text:
Why is kerosene destroying just my red LEDs?
Due to the nature of the problem, I've decided to post this on chemistry.stackexchange and not electronics.stackexchange, since I believe that the problem is chemical in nature.
Long story short - we have an electronic product that is submerged in fuels (kerosene being one of them) and uses an RGB LED (click here for datasheet). Due to a sealing problem in the enclosure, fuel has managed to get in and cover the PCBs. What's interesting is the effect that has had on the PCB. The PCBs functionality has been completely unaffected, apart from the fact that the red LED in the RGB LED module has completely stopped working. We've replicated this ourselves manually by submerging 2 new PCBs in kerosene for a day and then taking them out and powering them up and seeing that the red LED stops working entirely. The green and blue LEDs continue to work just fine.
Examination of the failed boards shows that there are no other electrical faults. It is just the red LED that completely stops illuminating.
Looking at the last page on the datasheet, the LED material is listed as AlGaInP / GaAs. Is there any obvious reaction between kerosene and these materials that would explain why just the red LED stops working?
Update After leaving the PCBs to dry, the red LED starts working again! So the problem is not permanent.

The issue probably isn't dependent on the diode material itself but on what the diode material is packed in. Typically diodes are encapsulated in some sort of resin (possibly a transparent epoxy or something like that). If kerosene penetrates that, it could damage the functionality. Moreover, the encapsulation material might differ for other LEDs (some might require higher standards of protection from air or light degradation than red LEDs).

@matt_black What confuses me is that the LEDs are all packed in the same package. If you have a look at the product page (http://uk.farnell.com/multicomp/703-1028/led-rgb-3528-plcc/dp/2112143) or the datasheet, the LEDs are all housed in the same package. I wouldn't have expected them to have different housing arrangements.

The device is probably not meant to be submerged in kerosene ;) Nevertheless, as the respective R,G,B LEDs are arranged in one part and have different semiconductor as support. I'd guess that they are "glued" on some support (silicon, ceramics) and connected with wires.  Could you inspect under microscope, if the part did not mechanically decomposed?
In all cases, I'd suggest to place some drop of epoxy or other resin on top of the PCB as a protection. Asphalt-like stuff is commonly used, but not transparent enough for LEDs ;)

@AmrBekhit You may still have 2 different types of packaging: the overall encapsulation is one but, if the device is made by assembling different individual diodes into a single device, each diode may have a different encapsulation on the original component.

@matt_black After drying out the PCBs, the red LED has started working again! So clearly the problem is not permanent.


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Why does electronegativity generally increase across a period?
I've been doing some research and the only answer I seem to be getting is that the increase of protons means electrons are more attracted to the atom.
I thought though it was because as you go across a period, an atom will slowly fill up its outer shell, and hence reduce its need to gain or hog electrons?? Or is this incorrect?

Electronegativity is the tendency of atom to attract the electron cloud towards itself in a covalent bond. Across the period, since the electrons are filled in the same shell(which means it should be of the same size without any electron) and the amount of positive charge increases and as a consequence the size of the atom also decreases(Each electron is pulled more strongly by a nucleus with higher positive charge). The electrons are hence more tightly held as you go across the period.

I thought though it was because as you go across a period, an atom
  will slowly fill up its outer shell, and hence reduce its need to gain
  or hog electrons?? Or is this incorrect?

A need to gain or lose electron is dictated by the nearest configuration of greatest stability(Usually Octet). And this is not closely related to Electronegativity either. Electronegativity only has to do with how strongly can an atom pull the electron cloud towards itself. Actually, as you go across the period, the tendency to gain electron increases(Indicated by increase in Electron Affinity) and not decrease as you thought.

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understanding redox reaction
I am a total noob in chemistry, so sorry in advance if this is a very stupid question... I am trying to solve this redox reaction:
$$\ce{KClO3 + FeSO4 + H2SO4 -> Fe2(SO4)3 + KCl}$$
$$\ce{K^{+1}Cl^{+5}O^{-6}3 + Fe^{+2}S^{+6}O^{-8}_4 + H^{+2}2S^{+6}O^{-8}4 -> Fe^{+6}2(S^{+6}O^{-8}4)3 + K^{+1}Cl^{-1}}$$
When I analyze this I can see the following things:

Fe lost some electrons, it went from +2 to +6
Cl went from +1 to -1: a gain of elecetrons
But
I see something happened to the SO ions as well: on the left side you have two times $S^{+6}O^{-8}_4$ and on the right three times so something changed here as well. What happened? Do I have to take this into consideration as well for the calculation? And on top of that $O_3$ on the left side disappeared.

My current guess is that you never have to look at the change in oxidation state of oxygen atoms. Is that correct?

First of all, when talking about oxidation states, normally you don't multiply. You just write $\ce{Fe2^3+}$, and people will get it (correctly) as two times $\ce{Fe^3+}$ with a total "charge" of +6; you don't have to actually write +6 to achieve that. Also, pay attention to the formatting, especially the \ce{} thing. As for the $\ce{SO4^2-}$ ion, nothing happens to it.

@IvanNeretin I thought smth did happen to $SO^{2-}_4$ as you have three of them on the right side and only two on the left side.

No, it doesn't work like that. Until the reaction is balanced, you have no idea how much of anything do you have. And once it is balanced, you'll have as much on the right as on the left.

yes you must balance the equation, meaning that there are the same number of atoms of each type on both sides . The ions present are Fe$^{2+}$, Fe$^{3+}$, ClO$_3^-$, SO$_4^{2-}$, K$^+$, Cl$^-$ and H$^+$

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Yellow precipitate in SnCl2 solution
I have a few small bottles of 0.5M $\ce{SnCl2}$ solution, however solution was prepared about 4 months ago, and there is a yellow precipitate on the bottom of each bottle. Working hypothesis is that this yellow precipitate is $\ce{Sn(OH)Cl}$, and reaction is:
$$\ce{SnCl2 (aq) + H2O (l) ⇌ Sn(OH)Cl (s) + HCl (aq)}$$
(from Wikipedia)
I added a few drops of $\ce{HCl}$ to push the equilibrium towards left-hand side, but so far only small part of yellow precipitate is dissolved. Should I add plenty of $\ce{HCl}$ then? If you have any idea on how much, please let me know.
Wikipedia says:

Therefore, if clear solutions of tin(II) chloride are to be used, it
  must be dissolved in hydrochloric acid (typically of the same or
  greater molarity as the stannous chloride) to maintain the equilibrium
  towards the left-hand side (using Le Chatelier's principle).

do I read it right that I have to add the same volume of $\ce{HCl}$ (0.5M) to my 0.5M $\ce{SnCl2}$? (well, they didn't say that, but sort of hinted)
Thank you very much in advance!
UPDATE: I added decent amount of $\ce{HCl}$, and I still have some yellow precipitate (I think less, than before, but still a rather large amount), and my solution is not transparent, it is whitish-opaque. $\ce{SnCl2}$ solution has to be transparent.

I'd rather suspect oxidation.

If you suspect oxidation by the air (oxygen), then I think it is  unlikely, as bottles were full of solution and closed tightly - there was no air (oxygen) access to the solution.

Maybe a silly question, but are you thoroughly agitating it? Increasing surface area always helps the speed of reactions.

Thank you for asking, @ringo! I spent some tie agitating it and was unable to dissolve all precipitate I had. However, both experiments I needed this solution for, went very well. So I guess it was OK. Experiments I conducted: (1) https://www.youtube.com/watch?v=Btaf9AUnjDo (2) https://www.youtube.com/watch?v=AklooXzB1w8. Here is a picture of my tin hedgehog: https://www.facebook.com/photo.php?fbid=331731090491714&set=a.107640622900763.1073741827.100009644765577&type=3&theater

Other than forming the basic salt, it will particularly form $\ce{H[SnCl3]}$,  hydrogen trichloridostannate(II)  if the concentration of HCl is in excess  or will form an aqua salt, $\ce{[Sn(H2O)Cl2]}$ if the conc. of SnCl2 is more than that of water. These are the undesirable  side products and will resist the precipitate to dissolve and show whitish opaque color.

https://chemistry.stackexchange.com/questions/54402/will-sno2-or-sno-form-from-this-reaction/54403

have you tried any other methods to speed it up


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Why does the concentration of hydrogen ions increase in the electrolysis of dilute sulphuric acid?
In the electrolysis of dilute sulphuric acid, I'm given these two half-equations:
anode: 4OH-(aq) -> 2H2O(l) + O2(g) + 4e-
cathode: 4H+(aq) + 4e- -> 2H2(g)
Why does the concentration of hydrogen ions increase?
Isn't the concentration of hydrogen ions decreasing here, rather than increasing as it is being discharged at the cathode?
N.B: (This question originates from a past AQA GCSE Question: see question 2 part (d) here.)

Replace the anode equation with your new equation. Then find the overall equation for the reaction. You are correct that the net number of $\ce{H+}$ is not changing as a result of this process.  However, something is being consumed. What is it? How does that change the concentrations of the other species present?

The sulphuric acid does not take part in any reactions, and is only present to ensure that water conducts electricity.
Only the water self-ionises to form hydroxide (OH- and hydrogen (H+) ions, and its concentration decreases as it decomposes to form O2 and H2.
Therefore, the concentration of hydrogen ions from sulphuric acid increases during the electrolysis of dilute sulfuric acid.

There does seem to be 'resistance' to be accurate on the underlying mechanics of electrochemical reactions, likely due to complexity and the presence of radicals and associated short-lived intermediates (see, for example, this discussion on the electrolysis of water presented on this forum).
In the case of dilute H2SO4 with H+ ions, expect the reaction:
$\ce{ H+ + e- -> .H }$
And, the kinetically possible self-reaction of the hydrogen atom radical:
$\ce{ .H  + .H  -> H2 (g) }$
So, the H+ ion is indeed being consumed and the dissociation of H2SO4 increases to re-balance the equilibrium:
$\ce{ H2SO4 <=> H+ + HSO4- }$
As support for my discussion I also note the electrolysis of dilute H2SO4 in organic reduction reactions (see this ebook).
To reiterate, electrolysis half-cell reactions do not relate to mechanics, more informative as to a possible reaction's likelihood only.

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Monitoring reaction progress?
One of the first thing that every one learns in organic chemistry lab is how to monitor a reaction progress with TLC.
My question is what methods do we have for visualizing TLC ( UV , Iodine vapor , etc ) and each one of them is suitable for what purpose and condition ? and is there other ways like TLC (fast , easy , relatively cheap , etc) for monitoring reactions progress?

