$ caught my attention. I thought that $\ce{H2SO4}$ was a strong dibasic acid, but a very low value of $K_{\mathrm{a}_2}$ implies that the second proton is not that easily donated. Why is the value of $K_{\mathrm{a}_2}$ so less?

This is the typical pattern. Roughly speaking, there are no strong polyprotic acids.

@IvanNeretin Actually some complex acid would qualify... But roughly speaking... :D

For the second deprotonation you are trying to remove a proton from something that is already negatively charged, a process far more difficult than removing a proton from something that is neutral.
Alternatively you can think that the $-2$ charged $\ce{SO4^2-}$ anion attracts really strongly the $\ce{H+}$ resulting in the equilibrium being towards $\ce{HSO4^-}$, making it a weak acid compared to $\ce{H2SO4}$.
It is similar thinking to why the second ionization energies are always much higher than the first.

There can always be more than one ways to explain an observed phenomenon. With this in mind, one might be tempted to believe that there is suppression of dissociation of the $\ce{HSO4^-}$ ion due to the common ion effect.  The reasoning goes along the lines of:

Note that even before $\ce{HSO4^-}$ dissociates, there is an equal concentration of $\ce{H+}$ ions already present in the solution. We know, by Le Chatelier's principle, that the increased concentration of the products favors the reactants. Hence for the second dissociation, since $\ce{H+}$ is one of the products, and is in large concentration, the second dissociation is actually favored in the reverse direction. Thus, deprotonation of $\ce{HSO4-}$ is not favorable.
Hence, the $K_\mathrm {a_2}$ of $\ce{H2SO4}$ is significantly lower than its $K_\mathrm {a_1}$.

However, this is a wrong reasoning. Note that $K_\mathrm{a}$ values are always a constant. The $K_\mathrm{a}$ value for the $\ce{HSO4-}$ ion would be the same whether we took $\ce{H2SO4}$ or $\ce{NaHSO4}$. However, in the case of $\ce{NaHSO4}$, there are no pre-existing $\ce{H+}$ ions in solution when $\ce{HSO4-}$ deprotonates. Hence, it would be wrong to invoke the common ion effect here.
Therefore, the common ion effect should not be invoked when explaining the low $K_\mathrm a$ of $\ce{HSO4-}$ ion as compared to $\ce{H2SO4}$.

The value of 2nd dissociation constant is less than 1st because in 2nd equation there is anion $\ce{HSO4-}$ which will not allow to release protons,but we know that $\ce{H2SO4}$ is strong dibasic acid.


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Why does the osmotic pressure method offer an advantage over other methods of molar mass calculation?

Measurement of osmotic pressure provides another method of
determining molar masses of solutes. This method is widely used to
determine molar masses of proteins, polymers and other macromolecules. The osmotic pressure method has the advantage over
other methods as pressure measurement is around the room
temperature and the molarity of the solution is used instead of molality.
As compared to other colligative properties, its magnitude is large
even for very dilute solutions. The technique of osmotic pressure for
determination of molar mass of solutes is particularly useful for
biomolecules as they are generally not stable at higher temperatures
and polymers have poor solubility.

This is the text from my Class 12 Chemistry Book Part 1 NCERT (India) (page 54). I didn't get why:

Molarity is used instead of molality. I think it is probably because we can measure the volumes of solutions easily using measuring cylinders, but mass of solvent in already prepared solution is difficult to measure.

It's magnitude is large even for dilute solutions. Why? I didn't get this one.

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Aromaticity of multi annuli systems
In high school books, Huckel's rule is used for determining aromaticity. We were taught this example

Huckel's rule is limited to monocyclic systems. We were taught to find aromaticity by -

looking for a closed loop. The two loops that were pointed out have been drawn (in pink) in the diagram above.
Applying Huckel's rule to it i.e counting the number of $\pi$ electrons and then applying the 4n+2 formula for aromatic compounds and 4n for anti.

No explanation was provided as to  why this concept works on the grounds of 'this is out of syllabus'.

I have multiple questions regarding this 'closed loop' concept

What is it really called?
How does this loop methodology work?
What would happen if the inner loop was aromatic and outer anti aromatic?
(the example has both loops anti)

Well, you put charges backwards...

https://chemistry.stackexchange.com/questions/78732/is-7%d1%81irculene-aromatic-or-anti-aromatic - pretty much same thing.

related https://chemistry.stackexchange.com/questions/37641/is-buckminsterfullerene-aromatic https://chemistry.stackexchange.com/questions/51484/does-aromaticity-require-a-co-planar-pi-system

@avatar if you want to separate charges to make the inner and outer rings look aromatic, the negative charge should be on the inner ring.

@Oscar the rings are anti aromatic

I see , the title is aromaticity that's Mithoron said that the charges are backwards

@AvyanshKatiyar The rule you stated is not very clear. Does all "loops" being aromatic or antiaromatic make the entire molecule the same? It would be great if you could provide some reference!

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Can we produce tritium from helium 3? If yes how?
I want to know the best and economic way of the manufacture of tritium from helium-3

I don't think this can be done

Then again, this might be a worldbuilding question (helium-3 and tritium are both used a lot in science fiction). However, at least for now it seems to ask a legitimate chemistry question.

From my knowledge, helium-3 is actually a by-product of tritium decaying. Tritium is usually manufactured industrially by bombarding lithium-6 nuclei with neutrons in a nuclear reactor, and is currently the cheapest way to manufacture tritium.

In fact, producing T from 3He is straightforward. A visit to the National Nuclear Data Files site gets you cross sections for the various reactions.
The He3(n,p)T reaction has a cross section in excess of $10^{5}$ barns at low (thermal) neutron energies.
The Li6(n,T)He4 reaction mentioned in another answer has a smaller cross section, in excess of $10^{4}$ barns at low neutron energies, so it is about 10x less efficient to make tritium this way (not even counting enriching to get Li6 instead of Li7).
Yet, tritium is indeed bred from Li6 in reactors. This is beacuse they use solid lithium compounds in the reactor. Trying to stuff as much He3 into the same volume would require high pressures, and you still would not get the same amount in a given reactor volume. So, they use Li6 and wait longer.

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Reaction of phenol with sulphuric acid and nitric acid
I'm confused a little bit about the products A and B .

Please identify A and B

Could you us what you already did?

Hi S. Dutta, welcome to Chem.SE! We require you to show your efforts on this problem. What formulae/steps did you try? What knowledge do you already have about this? Where did you get stuck? Please add this to your question. Thanks!

Actually, the reaction provided above is a better way to synthesize Picric acid(2,4,6-trinitrophenol).
First, when the phenol is reacted with concentrated $\ce{H2SO4}$ , it behaves like a normal sulphonation reaction at high temperatures. The major product at high temperature is p-hydroxybenzenesulphonic acid, although there is a small amount of o-hydroxy benzenesulphonic acid. So, you should write the major product in A which is 4-hydroxybenzenesulphonic acid.
But, when A is treated with conc $\ce{HNO3}$, due to high activating property of $\ce{-OH}$ group, the $\ce{-NO2}$ groups are substituted at both the ortho positions.If both the ortho positions are substituted by $\ce{-NO2}$ group, there will be extensive intramolecular Hydrogen bonding , which makes this product thermodynamically very very stable ($\ce{$\Delta$H << 0)}$and at the higher temperatures the thermodynamically controlled product is formed in more amount.
More over, the $\ce{-NO2}$ groups also substitute the $\ce{-SO3H}$ groups at para position easily, because $\ce{H2SO4}$ being a stronger acid than $\ce{HNO3}$ ,the leaving $\ce{-SO3H}$ is more stable  than $\ce{-NO2}$. So, the ultimate product gets all ortho-para positions of phenol substituted by $\ce{-NO2}$ groups.  So, B will be  PICRIC ACID(2,4,6 -trinitrophenol)


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Calculation of silver nanoparticle size using Mie Theory
I have synthesized silver nano-particles and I wish to determine the approximate particle size using the wavelength that was most absorbed according to its absorption spectrum.
I find out that Mie theory provides a good approximation for this, (Eq 4 in this paper). However, I am clueless as to how to use the equation. In other words, how should I use the $\lambda_{max}$ obtained through the experiment in the provided equation?

Wait isn't Equation 4 $\gamma (R) = \gamma 0 + (Av_f)/R$? Where's the $λ{max}$?

Yes, it is.  However, the first answer on the link suggests to use it. https://www.researchgate.net/post/Can_anyone_explain_how_to_calculate_particle_size_from_Uv-Vis_spectra

Oh, the application took pretty long to load. So I didn't see that at first! Thanks!

So any insights on my problem?

Not yet, but I’ll work on this in math class. Middle school is so easy :)

Just wondering, do you work in a lab? Because https://chat.stackexchange.com/transcript/message/43030181#43030181

Yes, I do work in a lab and I find it better to ask the question over here, then to reach those experts in the lab (which I think is allowed, since I have explained the question clearly and provided the research that I have done).

Let us continue this discussion in chat.

Plug it in to the math equation
Ok, so we have this equation: $γ(R)=γ_0+(Av_f)/R$
First, we substitute the values into the equation that we know.
$γ(R) = (5\cdot10^{12}) + (3/4\cdot1.39\cdot10^6)/R$. Simplified, this is:
$γ(R) = (5\cdot10^{12}) + (1.0425 \cdot 10^6)/R$
$γ(R)$ can be measured in your experiment. Then we can solve your equation for $R...$
$\large{R = \fbox{$\frac{1.0425 \cdot 10^6}{γ(R) - 5\cdot10^{12}}$}}$
Plug in your value of $γ(R)$, and you're there!*

*In other words, how should I use the λmax obtained through the experiment in the provided equation?
The equation only mentions $\gamma (R)$, I don't know how to help you there**

**Zhe: @JavaScriptCoder For these types of questions, the first thing to ask is "do you work in a lab?"
If not, why are you doing experiments outside of a lab, and if so, why is it that no one in your lab has any expertise on this?

Here you go. I don't know what $λ_{max}$ is.

I know the constants (provided in the paper) and I know enough math to manipulate the equation in the provided form. This answer does not answer my question at all.

Yes, I know, but I think $γ(R)$ can be measured by the scattering of the silver nanoparticles. I don't know how to find $γ(R)$ from  $λ_{max}$


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Is it possible to produce hydrogen sulfide with tap water and activated carbon or CuZn?
Please note I'm a bit out of my depth with chemistry.
The situation:
I have a KDF/GAC water filter that defies logic and releases $\ce{H2S}$ (assuming it by smell/taste) in the water after filtration.
Is it possible that some reaction could produce $\ce{H2S}$ with water and $\ce{CuZn}$ media (such as KDF) or GAC (Granulized Activated Carbon).
The more practical problem:
DIY Exchange - KDF Filter Question

On the link about those filters it says that they remove H2S not release it. The filter contains Cu and Zn so there is no sulphur source in there. The activated carbon is just carbon so no sulphur in there either and as you say your water is fine before it goes through the filter. There is no chance that the filter is generating H2S unless it has been used and it got saturated with H2S or other sulphur source. Just send it back, get a refund, and get another one.

@AMM Thanks for your comment, I agree it could be saturated, however I doubt it to be the cause. I was wondering if there is maybe some salt like PbSO4 and some other chemicals that could react in a way with the C, Cu or Zn as catalysts or as partakers to create H2S (or something that smells as bad). Also my question here is mainly for academic purposes and understanding the world we live in.

