text:
Correct way of drawing D-configuration with this stereochemistry?

Hello! I have completely memorized how to draw all 20 amino acids in the way seen in the picture (middle, between the "ball-and-stick" and fisher projection)
By my understanding this is in the L configuration, what is the correct way of drawing them in D- configuration? Is there a simple trick? Does anyone have an image of them all drawn in D-config in the same way?
Is it basically having the COO- and NH3+ switch places. Like here:

Thanks!

Take any two substituents at the asymmetric carbon and switch places; that would be another enantiomer.

I can see how that would work with the fisher projections but how about in the middle drawings? Can you explain further?

Are the charges a form of hypercorrection?

Middle drawings are no different from Fisher projections in this regard. Switch any two substituents, and you'll end up with another enantiomer.

So the way you say, how would some of the amino acids look? Thanks for the help!

Swap the bold wedge and the dashed wedge. That will reverse the stereochemistry. Remember to do it for every position. This particularly important for something like isoleucine.

So every bolded wedge into a dashed wedge and vice versa? Is everything else the same?

@Zhe Can you confirm?

Yes. For a tetrahedral center with 4 different substituents, you will change the relative stereochemistry by reversing exactly two substituents. You can prove this easily via the CIP rules.

@Zhe so for whichever of the 20 amino acids, if I want to turn them into D configuration I ONLY swap ALL bolded wedges into dashed wedges and ALL dashed wedges into bolded and it will be correct? And NH3+, COO- and the R groups are left just as they are? I just want to make completely sure.

Just someone say yes or no so I can rest assured.

YesYesYesYesYes

You don't have to memorize the whole structures. Easier is to memorize the general structure including the stereochemistry, and the residues, R (except from proline). The Fisher projection is useful, but on your depiction, it's not in the usual orientation. The carbon chain should be vertical, carboxy group at the upper side, residue at the bottom. Hydrogen and amino group (pointing towards observer) horizontal.

(non-ionized forms)
One can easily memorize that for L- amino acids, amino group is on the left, for D- amino acids the opposite, which is the case for your orientation as well (and don't forget to invert chiral residues for D- AA's, as Zhe has already noted in the discussion; it is the case for isoleucine and threonine).


text:
Why do osmium and iridium have the most oxidation states of all the elements?
Both osmium and iridium have 12 oxidation states.  Indeed, iridium has the highest oxidation state of all elements at +9.  see https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements
I'm just curious as to why these elements have the most oxidation states and whether this confers on them special properties or different reactivities or any special uses.  Thank you!

No, they just have a lot of d- and f-electrons to play with. One might suggest that elements one row down would be similarly flush in oxidation states, and perhaps even more so, if we had enough around to do much chemistry with.

I have another question:  The row below is the actinides and whilst they display similar positive oxidation numbers they don't display any negative oxidation numbers.  why is this the case?

Why do you ask another question after this: http://chemistry.stackexchange.com/questions/61346/which-element-has-has-the-most-oxidation-states-and-why ? You were already asking why.

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text:
Why is no MS-gas phase NMR coupling possible?
With ever stronger magnets coming out every year, LC-NMR coupling becomes more attractive and is now used frequently in the pharmaceutical industry as well. Considering the rapid progress that is made in that direction I wondered why no MS-NMR coupling has been reported yet, especially as gas phase NMR is possible already. So I wondered:
When will it be possible to do MS-NMR coupling? What are the biggest obstacles for achieving MS-NMR coupling?

Perhaps because the concentration of molecules is very much smaller than in LC. MS is very sensitive and works with very small numbers of molecules. The sensitivity would have to be incredibly high.

And why isn't it possible to just do mass spectrometry with a higher number of molecules, so adapt it to higher concentrations?

How do you keep the charged molecules in your NMR probe for hundreds of millicesconds, so you can record an FID? And why couple the two methods at all? Just run an NMR of the same sample! For separation according to size, use SEC/GPC, those have been coupled with NMR.