Slightly askew of what was asked, but still relevant.  TLC is not dependent on conjugation.  That reasoning suggests to me You are using UV light to visualize the TLC.  UV light is used because it is clean, quick and non-destructive.
There are destructive visualization techniques, such as vanillin stain.  In this case, after elution one dips the tlc slide into the stain, and then bakes the coated slide, and various compounds appear in vivid colours, that are normally somewhat related to the functional groups present in the molecules.  Myriad stains exist, vanillin was just my favorite when I was practicing.

(picture taken from http://www.ochemonline.com/images/thumb/5/5d/Tlc_stains.JPG/1000px-Tlc_stains.JPG)
These stains are not quite as clean or quick as UV visualization, but are still quite convenient.  There are other monitoring techniques, but nothing beats TLC for generality, cost and speed.


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How is heat physically released in an exothermic process?
Many non-covalent interactions (e.g. electrostatic interactions) result in decreased entropy and are driven by negative enthalpy. If the heat released in the reaction has its physical basis in an increase in the average kinetic energy of the degrees of freedom contributing to the heat capacity of the solution (i.e. surrounding water molecules move faster), how exactly does binding exert a force on these degrees of freedom? How does binding cause increased thermal motion of solution molecules? Is it simply that the two attractive moieties (e.g. a lysine and a glutamate residue) are accelerated towards each other as they convert potential energy into kinetic, and then release heat via collisions with solution molecules? Is all exothermic heat release the result of potential energy being converted into kinetic energy in reaction products and the subsequent collisions with solution?

I think that you are on the right track. I try to explain how I see things below.
As a bond forms the reactants must absorb some energy from their surroundings so that the activation barrier between reactants and products may be surmounted. (By the Boltzmann distribution there are always a few collisions with far more energy than the average). At the top of the barrier is the transition state (TS), for this to be crossed the excess energy has to be taken away from the TS more rapidly that it may cross back to reactants.
The transition state is effectively a very hot species; it has far more energy than the surrounding medium. This can be thought of as a species having a large number of vibrational and rotational quanta excited. As far as is known$^*$ a transition state lasts for only a few tens of femtoseconds to a few picoseconds because vibrational/rotational energy is lost very rapidly to the surroundings. This excess energy is carried away also as kinetic energy of the products if, say, two molecules of product are formed. As time progresses this energy excites more and more vibrational and rotational modes in the surrounding medium, which may be a solvent or protein's backbone and residues. Eventually this energy diffuses away from the reaction site (via anharmonic vibrational couplings) and the product eventually has vibrational and rotational modes excited with quanta in accord with the prevailing temperature.
$*$ Although transition states are universal there is a paucity of actual, i.e. time resolved spectroscopic data on their properties.
(A more detailed approach has been given by Kramers (see M. Daune ' Molecular Biophysics for a discussion). In the condensed phase 'friction' or 'drag' on the reaction coordinate slows the reaction (compare to gas phase) and the reactants 'diffuse' towards and across the transition state. 'Friction' here means vibrational and rotational energy affecting the reaction coordinate, by impeding or promoting, by turns, the progress of the reaction. The Kramers approach allows a consistent approach to  gas phase reactions from isolated molecules, then in the presence various pressures of inert gas right through to reactions in viscous solution.)

That clarifies a lot. I'm still unclear as to how a weak interaction such as a hydrogen-bond or electrostatic interaction could release heat. Is it just released via the acceleration of the species and collisional relaxation? Protein folding is said to involve many such electrostatic interactions, which must release heat to the surroundings in order to proceed (i.e. they must have negative enthalpy) because they necessarily decrease entropy by limiting conformational sampling (a source: http://www.ncbi.nlm.nih.gov/books/NBK22567/).

Or in the case of an electrostatic interaction, does the heat come from the photons that are released?

@jory, no photons need be involved. In any interaction (h bond, dispersion, dipole-dipole etc ) it is the potential energy that holds the two parts together. So when a bond breaks this energy has to go somewhere, usually it gets taken up by a combination of vibrational , rotational quanta and translation of the species. These constitute the heat released. Thus then diffuses away into the surroundings, (primarily via collisions) raising the temperature by a tiny bit.


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how do electromeric / inductive effect theories compare with resonance
Back in high school, long time and many seas away, I learned about electrons moving within organic molecules based on electromeric and inductive effect. Googling them today, I found that they are deemed obsolete by IUPAC, but still used, mainly in Russia and India.
True, today, for my limited chemistry needs, I can make do with resonance. But how did +E, -E +I and -I become obsolete? Is a mere statement of the various resonance structures enough to fully convey the information stated by describing those effects, or were they rejected for some inaccuracy?
Also, were these effects part of Chelintsev's "Marxist chemistry"? That would explain their historical and geographical extent.

Didn't heard earlier about electromeric, but inductive isn't obsolete, rather common in rationalising properties.

The concepts of electromeric and inductive effects are not obsolete.
Only the term "electromeric" is obsolete and has been subsumed in the terms "mesomeric effect" (+M and -M) and "resonance effect" (see iupac source). The term "isomeric effect" is still in use and I learned it not a decade ago along with "mesomeric" and "resonance".
Amusingly, your SE question is my first result in google for "Marxist chemistry". From the little I have just read in this book, Marxist chemistry had apparently more to do with refusal of the concept of resonance than anything else. So... maybe? Maybe not.

In the schemes describing electromeric effect, some arrows describe the movement of chemical species but standards now use arrows to describe the movement of a pair of electrons. The electromeric effect (on a scheme) is then implied.

Are you sure @SteffX? This page shows curved arrows "symbolizing the displacement of electron
pairs". I'm genuinely curious.

Sorry my comment was a bit misleading. I am not saying it was a convention but I have seen arrows from a proton to a double bond (and it actually makes sense in the context of electromeric effects). Maybe it was just a bad scheme.

Maybe =]. Anyway thank you for your elucidations.

@S.Chevalier Chelintsev's model could not deny resonance per se, because its measurable effects were undeniable. But he was uncomfortable with the thought that electrons move from place to place without a pulling /pushing measurable force. The absence of a force jarred with materialism - the basis of "Marxist science". So he called the effects "mutual influences", to make them seem force-y. I think.
In my understanding, re-wording is also the way to obsolescence for +E, -E.
Chelintsev was at the forefront for <6 years and failed to muzzle critics. "Marxist chemistry" is stretching it.

Indeed. Very interesting!


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Why can't a nucleophile attack the intermediate of an electrophilic aromatic substitution?
During an electrophilic aromatic substitution, the aromatic pi cloud interacts with an electrophile to form an arenium ion (AKA sigma-complex). The following image (from my organic chemistry class) suggests that aromaticity is always restored and that a nucleophile can't interact with the resonanced carbocation of the arenium ion.

If I were to replace the base in the above image with a nucleophile that is an extremely poor base (such as iodine), could the nonaromatic cyclodiene be formed?
Or would the reaction just halt instead?
Follow up question: If electrophilic aromatic subsitution doesn't work, how would one go about making the cyclodiene (with Nu and E substituents) from an arene? Is it possible?

The intermediate is non-aromatic arenium cation which is stabilized by resonance. Deprotonation of sigma complex is a fast process as it leads to much thermodynamically stable i.e., aromatic ring. Aromatization or deprotonation step is fast as Kinetic Isotopic effect is not observed. Hence even a weak base will extract the proton from the substituted site which is sp3 hybridized.
As for another nucleophile to attack the $\sigma$ complex, the positive charge is delocalized and deprotonation step is much faster. I think that it would be easier for the base to extract proton than to attack the Delocalized positive charge. The proton already has a $\delta$+ charge due to Hyperconjugation effect. I think that even if the base if a strong acid, it can dissociate to give H+ but the non-aromatic intermediate will not remain the solution as it is.

In the case of iodine, formation of HI will take place.

this answer would be better if it was explained in greater detail

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How do acids conduct electricity with H+ ions?
I know that ions can conduct electricity, but H+ ions has no electrons! How can it conduct electricity then. Like if we were talking about Cl- or Na+ then these ions do have electrons.
But how can something that has no electrons be used to conduct electricity?
Also, do acids conduct electricity only in water and not on their own?
And, do acids also stop conducting electricity after all the ions have been deionized? (like NaCl aqueous solution)

Electrons are very much not a magic substance that is involved whenever someone talks about electricity. It is electric charge that matters, and $\ce{H+}$ surely does have one.

so how can a positive charge conduct electricity?

Just like a negative charge does: by moving around.

so we can do the same thing with a proton too? If yes, then what about bases? They have a OH- ions. How can it conduct electricity?

What same thing? All ions have electric charge. Therefore they conduct electricity whenever they move. That's the reason for conductivity of electrolytes.

so we use molecules as electrons? Do we only need something to move?

we use molecules instead of electrons to produce electricity and we make them move due to their charge. That is what I got

Again, electricity does not mean that electrons are involved.

ok thx for your answer

Acids by themselves don't necessarily conduct electricity.
An example is boron tribromide, a Lewis acid.
What makes acid (aqueous) solutions good conductors, is the capability of the acid of interacting with water molecules, by creating positive and negative charged ions.

how does that help in conducting electricity?

Because the transmission of electricity involves the movement of charged particles (be them ions, such as in ionic solutions, or electrons, such as in metals).

ok, thx for your answer


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Nuclear spin isomerism in molecules other than H2
I was reading my teacher's material about ortho- and para- hydrogen, and I found stated that

"every homonuclear molecule, with nuclides with spin other than zero ($\ce{H2}$, $\ce{D2}$, $\ce{T2}$, $\ce{N2}$, $\ce{^{17}O2}$...) has nuclear spin isomers".

What is not clear to me is: are there any nuclides with spin zero? Is there, hence, any homonuclear molecule without spin isomers?

Nucleons of the same type tend to pair up and cancel out each others' spins, so nuclides with an even number of protons and an even number of neutrons will have zero spin. Some common nuclides with this property include $^4\ce{He}$, $^{12}\ce{C}$, and $^{16}\ce{O}$.


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How is Cu(OH)2 a base if it can't dissolve in water?
As the title says, how is $\ce{Cu(OH)2}$ a base if it can't dissolve in water?
I mean if it can't dissolve in water then it won't produce any $\ce{OH-}$ ions.
And if it won't produce any $\ce{OH-}$ ions then how can it be called a base?
Same question for $\ce{Fe(OH)2}$ and every other base which is not an alkali.
I feel like this question is quite obvious and don't know how to describe it more. So please help.

