text:
problem with loops over basis sets in psi4 using python
I've recently started using the psi4 quantum chemistry software. I wanted to compare a couple of different basis sets (mainly pople style) on an amino acid. So I've written a python code inside the input file to iterate over the different sets.But only the first loop runs and the second fails miserably giving an error code:
PSIO_ERROR: Attempt to write into next entry: 35, SO-basis Overlap Ints
PSIO_ERROR: unit = 35, errval = 18
PSIO_ERROR: 18 (Incorrect block end address)
Traceback (most recent call last):
File "<string>", line 42, in <module>
File "/usr/share/psi/python/driver.py", line 617, in energy
procedures['energy'][lowername](lowername, **kwargs)
File "/usr/share/psi/python/proc.py", line 1231, in run_scf
scf_helper(name, **kwargs)
File "/usr/share/psi/python/proc.py", line 1457, in scf_helper
e_scf = psi4.scf(precallback, postcallback)
RuntimeError: PSIO Error

here is my code:
basis_sets = ["6-31G", "6-31G*", "6-31G**"]
for basis_set in basis_sets:
    set basis =  $basis_set
    energy('scf')

I've tried to run it without the code and there was no problem.
I copied the code from the psi4 tutorial page for python scripting
Thanks for the help.

I guess this may be borderline off-topic here. However, please provide the complete input file which failed with along with the version and other parameters necessary to reproduce the problem. Have you tried using the original example with Dunning's basis sets? This error looks quite cryptic and I have no experience. Have you tried substituting the alias 6-31G* with the equivalent 6-31G(d)? I can't be sure, but maybe the asterisk is causing some kind of trouble. You can [edit] in more information, alternatively you might want to consider asking this on a more computational site.

I tried running these basis sets individually without the loop and they worked fine. The other part is just an .xyz file with my molecule nothing special.

Well, apparently there is something wrong with the loop then. Please provide the information I have asked for. This is already borderline off-topic and as it is not reproducible it basically is of no further use in the future. (Be aware that the community might decide to close this question.) It would also be helpful if you could try what I suggested, as I have a gut feeling that the python bit does not recognise the alias what the global program probably does.

I tried to run the different basis sets under different aliases and it recognizes them. I'll couldn't find the source of the problem, and I guess that for now I'll have to run them manually.

If it solves your problem, feel free to add an answer.

Here is an input file that will reproduce this error:
molecule {
    b      0.0000        0.0000        0.0000
    h      0.0000        0.0000        1.0000
}

basis_sets = ["6-31G", "6-31G*", "6-31G**"]
for basis_set in basis_sets:
    set basis = $basis_set
    energy('scf')

The problem is that Psi4 sees there are files remaining in scratch that it can use, starting with the spin orbital-basis overlap integrals, but because the basis set has changed, the file is the wrong size. You'll need to clean these up after the energy call by adding clean() within the loop.

So just by adding clean() after the energy ('scf') the problem will be solved?

Yes, try it for yourself


text:
What is the best way to introduce OH group into aromatics directly?
I will try reactions to hydroxylate a haloaryl compound containing fluorines and a NH group.
I consulted databases but could not find ways where yields are good.
If there are people good ways  to introduce a OH group into the compound directly, please tell me.

There is no universal and good direct methods, since electrophilic oxygen is not a nice thing to begin with and hard to come by.  The usual way to go around it is to utilized sequence of lithation (BuLi + ArH = BuH + ArLi) followed with synthesis of arylborate ester (ArLi + B(OMe)3 = ArB(OMe)2 + LiOMe) and than oxidation (ArB(OMe2) + H2O2 =  ArOH + HOB(OMe)2) in a basic solution).  Fluorine isn't a very mobile atom in transmetallation with organolithium compounds, so using extremely low temperature it might be possible to preserve it.

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text:
Element Song, updated with 118 names?
Tom Lehrer’s The Elements song (listen) mentions all elements up to Nobelium, and ends with the appology

These are the only ones of which the news has come to Harvard,
  And there may be many others, but they haven't been discovered.

Now that’s a bit dated. We can’t ignore trans-uranic elements when every bedroom in developed countries contains Americium-241, and now that official names have been ratified for all elements through period 7, I can hope that an updated song will be available.
Are there any such songs, whether a revision of the original or based on a new melody, or any known efforts underway to complete one?

I have downvoted. I personally really don't think this is on topic here, it is not an actual chemistry question. (But I'll hold back on the close hammer.) Would you consider coming to chat for these things?

While I believe this is on topic, downvoted because the answer is found by simply googling.

Americium is in the original song though...

I'm voting to close this question as off-topic because this isn't a question for the main site. This is the sort of thing which should be asked in chat.

Regarding close-vote: A good policemen/moderator/...  does not blindy enforce the rules, but allows an exception from time to time. In particular, when the rule violation does not really bother anyone, is especially popular and is a clear one-time exception.

@mhchem No, a good moderator sits back and only handles extremely clear-cut cases or those in which unresolved flags have been raised. It is the community’s job to decide whether this is on-topic or nur and it currently has accumulated three close votes (and two leave-open votes in the review queue). I think, it’s safe to say the community’s split on the subject.

The NEW Periodic Table Song (Updated)
AsapSCIENCE
https://www.youtube.com/watch?v=VgVQKCcfwnU

Whilst this song may technically be on similar subject matter, it's nothing like the original song. Much like many of his works, it was inherently comedic.

The Wikipedia page you linked to, links to the song's text which has an alternative ending containing 3 more elements.

[Alternate ending:]
Lawrencium and Hahnium and lastly Rutherfordium
  If there are any others, I'm afraid I haven't heardium

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text:
Bainite at room temperature?
I read that if, after a tempering from T>723°C to T*=[250°C,550°C], I do an isothermal transformation of an eutectoid steel, I'll get bainitic microstructure.
Does this microstructure exist only if temperature remains at T=T* also after microstructure was formed?
In other words, if I do a cooling until room temperature (after bainite was formed), will it exist at this temperature?
Thank you very much.

The microstructure is retained upon cooling. After bainite forms, you can even quench it and it will remain bainite. The only way to get martensite is to cool it so fast that bainite has no time to form. The only way to get bainite is to cool it so fast that bainite can form, but not so fast you get martensite, and not so slow that you get pearlite. After the microstructures have formed, they remain, unless you reheat it.

Although mechanical deformation or fatigue may give pearlite a shot at nucleating.


text:
Which metals hydroxides dissolve in excess NH4OH?
Some metal hydroxides like $\ce{Ag(OH)}$ and $\ce{Cu(OH)2}$ dissolve in excess $\ce{NH4OH}$ to form metal ammonia complexes.
I want to know which other common metal hydroxides dissolve in excess $\ce{NH4OH}$. Also, can it be somehow predicted ? I couldn't find much relevant information on the Internet.

There are a number of unknowns and a number of equations but this is a matter of the various equilibriums. Remember that you could also add NaOH so the solution could be more basic than you could get with NH4OH alone.

The Wikipedia article for "Metal amine complex" gives many examples.  According to the article:

Almost all metal ions bind ammonia as a ligand, but the most prevalent
  examples of ammine complexes are for Cr(III), Co(III), Ni(II), Cu(II)
  as well as several platinum group metals.

Specific examples include:
$\ce{[Pt(NH3)4][PtCl4]}$
$\ce{[RhCl(NH3)5]Cl2}$
$\ce{[Cr(NH3)4Cl2]Cl}$
$\ce{[Co(NH3)6]Cl3}$
$\ce{NH4[Cr(NCS)4(NH3)2]}$
$\ce{[Zn(NH3)4]^2+}$
$\ce{[Ag(NH3)2]+}$
$\ce{[Ni(NH3)6]^2+}$
$\ce{[Cu(NH3)4(H2O)2]^2+}$
Several others are listed just in that one Wikipedia article, along with a fair list of references.


text:
What is the effect of change in pH on precipitation?
What is the effect of change in pH on precipitation? I was curious as to how pH actually affects precipitation.

Now that's a really, really broad question... You realize there are quite a few different types of compounds that can precipitate, right?

What compounds are in your solution and what are you trying to / not to precipitate?

It actually has multiple effects.
First, most of the precipitates contain ions that are either acids or bases. For example, if you look at CaCO3, and decrease pH, the carbonate ion might get protonated:
$$\ce{CO3^2- + 2H+ -> H2CO3 -> H2O + CO2 ^}$$
So what you will see is that the precipitate is dissolved. It can also go the other way. If you have calcium-bicarbonate solution, and increase the pH, it might get deprotonated and precipitate:
$$\ce{HCO3- -> H+ + CO3^2- ->[Ca^2+] CaCO3 v}$$
Note that most of the cations can have similar reactions, but complexation might also play an important role. All in all, things can get really complicated, and this is why it is an useful test in determining compounds.
Second, the surface charge might also play a role. For example proteins can be denaturated — precipitated — by acids, because their surface charge, hence their solvataion state changes with adding more acid to it, making the solution instable. This is a whole research area in the chemistry of colloids.
Lastly there is an effect on ionic strength and ion activities. This is the most complex of all, and little is known about the exact nature of this effect.


text:
How can I find the ratio of molecular masses using the ideal gas equation?

A person evacuated a cylinder and filled in $4~\mathrm{g}$ of gas A at $25~\mathrm{^\circ C}$. Pressure was found to be $1~\mathrm{atm}$. Another person filled another $8~\mathrm{g}$ of gas B in the cylinder at the same temperature. The final pressure was found to be $1.5~\mathrm{atm}$. What is the ratio of molecular masses of A and B (assuming ideal gas behaviour)?

I tried to do it this way:
Taking the molecular mass of A as $x\ \mathrm{g}$ and that of B as $y\ \mathrm{g}$, I formed the respective ideal gas equations but the required terms ($x$ and $y$) just cancelled out.
$$N(a) = 4/x ;\quad N (b) = 8/y$$
So$$V (a) = \frac{4 \times  22.4}{x};\quad     V (b) = \frac{8 \times  22.4}{y}$$
So, ideal gas equation for A:
$$1 X \frac{4 \times 22.4}{x} = \frac {4 \times  R  \times  25} x$$
Here $x$ cancels out. Similarly $y$ cancels out in its equation.
This is the problem I am facing.
Another thing I want to know is that the total pressure (i.e $1.5~\mathrm{atm}$), will that be the sum of the individual pressures of gases A and B?

Could you edit your question to show the math whereby your terms cancel out?