Keeping charged molecules in NMR probe for hundreds of milliseconds $\Rightarrow$ could be possible with quadrupole ion trap (linear ion trap)
Why couple them? $\Rightarrow$ because MS is very powerful for large molecules and complex mixtures while NMR is better for small molecules, combination of those would be incredible.

An ion trap can hardly keep a macroscopic amount of sample, and in MS, there is no macroscopic amount of ions flying. Even with hyperpolarisation (which is likely not possible in an ion trap), NMR does not work with a few 10000 of particles. And "incredible" is a weak point. What actual problem could this solve? You can do HPLC-NMR, SEC-NMR, DOSY, etc. pp. today!

I see I get your point. I mean I already know that it is somehow technically impossible, and now I get that the problem is with the hyperpolarisation that isn't happening in vacuo. Thanks for pointing that out.  However,I still believe that MS-NMR, if it would be possible, would be an amazing technique as you would have the amazing resolution power of MS combined with the detailed structural information from NMR. Thinks like MS-IRMPD spectroscopy are already out there and provide amazing new possibilities but I do see now why that can't be done with NMR in the same manner.

(The magnets are not really getting stronger. The > 20 T magnets are still too big and expensive to be widely used. First 900 MHz was from 2001, the 1GHz Bruker magnet came out five years ago. Since then seemingly nothing... except this 1020 MHz hybrid prototype by JEOL et al. Looks a bit unwieldy. There is a reason it is a hybrid (inner coils made of Bi-2223, i think): The classic superconducting materials (Nb3Sn, NbTi) have an absolute upper limit of ~25 T. HTSC can go higher, but they are very difficult in manufacturing. )
Fundamental problem is T1 relaxation in the gas phase. It is much too slow. You need hyperpolarisation techniques, that you have to interface with the GC. Difficult. (Impossible with mass spectrometry, because there the sample is in vacuum.)
The amount of analyte would be tinytiny. Likely impossibly small for NMR, even with hyperpolarisation. Certainly too small with mass spectrometry.
Other point: Why? GC is not useful for complex molecules, because the have very little vapour pressure. For less complex molecules, GC-MS is fine. There  are a number of working NMR techniques to separate different analytes. HPCL-NMR, GPC-NMR, DOSY, etc.
I guess nobody has seen a problem that needs GC-NMR to solve it. Or at least he has not been able to convince anyone it would be worth financing it.
(I initially misread the original question to be about a GC-NMR combination. Hence all the GC in my answer.)

I see that GC-NMR would not be that much better than LC-NMR. But what about MS-NMR? I mean it would be a very powerful technique for e.g. proteomics.

MS-NMR? Not sure i understand that. NMR on single molecules?

MS doesn't consist of single molecules more like $10^6$ molecules something in that range.

But there will never be more than a few molecules in your probe at a time. Plus, they will never polarise, because they are in vaccuo.

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text:
Can a saturated hydrocarbon have side chains?

Saturated hydrocarbons: Each carbon atom is bonded to as many hydrogen atoms as possible. In other words, the carbon atoms are saturated with hydrogen (Source).

or

Saturated means that the hydrocarbon has only single bonds and that the hydrocarbon contains the maximum number of hydrogen atoms for each carbon atom Latech.edu.

But if a carbon chain has side chains, the carbon from the side chain takes the place of a hydrogen. So it doesn't have as many hydrogens as it could have right?
But then I guess I could say that anything except methane has less than the max (4) number of hydrogens.
So what exactly is meant by maximum number of hydrogens?

"Maximum number of hydrogens" is a poor definition, for it is ambiguous, just as you pointed out. Better stick with "no multiple bonds".

All saturated (and non-cyclic) hydrocarbons have the general formula $\ce{ C_nH_{2n+2} }$

@Karl You should just make that an answer...

So is something like methanol saturated compound (not hydrocarbon). What I mean is: if a compound has all single bonds but the carbon is attached to 4 groups other than hydrogen, is it still saturated; or is the term saturated specific to hydrogen?