You are using the Arrhenius definition of acids and bases, which only applies to substances that dissolve in water. The Bronsted-Lowry and Lewis definitions don't require water, so insoluble hydroxides can still be bases under these definitions.

oh, thx. Another question. My teacher gave us a question- "What happens when HCl gas is exposed to dry litmus paper". The answer was -"no reaction". Why is that so?

confusion-about-ph-of-a-substance - I had mentioned why things are acidic and why they aren't.

Dry HCl cannot dissociate and hence it won't be acidic. You need a medium such as water for HCl to dissociate. Acidic property shows up when there are free $H^{+}$ ions.

can it dissocciate when the weather is humid?

@Adi Yes it will but you do not even need a humid weather. Traces of water are enough.

Dry HCl is a nasty thing. It will suck water out of air, even if there is very little of it.

Copper (II) hydroxide has some small solubility in water, determined by its solubility product constant.  More importantly, it will dissolve freely and act as a base if a suitable acid, such as acetic acid, is is available to be neutralized.  See also the comment from @f".


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Why does multipying a scalar in both sides of a balanced chemical equation, change related quantities?
Stoichiometric coefficients appearing in a balanced chemical equation are unrelated to the actual concentration of reactant/products at any stage of a chemical reaction. They give information only about the ratio in which the reactants and products are consumed and produced. Multiplying by a scalar on both sides of the equation doesn't change that ratio.
Then, why does doing so result in changes in physical quantities related to the reaction, like equilibrium constant or Gibbs energy?

It is necessary to standardise things simply to remove tremendous confusion. As pointed out by @Ivan Neretin, the effect of not reducing the stoichiometry is to have, in effect , an infinite number of equilibrium constants /$\Delta G$ etc for the same reaction. Chaos would reign!

Maybe because the equilibrium constant, Gibbs energy, enthalpy, etc. are not the fundamental characteristics you think they are.
Really, what does enthalpy mean, for example? Say, a reaction $\ce{A + B -> C + D}$ has certain enthalpy. This means that running a reaction with 1 mole of A and 1 mole of B gives us certain amount of heat. Now let's multiply everything (including coefficients and enthalpy) by 2. That would be $\ce{2A + 2B -> 2C + 2D}$. In other terms, running a reaction with 2 moles of A and 2 moles of B gives us twice as much heat, just like we would expect.
Now think of the equilibrium constant. Say, we have a reaction $\ce{2NO2<->N2O4}$ and the constant just happens to be 4. This means ${\ce{[NO2]^2\over[N2O4]}}=4$. OK, and what if we write this reaction in a different way, like $\ce{4NO2<->2N2O4}$? Obviously, the equilibrium concentrations are still the same, but the expression for the equilibrium constant has changed to ${\ce{[NO2]^4\over[N2O4]^2}}$. That's a square of the previous expression, hence its numeric value would be $4^2=16$. So what? It's not like we can just up and measure this constant with some kind of "constantometer", anyway. It's the real concentrations that matter; and if we calculate the concentrations using either version of the constant, they will turn out the same.
This is also in accordance with $\ln K = {\Delta G\over RT}$. When you square the constant, its logarithm doubles, and so does $\Delta G$.


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How could you balance this redox equation involving Cr2O7 2- and H2O
Hi I know that balancing redox equations you first work out the separate half equations first and then combine them however I'm not sure how to balance this one?
$$\ce{Cr_2O_7^{2-} + H^+(aq) + e^- \rightarrow Cr^{3+}(aq) + H_2O(l)}$$
I've arried at $\ce{Cr_2O_7^{2-} + 3e^- \rightarrow Cr^{3+}}$ however for $\ce{H^+ \rightarrow H_2O}$ I'm not sure what to add because the oxidation for state of hydrogen is the same?

Why would you add 3e- in the first place? Also, welcome to Chem.SE, and pay attention to the formatting.

@IvanNeretin Hi I found the oxidation number for Cr and it was +6 on the left hand side so I thought that you would need 3e- to balance the difference in oxidation states?

That would be right if not for the fact that you have two of them Cr.

But wouldn't one Cr at the products be +6 so that both of them would give me +12? so Cr would be +6? @IvanNeretin

No, they both would turn from +6 to +3.

To balance a redox reaction (in acid),

First balance the atom being reduced or oxidized:

$$\ce{Cr_2O_7^{2-} \rightarrow 2Cr^{3+}}$$

Then balance the oxygens by adding $\ce{H_2O}$:

$$\ce{Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7 H_2O}$$

Then balance the hydrogen by adding $\ce{H^+}$:

$$\ce{Cr_2O_7^{2-} + 14 H^+ \rightarrow 2Cr^{3+} + 7 H_2O}$$

Finally, balance the total charge by adding $\ce{e^-}$:

$$\ce{Cr_2O_7^{2-} + 14 H^+ + 6e^- \rightarrow 2Cr^{3+} + 7 H_2O}$$


text:
Why does dilution reduce the concentration of hydronium and hydroxide ions?
So my teacher gave us the definition of dilution:

Mixing of an acid/base with $\ce{H2O}$ to reduce the concentration of $\ce{H+/OH-}$ ions per unit volume.

But doesn't mixing an acid/base to water increase the concentration of $\ce{H+/OH-}$ ions?
I don't understand this please explain.

Compared to the acid/base alone, it decreases the concentration.

How? I dont understand

He is talking about dissociated H+/OH-.
Consider, for example, a NaOH (strong base) aqueous solution:
You have a dissociation of Na+ and OH- ions.
Of course, the OH- ions give the solution its basicity.
By increasing the dilution of the solution, the concentration of dissociated OH- anions decreases, and thus the basicity of the solution decreases.

so we are adding in more water?

Yes, because you could consider (although it's not 100% correct, but let's consider the concept) as the OH- and H+ ions of water were not dissociated, but instead they were completely aggregated in the H2O molecule.
The H+ or OH- ions of the added acid or base, instead, are dissociated.
By increasing the dilution, you are decreasing the concentration of the ions that come from the acid/base, and so decreasing the acidity/basicity.

Let’s take an example base which can be said to be fully dissociated in water, e.g. $\ce{NaOH}$. You take $40~\mathrm{g}$ of the pure compound and dissolve it in $0.5~\mathrm{l}$ of water. The standard way to calculate the concentration of sodium ion is:
$$c = \frac nV = \frac m{MV} = \frac{40~\mathrm{g}}{40~\mathrm{g\cdot mol^{-1}} \times 0.5~\mathrm{l}} = 2~\mathrm{\frac {mol}l}\tag{1}$$
This builds on the fact that the amount of sodium ions is identical to the amount of $\ce{NaOH}$ added. The same thing can be done for hydroxide ions; within experimental error the concentration of hydroxide ions is also $2~\mathrm{mol\cdot l^{-1}}$ as calculated in equation $(1)$.
Now say you add another $0.5~\mathrm{l}$ to the solution you generated above. Obviously, you did not add any additional sodium ions, so their amount will remain the same. The new concentration can be calculated as shown in $(2)$:
$$c' = \frac n{V'} = \frac m{MV'} = \frac{40~\mathrm{g}}{40~\mathrm{g\cdot mol^{-1}} \times 1.0~\mathrm{l}} = 1~\mathrm{\frac {mol}l}\tag{2}$$
Thus, in the resulting solution, you are left with half the original concentration of sodium ions; similarly for hydroxide ions at first approximation. The concentration decreased by adding water.

Why did I always include a caveat with the hydroxide ions? Well, these are also determined by the autoionisation of water, defined by its equilibrium constant $K_\mathrm{w}$ of equation $(3)$ as shown in $(4)$.
$$\begin{gather}\ce{H2O <--> H+ + OH-}\tag{3}\\[0.6em]
K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = 10^{-14}~\mathrm{\frac{mol^2}{l^2}}\tag{4}\end{gather}$$
Since the concentration of added hydroxide is much larger than that present in pure water ($1 \gg 10^{-7}$), we can ignore the additional concentration of hydroxide introduced by the equilibrium. However, we can calculate a new concentration of $\ce{H+}$, see equation $(5)$:
$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} = \frac{10^{-14}~\mathrm{mol^2 \cdot l^{-2}}}{2~\mathrm{mol \cdot l^{-1}}} = 5 \cdot 10^{-15}~\mathrm{\frac{mol}l}\tag{5}$$
Thus, after dilution, the new concentration of protons is:
$$[\ce{H+}]' = \frac{K_\mathrm{w}}{[\ce{OH-}]'} = \frac{10^{-14}~\mathrm{mol^2 \cdot l^{-2}}}{1~\mathrm{mol \cdot l^{-1}}} = 1 \cdot 10^{-14}~\mathrm{\frac{mol}l}\tag{6}$$
Note how this concentration increased; this is reflected in the change of the solution’s $\mathrm{pH}$ value from $14.3$ to $14$.


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Is there a database on chemical reactions, similar to NIST, but far more complete?
The only thing I'm interested in is to be able to search all reactions involving that compound in the database. Equilibrium constant, reaction rate and energy of activation would all be very welcome, but in case these data are absent it's no big deal.
Doesn't need to be an online database either, I can download a software if it contains such data.

The best known databases are Beilstein and Chemical Abstracts with around 22 millions reactions each since the end of the 19th century but they are commercial.
Organic Syntheses has made available its database for free and it covers 6000 reactions. One of the greatest advantages is that each step is described in depth and each reaction has been thoroughly tested prior to publication.
Have a look here for a list of databases.

What used to be Beilstein is now Reaxys, I think it edges out Scifinder for organic. Most people I know use it

I have not used those databases for a while but I really liked Beilstein (now Reaxys) because the structure queries were more precise than in SciFinder.

I can't access any of those that are interesting :(

I checked out with the local university, and, because it is a Catholic institution, it has a special affiliation with my school (which is also Catholic, and run by a branch of nuns from the Franciscan Clarist Order). I can go there and research the Reaxys database. So, I'll be doing that as soon as possible.

SciFinder could be very useful in your case (Reaction option). However, you need to subscribe it first as it is not freely accessible.

The best thing that comes to mind is Scifinder (https://scifinder.cas.org/). You can search scientific publications by molecule or by reaction. I've used it for organic chemistry, I have no idea of how comprehensive will it be with other fields of chemistry. And you will need to sign up. Perhaps your school has the credentials.

A really nice database for reactions is Rhea. It is useful especially for biochemistry and has rather nice credentials such as support from the Swiss Institute for Bioinformatics. In my opinion, it is rather user friendly. Just type the reaction you want into the search bar at the very top.
Although it is a bit lacking in UX (User Experience and design), it has a lot of information with really nice diagrams.