To answer your last question, yes, that is a central concept of the ideal gas law; that the composition of the gases is irrelevant with respect to the total pressure of the system.  Regarding the rest of your question, don't forget units!  That will both further clarify what you are doing and help you make sense of  it yourself.  If the units don't make sense, your equation probably doesn't, and vise versa.

@airhuff units will only determine the quantity involved. But to solve we need the numerics.

I don’t quite get where and how parts of this interact with the ideal gas equation, and I don’t understand the last equation block (is there something missing where the $1$ is?) but at least you showed your work so it’s no longer homework.

@Jan Yes, it's not a part of my homework. The 1 is the pressure found when gas A is put in the cylinder. So that 1 is just being multiplied. It was a question based on the ideal gas equation so it has to be solved that way (according to me). But if u have any other simple method, please let me know.

The ideal gas equation is $PV = nRT$
But for this problem V, R, and T are constants. So we can collect terms as:
$\dfrac{P}{n} = \dfrac{RT}{V}$
So $\dfrac{P_A}{n_A} = \dfrac{P_B}{n_B}$
Now you correctly determined the number of moles of A and B.
$n_A = \dfrac{4}{\text{MW}_A}$
$n_B = \dfrac{8}{\text{MW}_B}$
But you need to consider the partial pressures of A and B.

The cylinder was empty before A was added, and after A was added the pressure was 1 atmosphere. So the partial pressure of A is 1 atmosphere.
Before B is added the pressure is 1 atmosphere and after B is added the pressure is 1.5 atmospheres.  So the partial pressure of B is 0.5 atmospheres.

$P_A = 1$
$P_B = 0.5$
Now substituting into our simplified equation
$\dfrac{1}{\frac{4}{\text{MW}_A}} = \dfrac{0.5}{\frac{8}{\text{MW}_B}}$
$\dfrac{8}{\text{MW}_B} = \dfrac{2}{\text{MW}_A}$
$\dfrac{\text{MW}_A}{\text{MW}_B} = \dfrac{2}{8} = \dfrac{1}{4}$

Everything's fine but I still can't understand why the volume is a constant.

Same cylinder. We're assuming that the pressure doesn't deform the cylinder. Obviously if the "cylinder"  was a balloon then the volume would change as the pressure changes.

Ideal gas equation is PV = nRT
Here R is constant temperature is constant mentioned in question and also the volume is constant as volume of gas equals to the volume of cylinder (volume of gas= space occupied by it)
So in ideal gas equation n and P are variables rest are constant equation can be modified as,
P1/n1 = P2/n2
P1=1
n1=4/x
n2
P2=1.5
By solving we get n2=6/x
It is total moles of gas filled in cylinder
i.e. 4/x + 8/y
6/x = 4/x +8/y
Multiplying equation by x we get
6=4+8x/y
x/y = 1/4

When you say that the volume of gas equals the volume of the cylinder, do you mean the cumulative or combined volume of both the gases A and B?

Moreover, the P2 which you have written as equal to 1.5atm, that is the combined pressure of both the gases when both of them are present in the cylinder simultaneously, and not only of gas B as you seem to suggest.

Cumulative as in Q. Cylinder dimensions aren't mentioned so for sake of question we must assume that both gases are filled in cylinder of same dimension and hence have same    volume

OK ok it's my mistake everything is ok but p2 will have different value

Ya that's why I was getting confused. Thanks anyways.

Is it correct answer saksham

Yes 1:4 is the correct answer. :)

Then vote my answer and ask more questions so I can clear your doubts if any


text:
Backbonding in molecules like BF3
Can anyone please briefly explain to me the concept of backbonding along with necessary conditions for backbonding with some example? As I understand, backbonding is a donation of electron from electron-rich to vacant orbital, but I don't know anything else.

Related 1, 2.

$\ce{BF3}$ has trigonal plannar structure all the three $\ce{B-F}$ bonds lie in plane and thus p-orbitals of boron and fluorine become parallel.
Boron has empty p-orbital and p-orbital of fluorine contains lone pair, and hence  boron acts as Lewis acid and fluorine as Lewis base. Fluorine donates its lone pair to boron and this bonding is called backbonding.

It is rough diagram not exact one

But why? Why does Flourine donate its lone pair to Boron? What good is it doing for fluorine?

@PianoLand,
Formation of an extra bond( even if partial) is exothermic i.e releases energy and stabilizes our system. Hence it is a sort of resonance amongst Boron and Three Fluorine atoms. Hope it helps.

@user206007 While your input is appreciated, please don't post your comments as answers — as soon as you have enough reputation (50 points), you'll be able to fully participate in comment section.

@user206007 From resonance perspective we should not expect a double bond because boron gains negative formal charge.

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text:
Salt and sugar surfactants?
How do daily compounds such as table salt ($\ce{NaCl}$) and sugar ($\ce{C6H12O6}$) affect adhesion or surface tension of water? Considering their polarity, I guess they both increase it.

Related: http://chemistry.stackexchange.com/questions/49570/does-nacl-reduce-the-surface-tension-of-water/49584#49584

Salt and sugar are not surfactants but they are solutes when dissolved in water.

The surface tension of water is increased by the dissolution of either glucose ($\ce{C6H12O6}$) or $\ce{NaCl}$.  The following are excerpts from this table:
Solution$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $  Temp ($\mathrm{^oC}$)$\ \ \ \ \ \ \ \ \ \ \ $  Surface Tension (mN/m)
Water$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $25$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 71.97
Water$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $0$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 75.46
6M aqueous $\ce{NaCl}$$\ \ \ \ \ \ \ \ $ 20$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 82.55
Sucrose (55%) + water$\ \  $20$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 76.45
(Water is included at two temperatures in order to bracket the temperature of the mixtures.)
According to this Wikipedia article:

Surfactants are compounds that lower the surface tension (or
  interfacial tension) between two liquids or between a liquid and a
  solid. Surfactants may [emphasis mine] act as detergents, wetting agents, emulsifiers,
  foaming agents, and dispersants.

The article further states that "surfactant" is a concatenation of the term "surface acting agent". Furthermore, according to the Encylopaedia Britannica :

Surfactant, also called surface-active agent, substance such as a
  detergent that, when added to a liquid, reduces its surface tension,
  thereby increasing its spreading and wetting properties.

So, by these definitions and the data above, the compounds in question do not qualify as surfactants as they raise the surface tension of water.  Additionally, as they have no amphiphillic (having both hydrophilic and lipophilic properties) nature, they would not be considered surfactants in the common context of chemistry jargon.


text:
Why aren't these two resonance structure same?

Draw the resonance structures of Fluorobenzene.

In the solutions the structures below were given, I wonder why these structures are not just same. Both of these structures have a negative formal charge on the ortho position.

Are the above structure of Fluorobenzene same ?

They are not "really" different, but there are two ortho carbons in fluorobenzene so you draw a resonance structure showing charge delocalization to each ortho carbon.  There is also a third resonance structure with the charge on the para carbon - 3 resonance structures in all.

@ron Does that mean that when we draw resonance structures we aim to show all the different places where electron can delocalize ? does not matter if it is resonance structures are similar or different.

Yes, that's correct.

@ron Thank you $\ddot \smile$. That instantly solved many of my questions about resonance theory.

Simply - When you're drawing resonance structures you effectively number the atoms to break symmetry.

There are called equivalent resonance structures since the number of pi bonds, lone pairs and sigma bonds are the same. On the other hand, resonance is the delocalization of pi electrons. These structures show how pi electrons are delocalised from ortho position to para position and then back to ortho(on the other side of the molecule). So both the structures you mentioned are different in this context and do have a purpose in explaining the resonance phenomenon.0
Note: This rule only applies to hydrocarbons. In other cases, the two resonance structures need to have the same structural formula to be equivalent.

"There are called equivalent resonance structures since the number of pi bonds, lone pairs and sigma bonds are the same. " Does not resonance structures ought to have number of pi, sigma and lone pair equal ? I maybe wrong.

Two resonance structures must obey this condition: they must have the same number of electrons involved in sigma bonds, pi bonds and lone pairs. If they also happen to have the same number of pi bonds, lone pairs and sigma bonds, they are called equivalent resonance structures. Every resonance structure need not be a equivalent resonance structures but every equivalent resonance structures must be a resonance structure.

What about these? Are they equivalent? They have the same number of pi bonds, lone pairs and sigma bonds, but they are most certainly not equivalent.

@orthocresol Oh yes I think you are correct. So what is the criteria for equivalence ?

That both resonance forms should have the same structural formula.


text:
Finding equilibrium temperature when two things are mixed

10 g of ice at −10 °C is added to 10 g of water at 85 °C. What is the final temperature and amount of ice left in the system? (System is kept inside an ideal insulator)

This question may seem quite easy for many of the readers, but please help me with such questions as I find them tricky. I understand that this is a question which involves the application of the $Q=mc\,\delta T$ formula and that we have to equate the $Q$'s of both ice and water in order to obtain the answer.
Another thing which is troubling me is that the answers to most of such questions that I have encountered is 0 °C, and this question is not an exception.
Please suggest me on how to go about such questions, and provide some conceptual clarity on the topic if possible.

Solve this in steps rather than looking for one equation to cut the Gordian knot. So how much heat does it take to raise the ice to 0 C, and when you extract that much heat from the water what is the temperature of the water? Then solve the next part of the problem.

I think the reason that the answer to these kind of questions are so often 0$^o$C is to 1) drive home the point of how much latent heat is involved and 2) to be sure that you see that any ice remaining means 0$^o$C.

Just from my memory remembering some numbers, it would result as all water 0 Deg C.

In this type of question, you should bring all the matter into one state and calculate net heat.
In this solution, we will consider heat released as positive heat, absorbed as negative, and the state to which we will bring all the matter be water at 0 °C.

Here we were left with 20 g of water at 0 °C and zero heat if heat was some positive quantity we would use the heat to raise 20 g water to a specific temperature.
By equation $20\times T\times1=\text{heat left}$
$T$ is temperature to which 20 g water is raised
If heat was negative quantity, we would convert part of water at 0 °C to ice at 0 °C by equation
$M\times80=\text{heat left}$
$M$ mass of water which is converted to ice

I'll point out that textbooks now tend to use Joules not calories.

It will just different in value of latent heat and specific heat process will be exactly same

I don't disagree with your method or answer. I'm just trying to point out that "old" books use calories for such problems and "new" books use Joules. I'm an old dog and would use calories too...