It still doenst make any sense because in order to get H2S you need to reduce SO4-2 from +6 to -2 which means that you need to go through the 0 oxidation state first meaning that you ll be having sulphur formation which will precipitate out of the solution, and anyway the filter shoul be oxidizing S-2 anions to elemental S. What does the supplier suggest is happening? This is odd.

@AMM thanks it makes sense. Yup, very odd. The supplier have not yet responded to my mail (they probably think I'm crazy haha).

What is "Ze"  ???

@Hexacoordinate-C Pretty sure he means Zn. It is a Zn/Cu filter.

@Hexacoordinate-C haha, fixed it thx. Meant Zn.


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How to calculate Absolute error under square root sign?
I was calculating standard deviation of a data set. Could someone help me find the square root of 15.13 with an absolute error of 0.005? Is there a rule for calculating absolute errors under square root sign?

I'm voting to close this question as off-topic because it's not about chemistry.

Sorry It was a Chemistry lab calculation. I was calculating the standard deviation for the volume of water delivered by a pipette using mass and density of water ( at given temperature). Thank you for the reminder! I'll change it then.

I don't understand how the error of a square root would result from the experiment you described.

I recommend this publication, which also correctly covers the common circumstance of correlated noises: J. Tellinghuisen, "Statistical Error Propagation", J. Phys. Chem. A 105, 3917-3921 (2001).

A slightly different approach to @Zhe's answer is to use the 'propagation of errors' formula. Your formula is $y=\sqrt{x}$  with $x=15.13$ and an error $\sigma_x=0.005$ then the formula becomes $\displaystyle \sigma_y=\frac{\mathrm dy}{\mathrm dx}\sigma_x = \frac{\sigma_x}{2\sqrt{x}}= 6.4\cdot 10^{-4}$.
[if you have a formula in more than one variable , say $x$ and $z$ then $y=f(x,z)$ and the equation is more complicated but still quite easy if done in steps and is $\displaystyle \sigma_y^2=\left(\frac{\mathrm dy}{\mathrm dx} \right)^2_z\sigma^2_x+ \left(\frac{\mathrm dy}{\mathrm dz} \right)^2_x\sigma^2_z$. Notice that the square of the final error is calculated in this case.]

This is much better for combining two uncorrelated errors than what I suggested.

Generally, you want to look at Taylor expansions when computing functions on small errors:
$$\sqrt{1+x} = 1 + \frac{x}{2} + O(x^{2})$$
This means that
$$\sqrt{15.13 \pm 0.005} = \sqrt{15.13}\sqrt{1 \pm \frac{0.005}{15.13}} = \sqrt{15.13}\left(1 \pm \frac{1}{2}\cdot\frac{0.005}{15.13}+O(0.005^{2})\right)$$

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How to make dichloromethane human-friendly and safe for storage?
I use dichloromethane to polish 3D printed object as postprocessing step.
Most of it is stored in darkened glass on my balcony (now at $\pu{-15^\circ C}$).
What chemical reactions I need to make it human-friendly (non toxic)? Is it ok to use non-darkened polypropylene/polyethylene box to store it?
What I can do during the summer season? (Dichloromethane is volatile liquid)

Dichloromethane is a suspected carcinogen. The way to not make it a carcinogen is to make it something else, but then it's not dichloromethane. Storing in a closed bottle where it could be exposed to direct sunlight seems like a terrible idea.

DCM might be suspected of being a carcinogen, but it has been tested a lot a there is no evidence it is carcinogenic in people. It is actually one of the safest solvents (it isn't flammable or very toxic.) But it might need to be stored in glass or appropriate metal containers. It is destroys a lot of plastics.

@matt_black Here is just a small section of the hazards of DCM: ''Warning!:
Causes eye and skin irritation. Causes respiratory tract irritation. Harmful if swallowed. May be harmful if inhaled.
May cause central nervous system effects. Methylene chloride is metabolically converted to carbon monoxide after systemic
absorption, which yields increased concentrations of carboxyhemoglobin in the blood. May cause kidney damage.''  Therefore i wouldnt say DCM is "safe" in general. Of course it also depends on what your idea of safe is as well. I am talking from a synthetic chemist's perspective.

@AMM I used to use it a lot in the lab. It was one of the more benign solvents I used. The biggest immediate danger of large quantities was asphyxiation caused from the vapour displacing air (which has happened in industrial reactors).

According to A Comprehensive Guide to the Hazardous Properties of Chemical Substances at pages 445-446:

Disposal/Destruction
...
Small amounts may be destroyed in the laboratory by hydrolyzing with ethanolic solution of potassium hydroxide

As in Zhe's comment, you cant make DCM human friendly and you shouldn't be handling it recklessly.
DCM, should be used in a fume hood to avoid breathing the vapors and should be stored in amber winchester bottles in a dark and cold place. It dissolves most polymers, so don't try to store it in a PP or a PE bottle as it will eat through it eventually and also get contaminated.
Lastly, if you want to dispose of it you shouldn't just bin it somewhere. Take it in a lab and they will do it properly. Reading the MSD of DCM would also be a good idea.


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Filtering conformation search: redundant elimination or clustering
In my project, I study reaction computationally. The initial step in my workflow involves a conformational search using a constraint to bring the two reactive parts on my molecule together (reaction forms ring).
This conformational searching generates many confirmations, and it can be difficult to tell which are different.
Previously I use redundant comformer search which narrows down before I pick which conformers to study by DFT  but today it was suggested to use a clustering method to group the conformers. In the suggestion i was told that the clustering would group into chair vs boat vs twist etc which would be useful. I tried this using maestros inbuilt algorithm but the results were not very good - all of the clusters contained multiple ring types.
Is there any better way of clustering in order to sort my results. Using external software for instance, or mathematically. I also wonder whether both are needed (redundant conformer elimination then clustering, or just cluster all results).

Welcome to the site! Great first question. If you haven't already, take the [tour] and if you have any other questions about the site check at the [help].

I'll preface this by saying I'm sure there are much better ways to do this, but a simple approach could be to DFT-optimize the structure of each group (1 boat, 1 chair, etc.) and then calculate the root-mean-square deviation in atomic positions of all the generated conformers from each of these prototypical structures. I'm sure Open Babel has some better method though, so this is just food for thought. (Tangentially related, but this seems like a good problem for machine learning-based multiclass classification.)

Thanks for commenting, but this doesn’t really simplify things at all - my current approach manually going through result gives this. I’m interested in clustering to do the sorting rather that sorting to be able to clusterering.

What type of clustering algorithm are you using? Most clustering algorithms have hyper-parameters, which influence which clusters are found and how many.


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Why is the N-N bond length greater in dinitrogen trioxide than in dinitrogen tetroxide?

Why is the N-N bond length greater in dinitrogen trioxide than in dinitrogen tetroxide?

If the lone pair on $\ce{N}$ in $\ce{N2O3}$ delocalises, then the $\ce{N-N}$ bond will have a double bond character, so it should be shorter.
From data on Wikipedia, the $\ce{N-N}$ bond length in dinitrogen trioxide is  $\pu{186.4pm}$ while that of dinitrogen tetroxide is $\pu{178pm}$.

I am thinking this is a very tough question. Here's a research paper on this topic. Read from "The $\ce{N–N}$ bond distance is exceptionally long $\pu{1.75–1.78 Å}$! while the..." second paragraph.

@GaurangTandon This paper highlights that N-N bond is exceptionally long in dinitrogen teroxide. But why is it longer in dinitrogen trioxide?

Indeed, it is. But, I do not know the exact answer. I was just linking to a related paper.

So is this an anomaly or is there an explanation to it?

@Parth there is no such thing as an anomaly , just things for which reasons haven't been found yet.


text:
Why hasn't chemistry brought about a greater level of human enhancement(as of yet)?
So, this question had me thinking, this as someone who comes from the field of machine learning. Why haven't chemicals enhanced our minds and bodies as much as was predicted in the early years?
From the 40s till about the 70s I would say, many scholars and general talking heads predicted that chemistry would not only revolutionize food, industry(which it has) but also completely redefine the limits of humans by extending life exponentially as well as give super-human like mental and physical attributes.
Seemingly, this has not happened yet. While chemistry has completely changed a lot of what humanity does, it has never quite lived up to expectations(much like mine). So what is the disconnect? Is it that we just need more time? I am suspicious of that as chemistry is an very old field by academic standards and from what I read has been seeing diminishing returns in research.
I ask this as someone with only a rudimentary understanding of organic chemistry. Thanks!

Ever heard of doping in sport? Ever heard of LSD? Chemistry did enhance human minds and bodies, but it comes at the cost of ugly consequences, hence it is generally forbidden, and rightly so.

Obviously, I know of substances like that. If you'll read, the question is "as was predicted in the early years?"

Well, machine learning did not live up MY expectations, so it is 1:1

It pretty much depends on what your expectations are, but life expectancy has increased a lot in the last 100 years, due in part to (synthetic) antibiotics and other drugs such as for treating cancers.  Now that antibiotic resistance is occurring apace chemistry will be needed again to  provide new compounds to tackle this huge threat.  The fact that Chemistry is old (but not as old as Physics/Astronomy) does not mean that there is nothing new to learn.

This is not a limitation related to chemistry, but related to the lack of understanding of the human body and mind.
There are a couple exceptions, such as the Cochlear implant (wich basically transforms a deaf person into a hearing cyborg), pacemakers, and some drugs that can enhance mental capabiliyties.


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What are the units that measure free energy?
Is it $\pu{kJ/mol}$? I have done problems and I always end up with these units, but is this correct?

In thermodynamics the change in free energy $\Delta G$ is given by,
$$\Delta G = \Delta H - T \Delta S$$
Hence by dimensional analysis, its units must be that of $\Delta$H, which is $\pu{kJ/mol}$.
It is logical because energy is usually measured in multiples of joules (SI Unit).

What are the units that measure free energy?

There are several different free energies, but I'll tackle Gibbs and Helmholtz only here:

Gibbs free energy: $G=H-TS$, for $H=U+PV$. Hence, by dimensional analysis, Gibbs free energy has the SI units of Joule.
Helmholtz free energy: $A=U-TS$. Again, by dimensional analysis, Helmholtz free energy also has the SI units of Joule.

Since both are only "free energies", it makes sense to assign them the SI unit Joule.
That said, however, almost everywhere they are specified in terms of $\pu{kJ/mol}$ because that makes the comparison across different substances easier. Since $G$ is an extensive function, its value can vary even with the amount of substance. Your calculated $\Delta G_\mathrm f^\circ$ can differ from mine even if we take the same substance, just because we had the amounts different. Hence, to ease comparison and computations, they are "specified" in terms of $\pu{kJ/mol}$, although their correct SI unit is Joule.


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Why does acetone have a lower boiling point than hexane?
Why does acetone have a lower boiling point than hexane? I thought that since hexane is non-polar then it should have weaker intermolecular forces and a lower boiling point, but it doesn't. Why?
For reference, the boiling point of acetone is roughly 56 °C  while that of n-hexane is roughly 68 °C.

Asphalt and parafin wax are just made of nonpolar molecules, too...