I checked that one out. But i's only use is in biochemistry. My objective has a focus on inorganic chemistry for now.

I use the www.reaxys.com most often, as it is really simple to use. See comparison of similar solutions at:
http://www.infotoday.com/OnlineSearcher/Articles/InDepth-Reports/Apples-and-Oranges-A-Chemistry-Searcher-Compares-CAS-SciFinder-and-Elseviers-Reaxys-91663.shtml
or
http://csulb.libanswers.com/faq/30611
Both scifinder and reaxys are paid, universities can have institutional access to them.

I've had a great time using the SMILES Chemical Reaction Database. It contains two million entries. It is simple to use. Contains reactant product pairs as a list of SMILES strings. Can be imported to programs such as ChemAxon's Marvinview or Mathematica. Searchable for structures. Advantages of the SMILES Chemical Reaction Database is that it includes stereochemical information. It is well known that the SMILES Chemical Reaction Database can be used for artificial intelligence algorithms in chemistry. It is available here. Download comes with an instruction book that teaches how to generate in silico combinational libraries and how to use AI to predict chemical reaction outcomes.


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Successive ionization energies (work functions?) for metals
Are there tables for successive ionization energies for metals (like for gases)? All I can find are work functions to remove one electron from the surface of a metal into vacuum (and to keep it there). However, I guess it gets harder to remove electrons the more electrons already have been removed, right?
Maybe there's a way to relate that energy to the average degree of ionization of the material. Is it possible at all?
Finally, is it harder to ionize a metal gas or a metal solid? Although most work functions are a little bit smaller than the (1st) ionization-energies of the same material in gaseous state but I still think ionization of solids is harder. Am I correct?

Someone knowledgable should verify this, but i think that work functions are not related to ionization energies in a simple manner (except thaat work functions are always greater than ionization energies), because work function includes the average energy lost by the electron in collisions with other electrons and nuclei. If this is true, then work function would heavily depend on the crystal structure of the metal etc.

Also, if there is no such table you can find, you can always open Excel and copy the IE values from the Wikipedia pages of each metal you want.

In an infinite metal solid, there are an infinite number of electrons, so each one removed will be at the exactly identical work function with no change in energy.
Intuitively, this may seem a bit strange because .. well, it should be harder to remove an electron once positive charges start to accumulate.
If I have even a micron-sized piece of metal, that's ~$10^{13}$ atoms. So you really have to do a lot of ionization before you're even going to get 0.1% of the metal ions.

Here is the link to the exact page you want: Ionization energy of elements (Data Page)
For a good measure, here is the one for electron affinities too: Electron affinity (Data Page)

Wikipedia defines ionization energy as:

The ionization energy is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation.

This means we cannot really talk about the ionization energy for solid metals. The closest interpretation we can get for solid metals is the work function, but i suspect they depend on the metal's crystal structure too, as i have written below.

I would say that metal solids are harder to ionize because at low temperatures, the average energy of metal atoms is low and so thermal collisions generally may not have enough energy to knock an electron of a metal atom. A more important reason is that electrons are locked away in metallic bonds with other atoms (for solid metals), so you have to provide sufficient energy to the electron to not only break the metallic bonds, but also release the electron from the atomic nucleus. (Please correct me if i'm wrong.)
Also, check out this ionization energy page on Wikipedia for more information.

thanks for your reply but actually that tables give molar ionization energies for GASES, not for solids....

@ODIUM Okay, i misread your question! The first line made it seem that you wanted ionization energies for gases... well i don't know if there are ionization energy tables for solids...

@ODIUM Also please note that the Wikipedia page for IE says that IE is defined for isolated gaseous atoms (and ions). So the thing you are looking for is most likely not called ionization energy. Work function is the closest you can get (AFAIK), and that you can easily find.

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


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Reaction mechanism of glycerol with hydrogen iodide acid
I was watching a chemistry video lecture today and I came across this:-
Basically the lecturer was illustrating the mechanism of reaction glycerol with $\ce{HI}$.See from right to left(follow the arrows :-P) in the screen shot.I have a doubt in the fourth step where it is shown by an arrow that the $\ce{I^-}$ ion shifts from terminal position to the central position therby forming a positive charge at the terminal position where another $\ce{I-}$ attacks.He says something like "The $\ce{I-}$ shifts to central position due to areal resonance(or something like that..hope I heard that word correctly) which increases stability".
Now that's my doubt!Why should the iodine ion shift its position at all?Isn't the positive charge more stabilized by hyper conjugation in the central position?And what on earth is areal resonance(I searched google too but no use)?
Can someone help me out ?Thanks in advance!

The iodine isn't changing positions, it's donating a pair of electrons to stabilize the charge by forming an iodonium ion similar to the bromonium ion formed during the halogenation of an alkene shown below. You can see that the bonded pair of electrons was not affected by the step. It looks like they confused the reaction by short-cutting the reaction step.

So the iodide ion attacks at the less hindered terminal position of the iodonium ion?Why? Is being less hindered the only reason? BTW thanks a lot for your answer!


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Does SCF energy mean same as HF energy?
I use Gaussian 09 and Turbomole to do a same calculation (B3-LYP/6-311G) for a same molecule. But the results are not same, Turbomole shows SCF total energy = -772.16125945927, while Gaussian shows SCF Done E(UB3LYP) =  -772.402241773, no UHF energy shows. I use UB3-LYP in Gaussian, while in Turbomole cannot specify that.
I use TmoleX to create the input file for Turbomole. It does not have UB3LYP or RB3LYP in the method part, I can only select DFT > B3LYP as my method. In previous part for molecular attributes, I can select multiplicity: UHF, then generate MOs.
The input file of Gaussian is very simple: UB3LYP/6-311g followed by cartesian coordinates of the molecule.
The energy is not the same in both of these cases. What causes the different results?

Answering your question from the title: no, SCF does not necessarily mean HF. Essentially the same self-consistent field (SCF) procedure is used to solve both the Hartree-Fock (HF) equations and the Kohn-Sham (KS) one, so for the case of DFT, SCF energy means not HF energy, but rather KS energy.
Regarding discrepancies in results obtained with different programs, there might be few reasons out there: different definitions of the B3LYP functional, different harmonics used for 6-311G basis sets (spherical vs. cartesian), different integration grids, different SCF convergence criteria, etc.

I checked the TURBOMOLE documentation and indeed found that to use the very same B3LYP functional that is used in Gaussian (with the VWN3 correlation) you need to use b3-lyp_Gaussian in TURBOMOLE, not b3-lyp. So, first of all, I suggest you try b3-lyp_Gaussian in TURBOMOLE instead of b3-lyp.


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What happens to a molecule while it is reacting?
I recently learned about the graph of activation energy that looks like this (source):

I was wondering, what actually happens to the reactants as time passes on the graph? If we were to look at the molecules when the graph was at its peak, what would we see? And lastly, can the reaction be stopped somewhere in between, say at the peak, or does it always have to stop either at one end as a reactant, or at the other as a product?

The horizontal axis on the graph is not time, but the reaction coordinate. Do you know what potential energy surfaces are?

In fact, this is related to the work of Ahmed Zewail: he devised methods (femtosecond-length laser flashes) to observe the so-called "transition state" of a reaction. A popular account by Zewail himself can be seen here.

@Hotoke-san Please do not try to circumvent the reputation requirements by leaving comments as answers.  I am in the habit of deleting such answers on principle, but this is a potentially useful bit of information.  Please do not attempt this in the future.

@jonsca That comment looks like a valid answer to me: maybe not a good one, but an answer nonetheless.

@OrangeDog I don't agree with that.  It doesn't touch upon the I was wondering, what actually happens to the reactants as time passes on the graph? If we were to look at the molecules when the graph was at its peak, what would we see? And lastly, can the reaction be stopped somewhere in between, say at the peak, or does it always have to stop either at one end as a reactant, or at the other as a product? at all and, even if it did in a direct fashion, requires the reader to go to an external link to determine that.

Well, a lot of things happen to the reactants. Some bonds stretch (and maybe eventually break), the others shrink, and your molecules morph into different molecules, which are the products.

(source)
As for staying at the very peak, that would be kinda unnatural, but luckily, not every peak looks like this; sometimes there is a tiny dent near the top, and with some effort you might be able to stop your reaction right there, that is, to isolate the activated complex. The discovery of its structure may shed some light on the reaction mechanism and earn you a lot of likes from your colleagues.

Weird -- it still doesn't work for me so the obvious explanation of a temporary server error doesn't apply. Anyway, thanks for checking: I don't see anything else we can do.

Does it at least point to www.visichem.thelearningfederation.edu.au/topic04.html for you?

Yes. But, for me, that gives a page whose source is literally the words "File not found" with no markup or anything. In fact, I get the same thing for any url within that domain, including www.visichem.thelearningfederation.edu.au/ .

Weird indeed. Looks like discrimination by IP on their end for some unfathomable reason.

Thanks! That animation worked fine for me, and really helped me picture what goes on in a reaction. I think that I was picturing bonds as physical, solid connections when they are really basically just attractions between two molecules. The activated complex sounds really interesting too :)

Please give one example where you actually can observe a transition state. The term activated complex just stems from an earlier version of the transition state theory. Whenever you have a local minimum you can observe it, a transition state, by definition, you cannot observe.

Cannot find the source file either, please make sure you can actually use it under the copyright our site provides. In any case the original copyright should be matured according to cc-by-nc-sa.

Well, like I said, the intermediate can't be isolated unless it corresponds to a local minimum on the potential surface, however minor (the mentioned "dent"), in which case the examples are plenty. (BTW, what is the right term for it? "Activated complex" does indeed sound sort of old-fashioned.) Of course you are right in that the system won't stay in the "top of the hill" state.

As for the copyright issues, sorry, I didn't get it at all. What is cc-by-nc-sa and what should I do besides leaving a link to the source page?

The top of the activation energy is called a metastable point.  In theory it could persist forever because there's nothing encouraging it to go one way or the other.  In practice, such a system is never perfectly balanced and the metastability is extremely hard to see.  Metastability is a bigger issue in other disciplines, like electrical engineering, where it's possible to create longer lasting metastabilities by accident, and they cause problems.

@IvanNeretin Can the activated complex return to reactants?

@ado Yes it can. Moreover, even the products can return to the activated complex and then to reactants.

@IvanNeretin When the activated complex is formed then the probability that it will go to reactants or products is the same (1/2)?