OK in place of 1 cal we can place 4.2J as 1 cal= 4.2J

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text:
Hydrogen reactions with catalysts
I have a question about reactions involving hydrogen using catalysts. I'm a layman, so forgive me if my questions seem elementary.
Reacting hydrogen plus oxygen and a flame reacts, unleashing energy (in the form of an explosion) and creating water.
My questions:

Why does it need a "flame or spark "? Is there something unique
about the fire that causes the reaction? Or is it merely the heat of the flame? In other words, if you put hydrogen and oxygen in, say, a
pressurized vessel and added energy in the form of heat (but not a
flame) would it still react and create water? Would it do it without an explosion?
I understand catalysts "speed up" the reaction, or put another way,
reduces the amount of energy required to cause it. I've read that in
this particular reaction ($\ce{H2 + O2}$) copper works as a
catalyst. Is there a way to determine by looking at the chemical
formulas if other catalysts would work as well or better or is it
just trial and error? For example, could you tell if silver would
also act as a catalyst just by looking at the formula? Or would you
need to actually add silver and see for yourself?
When a catalyst is used, is it inexhaustible or does it get "used
up"? If you were using copper to react these two, could you use it forever? If not, is there a way to "recharge" the catalyst?

Thanks in advance, I'm trying to learn about this particular reaction for a project I'm working on ($\ce{H2 + O2}$) but without a formal science background, some of the scientific papers I'm reading are a little hard to grasp.

Electricity works fine as a spark. And a flame is basically a chemical reaction giving off heat and light so it’s always either the light (not in the case of hydrogen) or the heat. I haven’t seen any experiments that try to ignite with hot air or equivalents yet. I’m imagining it’s hard since you need to transfer the heat to the hydrogen efficiently.

Catalyst finding is trial and error. In theory, efficient catalysts could be calculated computationally but we’re not there yet. And the definition of a catalyst implies that it exits the reaction unchanged so it does not get used up.

I’m leaving these thoughts here in comments for somebody else to flesh out into a nice answer.

Adding a bit of information for your question a). The reactin we are looking at is:
$$\ce{2 H2 + O2 -> H2O}$$
As you can clearly see, the final product water has one oxygen atom for 2 hydrogen atoms, while in the beginning, we only had diatomic molecules. This means that during the reaction, at least one fo the diatomics had to break to single atoms. In this case, this happens with the reaction:
$$\ce{H2 -> 2H^.}$$
From this step, we have a nice reaction scheme called a radical chain reaction. However, we first need to break the $\ce{H-H}$ bond somehow, and this is what flame is doing. If you break a significant amount, then you don't need to provide flame anymore, the reaction is self-supporting — the flame is ignited.
Now, a catalyst is doing exactly this. Many metals, including copper, can adsorb hydrogen molecules to the surface, and this also results in breaking the $\ce{H-H}$ bond — so the explosion can start this way too.

Why does it need a "flame or spark "? Is there something unique about fire that causes the reaction?

Yes and no. Hydrogen/oxygen flame is a branched chain reaction, so after injecting enough free radicals it may propagate on its own without heat applied. This, however, not the case for many flames, as they may be not branched (hydrogen/chlorine flame) or require heat to produce combustible gases (common wood/oxygen flame)

if you put hydrogen and oxygen in, say, a pressure vessel and added energy in the form of heat (but not a flame) would it still react and create water? Would it do it without an explosion?

Most combustible compounds have self-ignition temperature at which they ignite without flame. Hydrogen/oxygen mixture has self-ignition temperature too. However if heated below it it may react slowly and nicely... just very slow.

Is there a way to determine by looking at the chemical formulas if other catalysts would work as good or better? Or is it just trial and error?

Sometimes, but generally trial and error. In most cases a family of catalysts catalize a family (or several families) of reactions. Once families are established, an educated guess is possible. However, it is usually not possible to guess catalytic activity without previous experience.
For example. I know, that late transition metals usually are good catalysts for hydrogenation. Hence, I can guess that I can use platinum as hydrogenation catalyst for most alkenes. However, I the knowledge was obtained by trial and error (and explanations are not obvious even after it was found). Finding the best match for the given pair of process and catalyst is also mostly a trial and error, though typically for each family a good educated guess may be made for the several most likely candidates.

When a catalyst is used, is it inexhaustible or does it get "used up"?

Ideally it should be inexaustible. In reality various processes lead to gradual loss of activity.

It not, is there a way to "recharge" the catalyst?

sometimes. For example, catalytic cracking of hydrocarbons uses strong solid acidic catalysts. In the process, coke is formed on the catalyst, leading to loss of activity. Since the catalysts employed are usually pretty stable, it is possible to simply burn the coke and regenerate the catalyst. However, the catalyst may be too fragile for burning off the coke or it may loose activity in other irreversible way.

I always get a knee jerk reaction about SAFETY when I read such questions. Putting a spark in a mixture of hydrogen and oxygen is known to the FBI as detonating a BOMB.
(1) Given a mixture of hydrogen and oxygen:
(a) A spark would set off an explosion which would make water as the product and give off heat.
(b) If you "gently" heat a mixture of hydrogen and oxygen enough then the molecules would acquire enough energy to form water. There would be an equilibrium which would depend on temperature and pressure. Heating too vigorously could cause an explosion.
(2) As Jan pointed out there has been a lot of theoretical work on catalysts. However we still don't have the ability to calculate a priori how a particular catalyst will perform. Chemists rather use calculations to make educated guesses, then must actually do an experiment to determine the result.
(3) (a)In theory a catalyst will last forever. In practice catalysts tend to get "poisoned" for one reason or another - typically by contaminates in the reactants.
(b) It might be possible to recharge a catalyst. For noble metals like platinum you might just reclaim the metal and start over.

A few notes about this one. Firstly, you stated in point b) that vigorous heating can lead to explosion, while slow heating leads to equilibrium. This is not a good way to look at it, it is mixing up kinetics and equilibrium of the reaction. The EQUILIBRIUM of the system is always practically water, and this is independent of the heating speed. The explosion means the reaction is uncontrollably fast, which is again mostly independent of our heating speed. Also we CAN calculate catalysts efficiently. Finally, catalyst poisoning is NOT what you have described here - use the right words please.

@Ezze - Overall I disagree with all your points. We'll have to agree to disagree.


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Why does NMR deal with protons?
I ask this b/c I know the NMR deal with magnetic fields (MF) and it can change energy of proton’s magnetic energy. But my question is that electrons also have a spin and M F. Just like how IR and Mass deal with electrons, why don’t NMR also talk about electrons? Why just focus in the nucleus.

The energy gap between "up" and "down" spin states of an electron is much larger than the energy gap between the two spin states of a proton. So, the spectra don't quite interfere. It is possible to do what is essentially NMR, but on an electron: https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance

Your question and title are inconsistent. All nuclei aren't just a proton. $\ce{^13C}$ is also used for NMR studies as well as other isotopes.

Electron spins are used for spectroscopy. The expected energy gap is different so you would be looking in a different area of the EM spectrum (i.e., microwave versus radio). See: https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance

So can I know why electron spin are also not used to know a molecules structure in orgo? Why just focus on NMR? Does NMR have more usefulness?

EPR only works for unpaired electrons. Most organic compounds have only paired electrons. Organometallic complexes are fairly rare in the overall scheme of things.

NMR does deal with electrons, but not their spin. The nuclear environment that determines the NMR resonance of a nucleon is dependent on the electrons in the molecule and features like peak splitting are a result of transmission via electrons of features dependent on remote nuclei.

NMR deals with the nucleus of an atom (not just protons, but all nuclear isotopes with spin>0). Electrons also have a magnetic moment, and the technique used to probe this phenomenon is EPR/ESR. While NMR uses frequency in the MHz range (MHz), ESR uses higher energies in GHz range.
Lots of common references available on the internet to learn more, but you might be interested in:

https://physics.stackexchange.com/questions/171256/whats-the-difference-between-nmr-and-epr
https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance
https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance


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Spectroscopy pyrazine molecule
Considering pyridazine and pyrazine molecules, I am asked to say if I can distinguish one from the other in a sample.
I supposed that is possible because pyrazine belongs to D2h and so it has the inversion center. Thus, in IR spectroscopy I can appreciate the antisymmetric stretching while in Raman I can identify the total-symmetric ones.
Then I am asked to say if this spectroscopic technique allows me to predict quantitive results. and how?
This question floored me because I have never done spectroscopic experience!

You'd use relative peak intensities somehow. Very common technique in spectroscopy. Pyridazine has peak about 1300 http://webbook.nist.gov/cgi/cbook.cgi?ID=C289805&Mask=80#IR-Spec and pyrazine has peak about 900 for example http://webbook.nist.gov/cgi/cbook.cgi?ID=C290379&Mask=80#IR-Spec

How can identify which oscillation frequency is related to nth normal mode? For example in the pyrazine there are two total symmetric CH stretching: it is expected to find them in 3000cm^-1 region but I don't know how can I relate a particular frequency to a particular mode

Assigning particular peaks to  particular modes is the subject for a book. You have to use not only the position of the peaks but their relative intensity as well. Also you have to assign all the peaks to be sure what is going on.

Can you use any kind of spectroscopy? What about proton or carbon NMR?

You need to understand some details of using point group tables. The pyrazine  belongs to  $\ce{D_{2h}}$ point group and pyridazine belongs to $\ce{C_{2v}}$
In the third column of the point group are found x, y, z which are used to identify dipole transitions, e.g. vibrational symmetries that can exhibit  IR transitions and in the fourth $x^2, z^2, xz$ etc. which are used to identify Raman transitions. If there is a squared term in column 4 and in the top line of the table (totally symmetric representation) labelled $\ce{A_g, A_1}$ then a Raman transition is totally polarised. Raman transitions from other symmetry species vibrations are not fully polarised. (Ignore the $\ce{R_x, R_y,R_z}$ they do not refer to IR or Raman transitions)
The point groups are shown below

In $\ce{C_{2v}}$ all vibrational species $\ce{A_1}$ etc. can have Raman transitions but in the IR the $A_2$ does not have an IR transition, no x, x, z in column 3. The other three types of vibrations could have both IR and Raman transitions. In the $\ce{D_{2h}}$ you can see that, because of the centre of inversion, Raman and IR vibrational transitions are mutually exclusive (compare symmetry species of $x^2, xy$ etc. with x, y, z ) because there is no vibration with a symmetry species that can generate both IR and Raman transitions.
Thus, with your compounds, the one with $\ce{C_{2v}}$ symmetry will have IR and Raman transitions with the same frequency, but for the $\ce{D_{2h}}$ compound there should be no common IR and Raman frequencies.
(See the answer to this question Understanding group theory easily and quickly more details on using point groups)

Thank you very much! But I'd like principally to know if I can associate a particular frequency to a particular mode just looking the spectrum (without the help of the tables in book).