Why a comparison between acetone and hexane? Why not between 2- or 3-pentanone (bp 101-102 $\ce{^{o}C}$, MW 86) and hexane (MW 86)?

The real surprise is that the smaller molecule acetone has nearly the same boiling point as hexane.

Hexane is a larger, longer-chain molecule so there are more London dispersion forces between n-hexane molecule which outweighs the dipole forces in acetone.


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How does change in entropy influence the tendency of a reaction to occur?
I know that entropy is the measure of the randomness of particles within a system, but how would it affect the tendency of the reaction? Would it just increase or decrease the chance of the reaction occurring, based on what change in enthalpy is?

See Gibbs free energy.

Entropy in a sense decides how energy is distributed. More entropy, more even distribution of energy. Consider a reaction with high activation barrier. If that reaction is producing high entropy then, the chances of the reverse reaction are decreased as increasing entropy decreases the probability of the product molecules acquiring enough energy for the reverse reaction.

Entropy ($S$) is a measure of the disorder in a system.
In a closed system, entropy always increases over time. In an open system, energy can be added to a system to cause a decrease in entropy, but this is not necessarily a spontaneous reaction.
A reaction will be spontaneous if the change in free energy, $\Delta G$, is negative.
Quantitatively,
if entropy increases, and the reaction enthalpy is exothermic $(\Delta H<0)$ or weakly endothermic ($\Delta H>0$ and $\Delta H < T\Delta S$), the reaction becomes spontaneous according to the free energy equation.
Since the observable universe is essentially a closed system, entropy will continue to increase until the universe essentially ceases to exist.
This way an increase or decrease(sometimes) in the randomness can cause a reaction to occur.
Here is a table relating to spontaneity and signs of $\Delta G$:

Image Source


text:
Significance of Temperature in Gibbs Free Energy
Why is it important to know the temperature of both reactants and products of a reaction in order to determine change in the Gibbs free energy of the reaction?
Is it because temperature has an effect on both enthalpy and entropy? Or is it because of some other reason I am missing?
Any help on this would be great. Thanks!

Isn't that reason more than enough?

Change in the Gibbs free energy ($\Delta$$\ce{G}$) of a reaction at any temperature can be determined by the equation <$$\ce{$\Delta$G = $\Delta$G^0 + RTlnQ_c (Q_p if the reactants and products are gaseous)}$$ Here, $\Delta$$\ce{G^0}$ is the standard Gibbs-free energy change of the reaction which is determined by,$$\ce{$\Delta$G^0 = $\Sigma$a_iG^0_i_{products} - $\Sigma$b_iG^0_i_{reactants}}$$ where $\ce{a_i}$s and $\ce{b_i}$s are proper stoichiometric coefficients. Now, $\ce{G^0}$ of products or reactants are determined by producing 1 mole of ech of them from their corresponding constituting elements in their standard states and at a FIXED TEMPERATURE(generally at $\ce{298.15K}$). So,  $\ce{$\Delta$G^0}$ is determined also at that particular temperature,So, it is independent of Temperature.  But,when we have to calculate $\ce{$\Delta$G}$ for any reaction at any temperature($T$) by the above given relationship , we have to know the temperature of the reaction condition as you can see clearly, $\ce{$\Delta$G}$ is directly  dependent on $\ce T$.That's why, it's important to know the temperature.


text:
NMR Spectra Oddities
I've noticed that as the chain length of n-alkanes increase, their number of carbon environments tend to 5 peaks, and no more, no less. Why is this? I've looked on multiple NMR (C13 NMR, I might add) databases and they show that there are 5 carbon environments. The outer carbons are all distinct environments like you'd normally expect, but the inner part of it (e.g. the innermost carbons in tridecane, C13H28) is all one environment. I hope somebody could shed some light on it. It's stumped both me and my chemistry teachers.

The environments are the same only to within the limits of the spectrometer's resolution and other experimental details; remember that every measurement comes with uncertainties, some of which are baked into the very fabric of the universe.
If you had an "infinitely precise" NMR spectrometer, all the peaks would be distinct (except when symmetry demands they be the same). In fact, every single possible compound would be uniquely identifiable by its atoms' specific shifts, shared with no others. In principle, you'd be able to assign a carbon peak at 123.6207426[...]964109454164[...] ppm to riboflavin, whereas a peak at 123.6207426[...]964109454165[...] ppm would indicate monobrominated hexabenzocoronene intercalated into a strand of DNA, and so on. Of course this is just a thought experiment and would never be possible in practice, even though it would make characterisation of complex molecules and mixtures a dream...

Yes, I thought that may be possible. However, is that the only reason, do you think? Would there be another reason that it works too, apart from simply the spectrometer only able to be so precise? And why 5 peaks? Why not four, or six? Do you know why the number of peaks doesn't keep on increasing even after, say, dozens and dozens in the carbon chain? Thank you for the comment; it really helped. I'm just wondering if that's the only explanation for it.

I'm afraid that trying to quantify at what point the signals will start to overlap is probably quite hard from theoretical considerations alone. It makes sense for it to happen though; the further apart things are, the weaker they interact with each other. You might find it interesting to check the 13C NMR of a sequence of 1-functionalized alkanes, e.g. 1-hexanol, 1-heptanol, 1-octanol, 1-nonanol, 1-decanol, .... See, for example, here.

Specifically, to get such precise peak positions, you would need to have a lifetime of the excited state of the spin system in the order of days. I.e. a T2 of many hours. That's not a limitation of the spectometer, but of the very sample itself.

There is a very simple anwer: these methylene groups are practically chemically equivalent. It is impossible to make a specific reaction with one of them.
Chemically equivalent hydrogens have the same chemical shift, and that's it.
Additionally, when two protons with similar CS couple, they create a higher order spectrum instead of two doubletts. The closer the CS are, the more the outer peaks get suppressed and the inner coalesce. Classic transition from AB to AA', the roof effect becomes so pronounced that you only see one peak, perhaps with a funny shape.
With 13C+decoupling, you seem to be slightly better off because there are no interactions (line splitting) visible. Still each line has an intrinsic bandwidth, because the relaxation times are finite, and the magnetic field has a finite homogenity. And the temperature distribution over the sample has a finite homogenity. Etc.
And of course you have in fact still interaction with surrounding protons, only they cancel out in the spectrum. Doesn't mean they're not there. Because otherwise you'd get relaxation times in the order of days with decoupling, which is not the case.


text:
How to compare the boiling point of water, ammonia and hydrogen fluoride?
According to the values of boiling points that I found on internet the order is as follows:

$\ce{H2O}$ >  $\ce{HF}$ >  $\ce{NH3}$

I was expecting $\ce{HF}$ to have highest boiling point because F is the most electronegative atom hence should form stronger H Bonds.
Now, I think the factor here might be number of H Bond rather than strength of H Bond.

$\ce{H2O}$ have 2 lone pair and 2 H atoms, hence it can form 4 H Bonds with surrounding molecules.
Similarly  $\ce{NH3}$  is also capable of forming 4 H Bonds with surrounding molecules.
$\ce{HF}$ on the other hand is only capable of forming 2 H Bonds with surrounding molecules.

Still the above hypothesis doesn’t explain why the boiling point of $\ce{HF}$ is greater than $\ce{NH3}$. I’m not sure what am I missing here.

Not exact duplicates, but related: https://chemistry.stackexchange.com/a/35354/5026 https://chemistry.stackexchange.com/a/59820/5026

Possible duplicate of How many hydrogen bonds in these compounds?

Though ammonia has three H atoms, it is not capable of forming four H-bonds
An important difference in terms of hydrogen bonding between ammonia and water, lies in the ratio between how many partial positive hydrogen atoms and how many lone pairs of electrons each have.
A water molecule has two partial positive hydrogen atoms and two lone pairs of electrons located on the oxygen atom. This implies that each water molecule can potentially have both its hydrogen atoms and both its lone pairs involved in hydrogen bonding.
But in ammonia, there is a shortage of lone pairs on the central nitrogen atom.
Hence it is capable of forming only two H-bonds.
And the difference in electronegativity is higher in $\ce{HF}$ than in $\ce{NH3}$.
Hence boiling point order is $\ce{H2O} > \ce{HF} > \ce{NH3}$

The boiling point order is H2>HF>NH3
Water has the highest boiling point and melting point followed by hydrogen fluoride and lastly ammonia because water has two lone pairs on the oxygen atom which makes it highly electronegative and has two hydrogen atoms hence its capable of forming four hydrogen bonds with the surrounding molecules and hydrogen fluoride is highly electronegative than ammonia.


text:
Explanation for the trends in nucleophilicity—Orbital interactions or electrostatic attraction
The trend of halide nucleophilicity in polar protic solvents is $$\ce{I- > Br- > Cl- > F-}$$
The reasons given by Solomons and Fryhle[1], and by Wade[2] are basically as follows.

Smaller anions are more solvated than bigger ones because of their 'charge to size ratio'. Thus, smaller ions are strongly held by the hydrogen bonds of the solvent.
Larger atoms/ions are more polarizable. Thus, larger nucleophilic atoms/ions can donate a greater degree of electron density to the substrate than a smaller nucleophile whose electrons are more tightly held. This helps to lower the energy of the transition state.

Clayden et al.[3] give the reason that in SN2 reactions, the HOMO–LUMO interaction is more important than the electrostatic attraction. The lone pairs of a bigger anion are higher in energy compared to a smaller anion. So, the lone pairs of a bigger anion interact more effectively with the LUMO σ*. Thus, soft nucleophiles react well with saturated carbon.
What I don't understand is the trend of halide nucleophilicity in polar aprotic solvents.
$$\ce{F- > Cl- > Br- > I-}$$
In aprotic solvents, only the solvation factor is absent. Bigger anions are still more polarizable and can interact in a better way with the LUMO. Then, what is the reason for the trend reversal in aprotic solvents? Solomons and Fryhle[1] say

In these[aprotic] solvents anions are unencumbered by a layer of solvent molecules and they are therefore poorly stabilized by solvation. These “naked” anions are highly reactive both as bases and nucleophiles. In DMSO, for example, the relative order of reactivity of halide ions is opposite to that in protic solvents, and it follows the same trend as their relative basicity.

Shouldn't the orbital interactions matter more than the electrostatic attractions even in polar aprotic solvents? Why does nucleophilicity follow the basicity trend in polar aprotic solvents?
References:

Solomons, T. W. Graham; Fryhle, C. B. Organic Chemistry, 10th ed.; Wiley: Hoboken, NJ, 2011; pp. 258–260.
Wade, L. G. Organic Chemistry, 8th ed.; Pearson Education: Glenview, IL, 2013; pp. 237–240.
Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; pp. 355–357.

I have resigned to the fact that in aprotic solvent the trend is explained by orbital sizes.

What do you think of my answer @ApoorvPotnis ?

@TanYongBoon I really like it. I'll award you a bounty once I gain some reputation. So, can one say that the unequal solvation in polar protic solvents makes the trend reversal?

@TanYongBoon Also, sorry for the late reply.

Nope. Nothing to do with solvation. It is the neglect of the thermodynamics factor - bond strengths: C-F > C-Cl > C-Br > C-I . I have shown in my answer that thermodynamics does influence kinetics. When you make stronger bonds in your products, you ALSO have stronger bonds in your transition state. Consequently, Ea decreases for the reaction, increasing the rate of the SN2 reaction. Thus, we expect the natural order of nucleophilicity to be F- > ... > I-. This is essentially a summary of my answer.