No, otherwise all reactions would stop exactly halfway.

The normal way of drawing a reaction profile is to place a barrier between reactants and products, just as shown in the question. As the top of the barrier is the transition state. The reaction pathway is usually not well defined as such since there are many bond angles and distances that change during an reaction, nevertheless this idea of a pathways is very useful in understanding what is happening.  What the picture shows, and also the pretty animation, is that reactants approach, there is some sort of interaction and then the products are formed. However, the picture (or animation) does not tell us how this occurs and what are the important factors is left to the imagination.
First, the reactants have to acquire enough energy to reach the transitions state at the top of the barrier. The average energy of the reactants is usually depicted as the line on the left of the figure with the word reactants. This energy is usually far less than the barrier height (except for diffusion controlled reactions, see below), and this means that most of the times that the reactants collide with one another no reaction occurs. There is just not enough energy to react, whether or not the reaction is endothermic or exothermic.
Now since the molecule are in solution or in the gas phase at some reasonable pressure of inert buffer gas, many collisions per second are going to occur between reactants and solvent (or buffer gas. In solution collisions occur at a rate of at least $10^{12} s^{-1}$. Just occasionally, perhaps 1 in $10^9$ collisions one will be energetic enough to give the reactants enough energy to reach the top of the barrier. At the transition state the reaction can occur, provided the reactants are in the correct orientation, and that a collision does occur quickly enough while there to remove enough energy from the transition state and thus lower the energy making it impossible to react.
Although the average kinetic energy of the solvent is $3RT/2$ per mole, and is small < 4kJ/mol, there is a distribution of energies given by the Boltzmann distribution. The chance of having an energy $E$ is proportional to $exp(-E/(RT))$ so is very small if $E$ is, say, 100kJ/mol and this is why many reactions have small rate constants and why so many collisions are needed before a reaction can occur.
Once at the transition state, as the energy is here well above the average, then on average, a collision with solvent will remove a little energy, rather than add to it, and so the product becomes stabilised.
(The description given here is rather crude, a more detailed considerations lead to Kramers modification to transition state theory, and this involves considering ‘friction’ on the reaction coordinate, both from the solvent and from internal vibrations in the reactants hindering or supporting motion along the reaction coordinate.)
In some reactions the barrier to reaction is very small, which means that it is only slightly greater than the average energy at that temperature. In this case once the reactants collide with one another reaction occurs. The rate of reaction in now limited by how quickly the reactants can collide an this is determined by their diffusion constants, which in turn depends on the solution viscosity. These are ‘diffusion controlled’ reactions. In the same solvent they always have a larger rate constant than a reaction with a larger barrier.
It is important to realise that the experimental proof as to what transition states are is very, very sketchy indeed. In the nice animation we can see bonds stretch and break etc., but this is perhaps, based on calculation, certainly on chemical intuition, and on circumstantial evidence from chemical experiments but which are not time-resolved spectroscopic ones. The reason for not being able to spot a transition state is that it only lasts for approx $10^{-13}$ at most, and probably less, and even in reactions started with a femtosecond laser pulse it has not been possible to detect them even in photo-induced reactions such as the trans-cis photo-isomerisation of stilbene (diphenylethene) presumably because their concentration is minute.
In simple reactions such as O + H$_2$ = H + OH, which are performed in a molecular beam much more progress has been possible. These reactions are measured at very low gas pressure and a beam of atoms intersects a beam of molecules and the products are analysed spectroscopically and/or for their energy and spatial distribution.
It is possible to plot a potential energy surface as a series of contours, just like a map showing hills and valleys. Although the surface is empirical it is calculated by using known vibrational potentials for the diatomic molecules and the Heitler-London Valence bond approach. (The surface is called LEPS after London, Eyring, Polanyi and Sato.)
Figure 1 shows the potential energy plotted vs. OH+H separation (taken from centre of OH) as ordinate and the O+H$_2$ separation as abscissa. (Note that calculation assumed the collision to be linear (end on only) between O + H$_2$ and that H$_2$ rotations are not included in setting initial collision parameters)  The transition state is at the point approx (0.12, 0.09), yellow circle. This is at the ‘saddle point’, where the surface has convex curvature along the reaction path and concave curvature perpendicular to this. The dark blue colour shows lowest energy, yellow/green intermediate and red highest.

The solid line shows one trajectory of a co-linear collision between O+H$_2$ and starts at the far right of the figure. The dashed line contours describe the total energy, collisional (kinetic) energy plus the initial vibrational energy in H$_2$.  The oscillations in the right hand valley show the vibration of H$_2$ as it approached the O atom. In fig 1 although there is enough total energy, the vibrations are such that the molecule is at the wrong place on the potential, due to its initial speed and vibrational energy, to cross the transition state and so returns down the valley. No reaction has occurred.
In the second figure, the transition state is crossed, although the total energy is the same, as vibrations place the reactants on the right place on the surface. Vibrations in the product OH can be seen in the vertical valley. Thus the important point is that it is not just total energy but also orientation and vibrational energy that is important in causing reaction.

By observing rotational and vibrational energies in products and reactants in reactions such as these, the potential energy surface can be established. The connection with (thermal) rate constants as measured in normal chemical kinetics (non molecular beam) experiments is made by calculating the number of transition state crossings at a given energy and then averaging over collisional and kinetic energies.

I assume with the passing of time, you refer to traversing the reaction coordinate (the x-axis in your graph) from the left to the right.
We start at the left of the graph, with a reactant. This reactant gains enough energy to climb the hill of the graph. Let's wait there for a moment: what does it look like? It does not look like the product, because it's energy is still quite different. After all, it is on the hill and not in the valley at the right, so it also looks different. Further it also does not look like the reactant, because it's energy is different from the reactant.
But, according to Hammond's postulate it looks more like the reactant than like the product, because in energy it is closer to the reactant.
So, on the top, you see a molecule/structure that resembles the reactant more than the product. This is nicely shown in Ivan Neretin's animation.

And lastly, can the reaction be stopped somewhere in between, say at
  the peak, or does it always have to stop either at one end as a
  reactant, or at the other as a product?

The x-axis or reaction coordinate tells you how the transformation progresses from one form to another form. As the peak is what it is: a peak, it is not energetically favorable to stay there. At the moment the structure is at the peak, it will overshoot to the left or the right. So, this activated complex as it is called, can not be isolated in your reaction flask.

Thank you! Your answer contained some details that were not specifically stated in Ivan Neretin's. It was very helpful. I'm interested in why Hammond's postulate is a postulate. Is it because the actual molecules can't be observed in a reaction, or because there are other, opposing theories?@Lucademicus

In virtually all cases, there isn't just one distance or variable involved. Perhaps your chemistry classes have taught you that there are three 'types' of molecules? Monoatomic - ie atoms, they have the simplest motions basically just velocity and spin around an axis. So, (ignoring the specifics of the element's electron orbitals configuration) we have a velocity vector (3d) and a spin vector (also 3d). Di-atomic Add to the monos the vibration (variation in bond length) and maybe tumbling (end over end spin) and they aren't much more complex - another spin axis (maybe) and a periodic motion (but you have more orbitals to worry about as well). And then there's everything else. You got waggle, wobble, flapping, all sorts of motions of the nuclei. The energy landscape gets complicated. There is more than one way to get from reactants to products. A simple energy vs "reaction coordinate" (2d) diagram hides all that complexity. Reaction coordinate is like attempting to describe a fight using only impact force. It's not just the force, it's where you apply it. IOW, the 'energy' can be in various modes and still get the job done, while other even more energetic 'collisions/interactions' may have no effect. Best to look at it as an average (and an average which is context dependent, for instance changing solvent can dramatically effect activation energy). You can not really describe general properties of the reaction coordinate. They depend on the details of the reaction too sensitively for useful generalization over the entire subject of chemical reactions. It's a VERY useful and important concept, but it's very fuzzy. Trying to focus on the details is going to require very specialized considerations which aren't generally useful for "general" chemistry. The specifics don't generalize easily.

Any element or compound  react with  others to reach more stable state that is lower energy form. But before that we have to initialize the reaction (if it is not the deal then any reaction will occur automatically and they will reach the lowest energy state in no time and no further reaction you can see in your life) by giving some energy that is making it unstable where your graph is at its peak , then the existing bond will dissociate and the constitute elements will find the combination which have lowest energy and acquire that combination which is the  reactant of your reaction.     I think this is appropriate .**i m ready for further help ** if you want.


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The percentage of ingredients in a glass of tea
When a pot, pitcher, glass, etc of tea is brewed, how do I find the exact concentrations of tea and water in the end result. I realize it depends on the amount tea and water used when brewing. Is it still called tea, albeit weak tea, when the water is more prevalent in the water to tea ratio?

You can't really talk about the 'concentration' of tea, because 'tea' is not a well defined compound. It contains a lot of different chemicals, some of which are dissolved into water when brewing.
You could talk about the concentration of caffeine, epigallocatechin gallate, or manganese in a particular cup of tea, but they depend on the type of tea, the temperature of the water, the infusion time, etc.
The minimum strength of a cup of tea that still allows it to be called tea is a fuzzy question without an answer. Take a pile of sand. Remove grains of sand one by one. When does it stop being a pile? A reasonable limit for the tea would be when a normal human being can't taste the tea anymore.

Apart from tea being not a well defined compound, it is not even a well defined mixture. Even one type of tea is not even homogeneous as it depends on where it was grown, and the conditions thereof. || "A reasonable limit for the tea would be when a normal human being can't taste the tea anymore." Reasonable, but very subjective. I suspect there is no such thing as a normalised human being. || I agree with the sentiment of this answer, that it is too fuzzy to be answered rigorously.


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Separation of sulfur dioxide and sulfur trioxide in the contact process
When gases leave the catalytic reactor in the contact process of sulfuric acid generation, both $\ce{SO2}$ and $\ce{SO3}$ will be present. According to descriptions I have read, the trioxide reacts with the water spray, but the dioxide does not, and instead rises as a gas and is recirculated into the system.
I thought that sulfur dioxide was highly soluble in water. If so, should it not dissolve into the water spray and thereby contaminate the oleum?

$\ce{SO3}$ reacts rapidly and exothermically with $\ce{H2O}$, forming the liquid $\ce{H2SO4}$, sulfuric acid. This removes the $\ce{SO3}$ from the mixed gases, allowing yet more to react. In fact, the resulting acid can contain more $\ce{SO3}$ than $\ce{H2O}$. This is called oleum or fuming sulfuric acid.