Not really, but as a rule of thumb, the asymmetric stretches have highest frequency then symmetric stretches, then bends at lower frequency.  There are  rules also for particular groups such as C=O, CN, OH, NH etc. that I'm sure you will be familiar with and that are very useful when identifying compounds.


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Catalysist activation energy - the Brønsted-Evans-Polanyi relation
I have stumbled upon something called the Brønsted-Evans-Polanyi relation in a study about the design of catalysts (for reactions like those in hydrocracking fuel production.)
The relation shows that the transition state potential energy $E_a$ is linearly related to the CO dissociation energy (over transition metal surfaces):

Reproduction of Fig. 3B from J.K. Nørskov et al, Nature Chem. 2009, doi:10.1038/nchem.121
My question is simply: What exactly is on the axes?

I am not sure if the transition state potential energy $E_a$ is the same as the $E_a$ being described in e.g. the linked Wikipedia article in the first paragraph. Is it a term for the activation energies that belong to the intermediate steps, when the catalyst breaks down the original activation energy jump into more smaller jumps?
The CO dissociation energy is a bit odd for me. Is this the energy for adsorption of CO into the material surface?

According to the Wikipedia article, this relation in the graph should be given by the equation:
$$E_a=E_0+\alpha \Delta H$$
where $\alpha$ seems to be a constant and $\Delta H$ the enthalpy of reaction. This seems to say that my "CO dissociation energy" on the $x$-axis is the enthalpy (or the product of enthalpy with $\alpha$.) I would appreciate if someone could shed some light on the wording "CO dissociation energy".

Can you add a complete citation to the Norskov paper, or at least a DOI? In your second paragraph, do you mean "transition metal" instead of "transient metal"? I'm not sure where "dissipation energy" comes from. The x-axis of the figure is a CO dissociation energy -- is that what you mean?

Thanks @hBy2Py, those were some were good guesses. Two misspelled terms... Now fixed. The 2nd paragraph is actually the exact formulation.

Made some edits -- please feel free to undo anything you don't like, or roll back the whole thing if you want.  I would request you at least leave the DOI link, though.

First, a note on the Bell-Evans-Polanyi (BEP) equation:
The form you've written it in disguises what the constant $\alpha$ tells you. From the wikipedia article, it's said that $\alpha$ is a number between 0 and 1 which tells you how close the transition state is to the reactant state. Thus, this constant actually tells you quite a lot about the shape of potential energy plotted versus the bond coordinate.
To see why this is true, we should look at the derivation given in the article you linked. The BEP equation assumes that the energy of both the reactant and product vary linearly with the bond length of the atom being transferred. This is probably a very bad assumption for finding the actual shape of the well, but the intersection of these two lines seems to give a good approximation for the location of the transition state. Specifically, if one knew the slopes of these lines $m_1$ and $m_2$, we could find $\alpha$ according to $$\alpha=\frac{m_1}{m_1-m_2}.$$
See the wikipedia article you linked for the details on this, but hopefully this motivates the meaning of the constant $\alpha$.
So, what does this have to do with the article you're asking about?
Well, the reaction that is being catalyzed in the figure you have copied is,$$\ce{CO +3H2->CH4 +H2O}$$This seems very unlikely to happen in the gas-phase because the $\ce{CO}$ bond is very strong and we would have to have four molecules collide all at once and magically rearrange their atoms.
Thus, a transition metal surface is used as an adsorption site to get all these atoms close together.

What exactly is on the axes?

Well, now we can see that by the "CO Dissociation Energy" they mean the energy it takes to dissociate the carbon and oxygen atoms once they have been adsorbed onto the metal surface. So, a negative dissociation energy indicates that when $\ce{CO}$ adsorbs, it exists as two separated carbon and oxygen atoms.
Additionally, because they have shown the BEP linear relationship, the slope of that line is $\ce{\alpha}$ and the y-axis is the classical Arrhenius activation energy.
At first, it seems like the best catalyst should be the one which has the smallest CO dissociation energy, but as the article points out,

The reason is that in the limit of weak coupling (Ediss is only a little negative), the BEP relation gives that the barrier for dissociation of the reactants will be high and the rate low. For strong coupling (Ediss very negative) the activation energy of adsorption is small but now the barrier for forming the products will be large.

That is to say, if the reactants adsorb quite strongly to the surface (i.e. Rhenium in the plot) then reaction wil go really slowly because the activation energy (not the on you're seeing on the y-axis here) for forming the intermediates leading to methane will be very high.
On the other hand, if $\ce{CO}$ doesn't adsorb very favorably, the reaction will be slowed down because both $\ce{CO}$ and $\ce{H2}$ need to be adsorbed to react. When they do meet though, because they adsorb weakly, the activation energy to forming the products will be small.
This is why they say,

An optimal interaction strength must exist between these two limits.

The figure below on the right shows this intersection for pure metal surfaces. The closer you are to the peak, the better.

(source: nature.com)
Thus, either Cobalt or Ruthenium would be optimal for his reaction, but as the authors note, these are expensive, so nickel is used in the industrial process. Thus, some kind of alloy which behaves similarly to Ruthenium or Cobalt would be great if it could be found.
Finally, it should be noted that the BEP equation will only work in very specific circumstances where bond stretching is the dominant feature which determines if two things react. Thus, when reactants are adsorbed and thus all they can really do is vibrate, I would expect the BEP equation will do a pretty decent job.
I hope that clarified your questions. Let me know if I missed anything.


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Why is compressed air flammable?
I have a can of compressed air. The type you use to clean the insides of computer. It's "100% ozone friendly" and "pure compressed gas". So I assume its everyday regular breathable air. I used it to try blow out a candle. It had precisely the opposite effect.
Why?

Hint: what kind of chemical reaction is happening within the flame?

Hint2 - It isn't compressed air, it's compressed "air."

Hmm - can you feel something slosh around when you shake the can? Does the can get really cold when you let some "air" out? I really doubt the can contains compressed air ( 78 % nitrogen, 21 % oxygen etc...)

Most of the bottles of compressed gas used to clean computers are not actually compressed air, but other compounds. In the old days they were usually compressed CFCs (they are non-flammable but were banned by the Montreal protocol). Now they can be a variety of things but the mixtures often contain flammable components (labels often don't say what exactly but I have some that appear to be mixtures of hydrocarbons and HFCs and the labels warn that some components are flammable).

@Kevin. What is precisely "the opposite effect" ? What did you observe on the candle ?

Assuming the contents are what is written on the outside, compressed air is a mixture of about $78~\%$ nitrogen, $21~\%$ oxygen, $1~\%$ argon and about $0.04~\%$ carbon dioxide. None of these compounds are flammable. Oxygen, however, is able to support combustion. So increasing the amount of oxygen a flame gets will increase combustion.
The principle difference between blowing canned compressed air onto a candle flame and simply blowing into said candle flame is the focus of the air flow. These cans are designed to blow air in a rather defined way to a defined target so you can assume the air flow to be very focused. Blowing using your mouth would create a very unfocused air flow because there is only one constraint (your lips) and air can pass through at a variety of angles.
When reaching the candle flame, the focused air flow causes the mixing of gaseous wax (required for combustion) and oxygen from the air (both the canned and the uncanned) to improve, hence increasing the rate and heat of combustion. The small focused flow will create small whirls that help mixing. The mouth-blown air, however, will come as a strong wave, practically pushing the entire candle flame away. Therefore, the flame will go out. (That and only that is the mechanism of blowing out candles: blowing the flame away.)

Most probably some hydrocarbon like butane = big flame. A quick way to check is if the can makes a sound when you shake it. If it does then it's most likely hydrocarbon. It would have been less confusing if it said "compressed gas".

In former centuries the word "air" was used to descrive gases generally, not what chemists call "air" today.  Hydrogen used to be "fire air".  Sometimes this usage still creeps in, I've heard people use "air knives" for devices that wipe excess coating off a hot dip galvanized strip with nitrogen gas.
Don't rely on just the product name used to sell the product.  Check all the labelling.  If there is something (in)flammable, the manufacturer is required to report that information.

I would guess that because in the process of compressing air, we use oil to remove heat and lubricate the rotors in the air compressor. There is a separator in the tank that removes the oil from the air however they have been known to leave traces of oil in the compressed air and if there is excess oil in the reservoir then that's also going into the compressed air. They probably put flammable as a warning just in case a faulty separator is the cause of a can or two of having a percentage of oil in it.
This could be totally wrong tho, lmk


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How do I convert this ion exchange capacity into water softening capacity?
I am researching for an extended experimental investigation into methods of water softening. One such method is the use of ion exchange resin. I have contacted the supplier of the resin for its exchange capacity, but I'm not sure how their answer relates to the removal of magnesium and calcium ions from water.
They say the exchange capacity is "1.0 mEq/mL min of 99% in H form."

$\ce{H+}$ has charge of +1 but $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ both have a charge of +2. So the ion exchange bed capacity would be about 0.5 millimoles/ml for those cations. // Remember that random packing of spheres only occupies about 66% of volume. So a tube of 100 ml volume will only have about 66 ml of resin.

In the H form there are one plus charged ions ($\ce{H+}$) being exchanged in the water softening process. But the species being exchanged for the $\ce{H+}$ ions are doubly charged ($\ce{Mg^2+}$ and $\ce{Ca^2+}$), so two $\ce{H+}$ ions must be exchanged per $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ion.
So, "1.0 mEq/mL min of 99% in H form" could equivalently be read as:
The exchange resin has the ability to exchange $\pu{0.5mmol}$ of $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ions per $\pu{mL}$ of resin.


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Are there any Endothermic Pyrophoric Substances?
I've recently been reading about pyrophoric materials, specifically as they relate to rocketry, and this made me curious if there are any substances which are the complete opposite of pyrophoric. Essentially, are there any materials which react on contact with air (i.e. either gaseous nitrogen or oxygen), but endothermically? If so, do they react at STP, or only under special circumstances?

Pyrophoric is not the correct word; pyro means fire and such would very likely have to be exothermic unless you have very hot air.

@ChemBird, yes, I understand that's the case. What I'm trying to ascertain is whether there are any substances which will react on contact with air (like pyrophoric substances do), but endothermically rather than exothermically. I.e., they still react on contact, but consume energy rather than release it.