I really think you have asked a very good question. Very well thought out and very well referenced. Sometimes, we really can't depend too much on HSAB because its foundations, provided by Klopman and Salem, seem to be based on kinetic factors alone. Through writing my answer, I also learnt something really impt myself, that is, thermodynamics CAN influence kinetics. A very strange conclusion to come to since many separate these two considerations (e.g. we have thermodynamic vs kinetic control in org. chem) but yet is valid if we realise that the TS and reaction products are very related...

@ApoorvPotnis No need for the bounty. I am not so hungry for reputation points. I have benefited enough from your question through the thinking and writing processes. The purpose of my original bounty was to attract people to challenge/criticise my answer but seems like no one did that... Or maybe they thought that it was fine (but then, they didn't upvote if they thought so...)

@TanYongBoon What I am trying to say is that there are 3 factors which influence nucleophilicity: (1) Bond strengths of the resulting product. (2) The HOMO-LUMO interaction. (3) The solvation effect. Out of these three, (1) and (2) are the same for both protic and aprotic polar solvents. The resulting $C-X$ bond strengths and the HOMO-LUMO interactions are the same irrespective of the solvent. In the case of protic solvents, the energy required to break the hydrogen bonds is what reverses the trend.

Maybe we can discuss this in chat according to your conveience.

Yes that is what I'm saying. The cause of the trend reversal is the different solvation effects in the 2 different environments.

This is a rather intellectually-stimulating question and one that is also very difficult to answer. You have constructed a very good case for why the nucleophilicity order would not be expected to reverse in polar aprotic solvents, i.e. the order of intrinsic nucleophilicities of the halide ions should be $\ce {I^- > Br^- > Cl^- > F^-}$, based on the concepts of frontier molecular orbital interactions, as well as electronegativity and polarisability of the nucleophilic atom.
Affirming your viewpoint
Admittedly, the saturated carbon atom is a soft centre and based on electronegativity considerations, the energy of the $\ce {HOMO}$ of the halide ions should increase from $\ce {F^-}$ to $\ce {I^-}$, i.e. the hardness of the the halide ions as nucleophiles decreases from $\ce {F^-}$ to $\ce {I^-}$. Based on the hard-soft acid-base (HSAB) principle, we would expect the the strength of the interaction between the halide ion and the carbon centre to increase from $\ce {F^-}$ to $\ce {I^-}$.
We can provide more quantitative justification for this using the Klopman-Salem equation$\ce {^1}$:
$$\Delta E=\frac{Q_{\text{Nu}}Q_{\text{El}}}{\varepsilon R}+\frac{2(c_{\text{Nu}}c_{\text{El}}\beta)^2}{E_{\text{HOMO}(\text{Nu})}\pm E_{\text{LUMO}(\text{El})}}$$
The first term is indicative of the strength of the Coulombic interaction while the second term is indicative of the strength of the orbital interactions. In the reaction with the soft saturated carbon, it is actually expected for the reactivity to increase from $\ce {F^-}$ to $\ce {I^-}$ (ref 1, p. 117).
At the moment, all the evidence presented points us towards the conclusion that you have posited. However, we have overlooked the importance of a thermodynamic factor - the strength of the $\ce {C-X}$ bond.
A not-particularly-relevant add-on
Actually, if you think about it the considerations made in the Klopman-Salem equation are essentially kinetic considerations. We cannot conclude anything about reaction energetics from such considerations. The strength of the $\ce {HOMO-LUMO}$ interaction or that of the Coulombic attraction tells us nothing about how the electrons would organise themselves subsequently, what would be the strength of the resultant bonds etc.
A comment made on this by Olmstead & Brauman (1977)
On this topic of explaining the relative order of intrinsic nucleophilicity of the halide ions, Olmstead & Brauman (1977) had this to say$\ce {^2}$:

The intrinsic nucleophilicities follow the reverse order of the polarizabilities (e.g., $\ce {CH3O^-}$ > $\ce {CH3S^-}$ and $\ce {F^-}$ > $\ce {Cl^-}$ > $\ce {Br^-}$). This could be due to a stronger interaction between the more concentrated molecular orbitals of the anion with the carbon center. It could also be simply a reflection of the greater thermodynamic methyl cation affinities of the smaller anions.

It is important to note that Olmstead & Brauman are merely postulating the reasons. However, I do feel that the point on thermodynamics leads us on the right track to answering this question. On the other point of "more concentrated MOs", I am unclear of how it can be interpreted.
Let us focus now on this point.
How does thermodynamics even influence kinetics?
Well... It certainly does here. What are we concerned with when dealing with nucleophilicity? As it is a kinetic phenomenon, we are essentially concerned with the rate of the substitution reaction. This makes the consideration of the activation energy ($E_\mathrm{a}$) a particularly important one because the higher the $E_\mathrm{a}$, the lower the rate of the substitution reaction.
Here, the $E_\mathrm{a}$ can be taken to be the difference in energy between the reactants and the transition state. Thus, we can decrease the $E_\mathrm{a}$ by either raising the energy level of reactants or decreasing the energy level of the transition state. The reaction energy profile for an $\ce {S_N2}$ reaction is shown below, taken from ref. 3 (p. 394).

There are two ways to see how the strength of the $\ce {C-X}$ bond can affect the $E_\mathrm{a}$. I am not sure if they may be considered equivalent perspectives. Nonetheless, I will present both.
The first perspective is well-explained by Francis & Sundberg (2007) with regard to the $\ce {S_N2}$ reaction$\ce {^3}$:

Because the $\ce {S_N2}$ process is concerted, the strength of the partially formed new bond is reflected in the TS. A stronger bond between the nucleophilic atom and carbon atom results in a more stable TS and a reduced activation energy.

Another perspective (which may be somewhat similar to the first, or even possibly the same) is that the transition state structure bears resemblance to the product structure, based on the Hammond's postulate. However, this would be more applicable for an endothermic $\ce {S_N2}$ reaction where the transition state structure does indeed bear more resemblance to the product. When the product is more stable, i.e. has bonds that form more exothermically, the energy of the transition state would thus be lowered, and the $E_\mathrm{a}$ would also decrease, although not necessarily proportionally.
Conclusion
The argument you have presented failed to consider the important thermodynamics factor. Thus, it resulted in the predicted reactivity of the halide ions in the reversed order. The takeaway here is that the application of the $\ce {HSAB}$ concept alone does not always produce the correct prediction.
Additional information
More examples are cited for which the HSAB concept fails. These include:

The favourable combination of $\ce {H^+}$, a hard acid, and $\ce {H^-}$, a soft base (discussed here as well) (ref. 4).
The fact that the rate of reaction between $\ce {Ag^+}$, soft acid, and an alkene, soft base, is slower than that between $\ce {Ag^+}$ and $\ce {OH^-}$, a hard base (ref. 1, p. 110)

References

Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Student Edition). John Wiley & Sons, Ltd. United Kingdom, 2009.
Olmstead, W. N.; Brauman, J. I. Gas-phase nucleophilic displacement reactions. J. Am. Chem. Soc. 1977, 99(13), 4219–4228.
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.
Pearson, R. G.; Songstad, J. Application of the Principle of Hard and Soft Acids and Bases to Organic Chemistry. J. Am. Chem. Soc. 1967, 89(8), 1827-1836.

Upvoted for invoking Salem-Klopman.

In polar aprotic solvent, there is no possibility of formation of a hydrogen bond, to reduce the nucleophilicity of a nucleophile.
The order of basicity of $\ce{F-,Cl-,Br-,I-}$ are,
$\ce{F- > Cl- > Br- > I-}$ (since the order of acidity of the hydrogen halides are $\ce{HI > HBr > HCl > HF}$).
So, among halide ions, $\ce{F-}$ has the highest ability to donate its electron pair, and thus, act as a nucleophile.
Also, $\ce{F-}$ being the smallest faces less steric hindrance as a nucleophile.
So, the order of nucleophilicity in polar aprotic solvent is, $$\ce{F- > Cl- > Br- > I-}$$

Why does $\ce{F-}$ have the maximum capacity to donate its electron pair? Larger ions should be more polarizable and polarizability favours $\mathrm{S_N 2}$ reactions.

If you consider the Bronsted-Lowry definition of bases, the stronger bases have higher ability to donate electron pair

But basicity doesn't always parallel nucleophilicity; e.g. in polar protic solvents. Does it in polar aprotic solvent? If yes, why?

First of all, the basicity judgement by the ability of donating electron lone pairs  is Lewis definition. Secondly, being most electronegative shouldn't $\ce{F^-}$ try to keep electrons on itself rather than donating (according to the concept of electronegativity). How would you account for that ?

@ApoorvPotnis Generally, in polar aprotic solvents, nucleophilicity follows basicity.

@SoumikDas Although F has the highest electronegativity(attracts electron pairs in covalent bond), the small $\ce F^-$ has less stability than other halide ions due to high repulsion between electrons. For this reason, F has less electron affinity than Cl. Note that $\ce F^-$ is even smaller than the hydride ion.

@SoumikDas This suggests that $\ce{F-}$ is a 'hard' nucleophile. But 'softer' nucleophiles react better with saturated carbons, don't they?

@ArchismanPanigrahi , what electron affinity has to do with this? Electron affinity of $\ce{I}$ is even smaller than $\ce{F}$, Then what do you conclude from electron affinity ?

I mentioned electron affinity to give an example to illustrate that high electronegativity of Flourine does not indicate it will not donate electron

I've edited the question. Please see it. Thank you.

You are definitely mistaken. Electron donation looks at Lewis acidity. Clearly, I- is the strongest Lewis base among the 4.

In polar protic solvents since cations and anions are well solvated , attractive forces among the ions rupture due to which anions with higher solvation , i.e. high charge density have lower nucleophilicity than the anions with lower solvation, i.e. of lower charge density.
e.g., in $\ce{H2O}$ order of nucleophilicity of halides is  $\ce{F− < Cl− < Br− < I−}$.
In polar aprotic solvents, only cations get solvated but not anions hence anion which forms stronger bond with carbon has higher nucleophilicty.
e.g., in $\ce{DMF}$ and $\ce{DMSO}$  order of nucleophilicity of halides is  $\ce{F− >Cl−  >Br− >I−}$.

Acetone is also an aprotic polar solvent, then why is the order $\ce{Cl^-< Br^-<I- }$ in it?

I don't think your answer is answering the question. You are essentially answering a different question, which Apoorv Potnis has already answered in his question write-up. Your point on solvation is the bullet point 1 in the write-up.


text:
What is non PV work? How is it related to Gibbs energy?
Recently, I was trying to understand more about Gibbs free energy and got stuck at this point. Some books write that if $\Delta G=0$, then the reaction is reversible while some other sources say that a reaction is reversible when  $\Delta G$ = Non-$PV$ work.
I am not able clarify things here and am unable to understand the core concept of non $PV$ work.