Nobody knows why $\ce{SO2}$ is not soluble in oleum. The prediction of the solubilities is one of the yet unknown questions to be solved in chemistry. There are plenty of theories for explaining the solubilities of a group of substances in a group of solvents, using different parameters like polarity, dipole moments, electronegativities, entropy, etc. But no theories are general enough to explain the solubility of any solute in any solvent.

You are true that $\ce{SO2}$ is very soluble in water but it depends on the temperature. Also, $\ce{SO3}$ reacts with water then dissolves in it in a very exothermic way. My understanding is that, if the quantity of water is well adjusted, you will actually get hot concentrated sulfuric acid and sulfur dioxide is not quite soluble in it (ca. 8 g/kg at 120 °C). So $\ce{SO3}$ reacts with water and sulfur dioxide gas can be recirculated.
HTH

When gases leave the catalytic reactor The hot gas mixture is then passed through concentrated sulfuric acid , Sulfur trioxide  readily reacts with concentrated sulfuric acid to form oleum but other gases like O2, N2 don't react and pass through
The reaction is as follows:
SO3(g) + H2SO4(l) → H2S2O7(l) (Oleum)
In this step, the SO2 and unreacted O2 pass through the absorption system relatively unaffected, while the SO3 is captured and converted into oleum.


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How can a bond be more stable and more polar at the same time?
When comparing the acidic strength of the halogen acids, we see that the acidic strength decreases as $\ce{HI>HBr>HCl>HF}$. It is said that the reason for this is the bond strength of $\ce{HI}$ is the least, while that of $\ce{HF}$ is the greatest. But, the polarity of the bond in $\ce{HF}$ is the greatest, how then is the strongest bond?
I understand that fluorine has smaller valence atomic orbital (2p) than, for example, iodine (5p), and so has more effective overlap with the s-orbital of hydrogen. But how do we reconcile this fact with the fact that the fluorine hydrogen bond is more polar too? The electrons remain more with fluorine in $\ce{HF}$ than with iodine in $\ce{HI}$, so intuitively it seems that the hydrogen fluorine bond may be more easily broken by a base than the hydrogen iodine bond. So what is going on?
In short, how can a bond be stable when it is highly polar?

Edit
My question is regarding bond strength, and not the acidic order. For example, in organic compounds, the leaving group ability of the halide ions decreases as$^1$:
$$\ce{I^{-}>Br^{-}>Cl^{-}>>F^-}$$
This means that the fluorine-carbon bond is the hardest to break. But fluorine has very high electronegativity, and so the bond must be highly polar. But inspite of this, fluorine has the highest bond strength!
So is this just a misconception I have? If it isn't a misconception, then why does the highly polar bond have the maximum bond strength?
Thank you!
$^1$: https://chem.ucr.edu/documents/curricularmaterials/neumantextbook/Chapter7.pdf

Why wouldn't it, really? When you break a highly polar bond, you're left with two ions which are attracted to each other with great force.

@IvanNeretin Hmm, then when comparing acidities of compunds across a period, we say that because of increasing polarity of the bond with hydrogen (moving from right to left), the compound becomes more acidic? ($\ce{HF>H2O>NH3>CH4}$)

Sounds true to me.

@IvanNeretin I agree that what you say makes sense, but we use exactly the opposite explanation while comparing acidities in a period! What is the correct explanation for that, then?

@Mithoron I believe my question is different (I have checked your suggested duplicate). Please check my edit and guide me if I can put it in any better words!

The important thing here to keep in mind is that the bond strength is not a spherical chicken in vacuum, meaning that we need to keep the entire process in mind when analyzing all the factors that can influence it. Although the fluorine has a higher electronegativity, it is significantly smaller than Cl, Br, or I (same goes to their corresponding ions). And without going too deep into intricacies, $\ce{F-}$ is $way$ less stable than $\ce{I-}$ and thus, using the OChem language, a much better nucleophile. Due to it's inability to stabilize the charge, it holds onto the hydrogen with all it's might (pardon me for anthropomorphizing the atoms).
SO, when you're thinking about breaking $\ce{H-F}$ bond, you gotta take into consideration what happens with the products. In the example with organic chemistry and leaving groups that you have pointed out, fluorine is the worst leaving group for the precisely the same reason--it cannot stabilize the negative charge well, so it doesn't "want" to leave. However, if you try to do a, say, $S_N2$ reaction in a protic solution (not the best way to go around, but doable), then $\ce{F-}$ will be a much better leaving group than $\ce{I-}$, b/c $\ce{F-}$ will instantly "catch" a hydrogen from the solution and stop being nucleophilic, which according to Le'Chatelie'r principle shifts the equilibrium to where $\ce{F-}$ is a leaving group.

Thank you for your answer! You mentioned that $\ce{F}^-$ ion is less stable. Is it because of the repulsion between its lone pairs, or due to high charge density (making it attractive to electrophiles)? Or is it something else entirely?

@FreezingFire: if explaining it "on fingers" without going into the depths of quantum theory: yes, and yes. Having a high density of a charge (either positive or negative) is never a good thing, which make the corresponding species unstable.


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How do electron donor groups affect the polarity of an O-H bond in alcohol?
My textbook (Chapter Alcohols) , says:

An electron releasing group ($\ce{-CH3}$, $\ce{-C2H5}$) increases the electron density on oxygen tending to decrease the polarity of $\ce{O-H}$ bond.

Now, my question is that if the electron density increases on oxygen, then shouldn't the polarity of the bond increase? Why does it decrease instead?

Oxygen tries to draw electron density from the two moieties it's bound to, thus making both bonds more polarized than the normal carbon-carbon bond. Some functional groups offer more resistance than others to this polarization depending on their electronegativity. For this reason, if one of the groups "puts up more of a fight" than the other, oxygen will draw more electron density from the weaker one.
In your case, hydrogen is always the same since we're always talking about alcohols. Alkyl substituents donate electron density to the oxygen atom, so it is less eager to draw electron denstity from the hydrogen bond and will make it less polar. Change the alkyl for a primary alkyl chloride for example and the effect will be reversed.


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Why is gauche conformer of 2,3 dimethyl butane more stable than its anti conformer?
Why is gauche conformer of 2,3-dimethylbutane seen along 2,3 bond more stable than its anti conformer? It seems to be anomalous to me as in most other butane like compunds anti form is more stable than gauche form.Can someone explain me the reason why the compound shows Gauche effect?
Also does this anomaly hold in case of 2,3-diphenylbutane and 2,3-dimethylpentane?

Related: Conformations of butane and How can the gauche-effect be explained?

@Loong I saw those links.They do not explain the case of 2,3-dimethylbutane.

I am also confused by this and asked a more general question here. When it's answered, if it doesn't also answer yours, I'll be happy to come back here.

@S.Chevalier Your question is different from mine.Your one is answered in Wikipedia page I linked named Gauche Effect in my question above.

Could it be possible that the conformation where the hydrogens are anti is called gauche?

No.Even the hydrogens are in gauche position.And I found out the reason for stability  today.Will answer the question soon.


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Curtin-Hammet principle
Define the Curtin-Hammet principle and use it to predict the expected trans/cis product ratio in the following example:
       PhCH2CH(Ph)OCOR    strong base     PhCH=CHPh
                              100C

This part-Q (12-marks) did not include any numerical data. The student is, presumably, required to derive the Curtin-Hammet equation; which, establishes the trans/cis product proportions in terms of their respective transition-state (free energies of activation) energies. But without numerical data how is the trans/cis ratio to be determined?
It may be interesting to note that stilbene, in uv light, undergoes a cis-trans isomerism-equilibrium. So, given such circumstances, the trans/cis ratio must be 50:50? Such a convenient short-cut may not have been required; nor, would it earn the 12-marks.


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Confusion about gauche effect
This recent question about the gauche effect led me to read this rather older answer. The problem is that I notice I am still confused.
To take the case of 1,2 difluoroethane, I drew these orbitals:

Figure1: orbitals of $\ce{C2H4F2}$ in the gauche (1,2) and anti (3,4) confromations
Now I understand that 1 (gauche) leads to more stabilization than 3 (anti). But 2 (gauche) leads to less stabilization than 4 (anti). I don't know which effect is strongest: the destabilization in 2 or the stabilisation in 1. So I don't know why the gauche conformation (1,2) leads to more stabilization than the anti (3,4).
Did I draw something wong? Did I make a mistake in reasoning? Did I miss something else? Is this one too many error-acknowledgement questions?

The most important interaction in depicted in image 1. The energetic match of these oribtals is the best from all the listed.

Yes, but how do I know that it's enough to compensate for the loss of stabilization in 2 vs. 4?

In figures 2, it looks like you're saying the sigma of the C-F will be donating into the sigma* of the C-H, and that causes stabilization. However, a C-F sigma bond is a much poorer donor than a C-H due to the high electronegativity.
Since there is little to no donation actually occurring in 2 (or 4, for the matter) those interactions are much less relevant than depicted in 1/3 where there is significant donation for C-H sigma into C-F sigma*. Frankly I would never have even thought to draw 2/4 since there is so little electron density in the H lobe of the C-F sigma, but cool!

Wow I'd given up hope on this one.


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When we blow on/stir a hot drink, why aren't we making it hotter?
This may have already been a question on Chemistry StackExchange (in which case I will delete it if someone points me to the question), but I was wondering why providing even more kinetic energy into the system (stirring a cup of hot coffee) why are we expecting the coffee to become cooler?
If I understand correctly, heat is caused by molecules in a system moving rapidly and "rubbing each other" -- creating heat. So the more movement --> the more heat.
Doesn't this not make sense with the idea of stirring a cup of hot coffee?
Edit#1
Excuse me for the title and the body implying different things, I took blowing on a hot drink and stirring it as the same thing -- stimulating movement in the drink.

Open your mouth and only exhale to your hand - it's getting warmer. Close your mouth to form a small o and blow  - your hand is getting colder.

Is there a reason why the title says "blowing" while the body says "stirring"? These are not the same question (though they may have an equivalent answer).

You're right, @Nij, I have fixed this. Sorry about that.

(Very) relevant XKCD (What if?).