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Law of mass action & Collision Theory
I don't understand why the stoichiometric coefficients of reactants and products are expotentialized in law of mass action.
So I checked on Wikipedia. And this popped out:

In the 1879 paper[9] the assumption that reaction rate was proportional to the product of concentrations was justified microscopically in terms of collision theory, as had been developed for gas reactions. It was also proposed that the original theory of the equilibrium condition could be generalized to apply to any arbitrary chemical equilibrium.

How does collision theory explain this?

The law of mass action only rigorously applies to elementary reactions (in that otherwise the exponents do not match stoichiometric coefficients). These are reactions whose equations match which particles are colliding on the microscopic scale. Consider the reaction
$$\ce{$r_1$R1 + $r_2$R2 + \ldots + $r_n$R_n <=> $p_1$P1 + $p_2$P2 + \ldots + $p_m$P_m}\tag1.$$
Reaction $(1)$ is elementary if and only if it proceeds in a single step. At some time $t = t_c$ all reactant particles $\ce{R1, R2, \ldots, R_n}$ collide. Quantum mechanically this collision is not too well defined, so view it to be Newtonian in essence. Not to mention: the probability $P$ of the simultaneous collision goes to zero as $n \to \infty$. This is why in actuality most elementary steps involve at most three individual constituents.
We therefore sacrifice some generality and instead consider the easier to handle
$$\ce{R1(g) + R2(g) <=> P1(g) + P2(g)} \tag{1'}.$$
Energies
Collision theory postulates that particles only initiate a transformation when they have enough energy. This is called activation energy and denoted by $E_a$. Maxwell-Boltzmann distribution
$$\frac{\mathrm{d}N_v}{N} = 4\pi\left(\frac{m}{2\pi k_bT}\right)^{3/2}v^2\exp\left(-\frac{mv^2}{2k_bT}\right)\mathrm{d}v\tag2$$
allows us to estimate the fraction of particles that have the required amount of movement. Note we are technically assuming $\mathrm{total\ energy} = \mathrm{kinetic\ energy}$. Similarly relativistic effects are ignored, and no cap is placed on maximum velocity. With these simplifications, the fraction of interesting particles is
$$f = \frac{N^*}{N} = \frac{1}{k_bT}\int_{E_a}^\infty\exp\left(-\frac{E}{k_bT}\right)\mathrm{d}E\tag3$$
which evaluates to
$$f = \exp\left(-\frac{E_a}{k_bT}\right)\tag{3'}.$$
Number of interesting collisions
Say $Z^*$ is the number of collisions that matter. We will look for a function $Z$ such that
$$Z^* = Z\cdot f.\tag4$$
The Maxwell-Boltzmann distribution $(2)$ gives for a single kind of particle (say two $\ce{R1}$'s) that there are
$$Z = 2N^2\left(\sigma_\ce{R1}\right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\tag5$$
collisions in $1$ second per $1\ \mathrm{cm^3}$ if $N$ is the number of particles per square centimeter. (This is the traditional starting point unit.) Hence the rate of such a reaction would be
$$v_r = \frac{2Z^*}{N_A} \cdot 10^3\ \ \ \left(\frac{\mathrm{mol}}{\mathrm{s\cdot dm^3}}\tag6\right).$$
Reaction rate $v_r$
If you like, you are now able to substitute $f$ from $(3')$ and $Z$ via $(5)$ into $(4)$. Then insert $(4)$ into $(6)$.  After necessary manipulation, this yields $(7)$
$$v_r = \underbrace{4 \cdot 10^{-3} N_A\left(\sigma_\ce{R1}\right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot  \overbrace{\left(\frac{N}{N_A}\cdot 10^3\right)^2}^{c^2}.$$
Coming back to our actual reaction $(1')$ involves more relative quantities. For example, $\sigma_\ce{R1}$ becomes $$\sigma_\text{aver} = 0.5\left(d_\ce{R1} + d_\ce{R2}\right).$$
So it is a bit more difficult. But the result is analogous to $(7)$.
$$v_r = \underbrace{2\sqrt{2} \cdot 10^{-3} N_A\left(\sigma_\text{aver}\right)^2\sqrt{\pi RT\left(\frac{1}{M_{\ce{R1}}} + \frac{1}{M_{\ce{R2}}}\right)}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot c_{\ce{R1}} \cdot c_{\ce{R2}}$$
More compactly, if $\ce{R1} = A$ and $\ce{R2} = B$
$$v_r = k[A][B]\tag8$$
which is what we set out to prove.

This derivation assumes that every active collision leads to a reaction. When theoretically computed $k$ were compared to experimental, an extra factor was introduced, $P$. This is called the steric factor, and in classical collision theory $P$ remains empirical in essence.

Collision theory is for elementary steps
As your quote suggests, collision theory is a first theoretical explanation for the proportionality to products of concentrations. It does not generally explain various exponentiation. It does not have to either. The higher (or non-integer or negative) exponents usually derive from the mechanism itself. In other words, the fact that common reactions are not elementary comes into play.
For instance, the transition
$$\ce{2Br- + H2O2 + 2H+ -> Br2 + H2O}$$
is experimentally found to follow
$$v_r = k\ce{[H2O2][H+][Br-]}\tag{a}.$$
Mathematically, we can verify that one possible mechanism is
$$\ce{H+ + H2O2 <=>[K] H2O+-OH,} \tag{fast equilibrium}$$ $$\ce{H2O+-OH + Br- ->[k_2] HOBr + H2O,}\tag{slow}$$ $$\ce{HOBr + H+ + Br- ->[k_3] Br2 + H2O.}\tag{fast}$$
Indeed,
$$K_c = \frac{\ce{[H2O+-OH]}}{\ce{[H+]}\ce{[H2O2]}} \implies \ce{[H2O+-OH]} = K_c\ce{[H+]}\ce{[H2O2]}\tag{b}.$$
Applying the method of stationary concentration gives
$$v\left(\ce{HOBr}\right) = 0 \implies k_2\ce{[H2O+-OH]}\ce{[Br-]} = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]}\tag{c}.$$
The overall rate of the reaction is characterised by the rate of formation of bromine $\ce{Br2}$. So,
$$v_r = v\left(\ce{Br2}\right) = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]} \overset{(c)}{=} k_2\ce{[H2O+-OH]}\ce{[Br-]} \overset{(b)}{=} k_2K_c\ce{[H+][H2O2]}\ce{[Br-]}$$
or more briefly using $k_2K_c = k$
$$v_r = k\ce{[H2O2][H+]}\ce{[Br-]}\tag{d}.$$
I repeat: this only shows that it might be a valid mechanism, not that the reaction actually follows such a pathway. Therefore, while we can use collision theory to derive the law of mass action as a first approximation, the exponents for non-elementary reactions are determined via experiment.

Lets start off with the Wikipedia definition of the law of mass action:

"In chemistry, the law of mass action is the proposition that the rate
  of a chemical reaction is directly proportional to the product of the
  activities or concentrations of the reactants. It explains and
  predicts behaviors of solutions in dynamic equilibrium. Specifically,
  it implies that for a chemical reaction mixture that is in
  equilibrium, the ratio between the concentration of reactants and
  products is constant.
  [...]
  When two reactants, A and B, react together at a given temperature in a "substitution reaction," the affinity, or chemical force between them, is proportional to the active masses, [A] and [B], each raised to a particular power"

The "affinity", also describes the rate at which the reaction will happen.  From this statement and the above definition we can also get the rate law for an irreversible chemical reaction as:
$$\ce{A + B -> C}$$
$$\ce{rate = k [A]^a[B]^b}$$
Here k is the rate constant for the particular reaction. For the simple reaction given above, the exponents are just one.
Collision theory ties into this in a very basic manner. Considering the simple reaction above, the frequency of collisions should be directly proportional to the concentrations of each reactant (as stated in the definition of the law of mass action above), thus the exponents should be one for the above reaction. Now consider this reaction:
$$\ce{2A -> B}$$
which we can write as:
$$\ce{A + A -> B}$$
Now the reaction rate (which is proportional to the "affinity" of the reactants for each other) is given by:
$$\ce{rate = k [A]^1[A]^1}$$
or
$$\ce{rate = k [A]^2}$$
I hope this description sufficiently ties together the law of mass action with collision theory, I realize in a fairly simplistic manner, in a way that also explains why the reaction coefficients represent the exponentials in the law of mass action. Please let me know of anything requiring clarification.

But why should Kc be proportional to the active masses?

The collisions of the reactants is proportional to their concentration. A reaction cannot take place without a collision. Though not every collision of A and B reacts to make C, some fraction of them will.  If you double the concentration of A, you will double the number of A-B collisions, increasing the collisions and thus the products by a factor of 2. Did I understand your question correctly?

Great! Don't hesitate to ask if you have any other questions.

I don't feel this is a full answer, but it may allow you to clarify your question.
The reactant molecules must collide in order for a reaction to take place. Therefore we must know the collision frequency in order to determine the reaction rate. This rate must be the product of some probability and the collision frequency.
Collision Theory gives a way to calculate this collision frequency. In doing so Maxwell-Boltzmann distributions are used to calculate the ratio of atoms that can react within a given concentration. Most of this 'stuff' gets swept up in the reaction rate constant. After multiplying all these probability ratios together you are left with the product of the concentrations of each reactant.
Since every reagent must be treated in this way, identical components end up with exponents. That is, stoichiometry goes to the exponent.
If you are still confused, or not confused enough: this approach assumes that this reacting system is deterministic and continuous. While many systems can be modeled by treating them this way, they are not deterministic or continuous. They are discreet, stochastic events.

But whether 2 moles of product is produced or any number of mole why should they be considered different?

Need to think in a probabilistic way?


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Unknown white stuff from eggshell reaction with acids and bases
I have just finished some experiment at home using tea eggs.