At constant temperature and pressure, the maximum non expansion work is given by the change in the Gibbs Free Energy.
$$dW_{\text{max}}= dG$$
Proof:
At constant temperature and pressure,

from the definition of enthalpy: $H= U+pV$
first law of thermodynamics: $dU = dq+dw$

$\implies dH = dq+ dw + d(pV)$
Now,
$dG = dH - TdS -SdT \\\implies dG = dq + dw + d(pV)- TdS - SdT$
For isothermal process $dT =0$
$\implies dG = dq+dw+ d(pV)+ -TdS$
When the change is reversible,
$dq = dq_{\text{rev}}= TdS$ and $dw= dw_{\text{rev}}$
$\implies dG = dw_{\text{rev}}+d(pV)$
The work consists of expansion work, which for a reversible change is given by
−pdV, and possibly some other kind of work (for instance, the electrical work of
pushing electrons through a circuit or of raising a column of liquid); let's denote it by $dw_{\text{add},\text{rev}}$
$\implies dG= dw_{\text{add}, \text{rev}}+ Vdp$
For isobaric process, $dp =0$
$\implies dG = dw_{\text{add}, \text{rev}}$
You must be aware the the maximum work is obtained when the process is reversible, if not you may refer to this answer.
$\implies dG = dw_{\text{add}, \text{max}}$

Some books write that if $\Delta G=0$, then the reaction is reversible while some other sources say that a reaction is reversible when  $\Delta G$ = Non-$PV$ work.

$\Delta S_{\text{total}}= \dfrac{-\Delta G_{\text{system}}}{T}$
Thus, for a reversible process $\Delta G_{\text{sys}}= 0$
And I have also proved to you above that $\Delta G_{\text{sys}}=$ maximum non expansion (non PV) work.

I am not able clarify things here and am unable to understand the core concept of non $PV$ work.

PV work is only the work done during the change in volume. There can be different types of works that a system can do during a process, for instance: electrical work, work against gravity, work against friction, magnetic work, shaft work, etc. This is the concept of non expansion or non PV work.
Source:

Physical Chemistry by Atkins (8th ed. )
Wikipedia article on work in thermodynamics

You proved $\Delta G_{sys=0}$ but also proved that $\Delta G_{sys}=$ maximum non expansion work , does that imply maximum non PV work=0?


text:
Is there enough information in this question?
Here is the question:

Some countries still measure liquid volumes in gallons. (1 gallon = 3.785 L). A test in America recently found that a certain car's engine produces on average 9.50 kg of $\ce{CO2}$, which is a Greenhouse gas. Calculate the annual production of $\ce{CO2}$ in kg if there are 40 million cars in the USA and each car covers a distance of 12 000 km at a consumption of 20 miles per gallon. (1 mile = 1.609 km)

I don't think there is as I work out the annual fuel consumption, but am unsure about how to proceed from there. I think this is because it is not actually stated what the 9.5 kg figure for the production of $\ce{CO2}$ refers to (e.g. per year, per gallon of fuel, etc.)

It seems that you are correct :)

A gallon of gasoline weighs about 6.2 pounds, and is approximately (CH2)n. If combined with O2, the products would be 19.48 pounds (8.85 kg) of CO2 and 7.97 pounds of H2O. OK, so maybe the problem used slightly denser gasoline or it had less hydrogen in its composition. Close enough!
It seems that the problem assumes a familiarity with all the factors in the situation. Sometimes you have to read minds!


text:
How does cullet reduce the production temperature in glass production?
In my book, I read that broken glass pieces are reused in the production of glass, known to be cullet. This on one hand saves the production cost and also reduces the production temperature.
My question is: how does adding cullet reduce the production temperature? I see no reason why the materials would react any differently with or without cullet.

No expert here, but cullet would react faster than sand with the alkali, developing into a more uniform fluid that can be mixed at a (slightly) lower temperature so as to eventually dissolve the sand.
It is like beginning the reaction to produce glass at 25% (or however much cullet is used), rather than 0%, so that raising the temperature is not needed to to speed up the reaction. Lower temperatures can be used without needing to increase the time, and thus wear and tear on the equipment can be reduced.
So it's not that the cullet causes a reduction in temperature, but rather allows it.


text:
How are ions formed?
I know that $\ce{H2O}$ is neutral.
So how does adding one more hydrogen atom to it make it positive ion ($\ce{H3O+}$)?

Hi Hark, welcome to Chem.SE! Where did you read that addition of an hydrogen atom forms a hydronium ion? In fact, it doesn't. Hydronium ions are formed by addition of a proton to a water molecule.

So all ions are formed this way?

Ions are always charged. You cannot hope to form them by combining one atom with another atom.

How is sulphate ion formed?

Deprotonation of sulphuric acid. Twice.

Is it possible to make neutral SO4? How?

A "neutral ion" does not exist. Neutral gases of a similar formula - SO3 and SO2 - do exist, though they are not ions.

The flaw in the question is exactly what Gaurang Tandon said: $\ce{H3O^+}$ is formed by a water molecule and a proton ($\ce{H^+}$), not a hydrogen atom.
Compare the number of electrons in both sides of the following two equations:
$$\ce{H2O + H^+ -> H3O^+}$$
$$\ce{H2O + H^. -> H3O^+}$$
The last equation is wrong: there's an extra electron in the left hand side.


text:
finding out an known gas by determining its molecular weight through an experiment

A large evacuated flask initially has a mass of 134.567g . When the flask is filled with the gas of Unknown molar mass to 735 torr at 304K , it’s mass is 137.456g . When the flask is evacuated again and then filled with water at 304K , it’s mass is 1067.9g . Density of water at this temperature is given to be 0.997g/mL .

I am trying to find out what is this gas through finding the molecular mass of this gas derived from the ideal gas law-
Molecular weight = $\frac{Density \times R \times T}{P} $
To determine density of the gas, use mass of the gas divide by volume of the gas.
Mass of the gas = $137.456 - 134.567 $
Finding volume of the gas is where I started to struggle. I know it’s just $1067.9 - 137.456$ because total mass when gas added into the flask minus weight of the flask,  but I don’t understand the concept behind it.
Is there any diagram to understand the question statement of “ When the flask is evacuated again and then filled with water at 304K , it’s mass is 1067.9g .”

When the flask is evacuated and filled up with water again, its mass increases by $(1067.9 - 134.576)$g. or $933.324$ g. not $(1067.9 - 137.456)$g. because the gas was not there when it was being filled. It had the weight of an empty flask.
So the volume of the flask is $933.324/0.997$ mL. or $936.132$mL. which will be the volume of the gas when the flask is only filled up by the gas .

How does this answer the question?

Apply $$\ce{PV =nRT}$$. $P$ is known , $V$ is known now, $ R$ and $T$ is known . Find $n.$  $n =W/M$ ,$W$ is the increase in mass of the flask in presence of gas. Find $M$.  Think a little before adding a comment.


text:
Directional nature of the acetamido group in electrophilic substitution reaction
Here, in $\ce{CH3CONH2 -}$ group, the nitrogen has a lone pair and can thus activate the benzene ring along with having an ortho-para directive effect in the following manner:

However at the same time, both nitrogen and oxygen are electronegative. Hence, they should attract (roughly speaking) the electron cloud of the ring to deactivate the ring and may result in a meta directive effect in the following manner:

So which effect is applicable here and why?

What is chlorine , fluorine ? What sort of director.

Halogens are, as far as i know, ortho para directing even though they are deactivating

I believe the same principle is applicable here but not a deactivator as as halogens have the strongest $\pu{-I}$.

First of all, your second diagram concerning canonical structures are INCORRECT. There $\ce{N}$ has formed five bonds. $\ce{N}$ can't form five bonds because of unavailability of valance $\ce{d}$ subshells. It can only have the maximum covalency upto $4$ and how do you even think of keeping $\ce N$ atom neutral after forming five bonds which is anyway wrong. So, please make that correction.
There will be only one resonance structure:

(source)
Due to that delocalisation with carbonyl carbon, $\ce N$ will acquire a $\delta$+ charge which will increase its electron withdrawing effect (similar to $\ce{-I}$ effect) to some extent. Still, due to large no. of canonical structures of delocalisation in the benzene ring from the $\ce N$ atom, its $\ce{+R}$ effect is much stronger (strong $\ce{2p$\pi$-2p$\pi$} $ overlap) than its $\ce{-I}$ effect and thus the $\ce{-NHCOCH3}$ group still acts as an ortho-para directing group. But, its activating ability is slightly decreased from $\ce{-NH_2}$ group due to that $\ce{-I}$ effect.

Procedure for the nitration of acetanilide, noting that the products are primarily p with a small percentage o.
http://jscienceclass.blogspot.co.uk/2012/01/nitration-of-acetanilide-lab-report.html


text:
Do alkali metals not form diatomic molecules?
I know hydrogen forms a diatomic molecule $\ce{H2}$, where the electronic configuration of hydrogen is $\ce{1s}$.
But why doesn't lithium also form a diatomic molecule? Its electronic structure is $\ce{1s^2 2s}$, so can't two lithium atoms come together, share their outer electron and form $\ce{Li2}$? Same with $\ce{Na}$, $\ce{K}$ etc.
Am I missing something obvious?

Note that this is one of the properties that is usually taught at elementary level to differentiate hydrogen from alkali metals, even though they occupy the same group. In fact, some periodic tables depict hydrogen with a "dual position" - in both group 1 and group 17, since it shares several properties of both alkali metals and halogens. I'll spare the details for a high school textbook and this quora page.

Strongly related: https://chemistry.stackexchange.com/questions/32686/is-it-possible-to-have-a-diatomic-molecule-of-sodium-in-gaseous-state

Diatomic molecules of alkali metals are detected in the gas phase. However, it's so happens that the bond in them is very weak and at the temperature the alkali metal vapors develop only a few percents of the metal in the vapors exists as diatomic molecules. It so happens, that metal bonding allows to achieve an overall more energetically favored state.
Outer orbitals of the alkali metal atoms are very diffuse, so their bonds are weak. Only Lithium, the smallest of the alkali metals, has practically meaningful covalent chemistry to speak of.

for completeness $\ce{Li2}$ has a bond length of $0.267$ nm, a vibrational frequency of $351;\mathrm{cm^{-1}}$ and a dissociation energy of $\approx 8307,\mathrm{cm^{-1}}$. Na and K dimers as well as mixed dimer such as LiK are known.


text:
Molpro wavefunction symmetry specification
I'm beginning to use Molpro software for some computations. But I got completely confused by its description of specification of wavefunction symmetry.
It tells us, that the symmetry of wavefunction is described like
$$\text{wf},\text{nelec},\text{irrep},\text{spin},$$
where $\text{nelec}$ is the total number of electrons in the molecule, $\text{irrep}$ is the number of the irreducible representation (list of numbers in Molpro docs) and $\text{spin}$ is the total spin number.
I know, that water belongs to the $C_{2v}$ point group, and I found its character  table:

(source: uscupstate.edu)
I know, that 1/-1 values correspond with variance/invariance of orbitals, i.e. $A_1$ representation corresponds to the fact, that $\ce{s}$ and $\ce{p_z}$ orbitals are
not inverted by any of the four symmetry operations.
I also know, that the electron configuration of oxygen is $\ce{1s^2 2s^2 2p^4}$ and both hydrogen atoms have $\ce{1s^1}$.
But I'm not sure how am I supposed to choose the irreducible representation of the whole molecule.
In this answer the similar problem is described and $\ce{H2O}$ molecular orbital diagram is given:

The answer tells us, that molecular irreducible representation can be obtained like this:
$$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2 = (A_1)^2 \otimes (B_2)^2 \otimes(A_1)^2 \otimes (B_1)^2 = A_1,$$
which is in accordance with the first example in the Molpro docs, where $\ce{H2O}$ wavefunction is specified like WF,10,1;.