Ha, that was relevant enough to be a bit scary. Thanks for the link!

see this:-http://physics.stackexchange.com/questions/5265/cooling-a-cup-of-coffee-with-help-of-a-spoon

This is related to a concept called "vapour pressure". Liquid water (and any other liquid for that matter) always has some amount of water vapour that has to be in equilibrium with it. For example, the air above water of 5 °C has about 1% of water vapour, 40 °C has about 7% of water vapour and at 90 °C it has about 70% of water vapour.
If there is too much vapour in the air for the temperature it is, then water condensates. That's why you get condensation on your glasses when you walk from a hot humid room outside into the cold at winter. If there is not enough vapour in the air, water evaporates to fill in for the missing vapour. That's why taking off your shirt after you've exercised and sweated makes you dry faster. The air trapped between your skin and clothes is very humid and in equilibrium with the sweat on your skin. Once you take it off, dry air comes into contact with the skin, forcing evaporation of your sweat (even though you're below the boiling point of water).
This process of evaporation is endothermic, meaning it consumes heat. Whenever water is evaporating, the temperature of the water goes down. When you're blowing air on a hot drink, you're basically replacing the humid air on top of the drink with dry air. This allows the water to evaporate more easily, reducing the overall temperature of the drink. This is also why you're feeling cold when you got wind blowing on your when you're sweating.


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Is LiF a weak electrolyte in acetic acid?

The vapor pressure of acetic acid is 0.033 atm at 25°C. If 5.00g of $\ce{LiF}$ is added to 100g of acetic acid, what would be the vapor pressure of the solution at 25°C?

The lowering in vapor pressure (VP) can be calculated with this equation:

${∆P = X_{solute} * P°_{solvent} * i}$
${∆P}$ is the change (lowering) in VP.
${X_{solute}}$ is the mole fraction of the solute.
${P°_{solvent}}$ is the VP of the pure solvent.
${i}$ is the van't Hoff factor of the solute.

What is the "correct" van't Hoff for this problem, from the perspective of a general chemistry student? I know that $\ce{LiF}$ is not very soluble in water, and I know there are papers on lithium fluoride and acetic acid, but that's all beyond the scope of a general chemistry class.
Therefore, I think it's 2, because $\ce{LiF}$ is expected to be a strong electrolyte. It's ionic, and the solvent is similarly polar (acetic acid has hydrogen bonds). The solvent even resembles water in that both have hydrogen-bonding. So I would expect $\ce{LiF}$ to dissolve fully in acetic acid.
Remember, this is a general chemistry class, and the students aren't expected to know much beyond "like dissolves like." $\ce{LiF}$is polar, and so is acetic acid. So to them, $\ce{LiF}$ should dissolve in acetic acid.
Problem is that I'm in trouble with my supervisor since I told someone that i = 2. My supervisor is telling me that the van't Hoff factor should be 1. I looked over his work to see if he accounted for the dissolving of $\ce{LiF}$when calculating the mole fraction of the solute but he didn't - he just found the mole fraction of $\ce{LiF}$, not the combined mole fractions of $\ce{Li+}$ and $\ce{F-}$.
So, what's the deal with this problem? Is there some weird exception I'm not aware of regarding $\ce{LiF}$ and acetic acid? The van't Hoff factor should be 2, correct?

According to Acid-Base Equilibria in Glacial Acetic Acid. III. Acidity Scale. Potentiometric Determination of Dissociation Constants of Acids, Bases and Salts J. Am. Chem. Soc., 1956, 78 (13), pp 2974–2979:

It is generally agreed [references 3 and 4] that acids, bases and
  salts are only slightly dissociated in glacial acetic acid.

While the article does not specifically address lithium fluoride, it has quantitative data for lithium chloride.
Lithium Chloride dissociates with a pK of 7.08 +/- 0.02 in acetic acid.
So considering the high concentration of LiF in the problem (5 in 100 grams), if LiF is similar to LiCl and the other salts studied, 1 rather than 2 is correct.
This knowledge is not part of a standard general chemistry curriculum, so unless a specific book or lecture provided relevant information, I wouldn't expect students to know this.
Furthermore, the solubility of LiF in acetic acid is only 0.84 grams per kilogram according to Glacial Acetic Acid as a Non-aqueous Solvent for Metal Fluorides, so the problem is not realistic. (Only 0.084g of the 5.00 g would dissolve).  Also, HF is a strong acid in acetic acid according to that paper, so LiF might even be an exception to the general observation that salts do not dissociate in acetic acid.
Also, keep in mind that in the liquid phase, acetic acid exists as mostly cyclic dimers, which gives the solvent some non-polar characteristics.


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What does the amount of acetic acid in vinegar depend on?
I have different types of vinegar and I'm wondering what factors influence how much acetic acid is present in different vinegars (e.g wine vinegar/white vinegar/apple cider). I was thinking maybe it has something to do with different fermentation processes. Any thoughts?

You know that in many cases it's diluted to suitable concentration?

See this http://www.hindawi.com/journals/tswj/2014/394671/

Vinegar comes from French "vin aigre", aka sour wine. The transformation of wine into vinegar happens when some (possibly unwanted) aerobic bacteria transform the alcohol into acetic acid by oxidation. So any alcoholic mixture which allows growth of those bacteria can be turned into vinegar, like cider, sake...
The amount of acetic acid depends on the alcohol content which in turn depends on the initial concentration of sugar. It is also dependent on the time allowed for fermentation.
In fact it is much more complicated than that. Have a look at Wikipedia for more info.

My 3.785 L, extra strength 6% "Acitity" vinegar (V), means (1), and I just found this out myself. Vinegar produced by the Generator Method, which produces anhydrous acetic acid. The pure Acetic acid (Ac) crystalizes to a solid at less then 16.7 C. (2). You can imagine how easy it is to make any concentration you want, when you have a powder to mix with water = distilled vinegar. That means that in that 3.8 L bottle only about 270g is actually Acetic Acid. However because Acetic Acid is a weak acid in water with a pH of ~ 2.2 (= ) for 6% acidity. Some of that Acetic Acid (moles of Ac - mole of protons = moles of acetic acid left)
dissociates so, there is some (equal to ~moles of protons) Acetate in there as well.
Link 1 seems pretty interesting regarding the making of vinegar
1 - http://www.madehow.com/Volume-7/Vinegar.html
2 - http://antoine.frostburg.edu/chem/senese/101/organic/faq/ethanoic-acid.shtml

Fun fact: you might be able to mix something that tastes like vinegar but (in the UK at least) you can't legally call it vinegar; it's 'non-brewed condiment' (https://en.wikipedia.org/wiki/Non-brewed_condiment).

White vinegar is produced by distillation from any source: fermentation of apples, grapes, potatoes, or just sugar with some yeast nutrients. However, it is usually standardized at 5% acetic acid by adding water. Putting glacial acetic acid on your fries would not be healthy ;-(


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Iron Filing Catalyzed Toluene Reaction with Bromine Solution
When toluene and a bromine solution react, there is no reaction. That is, the solution turns a yellow/brown colour, the same as the bromine solution.
I conducted an experiment where toluene and a bromine solution were meant to be catalysed by iron filings. However, instead of the solution losing its colour (like you'd expect in an alkene/alkyne), it became rustic in colour. If the iron filings were intended to catalyse the reaction, why did the solution become darker? The reaction that took place seems entirely counter-intuitive.

Your black material is likely from impurities which is a common occurrence with organic reactions especially aromatic ones.

The iron and bromine must first react to form the catalyst (FeBr3) as the iron does not actually catalyze the reaction (this is known as in situ synthesis of the catalyst).  The FeBr3 catalyzes the reaction between toluene and Br2 to form ortho and para bromotoluene.  Catalysts are NOT consumed in the reaction though, which means the culprit of the rusty color is the catalyst (google iron(iii) bromide to view the color).
Also, ensure that you use a larger excess of the bromine.  The FeBr3 interacts with bromine to make it more electrophillic and able to react with the benzene ring of toluene.  If there is no bromine left after the catalyst has been formed, no reaction will occur.


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Nucleophilic substitution with chloroethane: hydrogen cyanide vs. potassium cyanide

Chloroethane can be used to make sodium propanoate.
$$\ce{chloroethane -> Q -> sodium propanoate}$$
The Intermediate, $\ce{Q}$, is hydrolysed with boiling aqueous sodium hydroxide, to give sodium propanoate.
Which reagent would produce the Intermediate, $\ce{Q}$, from chloroethane?
A concentrated ammonia solution
B dilute sulfuric acid
C hydrogen cyanide
D potassium cyanide

What is the difference between using hydrogen cyanide or potassium cyanide? Why would the answer be D?

Because hydrogen cyanide is not a very strong acid.
Potassium cyanide exists as potassium and cyanide ions, so here the cyanide ions can act as strong nucleophiles. Hydrogen cyanide is only a weak acid and so it will not dissociate to produce a significant amount of the cyanide ion nucleophiles. So it is a better idea to use potassium cyanide.

There's also the safety issue. Hydrogen cyanide is a toxic gas that can be lethal even at really small doses.

For cyanide to act as a nucleophile, we need the carbon-centred lone pair to be accessible as this orbital corresponds to the nucleophilic HOMO. In hydrogen cyanide, this is not a lone pair but occupied by a hydrogen atom forming a covalent bond. See the structures of hydrogen cyanide and the cyanide anion below.
$$\begin{array}{cc}
\ce{H-C#N:} & \ce{:\overset{-}{C}#N:}\\
\text{hydrogen cyanide} & \text{cyanide anion}\end{array}$$
This immediately tilts the balance in favour of the cyanide anion. But hydrogen cyanide is also an acid (e.g. compare its German name Blausäure or blue acid). Can it not deprotonate to give cyanide? Well, it can, but it is a very weak acid. The Evans $\mathrm{p}K_\mathrm a$ table gives its $\mathrm pK_\mathrm a$ value as $9.4$ in water and $12.9$ in $\ce{DMSO}$. Therefore, relatively strong basic conditions are required to generate substantial amounts of the anion. The reaction conditions, however, do not mention any added base. It is therefore reasonable to assume neutral conditions, which means that the vast majority of cyanide species when using hydrogen cyanide will indeed be the free acid $\ce{HCN}$ which is not able to attack.
This is avoided by using the already deprotonated ionic compound potassium cyanide.


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Why are protein hydrophobic cores denatured by heat when entropic forces should be favoured at high temperature
The hydrophobic interaction is thought to be driven by an entropic force. If it is, shouldn't hydrophobic interactions be stronger at higher temperatures, where states of higher entropy are favoured? Why, then, are protein hydrophobic cores denatured by heat?

I'll just guess a bit... Is it maybe that the hydrophobic effect requires both entropic stabilization as well as van der waals interactions between non-polar moieties? Maybe with the van der waals forces overcome, the entropy of the hydrophobic moieties themselves becomes more stabilizing than water expulsion from hydrophobic surface area?