First I peel off the shell of the tea egg and place it in a bowl of white vinegar (of course, I consumed the rest of the egg). Not surprisingly, bubbles were seen as the carbonate in the eggshell reacted with the vinegar.
After a day, part of the mixture was decanted into a small glass jar. Sodium carbonate (baking soda) chunks was then added which give off a lot of bubbles of $\ce{CO2}$ plus thick foam of undissolved eggshell mixture. This is repeated until no more reaction occurred (meaning the acid is completely neutralised with the baking soda in excess). After it is settled, a simple filter is made using tissue paper (Coles White Facial Tissue) as the mixture is then poured on top of the tissue into a clean glass cup. This is then left for 7 days, sealed by the tissue paper.
More white vinegar is added to the same liquid level. This is then left to evaporate a bit (and part of the liquid is decanted into another glass jar. Step 3 is then repeated until all eggshell has dissolved (leaving behind brown film like residue which is the membrane of the egg and also the tea leaf related residues
After that,the clear solution is transferred by capillary action using a rolled tissue paper (same brand) to a new glass jar. A few hours later, some white powder is formed on the rolled tissue paper. This white solid is then scrapped off to be stored in another glass jar.
A few days later after realising the colorless solution in the glass jar still does not evaporate significantly, it is poured onto a spread out tissue placed on a clean flat surface. After the solution evaporated, similar white powder was deposited near the edge of the tissue. This is then scrapped off to the same glass jar that contain the white powder.

Now when I prepared to wash everything, I noticed some strange slimy looking sediment (which look like tangled threads and web like in solution) formed that was not there 7 days ago.
I have a couple of suspicions:

Dissolved egg proteins that precipitated
Some unknown reaction between sodium carbonate and eggshell solution
A biofilm of some unknown microbes

Any idea what that white stuff could possibly be?

The white powder is likely either sodium or calcium acetate, likely a mixture of the two. Egg shells are mainly calcium carbonate and baking soda is sodium bicarbonate, which each react with the acetic acid (vinegar) to form acetates releasing carbon dioxide bubbles.
As far as the the slimy stuff goes, your protein guess seems pretty plausible.

Possible duplicate of Why do the physical properties of an egg shell change when the egg shell is exposed to vinegar for a week?

Hmmm... I am not sure if the suggested link resolves it fully, given how I actually kinda separated the egg membrane from the eggshell itself by the filtration, unless part of the protein dissolved in the baking soda and then deposited afterwards after standing for a week


text:
Entropy change in mixing a gas at constant temperature and pressure
The expression given on Wikipedia is (https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwj-_pDA2IfSAhUFto8KHa2kBAwQygQIIDAA&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FEntropy_of_mixing%23Mixing_of_ideal_materials_at_constant_temperature_and_pressure&usg=AFQjCNFVf8mAoquhiL05HeVXC1H3cw54xA)
But If I evaluate it as following I get a different answer .
$$\Delta S=\int q_{rev}/T=\int pdV/T $$
as p and T are constants $$p(V_1+V_2)/T $$
What is wrong with my method ?

The reference you gave involves mixing two different chemical species together at constant temperature and pressure.  The problem you seem to be solving is mixing the same chemical species with itself at constant temperature and pressure.  For that process, the change in entropy is zero.  If you really intended to determine the entropy of mixing two different chemical species, then the path you identified is not a reversible path for transitioning between the initial and final states of the system.


text:
Lennard-Jones diagram - what is on the x-axis?
I have a Lennard-Jones diagram here where I can't figure out, what is on the x-axis.

G. Ertl, Angew. Chem. Int. Ed. 2008, 47
We are looking at a diagram for a molecule $\mathrm A_2$ composed of two $\mathrm A$ atoms that comes into close contact with a surface.

If the molecule $\mathrm A_2$ forms a bond with the surface, a shallow minimum ($\mathrm A_{2,ad}$) is reached.
$E_{diss}$ is the dissociation energy needed to seperate/dissociate the two $\mathrm A$ atoms from each other (breaking the molecule). In this form (as two individual atoms), they could both seperately bind to the surface, resulting in much stronger bond formation $2 \mathrm A_{ad}$. This would be called chemisorption, if I'm not wrong.
Now combinding these two processes, we would have dissociative chemisorption. The crossing point between the two curves markes the activation energy of this.

This was just an explanation of the graph. My question is, how I should understand the horizontal dimension here - the x-axis. On a google search many diagrams pop up, but never with the dissociation included. Many of those as well as the Wikipedia article about Lennard-Jones potential show the x-axis as a distance between atoms - the interatomic distance $r$.
I cannot make sense of that on the diagram I am showing above. The distance between atoms should be smaller for the molecule than for the seperated atoms. What is on the x-axis in this case?

Ahh, silly me. The x-coordinate is of course just the distance between the $\mathrm A$ atoms / $\mathrm A_2$ molecule and the surface, to which they bind. This diagram doesn't show the interaction between $\mathrm A$ atoms but between them and the surface. I will close this question as solved.


text:
Do enzyme-substrate reactions show reversibility(to a considerable amount)?
I was studying about the catalysts of enzyme-substrate reactions. My biology book says that if the density of product gets high the enzyme-substrate reactions get slower. It quite sounded like Le Chatelier's Principle. But Le Chatelier's principle only works for reactions in which backward reaction happens to a considerable amount. But I am not sure if it is the normal case for enzyme-substrate reactions.
So, Does enzyme-substrate reactions show reversibility(to a considerable amount), or my reasoning is somewhere wrong?

There are many enzyme catalyzed reactions that are reversible. In such cases, buildup of product may slow and eventually stop the net reaction as you reach equilibrium conditions. Product binding to the active site or to a secondary allosteric binding site may also inhibit the catalytic reaction. This can occur even in the case of irreversible enzymatic reactions.


text:
Reactivity of lithium in water
Why lithium is least reactive in water among group one elements though it has the most negative Electrode Potential value?

Potentials only tell you about thermodynamic stabilities. They do not tell you anything about the kinetics of the system, which would correspond to how quickly the system reacts.

Lithium is the least reactive because it is the one with the least electrons. That means the electrons are closer to the nucleus and therefore more attracted to it. AS you go down the elements in group 1 the electrons are able to react more violently with water because less energy is needed for the electron to leave the metal.


text:
Force Fields and units
I am a math student trying to implement a simple force field in python for determining the energy of a small molecule and trying to minimize it in order to find a stable configuration. I was wondering if there is a prefered system of units to be used am currently using the SI system of units however when i calculate Van der waals energies the numbers are quite difficult to deal with . Any reference of basic force fields would be much appreciated

You usually want to use atomic units (Hartree are very common) for such calculations, since it makes the math easier and there are less possibilities for mistakes, bugs and stuff like that. See: http://muchomas.lassp.cornell.edu/P480/Notes/dft/node13.html You can always translate those into SI after the calculation.

Atomic units are the standard in almost any ab-initio (LAMMPS, NWChem, Orca, QChem) and/or classical force field (LAMMPS, MPMC, CHARMM, AMBER, GROMACS etc.) modeling code because you end up computing numbers much closer to 1.
Things can get bad very quickly for small values in python. To prevent this, I would use atomic units, and wherever arithmetic error occurs, try using the decimal type, instead of default float, via float -> string -> decimal:
from decimal import *
energy = Decimal(str(3.4280284e-19))
Here is a sample file I've used for atomic units in python:
pi = 3.14159265359
kb = 1.38064852e-23 # boltzmann / J/K
ab = 5.2917721092e-11 # bohr radius / meters
hbar = 1.054571726e-34 # planck / Js
h = hbar*2*pi
me = 9.10938291e-31 # electron rest mass / kg
e = 1.602176565e-19 # electron charge / C
ke = 8.9875517873681e9 # coulomb constant / kgm^3 s^-2 C^-2
Eh = 4.35974417e-18 # hartree energy / J
c = 299792458 # speed of light, m/s
cs = 340.29 # speed of sound, m/s
t = hbar / Eh # time / s
v = ab*Eh/hbar # velocity m/s
F = Eh/ab #force
T = Eh/kb # temp
P = Eh/(ab**3) # pressure
Ef = Eh/(e*ab) # electric field
Ep = Eh/e # electric potential
u = e*ab # electric dipole moment

This very much depends on whether you're trying to calculate the total energy, or whether you want to know the relative stability of different structures. The most convenient choice would be akin to Lennard-Jones dimensionless units (https://en.wikipedia.org/wiki/Lennard-Jones_potential#Dimensionless_.28reduced.29_units), if you're using a relatively simple model. If you're after real values, and have some empirical potential, you may want to consider atomic units (Hartrees for energy, Bohr radii for length, atomic time units for time), or alternatively electron volts, angstroms, femtoseconds, etc.


text:
Basicity of hypoboric acid
Why is the basicity of hypoboric acid $\ce{H4B2O4}$? Is it $2$ even though it has $4$ hydroxyl groups?

My teacher told us that the basicity of this acid is $2$ but I am not remembering the exact reason for it.

Note that boric acid is not a Brønsted acid.

Adding to @Loong comment - Boric acid is  a Lewis acid - H2O forms adduct with eg. B(OH)3 --> B(OH)4- + H+. Similar will happen with your structure - we are not losing H+ from the structure above to form an anion, we are losing H+ from an adduct of the above in water, which we typically measure pKa in - thus pKa has a degree of solvent dependency.

@Loong Boric acid can act both as Bronsted and Lewis acid.

This is a Lewis acid, not a Brønsted acid. Like boric acid $\ce{B(OH)3}$, which form an adduct $\ce{B(OH)3(H2O)}$ which then lose a proton from the water molecule of the adduct: $$\ce{B(OH)3(H2O) -> B(OH)4- + H+ }$$
Above molecule will do this at each B atom, and so lose 2 $\ce{H+}$ in total, not 4. IE will have:
$$\ce{B2(OH)4 + 2H2O -> B2(OH)6^{2-} + 2H+}$$
as the overall reaction.
Note $\ce{B(OH)3}$ acts as a Lewis acid due to vacant B p orbital, which can accept lone pair of $\ce{H2O}$ and so form an adduct. From this adduct lose proton, and hence acidic.

I am not familiar with this interesting-looking compound, but (assuming the structure you gave is correct) my guess is that once you have created a dianion from this acid, going to a trianionic species becomes very unfavorable and you would need a base much stronger than hydroxide to do it. For all practical purposes it would behave as a dibasic acid in water solutions.

any reasoning why it will become unfavorable for trianion?  I was thinking about Hydrogen bonding that will take place between the H atom and anionic O atom is that correct?

I think the main reason it is so unfavorable to make the trianion is that you are putting a lot of negative charge into a small volume. Fewer opportunities for hydrogen bonding may also be a factor.