Questions

So, is this the correct way to find the irreducible representation of the whole molecule? And if it is, could you, please, explain, how to get the molecular orbital representation?
How can I obtain the total spin number?

Related: https://chemistry.stackexchange.com/q/58229/41556

I have used Molpro a decent amount and none of these inputs are actually directly required. Basically the only cases you would want to use them are for open shell systems and charged systems. Charged, you will specify nelec so that it does not simply create a neutral system. For open shell systems or transition metals, you may wish to choose a particular high-spin complex, and so use the total spin input.
The symmetry is much easier to specify using the symmetry directive where you just type in the point group.

@jheindel I was under impression, that in some cases it's not enough to just specify the point group and that I have to explicitly specify even the irreducible representation of the molecule.

You might be right... I have never encountered these situations. I guess we can wait and see.

You are using the term "irreducible representation of the whole molecule", but to me this does not make a lot of sense. So I would like to clarify something about this:
Irreducible representations are used to classify different electronic states (or electronic wave functions) of the molecule. The irreducible representation indicates sign changes (and therefore nodal planes) in the wave functions with respect to the symmetry operations of the molecule.
The "whole molecule" does not have sign changes, there are no negative atoms. Therefore it does not make sense to talk about "irreducible representation of the whole molecule". You may argue now whether the "whole molecule" always belongs to the totally symmetric irreducible representation ($A_1$ in case of $C_{2v}$) or does not have one at all. I would just stick with following statement: The molecule has a point group and its electronic states can be classified by irreducible representations.
Answering your questions
1) So, is this the correct way to find the irreducible representation of the whole molecule? And if it is, could you, please, explain, how to get the molecular orbital representation?
Which irreducible representation you need depends on the electronic state you want to calculate. What you showed would be the correct way to get the irreducible representation of a given electron configuration (in this case the ground state).
The molecular orbitals can be obtained from a calculation (e.g. HF or DFT), or by symmetry considerations. But the later will only give you qualitative results, no orbital energies.
2) How can I obtain the total spin number?
Just as for the irreducible representation, this depends on the electronic state you are interested in. Usually this is the ground state, and for a closed shell system (like $\ce{H2O}$) we have $S=0$. But in principle it is possible to split up one, or more electron pair(s) and get $S=2,4,\dots$, but this will results in higher total energies (= excited states).

Yes, this is the correct way to find the irreducible representation of the molecule. To obtain the energy ordering of the molecular orbitals (which determines the order in which they are filled for the ground state molecule), you need to run a molecular orbital theory computation (e.g. Hartree-Fock or density functional theory).
The spin number in Molpro is not particularly intuitive. For a molecular term symbol such as $^1B_2$, spin information is conveyed as $2S+1$, where $S$ is the total spin quantum number. In the specific case of a $^1B_2$ state, the molecule has a total spin quantum number of $0$. However, Molpro users must specify spin as $2S$.
As noted in the comments, however, specification of the irreducible representation or spin is generally unnecessary for Molpro computations. For Z-matrix inputs, Molpro determines the symmetry automatically, and in most cases this will be the symmetry the user desires. Molpro also automatically sets the spin if the user doesn't specify it. By default, Molpro uses $S=0$ for molecules with an even number of electrons and $S=1$ for those with an odd number of electrons. Since a large number of molecules that are kinetically stable under typical terrestrial conditions have an even number of electrons, the default spin should work well for many users. Molpro users who work on excited electronic states or molecules with unpaired electrons are much more likely to need to specify the spin.


text:
Phase diagram of water using Gibbs free energy instead of pressure
From my lecture course it was indicated that this is what a phase diagram looks like.
Is this diagram applicable to water? All the ones I can find online seem to use pressure on the y axis as oppose to G. If if it is applicable to water, where is the triple point on the graph? I understand it's where the 3 phase lines intersect but this doesn't appear to occur on this type of graph.

I'm going to get out my steam tables and calculate the Gibbs free energy as a function of temperature along the liquid-vapor saturation line. ( The steam tables give values for the specific entropy and specific enthalpy for each state.  From that, you can calculate G. ).  I suggest you do the same.  Sounds interesting.  The steam tables for superheated states can also be used to determine which side of the saturation line the vapor and the liquid reside.

I made the calculation, and, unlike the figure in the post, the saturation line is a smooth  curve.  But the other features are the same.  The superheated vapor lies below the saturation line, and the compressed liquid and solid are above the line

Phase diagrams are not limited to temperature vs pressure. They can be made with any set of thermodynamic quantities. The diagram in your question is for a particular pressure at which the liquid phase exists. If you changed the pressure, the lines would shift. A triple point will be represented by all three lines intersecting through the same point.


text:
Influence of adding an acid to an equilibrium reaction that is not part of an equilibrium reaction
Question:

The equation of chemical equilibrium reaction is the following: $$\ce{[Cu(OH2)4]^2+ + 4Cl- <=> [CuCl4]^2- + 4H2O}$$
  The $\ce{[Cu(OH2)4]^2+}$ ions are of blue colour, while the $\ce{[CuCl4]^2-}$ ions are of yellow colour. The mixture at equilibrium is of green colour (meaning that there are both types of ions inside it). How does the colour of the mixture change, if $\ce{HCl(aq)}$ is added?

I know that, if a reactant is added to the equilibrium reaction, the chemical equilibrium is "moved" towards the products, vice versa.
However, the hydrochloric acid $\ce{HCl(aq)}$ is not a part of the equilibrium equation, so I don't know how it could in any way influence the chemical equilibrium. I need explanation for this problem.

HCl splits into H+ and Cl- in solution.

The equation as written has two typos: one is the lack of a + sign on the left and the other is H20 instead of H2O on the right. They distract from seeing the beauty and intricacy of the problem.
You have cupric ion (hydrated) plus four chloride ions on the left in equilibrium with tetrachlorocupric dianion and water on the right.
The key concept is that the addition of hydrochloric acid is equivalent to adding hydrogen ions and chloride ions. In a way, adding "HCl" is a mental typo, because it disguises the fact that you are really adding H+ and Cl- ions.
The addition of more chloride ions forces the equilibrium to the right.
And thanks to user32369 for the suggestion to edit.


text:
Why do frozen whipped cream remain "frozen" when thawed?
I froze some pastry with whipped cream on the other day (by mistake, I placed it on my balcony and expected it to be a few degrees above zero, but it wasn't). When I unfroze it again I was expecting the whipped cream to become soft again (or, I was actually expecting it to "collapse", that is lose all the air whipped into it) but instead, even when it was at room temperature, the whipped creamed remained basically as firm as it was when it was -10°C. Leaving it for a few hours didn't change that.
Milk, "unwhipped" whipped cream, butter and so on all can be frozen and then thawed and the result is more or less its original texture and consistency (just a quick shake and your fine). Why does whipped cream behave so differently?
Edit: I can add that I have googled this and find a lot of articles that confirm freezing whipped cream is a bad idea but no one explains why.


text:
Is the haloform reaction possible with fluorine and pseudohalogens also?
The haloform reaction is:
$$\ce{RCOCH3 +3X2 +4OH^- -> RCOO^- +CHX3 + 3X- + 3H2O}$$
where, $\ce{X}$ is $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$. However, I was wondering if this reaction works with fluorine and pseudohalogens? For instance would the following reactions occur?
$$\ce{RCOCH3 +3N2O4 +4OH^- -> RCOO^- +CH(NO2)3 + 3NO2- + 3H2O}$$
$$\ce{RCOCH3 +3(CN)2 +4OH^- -> RCOO^- +CH(CN)3 + 3CN- + 3H2O}$$

Nitroformate would be fine as a leaving group, but you would probably need a source of NO2+, as the electrophile. This is usually done in acidic conditions (concentrated sulfuric acid), so it probably wouldn't work (at least with the example you gave).

TL;Dr: Highly unlikely to be made using the general haloform reaction.
Long answer:
The mechanism of haloform reaction involves hypohalite ions resulting from the reaction between the base and the halogen.
$$\ce{X2 + NaOH -> X- + XO- + H2O}$$
If we use fluorine as halogen, the resulting reaction will produce hypofluorite ion but this ion is extremely unstable and as such fluoroform can't be made using this reaction.  Trifluouromethane is known and used industrially in the manufacture of polytetrafluoroethylene ("Teflon"), but it is commonly made by anion exchange with another haloform, typically chloroform plus $\ce{HF}$.

For many years, chemists have been unable to isolate cyanoform as a neat, free acid. However, in September 2015, it was successfully isolated by a team of scientists at Ludwig Maximilian University of Munich. It is stable at temperatures below – 40 °C. So, it requires drastic conditions to synthesize cyanoform and is only stable at low temperatures. So, it is highly unlikely to be made at ambient conditions using laboratory grade reagents.

Nitroform has been made in the laboratory by the hydrolysis of tetranitromethane under mild basic conditions. But can it be made using general haloform reaction? Yes, it can technically be made but there are face several problems. Let's start with the starting reaction i.e. reaction between base and nitrogen dioxide which produces nitrite and nitrate ions:
$$\ce{2OH- + 2NO2 -> NO2- + NO3- + H2O (+other nitrogen oxides)}$$

The nitrite that forms is not that great reactive oxidizer.
There might be an undesirable side reaction which is kind of a Meyer reaction to form undesirable nitrolic acid/psudonitrol which in basic conditions gives blood red/blue colored solution.
The reaction involves huge amount of nitrogen oxides in general which is not very economical to deal with.
Multiple SN1 reactions occurs that generates a lot of isomers and other nitro compounds of compounds along with a whole lot of nitrite esters.
Steric hindrance might play a role and the reaction would not proceed properly.
The nitroformate ion ($\ce{C(NO2)3^-}$) might ionize to form different ions: $\ce{CH(NO2)2^-}$, $\ce{CH2NO2^-}$.
Nitroform is itself a strong oxidizer.

There are discussions in forums regarding trinitromethane synthesis using general haloform reaction but this ended up being fruitless. It even discussed using copper salts as catalyst and even employed haloform itself to react with $\ce{Ag/NaNO2}$ in DMF/ether but these syntheses posed so much problems and the yield was generally very, very poor.


text:
H1 NMR Spectra of inorganic complex, Cu4OCl6(2-mepy)4
I'm having difficult interpreting the following H1 NMR I have been given for Cu4OCl6(2-mepy)4.

The NMR spectra obtained with integration trace beneath. I'm not too sure how to rationalise the 4.5:1 ratio between the peaks nor the cause of the small broad peak at 115ppm (well to the left of the spectra).