You are correct that the hydrophobic interactions are stabilized at higher temperatures. However, I believe the addition of heat mainly disrupts protein structure by breaking hydrogen bonds through increased thermal motion.
Once the secondary structure starts to go, any tertiary interactions will get scrambled, likely resulting in the separation of hydrophobic residues. The refolding of the polypeptide backbone must be stronger than the enhanced hydrophobic interactions.

I like this answer. I guess the energy of the hydrophobic interaction alone is not sufficient to conformationally restrict core residues in an efficiently packed structure. So just to be clear, if you heat oil in water (or any hydrophobic solute in water), you would, in fact, be favouring coalescence of the hydrophobic contents, right?

Yep, exactly, but there are exceptions. When an organic solute has a polar group (e.g. the alcohol group in octanol) increased temperatures can also increase solubility, for much the same reason hydrophobic cores of proteins can be denatured by heat.

To add to the answer of Dan Burden: the interactions might be stronger, but you're still making it easier for the protein to cross the energy barriers towards unfolding if you heat the protein.
Also, one thing that helped me understand protein physics is the following: an interaction might be worth 2 kcal/mole in a vacuum, but if you break this interaction you probably don't go to 0 kcal/mole. It's likely the protein forms another hydrogen bond worth only 1.8 kcal/mole with water, or forms a similar hydrophobic interaction with a non-fitting patch of the protein that's almost as favorable.

The energy barrier is a good point. After all denaturation involves transition to a kinetically trapped state.  In terms of such replacement interactions, I'm familiar with that idea in terms of electrostatic interactions such as when secondary structural h bonds are replaced by ones with water with little energy difference. Where do the replacement interaction partners come from for the hydrophobic moieties in the core though? Are you thinking it's more just intramolecular interactions with other core residues, but in a way less tightly packed and organized manner? Or are you thinking more th

I think in the end the hydrophobic residues will also interact with the residues of other proteins that are starting to unfold and aggregate.

As has been pointed out by @VonBeche adding energy (heating) is eventually going to unfold/denature a protein. Intermolecular (or other) forces cannot ultimately withstand thermal energy.
However, the use of 'entropic forces' while common is misleading since no forces are involved due to entropy. Entropy has units energy/temperature and so $T\Delta S$ units of energy not force, which is energy/distance. I try to explain, below, why it appears as a force.
When placing a protein into water, a 'cavity' in the water  has to be made which means breaking some hydrogen bonds (expensive in energy terms) and thus also some loss of entropy and the water has to bridge the cavity, also expensive. If the protein is unfolded and in water it has very many different configurations, possibly all of about the same high energy, relative to its folded state.
The water will try to hydrogen bond to polar residues but for non-polar residues hydrogen bonds will have to be broken and then remade as water tries to bridge around the residue (Remaking bonds recoups some of the energy lost and bridging recoups some of the entropy). In thermally induced, random diffusion of the protein chain, at some instant residues will come close enough to one another for relatively long range intermolecular forces can act between them. If water can be pushed out from between residues the intermolecular forces can increase, first, because there is no longer a polar molecule in the way (high energy interaction) and second because the residues can become closer.
In this manner in a globular protein the hydrophobic core can form, with polar residues on the outside hydrogen bonded to the water. So its not that entropy is a driving force but that by expelling water internal energy (enthalpy) can be lowered and water can now join with other water molecules making hydrogen bonds (enthalpy lowered) and increasing entropy in the water, thus lowering $\Delta G$ overall. The reduced entropy of forming a hydrophobic core is more than offset by removing water and allowing it to enter the bulk.
There is a decrease in free energy by forming a large cluster of particles out of several smaller (hydrophobic) ones provided that the combined particle is large enough; calculation indicate a volume/surface area ratio > 1 nm.  The decrease in free energy is largely enthalpic, whereas for smaller particles free energy change is largely entropic.
(Some other points about globular proteins, hydrophilic residues  do hydrogen bond with water but there is still an entropic cost since now these water have restricted rotational & translational entropy vs. bulk water. When a hydrophobic residue is at the surface, waters have to bridge round it at large entropic cost that is roughly proportional to surface area. If a polar group is in the hydrophobic core (for functional reasons) its energy must be controlled by neutralising its charge or dipole with other residues.)

Right and all that is exactly why I thought heat would favour the interaction. If it is driven by the entropy difference of water expulsion, the entropy term should be an even more negative contribution to the Delta G at higher temperature.


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A and B irreducible representations?
In character tables, I know that A and B are one-dimensional irreducible representations, but what is the difference them?
And why do some character tables list them as A1/A2, B1/B2 and others have Au/Ag, Bu/Bg?

g|u are from the German terms for even|odd, 1|2 are for people who doesn't care to remember those, and the difference is all there in the tables. Also, welcome to Chem.SE.

The traditional labelling of irreducible representations is the following:

A, when rotation around the principal axis is symmetrical
B, when rotation around the principal axis is asymmetrical

You can see, for example, that $C_n$ groups of even order have both A and B representations, while those of odd order cannot feature B representations.
As others have commented, there are additional rules… E for doubly degenerate representations, T for triply degenerate, u and g for parity under inversion (odd and even, respectively), greek capital letters Σ, Π, Δ for infinite groups (C∞v and D∞h).
Finally, if there are several irreducible representations that are not distinguished under those conventions, they are numbered.

@ F'x, its a small point but it would it not be better  to say for A, that after rotation about the principle axis the molecule is indistinguishable  compared to that before rotation, (or words to this effect) and in B it is distinguishable. I suggest this because symmetrical/asymmetrical are either not exact enough or confusing in this context.


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About ionic bonds (and ionic compounds)
I have a few questions:

All ionic bond occurs between a metal and a non-metal. Is this true? In the definition of metal and/or non-metal are the metaloids included? In the definition of non-metal are all diatomic and polyatomic non-metals included? Meaning all of those elements form ionic bonds with metals?
When 2 charged molecules form an ionic bond, how is the compound classified?
For ionic compounds formed by only 2 different elements, knowing the metal and the non-metal in question, is it possible to have more than one formula for the compound? I think this is probably possible and maybe even common for the transition metals, because we usually see those roman numerals in their formulae. Am I correct in assume transition metals can form different ionic compounds (meaning different formula) with one same element? If yes, why does that happen?
does item 3 extends to charged molecules?

Well, this is stark example of too broad. Didn't I already tell you not to ask like this?

how can I add something to this question to make it more narrow? I'm beggining to think some people simply don't like me around here. A person managed to answer the question with ease, and even added extra information.

Finn this is what you should add as a comment to his answer, not as an edit to your question. Questions without noise and clutter get higher scores too.

I don't care about the scores, I care about learning. I wanted this question to have a high score because of the great answer I received. I'm not a Chemist, I have no intention to amass a good reputation with Chemists online, since they seem to me like a very judgemental bunch. If I'm asking a broad question, it means I don't know enough to ask a specific question. So, they judge my question unfit for the forum in an elitist basis. I got my answers, that's all I care about. They can downvote this to oblivion as far as I'm concerned.

You asked four different questions in one post. That's why it's "too broad".

@f'' Thank you, I would never have guessed that. The questions are related by a theme, and if someone would bother to read them carefully they would maybe notice that they are put in the form of 4 separate items, but they are all the same thing: I'm trying to make better sense of ionic bonds, and by starting with these questions clarified I can then ask more specific ones. Apparently, people in this community have forgotten what does it mean to "not know" something.

Not exactly. In ammonium chloride ($\ce{NH4Cl}$), the bond is between a cation ($\ce{NH4^+}$) that has a non metal as its central atom. So the condition that ionic compounds are formed between metal cations and non metal anions is not necessary. Additionally, you will find many organic compounds where the anions or the cations are made up of many atoms, that is the cation itself consists of many atoms, and likewise for the anions. (I guess this was what you meant by charged molecules!). Organic cations (where the cation is not a metal cation) usually have their positive charge centered on a nitrogen atom. Secondly, almost all the non metals are capable of forming ionic bonds, regardless of whether they are diatomic or polyatomic in their elemental state.
It is called an ionic compound.
Yes, in fact transition metal compounds can have different formulae with the same metal and non metal elements! The best example would be $\ce{FeO}$ and $\ce{Fe2O3}$. So why do these different formulae exist? Here is a hint: $\ce{FeO \;=\; Fe^{+2} + O^{-2}}$ and $\ce{Fe2O3\;=\; 2Fe^{+3} + 3O^{-2}}$. Did you spot it? In the first compound, there are $\ce{Fe^{+2}}$ cations, while in the second one, there are $\ce{Fe^{+3}}$ cations! Why does this happen? Well some metal atoms can show two or even more different valencies. For example, once iron has lost the first two electrons, it has the possibility to lose the third electron as well, so it can show both types of positive charge on it, +2 as well as +3. Same is with other cations.
Not sure if anything like that happens. The problem is that generally only metals can lose an extra electron without changing the cation structure, charged molecules may require to change their structure to change the amount of charge they hold. The closest I know is transition metal complexes. The cation does change its compostion in various compounds (as a whole cation), but the central metal atom can remain the same. That is pretty advanced stuff though. You will learn about it in higher classes (in class 12).

Yes, by "charged molecule" I meant something like $NH_4^+$. Something that has covalent bonds but has a charge. About 2, even if it's a big molecule (covalent bonds) with just one ionic bond somewhere, it is still an ionic compound? About number 3, teachers talk about atoms as if they have a will, they say "the atom wants to have a full outer shell so it gives away/accepts electrons (depending on the atom)". Why transition methods "don't care" about their outer shell so much like the other atoms? About 4, ok, I don't want to deal with nothing extra advanced.

About 2, technically maybe yes, but by then we wouldn't bother classifying it like that! For example, charged amino acids may be called ions, but it is uncommon to read about ionic proteins or DNA ion! About 3, you should read about chemical bonding on sites like Wikipedia and chemguide.co.uk. The essence is that all atoms and molecules "want" to decrease their total potential energy. Some have the ability to show different charge on them to suit the situation. But some might do it only because they have no choice.

For example, if you were trapped in a castle on a hill. You want to go to the valley, but you do not have enough energy. Then you are stuck there. Now think of your height above sea level as your energy. You cannot lower your energy till someone gives you a boost up the wall! Same is with molecules. Iron may be trapped in +2 state till someone gives it a boost over the wall and sends it to the +3 state!

For transition metals, these walls are smaller, and it has a possibility for going lower, so they show multiple oxidation states (look that up). Most of others do not have multiple energy levels to choose from, or have too high walls to climb over.