Boric acid is a Lewis acid, not a Bronsted acid - we form an adduct of B(OH)3 with water and then lose a proton from the adduct, not from B(OH)3 directly - we have a different mode of reactivity and acidity


text:
Difference between Force Field and topology, and other related questions
I am performing some Molecular Dynamics computer simulations and I am learning about the various Force Fields (FF) such as OPLS-AA, OPLS-UA, AMBER, etc.
Many things are still unclear to me. I put these questions together because I think they are strictly related.
1) What is exactly the difference between "Force Field" and "topology", since they define similar parameters?
2) Which parameters can be defined as "belonging" to the FF, and which not?  It seems to me that sometimes topology files recall atoms from the force field and edit some parameters such as charge, mass, bond lengths, elastic constants.
3) Why is water model usually considered separately from the FF, and why is it not included in it?
4) When a molecule does not belong to a known topology and we have to model it, how do we establish the compatibility between our model and the FF used for the other molecules in the same simulation?
Thank you.

1 - "Force Field" (FF) means the set of equations that describe the interactions between atoms and the set of numerical parameters that people using those particular functions have come up with to describe some molecules (typically proteins, Nucleic Acids, lipids, and a few others). Many FFs exist because of different choices in functional forms, in the procedures adopted to obtain the numerical parameters etc etc.
A topology is just the description of a molecule in terms of parameters that can be replaced by the numerical ones contained in the FF of your choice, so that the equations of the FF can be worked out numerically.
2 - see 1.
3 - Water is described separately because there are specific models for water (TIP3P and so on) that work reasonably well, so people stick to these models indipendently of their choices for the rest of the system.
What do you mean by "it is not included"? I can see water parameters in all MD programs I have used.
Usually, FFs are always divided by "optimisation" let's say, meaning that a set of parameters have been optimised for a particular task/family of molecules, and those are given together.
4 - if you want to model a molecule for which you don't have the parameters, then you should fit them according to the procedures that have been used to obtain the other parameters belonging to the FF (good luck). If parameters exist for a similar molecule, you can try to change those to represent your molecule.
In general, it is not a good idea to use mixed parameters, meaning e.g. AMBER for the protein, CHARMM for a DNA in the same simulation.
Choose one FF and if you need parameters for some new molecule, you have all my sympathy (you can find tools to obtain these "pseudo-automatically", though).


text:
Electrolysis of water - why not pure water?
In the electrolysis of water, $\ce{H2O}$ molecules are being reduced at the cathode (forming $\ce{H2}$ and $\ce{OH-}$) and oxidized at the anode (forming $\ce{O2}$ and $\ce{H_3O+}$).
It is said that:

The electrolysis of water usually involves dilute, or moderately concentrated, salt solutions in order to reduce the power loss driving the current through the solution, but the presence of salt is not a requirement for electrolysis.

What is the reason for "power loss"? How does the presence of foreign ions change that?
The electrolysis seems possible to happen in pure water alone (as the quote also says).

Steeven, notice that I changed the mathrm formatting to ce (chemical equation). That is the more proper format, it's simpler to use for chemical equations and has a slightly different appearance upon rendering.

The terrible conductivity of water can explain this.
Pure water is a really bad electric medium, since it has a low tendency to ionize by itself (we know the low concentrations of ions involved by looking at the Kw).
This means that the voltage needed to get a current flow from anode to cathode would be really high, with consequences both in term of safety and energetic waste.
A salt dissolved provides ions, which make the diluted solution a good electric medium, and allow a current to easily flow between anode and cathode.

Water molecules reach the cathode > They split into $H_3O^+$ to receive the $e^-$ > The result is $H_2$. I understand that water is very reluctant to split into it's ionic components $H_3O^+$ and $OH^-$, but why does the salt ions do any difference to that?

Why would water molecules reach the cathode in the first place?

@IvanNeretin Is the purpose of other ions to "carry" the water ions away from the electrodes after a reaction?

Pure water is a very good insulator. Due to it's very low self ionization, there is not a sufficient amount of ions to carry current unless a significant overpotential is applied in order to overcome this kinetic barrier.
Sea water, which has a salinity of about 3.5%, is about one million times as conductive as pure water. According to this Wikipedia article:

Electrolysis of pure water requires excess energy in the form of
  overpotential to overcome various activation barriers. Without the
  excess energy the electrolysis of pure water occurs very slowly or not
  at all. This is in part due to the limited self-ionization of water.

The article sited by the OP states that the resistivity of pure water is 18 MΩ cm while that of sea water is  20 Ω cm.  This implies that a relatively huge overpotential is required to electrolyze pure water.  At these high potentials more high-energy side products like ozone are likely to be produced, as well as excess heat.  This results in power lost to the production of both chemical and heat energy.


text:
Alphabetical priority of the isopropyl group
This passage is contradictory. I've boxed the section that confused me.

Why would the book tell me to alphabetize substituents, only to clearly ignore its own rule about alphabetizing substituents?
Seems to me that "isopropyl" should come before "methyl," as the letter "i" is most definitely before the letter "m" in the alphabet. And the book makes no mention about ignoring the "iso-" prefix.
So:
1) Should methyl come before isopropyl?
2) If so, why?

Related: Alphabetical order in nomenclature

The iso prefix is included in the alphabetic order.
For eg:- The compound Thymol has the following IUPAC name :- 2-Isopropyl-5-methylphenol and not 5-methyl-2-Isopropylphenol.
Therefore the name of your compound should be 4-isopropyl-2,3,3-trimethylheptane.


text:
derivation of partial differential equations for changes in U

Changes in internal energy at

At constant temperature, $$d U = \left({\partial U \over \partial V}\right)_TdV$$
At constant volume,
  $$d U = \left({\partial U \over \partial T}\right)_VdT$$
When both temperature and volume is changing,
  $$d U = \left({\partial U \over \partial V}\right)_TdV +\left({\partial U \over \partial T}\right)_VdT$$

I want to know how do we get these equations.

I think I can speculate a bit about how do we get the first and second equation.
For the first equation, since the temperature is kept constant, we can plot a U-V graph.
Tell me if I am wrong here but for that graph the slope at a given point is $\displaystyle \left({\partial U \over \partial V}\right)_T$.
Now since the change is a infinitesimal change, the slope of tangent at that point is given by $\displaystyle \left({\partial U \over \partial V}\right)_T$.
So $\displaystyle {\Delta U \over \Delta V} = \left({\partial U \over \partial V}\right)_T$, since the change is infinitesimal, $\Delta U = dU, \Delta V = dV$ and substituting this we will get out equation 1. Same can be said for equation 2.
I have no idea how do we get the equation 3, I think it is a clever use of  product rule or something like that.

Is my understanding for the equations 1 and 2 correct ?
How do we get equation 3 from 1 and 2 ? because simply adding 1 and 2 does only give the RHS of 3 not the LHS.

One definition of a state function such as internal energy $U$ is that its infinitesimal change is equal to an exact differential. Thus in general
$$\mathrm{d}U=\sum_{i=1}^n\left(\frac{\partial U}{\partial x_i}\right)_{x_j} \mathrm{d}x_i.$$
There are various variables with regard to which one can apply the partial derivative. Entropy, volume, temperature to name a few. Usually we assume that our thermodynamic state functions are two-dimensional but this is only if the number of moles is constant.
$$\mathrm{d}U= \frac{\partial U}{\partial T}\mathrm{d}T + \frac{\partial U}{\partial V}\mathrm{d}V +\sum_{i=1}^k\frac{\partial U}{\partial n_i}\mathrm{d}n_i \overset{\mathrm{d}n_j\ =\ 0}{\implies} \mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V.$$
A good way to think of this assumption is that other functions are also simply functions of $T$ and $V$, e.g., $S = S(T,V)$. From the equation
$$\mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V$$
you can reach the other two equations by imposing $\mathrm{d}T = 0$ and $\mathrm{d}V = 0$, respectively.

So my way of thinking of how first and second equation is incorrect ?

Also can please you explain a bit more about how did you get the second equation. I did not understand that.

@A---B You are right that if temperature is constant, then you are essentially in a sliver where $U$ varies as $V$ is changed. And yes, the derivative $\left({\partial U \over \partial V}\right)_T$ will thus be the "slope" in that particular direction. Using $${\Delta U \over \Delta V} = \left({\partial U \over \partial V}\right)_T$$ is not a rigorous way of thinking about it though. I would focus on the general case (the total or exact differential) and derive equations $1$ and $2$ from there.

Mostly likely you are correct. But I did not understand from where did you get  $$\mathrm{d}U= \frac{\partial U}{\partial T}\mathrm{d}T + \frac{\partial U}{\partial V}\mathrm{d}V +\sum_{i=1}^k\frac{\partial U}{\partial n_i}\mathrm{d}n_i$$ and what does $x_i$ mean here ?

@A---B I assume that $U = U(T, V, n_j)$ and calculate a total differential. My point was that there is no need to add partial derivatives with respect to entroy etc. because $S = S(T, V, n_j)$ is already included in the $k + 2$ partial derivatives.

@A---B It has $n_j$-s, (number of moles) it is indeed smally written and I was afraid that it is not easy to see :)

I did not see that. I was thinking about $x_i$ in $$\mathrm{d}U=\sum_{i=1}^n\left(\frac{\partial U}{\partial x_i}\right)_{x_j} \mathrm{d}x_i.$$. What is x here ?

@A---B Oh, just a random variable. Later on I took $x_1 = T, x_2 = V, x_3 = n_1$ and so on.

@A---B Glad I could help!


text:
What is the physical interpretation of "internal pressure"?

$$ \require{begingroup}
\begingroup
\newcommand{\md}[0]{\mathrm{d}} \md U = \pi_T\,\md V + C_V \,\md T$$
  where  $\pi_T$ is the internal pressure and is given by $\displaystyle {\left(\partial U \over \partial V\right)_T}$.

What does the internal pressure actually physically mean? I can see it is the slope of the $U$-$V$ graph at constant temperature but that does not clear things up.
Also why is this quantity called the internal pressure and not something else? $\endgroup$

I'd assume that the phrase is trying to acknowledge the wall effect. Pressure at the wall of the container isn't the same as pressure away from the wall.

@MaxW Outside the container or inside the conatiner ?

Inside of course. what would the pressure outside of the container have to do with anything?

I also thought that but I did not understand where that pressure will be acting ?  on other molecules ?

I don't understand your last remark. Could you please clarify what "that pressure" is?

@MaxW I am really sorry. The pressure that is away from the wall.

Let's look at it this way. I am on the starship Enterprise and contain a 1 meter sphere of gas in a force field. I then beam the sphere into deep space and then release the force field. What happens? The gas dissipates in a spherical direction since there is nothing pushing the gas back together.

@MaxW So that force field is internal pressure ?