I'm also not too sure why the H1 NMR range is so vast. I appreciate being bonded to metals the electron density is changed across the relevant proton environments but by that much?
Thanks a lot

What you have there is a typical example of an H NMR spectrum of a paramagnetic compound. The formal oxidation state of the Cu here is Cu(II) resulting in d9 configuration. This means that there will be an unpaired d electron making the Cu atoms paramagnetic.
The lone electron in the Cu atoms disturbs the magnetic field the Hs' feel resulting in extreme chemical shifts and also provide a very efficient relaxation mechanism resulting in extreme peak broadening (in some cases you cant even see the resonances at all). The highest paramagnetic shift I have encountered so far was around 250 ppm for an Fe(II) complex I made. In another occasion we had resonances at 1200 ppm!
As for the peak broadening I had peaks of ν1/2 of a few thousands of Hz. The integration wont always make much sense with paramagnetic compounds either.


text:
How to give an evidence in support of the arenium ion mechanism?

Outline the mechanistic steps involved in the arenium ion mechanism for
  aromatic electrophilic substitution. Draw the energy profile diagram for the proposed mechanism. Give one evidence in support of the proposed mechanism.

I can solve the first two questions but I got confused about what the last question exactly asking for. Any help will be appreciated.

You say you've already drawn the energy profile diagram for your proposed mechanism , now you want to make sure that your reaction mechanism proceeds through the most stable transition state , ( Least activation energy ) and therefore , on the energy profile diagram , the peak occuring at the transition state should not be too high .Also you wanna make sure that your final products have energies lesser than the  reactants , because in that case , your $\Delta H$ would be $-ve$ and from the equation $\Delta G=\Delta H-T\Delta S $, there would be a greater chance your reaction moves forward.
You also should have a look at this

Thanks But the last question that I am asking for carries only one mark. How can I answer it precisely?

@Sanu_012 You could explain it from the energy diagram , like I have , or you could point out in the mechanism that the carbocation formed during the mechanism is resonance stabilised and hence favours the forward reaction.

Thanks a lot.  I gonna explain it in the carbocationic way.  It is worth one mark.

I don't see how this is supposed to answer the question...


text:
Why does free-radical chlorination of isopropyl bromide give 1-bromo-2-chloropropane?
Source: Morrison and Boyd - Chapter Alkanes - exercise no. 15

Free-radical chlorination of either n-propyl or isopropyl bromide gives 1-bromo-2-chloropropane, and free-radical chlorination of either isobutyl or tert-butyl bromide gives 1-bromo-2-chloro-2-methylpropane. What appears to be happening? Is there any pattern?

I have really no idea how this can be happening.


text:
Rendering software for publication
Does anyone know how I can render chemical structures of small organic molecules to look like the below images?

I have been looking around, and I cannot seem to find an answer. The closest lead I had was this link saying you can "copy" your ChemDraw then "paste" into Chem3D. I looked through Chem3D's manual and found no such rendering style as the pictures linked above.
I know Chimera can do Simple Line Drawing but it seems to render small molecules only within the plane. I have seen plenty of examples of proteins, but I am trying to find something for small organic molecules.
I have also found references to the style named after the artist David Goodsell. I have seen this style referred as a non-photorealistic rendering referenced in this sci-hub link, page 11, figure 9.  Again, I have seen this shell/toon shading for large biomolecules, but not for small molecules.
Any help would be appreciated. I linked to a few sources above hoping it might help someone else trying to find similar rendering styles.

Have you tried ORTEP? These pics look like the output of one of the ortep options but you need to have the molecular structure of the compound which you might be able to find on conquest (CCCD) if it is known. Then take the cif file and load it to ORTEP. Personaly i (and most people) prefer Olex2 though. Alternatively, i think you could generate a coordinate file from a computational software (e.g. Gaussian) and get a similar picture with computed bond lenghts.

If you actually have the 3D coordinates for the atoms then you can use PyMOL for even prettier - but highly customisable - pictures, e.g. here (see here for parametrisation).

@AMM I see, these structures do look standard among the crystallography community. I will consider trying Olex2. I am concerned if this style is a "standard," when presented in a presentation or publication the structure will infer as s crystal structure and not a result of an optimized electronic structure theory calculation.

@lemon I do have all the 3D coordinates so I will have to try PyMOL. I have never used the software, and the learning curve seemed intimidating. None the less, I will give it a try.

If the compound is known and the structure exists then most computational people will use it to optimize their geometry. If not they start from something similar and add the missing atoms. In either case the the software used for optimization (e.g. gaussian) can generate nice pictures. It is not very clear to me why you need the pics to look like above which looks quite old (hence why i sugessted ortep).

@user3587374 If you get stuck then feel free to ask me for guidance - I'm a PyMOL user.


text:
Theoretical comparison of critical constants of real gases
We know that for a real gas, its critical constants are: $T_\mathrm c=\frac{8a}{27Rb}$, attempt to $P_\mathrm c=\frac{a}{27b^2}$ and $V_\mathrm c=3b$. Now, let's compare the value of these critical constants for three real gases, say, $\ce{CO2}$, $\ce{NH3}$ and $\ce{H2O}$.
Knowing that $V_\mathrm c=\frac38\cdot\frac{RT_\mathrm c}{P_\mathrm c}$ (for one mole), experimental data from Wikipedia is:
$$
\begin{array}{|c|c|c|c|}\hline
&T_\mathrm c&P_\mathrm c&V_\mathrm c\\\hline
\text{Carbon dioxide}&\pu{304K}&\pu{7380 kPa}&\pu{0.1284L}\\\hline
\text{Ammonia}&\pu{406K}&\pu{11280 kPa}&\pu{0.1122L}\\\hline
\text{Water}&\pu{647K}&\pu{22060kPa}&\pu{0.0914L}\\\hline
\end{array}
$$
We will now attempt to justify this regular trend in the values qualitatively.
For $T_\mathrm c$, we know that it is the maximum temperature above which a real gas cannot be liquefied. If a gas has stronger intermolecular forces (IMFs), its molecules possess greater attractive forces to be able to overcome separatist forces that might separate them, and thus they'll still be able to liquefy. Thus, if a gas has higher IMFs, its $T_\mathrm c$ would be higher.
We know that H-bonds formed by water are much effective than those by ammonia (read), and carbon dioxide is a non-polar molecule with only London dispersion forces. Hence, we would expect $T_\mathrm c(\text{water})>T_\mathrm c(\text{ammonia})>T_\mathrm c(\text{carbon dioxide})$ by our qualitative reasoning, which agrees with the experimental order.
By definition, the critical pressure is the pressure needed to liquefy a real gas at its critical temperature. Now, since water molecules have the highest attractive forces, one would expect that they require the least external "support" (aka force per unit area) to be able to liquefy themselves (by bringing the molecules closer). Hence, we would expect that $P_\mathrm c(\text{water})<P_\mathrm c(\text{ammonia})<P_\mathrm c(\text{carbon dioxide})$. However, this does not match with the experimental trend! (which happens to actually be in the reverse order)
Moreover, I am unable to come up with a way to predict the trend in the critical volume at all.
Hence, my questions are:

Since $V_\mathrm c$ depends on the ratio of $T_\mathrm c$ to $P_\mathrm c$, am I correct in saying that its trend cannot be determined without actually being given the values for the other two constants? Or is there some clever way that I am overlooking?
Is there any other way to justify the trend in $T_\mathrm c$?
What is the fault in my reasoning for the trend in $P_\mathrm c$? And what is the correct reasoning?

In essence regarding this problem
$$T_\mathrm c \propto \frac{a}{b}$$
$$P_\mathrm c \propto \frac{a}{b^2}$$
$$V_\mathrm c \propto {b}$$

Values of $a$:
$\ce{NH3} \to 4.225\ \mathrm{l^2\ bar/mol^2}$
$\ce{CO2} \to 3.640\ \mathrm{l^2\ bar/mol^2}$
$\ce{H2O} \to 5.536\ \mathrm{l^2\ bar/mol^2}$.
This trend as justified by the OP through hydrogen bonding is absolutely correct.
Values of $b$:
$\ce{NH3} \to 0.0371\ \mathrm{l/mol}$
$\ce{CO2} \to 0.04267\ \mathrm{l/mol}$
$\ce{H2O} \to 0.03049\ \mathrm{l/mol}$.
$b$ or effective molecular size is essentially calculated by multiplying the size of a molecule by four. The trend over here is obviously determined by molecular size only.

Now:

The critical temperature and critical pressure are in direct /inverse proportion with $a$ as well as $b$ .An interplay of two factors. Predicting the orders of both of these cases become extremely difficult  without data. The opposing orders of $P_\mathrm c$ and $T_\mathrm c$ are seen as the order of proportion changes i.e $P_\mathrm c$ is inversely proportional to the square of $b$ and $T_\mathrm c$ is inversely proportional to $b$.
Trends in $V_\mathrm c$ on the other hand can be safely predicted by estimating the size of molecules as it only dependent upon one variable that is $b$.

All data is from this Wikipedia page


text:
Le Chatelier’s principle when pressure of gases increased with the same stoichiometric coefficient

$$\ce{C (s) + H2O (g) <=> CO2 (g) + H2 (g)}$$
What will happen to the equilibrium when the pressure of gases is increased? What will happen to the yield of $\ce{H2}$?

Since pressure increased means the volume is decreased, equilibrium will shift to the direction with the least number of moles. But since the stoichiometric coeffients are all the same, am I right to say that it will not shift?
But I doubt my answer as the question states - "pressure of gases increased". The reactants include solid carbon, meaning I do not count that in. Does this mean that equilibrium shifts to the right?

Your thinking is on the right track. However, what you really need to look out for is number of gas moles.
In order to understand this, think in terms of the equilibrium constant. The equilibrium constant for this reaction is:
$$K_\text{p}=\frac{P_{\text{CO}_2}P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$$
Remember, since carbon is a solid, changing the volume/pressure is not going to change anything about it, so it is not in the equilibrium constant.
Since we are introducing a supposed change into the system, we have to write its reaction quotient. The reaction quotient, $Q$, is just a way to determine which way the reaction will have to shift in order to return to equilibrium. We write it in pretty much the same way:
$$Q=\frac{P_{\text{CO}_2}P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$$
Even though it looks the same, remember that we are calculating this after introducing a change. So what we really plug into this equation is the following, where $x$ is the factor by which we increase pressure (suppose that $P_{\text{CO}_2}$, $P_{\text{H}_2}$, and $P_{\text{H}_2\text{O}}$ are the equilibrium pressures):
$$Q=\frac{(P_{\text{CO}_2}\cdot x)(P_{\text{H}_2}\cdot x)}{(P_{\text{H}_2\text{O}}\cdot x)}$$
$$Q=\frac{x^2P_{\text{CO}_2}P_{\text{H}_2}}{xP_{\text{H}_2\text{O}}}$$
Since we are increasing pressure, $x$ is obviously positive. Thus, $Q>K_\text{p}$.
Due to Le Chatelier's principle, since we introduced a stress into the system, the system will shift away from that stress. $Q$ and $K_\text{p}$ are a good way of mathematically representing this stress. $Q$ is essentially a representation of the system's current state, and $K_\text{p}$ is the representation of the system's ideal state.
So, how do we reach this ideal state? It's simple; we want $Q$ to equal $K_\text{p}$. Obviously this will happen through a change in pressure. Since $Q$ is too big, we need to decrease the numerator in order to make it equal to $K_\text{p}$. The numerator represents the pressure of the products, so the reaction will shift away from the products, using up more products and creating more reactants, until it reaches equilibrium.
Hope that helps.