No, the force field acts against the internal pressure.

@MaxW Yes sorry, my mistake I thought the force field is inside the sphere of gas, whereas it is the exact opposite.

See orthocresol's derivation of an internal pressure $\pi_T$ relation $(1)$ here.
$$\pi_T = T\left(\frac{\partial P}{\partial T}\right)_V - P\tag{1}$$

Internal pressure $\pi_T$ measures what you would expect from its definition:
$$\pi_T \equiv \left(\frac{\partial U}{\partial V}\right)_T.$$
It shows how internal energy changes when volume is changed and temperature is constant. For ideal gases the change is zero, something that also follows from the equipartition theorem.
$$\mathrm{d}U_{\text{ideal gas}}= \frac{\nu}{2}nR\mathrm{d}T \implies \left(\frac{\partial U}{\partial V}\right)_T = 0 \ \  \text{for an ideal gas}$$
In other words, the total energy an ideal gas has is not dependent on volume. There is no repulsion nor attraction amongst ideal gas particles.
As a first approximation to model real gases, the van der Waals equation $(2)$ is used.
$$\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT\tag{2}$$
Hence,
$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$
Using relations $(1)$ and $(3)$
$$\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V-nb} \overset{(3)}{=} \frac{1}{T}\left(P + a\frac{n^2}{V^2}\right).$$
Again via formula $(1)$
$$\pi_T = T \cdot \overbrace{\frac{1}{T}\left(P + a\frac{n^2}{V^2}\right)}^{(\partial P/\partial T)_V} - P = a\frac{n^2}{V^2}\tag{4}.$$
Result $(4)$ $\pi_T = a\frac{n^2}{V^2}$ demonstrates that for real gases internal energy will change when volume is decreased or increased. Real particles have interparticular forces that vary with distance. A hint as to why $\pi_T$ is called internal pressure will be given when $(4)$ is substituted into $(2)$:
$$\left(P+\pi_T\right)\left(V-nb\right)=nRT,$$
or even better, $(3)$:
$$P = \frac{nRT}{V-nb} - \pi_T.$$

If $\pi_T > 0,$ then $P$ will be smaller. Therefore, attractive forces are dominant.
If $\pi_T < 0,$ then $P$ will be bigger. Therefore, repulsive forces are dominant.

I don't think this is what the OP is asking about. I think question is more about why "internal pressure" vs. just "pressure." In other words what does "internal" add?

@MaxW I did always think about it as 'a correction to pressure that derives from intermolecular forces'. As opposed to traditional (ideal gas) pressure which can be modelled as collisions with container walls. To recap, real gases occupy a finite volume, and if one decreases the volume of the system (via an external pressure), the molecules are forced more close together. How these particles respond is what internal pressure measures. This is, in turn, to be differentiated from the traditional pressure (of which internal pressure is a component).

@MaxW Whether this is a good way of thinking about it, remains to be seen. :=) (Feel free to downvote accordingly if this is wrong.)

@LinearChristmas Wow, this is easily one of the most complete answers I have ever got. Crystal clear. Thanks for the answer.

That is pretty much the significance of $\pi_T$, it is a measure of intermolecular (IM) forces. As V changes, the average IM distance will change, and then the difference in the IM potential energy will affect U. For an ideal gas, there are no IM forces; hence there's no effect on U and $\pi_T = 0$. (And I realise I am just repeating what you said... whoops)

Assume that the gas is enclosed in a cylinder with a movable piston but the piston is not moving now. If the piston - wall interface is frictionless then the internal pressure (the one you are asking about) is the same as the external pressure (the one that can be controlled by mechanical means, e.g., by pushing or pulling on it.) If there is no friction and the process is reversible then the internal pressure is equal to the eternal pressure (the control variable) and is a state variable one that can be described by some other equilibrium parameters, such as $U$ or $T$ or $V$. If there is friction between the piston and the wall then depending on whether one pushes the piston in or pulls it out the internal pressure maybe larger or smaller than the external one and there is no such functional relationship among these parameters.


text:
What is meant by equal chemical environment in NMR?
I am studying NMR and the concept of chemical equivalence is confusing. What does it mean exactly? Does it involve having different atoms to the left, right, up and down of a particular hydrogen in question? Please explain. Also I don't understand diastereotopic hydrogen give two shift? Like wouldn't rotating the bonds give same chemical environment to the two hydrogen giving them one signal. I am confused.

Homotopic, Enantiotopic, Diastereotopic - http://www.masterorganicchemistry.com/2012/04/17/homotopic-enantiotopic-diastereotopic/

In reference to NMR, chemical environment would be something that causes H peak to shift or split.

Chemical and Magnetic Equivalence - http://u-of-o-nmr-facility.blogspot.com/2008/08/chemical-and-magnetic-equivalence.html

The two questions are only mildly related; you should separate them into two different questions. Incidentally, I’m sure the diastereotopic-hydrogens-two-signals-question has been asked before on here.


text:
How to redefine internal pressure in terms of κ and α?
The formula for internal pressure of a system is:
$$\pi_T=T\left(\frac{\partial p}{\partial T}\right)_{\!V}-p$$
My professor said we can redefine this equation in terms of either the

thermal compressibility, $\kappa$
expansion coefficient, $\alpha$

How do we do this?
I was able to derive for a van der Waals gas:

$\Large\kappa = \frac{V-nb}{pV-\frac{an^2}{v}+\frac{2abn^3}{v^2}}$
$\Large \alpha= \frac{nR}{PV-\frac{an^2}{v}+\frac{2abn^3}{v^2}}$

See equations (5) through (9) of getafix's answer here: Relation between constant-pressure and constant-volume heat capacities: Cp - Cv = nR

@orthocresol:  This is not correct for a non-ideal gas.

@ChesterMiller Do you mean the answer that I linked is wrong? (Edit: The answer that I linked derives the case for a non-ideal gas. Right before equation (9) it is written that $(\partial p/\partial T)_V = \alpha/\kappa_T$, which is the same as what you have in your current answer. The title of the question does, indeed, only apply to an ideal gas.)

@orthocresol:  The result you gave after the : is only valid for an ideal gas.  But other parts of the development, particularly the result that matches the final equation I gave are correct.  Still, neither the development you gave nor mine substituted back into the equation to get $\pi$.  Please advise me whether you think it would be best if I removed my answer?

@ChesterMiller I thought your answer was OK (and upvoted it). It was what I had in mind anyway when I linked the answer above.

$\def\d{\mathrm{d}}$I thought I might include the slightly longer approach because the non-standard treatment of partial differentials can be confusing.

Begin with the differential of volume assuming a constant number of particles.
$$\d V= \left(\frac{\partial V}{\partial T}\right)_P\d T + \left(\frac{\partial V}{\partial P}\right)_T\d P \tag1$$
Similarly, assuming that $P = P(T,V)$ gives us
$$\d P= \left(\frac{\partial P}{\partial T}\right)_V\d T + \left(\frac{\partial P}{\partial V}\right)_T\d V.\tag2$$
Substitute $(2)$ into $(1)$ and rearrange the differentials:
$$\left[1-\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial P}{\partial V}\right)_T\right]\d V = \left[\left(\frac{\partial V}{\partial T}\right)_P+\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial P}{\partial T}\right)_V\right]\d T.$$
This equality will hold in general if and only if the corresponding coefficients of the differentials are both equal to zero. We are interested in the coefficient of $\d T$, namely
$$\left(\frac{\partial V}{\partial T}\right)_P = -\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial P}{\partial T}\right)_V.$$
Assuming the partial derivative of $V$ with repect to $P$ is non-zero we can solve for $\left(\frac{\partial P}{\partial T}\right)_V$.
$$\left(\frac{\partial P}{\partial T}\right)_V = -\frac{\left(\frac{\partial V}{\partial T}\right)_P}{\left(\frac{\partial V}{\partial P}\right)_T}$$
This result is identical to what Chester Miller reached.

 Hint: multiply by $1$ and apply definitions of expansion coefficients to finish the derivation.

Great answer! I have a question though. Why do you say that "this equality will hold in general if and only if the corresponding coefficients of the differentials are both equal to zero"? Why must the coefficients be equal to 0?

@Nova It's similar to how can e.g. $ax^2 = by$ be equal for all $x$ and all $y$. We haven't really made any assumptions about the partial derivatives, hence the LHS and RHS equality "in all cases" requires both coefficients are equal to zero. Thanks for your kind words!

$$dV=\left(\frac{\partial V}{\partial T}\right)_pdT+\left(\frac{\partial V}{\partial p}\right)_Tdp$$So at constant volume dV = 0 and $$\left(\frac{\partial p}{\partial T}\right)_V=-\frac{\left(\frac{\partial V}{\partial T}\right)_p}{\left(\frac{\partial V}{\partial p}\right)_T}$$
The rest is straightforward.


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Mole Concept: how it applies to alloys
I was recently asked to find the total molar mass of a gold and palladium alloy, given the total number of atoms in the compound.
The method was to simply convert the total number of atoms to the total number of moles by using Avogadro's number as the conversion factor.
But as I understand it,  Avogadro's number and his concept only works if all the atoms are the same.  If multiple types of atoms are present in a group of $6.022\times10^{23}$ of them,  how could the different masses add up to a mole?

22.4 liters of air (nitrogen + oxygen mostly) at STP would have Avogadro's number of molecules. The mass of the 22.4 liters would be the "molecular weight" of air.

One mole is a number, not a mass.
See the following figurative example:
Let's say we are dealing with a sample of pure chemistry books.

A book's weight is 1 kg.
One mol of books would be 6,022 × 10²³ books.
The mass of one mol of books (the molar mass of books) would be as follows:

1 kg (mass of one book) × 6,022 × 10²³ (# of books)
Now let's say we're dealing with a mixture (alloy) of 60% chemistry and 40% history books. What is the mass of 1 mol of this mixture?

A history book's weight is 2 kg

So, our entire collection of books, sum up to Avogadro's number = 6,022 × 10²³ units. And we know that it is 60% Chem. and 40% Hist. So the mass of the entire collection would be:
m = N × 0,6(1kg) × 0,4(2kg)
where N is the Avogadro's number.

Enough of analogies. You have the mass of the gold and palladium atoms from the periodic table. I am assuming you also have the proportion Au/Pd.
You also know that there's 1 mol of atoms. The exact same equation would apply:
m = N × P.M(Au) × P.M(Pd)
where N is the Avogadro's number, P and M are the percentages and atomic mass, respectively, of gold and palladium.