First of all, How can you practically increase the pressure of the gases individually when they are all present in the reaction vessel in some definite proportion and they are all in the equilibrium state? You can only increase the pressure of the  whole system. That's the only thing possible.
Here $\Delta n_\mathrm{g} >0$. According to Le Chatelier's principle, increase in pressure favours the backward reaction. So, equilibrium shifts to the left.
If you have any doubt in the statement "If $\Delta n_\mathrm{g} > 0$, pressure favours the backward reaction. See my answer in the opposite analogous statement  of If $\Delta n_\mathrm{g} > 0$, pressure favours the
 forward reaction.
Pressure dependency in Haber Bosch ammonia synthesis


text:
Why is the rate of nitration greater for thiophene than toluene?
Toluene has a $\ce{CH3}$ group attached to the benzene ring and as a result due to the +I effect of $\ce{CH3}$, the electron density in the ring increases and it undergoes nitration faster than a normal benzene ring.
But what causes thiophene to have an even greater rate of nitration ?

Related answer https://chemistry.stackexchange.com/questions/89471/difference-between-tendency-of-benzene-and-thiophene-to-undergo-sulfonation/89576#89576

I would suggest two reasons:

Thiophene is an electron-rich compound because of presence of two lone pairs on the sulphur atom. Hence, it attracts electrophiles readily.
Though its aromaticity is lesser than that of benzene, the delocalisation is among five atoms in thiophene whereas in benzene it is among six atoms. Hence electron density in thiophene is more.

I am not sure if this answer is correct. (1) Yes sulphur has lone "pairs", but only one of them is involved in resonance with the ring (the other is out of the place of the ring) (2) sulphur's 3p orbitals have weaker overlap than that of carbon's 2p orbitals with the ring (3) you seem to be comparing thiophene with benzene instead of toluene (as asked by OP), and have completely ignored the hyperconjugative increase in electron density (due to the methyl group in toluene which is absent in thiophene)

Yes, (3) I didnt notice toluene, my bad. (2) I don’t get why. (1) is again a drawback. My bad! Just realised how poor my answer is. Can you explain me (2)? And should i remove the answer?

(2) same size $\ce{2p\pi-2p\pi}$ overlap is better than different size $\ce{3p\pi-2p\pi}$.  "And should i remove the answer?" I am not sure what to do in this case. Lemme think.

I think you should remove it. It's going to be misguiding for other visitors. No offence but please remove.

It says I can’t delete an accepted answer :(


text:
Reactivity order of electrophilic addition

Find the order of the ease of electrophilic addition of the following:

$\ce{H3C-O-CH=CH2}$
$\ce{ F-CH=CH2}$
$\ce{Cl-CH=CH2}$
$\ce{NO2-CH=CH2}$

According to me, it should be 1 > 4 > 2 > 3 but in my textbook, it says 1  > 2 > 3 > 4.
If we take $\ce{F-CH=CH2}$ and $\ce{NO2-CH=CH2}$, the result after electrophilic addition would be $\ce{F-CH+-CH2E}$ and $\ce{NO2-CHE-CH2+}$ where E is the electrophile. Since both F and $\ce{NO2}$ are -I groups, further the $\ce{C+}$ carbocation is situated, more stable is the compound. The latter should be more reactive than the former. So why is my textbook going against this logic?

"Since both F and NO2 are -I groups, further the C+ carbocation is situated, more stable is the compound." This is wrong. F is also a +M effect group, which NO2 is not. Thus the former is more stabilized than the latter, which is also what your textbook says. Mr pea's answer below is correct. What is the cause of the reluctance you have in accepting that answer?

I verified with my professors and they say I am correct. Reactivity among halide and other groups is made only on the bases of -I effect and not +M. Only the orientation is made on the bases of +M not reactivity. @GaurangTandon

What do you mean by "orientation"?

Meaning, where the carbocation will form after the electrophile just attacks. @GaurangTandon

Which book stated that order?

For the electrophilic addition to carbondouble bonds, the $\pi$- electron cloud double bonds first attack the electrophile and then a carbocation is created adjacent to the carbon where the electrophile is added. Then that carbocation is attacked by the nucleophile to form the addition product. $$\ce{R-CH=CH2 + E -Nu -> R-CH^+-CH2-E -> R-CH(Nu)-CH2E}$$The above path is the general electrophillic addition reaction in the case of alkenes. The carbocation position may vary depending upon the $\ce{R}$ group.

When the $\ce{R}$ is $\ce{CH3-O}$, there is a positive mesomeric effect (+M or +R effect) of the group due to which  when the carbocation is formed in the adjacent carbon, it gets  high stabilisation due to the +R effect, which in turn decreases the energy of the transition state of carbocation heavily, and thus the reaction rate becomes very high. That's why, the rate of electrophilic addition is very high in this case is very high, i.e it is the most easy.
When the $\ce{R}$ is $\ce{F}$, there is a significantly stronger +R effect due to $\ce{2p\pi -2p\pi}$ overlap. So, the carbocation gets a slight stabilisation due to this +R effect but there is also a -I effect in case of $\ce{F}$ which is the factor of destabilisation of carbocation. That's why, ease is relatively lower due to significant +R but stronger -I effect.
When the $\ce{R}$ is $\ce{Cl}$, there is a very weak +R effect (vinyl chloride has a weak  effect of +R, which is observed from the lesser dipole moment due to lesser charge separation  than ethyl chloride, where there is no +R), but dominant -I effect. So, in this case carbocation is more destabilised. So, the ease of addition is decreased even further.
When the $\ce{R}$ is $\ce{NO2}$, there is -I effect and also a dominant -R effect or -M effect. So the $\pi$ -electron cloud gets delocalised with the nitro group and thus due to decreased $\pi$-electron cloud density, it becomes difficult for it to even attack the electrophile. So, fractions of molecules forming the carbocation is extremely less, which actually increases the activation energy of the reaction. Thus, the reaction becomes very difficult with nitroethylene.

Consideering the overall effects described above, the ease of electrophilic addition should be,$$\ce{CH3 -O - CH=CH2 > F - CH=CH2 > Cl - CH=CH2 > NO2 - CH=CH2 }$$

"significantly stronger +R effect" stronger than what?

@GaurangTandon I think it is as compared to chlorine. Would have made more sense if he answered fluorine after chlorine, or mentioned that resonance is less significant with chlorine, in the 3rd part.

There will be +R(or +M) in the cases involving F and Cl, +R being better with F than Cl due to better orbital overlap with C.
Fluorine donates a lone pair to carbon, leading to a positive charge on F. Although F has a positive charge, this resonating form has special stability since this resonance form has more number of bonds, and since octet of all atoms and duplet of $\ce{-H}$ are complete.*
There is no lone pair on N in $\ce{-NO_2}$ so there is no +R effect in this case.

Yes, your assumption is correct, I edited my question so it's easier to understand. I know the cases involving F and Cl. My question is, which would be more reactive between the halogen group(II and III) or nitro group(IV)? Could you explain that? Thank you.

@raajsuriya both II and III will be more reactive than IV due to more stability of the carbonation intermediate. -I effect of nitro group is much stronger compared to that of fluorine. Effect at second carbon of nitro group will be comparable to that of fluorine group at first carbon(not completely sure about them being comparable, but there is a considerable difference in their inductive effects)

Since F and Cl are both deactivators (in the context of EAS) with F being more deactivating than Cl, it would thus make sense to say that F must be more electron-withdrawing from the pi bond than Cl, making the bond less nucleophilic. The + M/R effect is less significant than -I effect, as evidenced by the rates of reaction in EAS of F-substituted arenes and Cl-substituted arenes. Do you have any references to support what you are saying?

@TanYongBoon I’m sorry but I don’t have any reference to support what I am saying. Although what I am saying is what my professor has said at least a couple of times and according to him, in this case the extent of orbital overlap between F and C is the deciding factor.

@TanYongBoon I guess Mr_Pea is right...I've heard the same thing from my teacher a decent number of times. +M of F>+M of Cl and -I of F<-I of Cl

In electrophilic addition of a hydrogen halide ($\ce{HX}$), the $\ce{H+}$ attacks first. Reactivity in this case is measured by the stability of the carbocation. The reason for this is more stable the intermediate/carbocation, greater the time for the subsequent attack by the anion.
Pertaining to the question in hand we will be forming a two degree carbocation in all the cases as the only other option is a very unstable three degree carbocation. Let's analyse the carbocations one by one.
First let's rank  the cases according to the mesomeric effect.

Oxygen(+M): Good overlap due to matching orbital size.
Fluorine(+M): Good overlap due to matching orbital size
Chlorine(+M): Poor overlap due to a larger orbital size than carbon.
Nitro (-M): Actively destabilizes the intermediate

From this we can deduce the order to be:
$$\ce{O} \overset{?}{<=>} \ce{F > Cl > NO2}$$
The deciding factor between oxygen and fluorine is the $\ce{-I}$ effect.
Fluorine has a greater pulling tendency therefore it destabilizes the carbocation more than oxygen. Thus, the final order is
$$\ce{ O > F > Cl > NO2}$$
which is also the order given by your book.

$\ce{-OCH3}$ is a strong activating group which activates or increase the rate and $\ce{-NO2}$ is a strong deactivating group which decrease the rate. $\ce{-F}$ and $\ce{-Cl}$ are weak deactivating groups; they also decrease the rate but less in comparison with $\ce{-NO2}$. $\ce{-F}$ is a weaker deactivating group than $\ce{-Cl}$.
So the order is $1>2>3>4$


text:
Identifying covalent character in ionic compounds when varying multiple ions
By Fajans' rules, we can easily find out which compound shows covalent character.
Example: Among $\ce{NaCl}$, $\ce{MgCl2}$, $\ce{AlCl3}$ which one is more covalent?
Answer: There is a point in Fajans' rules that the compound which has more charge number that compound will show covalent character. In $\ce{AlCl3}$, Al has the highest oxidation state (+3) so it will show more covalent character than $\ce{NaCl}$ and $\ce{MgCl2}$.
The point is that since all three compounds I have mentioned have chlorine in common, it was easy to catch the solution but if the compounds were given as $\ce{NaF}$ and $\ce{LiCl}$, which one shows more covalent character, and how can we apply Fajans' rules?

You will have to consider two factors here: polarising power of the cation and polarisability of the anion.
There is more than one part of Fajan’s rule(Source-J. D. Lee, Concise Inorganic Chemistry):

A small positive ion favours covalency.
A large negative ion favours covalency.
Large charges on either ion, or both ions, favours covalency.
Polarisation and hence covalency is favoured if the positive ion does not have noble has configuration.

According to these, the $\ce{Li+}$ ion is more polarising than $\ce{Na+}$ ion, and $\ce{Cl-}$ ion is more polarisable than $\ce{F-}$ ion.
Since both factors support it, $\ce{LiCl}$ is more covalent than $\ce{NaF}$.
Somtimes, like in your question, both the factors support each other. Sometimes, when they are opposite, like in the case of $\ce{LiF}$ and $\ce{NaCl}$, it is tough to determine relative covalency using Fajan’s rule.

does that mean that d block elements (Transition elements) are more ionic than alkali metals and alkaline earth metal compounds ?

@Tanuj actually it means the opposite; the d-block elements are generally smaller than the corresponding alkali metals/ alkaline earth metals.

Okay , got it .