Standard organic chemists’ understanding is that saturated/unsaturated is a word pair entirely independent of branched/linear/cyclic. Thus, all six combinations are possible:

branched/saturated (e.g. neopentane or 2,2-dimethylpropane)
branched/unsaturated (e.g. 2,3-dimethylbut-2-ene)
linear/saturated (e.g. n-pentane)
linear/unsaturated (e.g. hex-3-ene)
cyclic/saturated (e.g. cyclohexane)
cyclic/unsaturated (e.g. cyclohexene)

I chose these examples for a reason. The central carbon atoms in both neopentane and 2,3-dimethylbut-2-ene are only bonded to other carbon atoms while all other carbons in the branched examples are methyl groups ($\ce{CH3}$). Yet, no organic chemist will classify neopentane as ‘unsaturated’ — so clearly there must be an additional criterion other than ‘maximise the number of bonds to hydrogen atoms’.
On the other hand, hex-3-ene has almost maxed out its number of hydrogen atoms: two carbons with three, two carbons with two and only two carbons which are lacking a hydrogen each. But that does not make it any less unsaturated. Note that cyclohexane and hex-3-ene have the same formula $\ce{C6H12}$ and thus the same carbon/hydrogen ratio. Yet, the cyclic compound is considered saturated and the linear isn’t.
Thus, forget about the maximum hydrogen definition. A saturated molecule is one that is made up of single bonds only.
Note: Of course, cyclic compounds can also be branched or ‘substituted’. But that doesn’t add too much.

It is better to stick with the definition that describes a saturated hydrocarbon as one that does not contain any double bonds, and an unsaturated one as one that does. Using this definition will allow you to clearly see whether a hydrocarbon is saturated or not. The other one is better used in order to understand what something 'saturated' is, and how something can go from being unsaturated to saturated, for example by addition of hydrogen.

The only hydrocarbon that has "as many hydrogen atoms as possible" is methane, with four H per C. For all other hydrocarbons, a C replaces at least one H. Whether in a straight or a branched chain, in a non-cyclic saturated hydrocarbon or alkane the number of hydrogen atoms is given in the formula @Karl cited, $\ce{C_nH_{2n+2}}$. For cycloalkanes such as cyclopropane, the formula is $\ce{C_nH_{2n}}$, since two end bonds go to carbons.


text:
B3LYP Explained
In my inorganic chemistry class, we were introduced to computational chemistry. We are told to use B3LYP theory in the program, but we never actually learned any of the computation yet. Could anyone explain simply what B3LYP is and how it fits into the DFT equation that looks like
$$
E_{\text{DFT}}[\rho] = T_{e}[\rho] + V_{ne}[\rho] + J[\rho] + E_{xc}[\rho]
$$

Uh, I think you should grab a tutorial or something. This's got to be too broad.

B3LYP gives you an expression for the exchange-correlation functional in the DFT energy equation. For a few more details this might be of interest for you.

I don't really think it's too broad. It's quite possible to give a relatively short answer explaining what an exchange-correlation functional is, pointing out that it refers to the last term in that equation above, and talking a little bit about B3LYP itself.

'The level of theory' is a fancy word for scheme used to calculate energy of a molecule. There are quite many ways to do it, most very computationally expensive. However, quite recently it was proven that state energy depends on electron density distribution only, and details of correlation of electron movement are derivable from said distribution in all entirety. This gave rise to density functional methods, where instead of considering all details of electron movement, only electron density is considered. Theoretically this allows to lower computational requirements from fourth-seventh (depends on the method used) power of number of basis functions to only third power of said number. This is a big deal.
The problem is that exact and universal way to derive energy of a system of electrons from their distribution is unknown. Thus, a lot of various ways were tested and most successful ones made it into available software.
Most of such ways - functionals - split energy of the system into several parts. Some of them may be known exactly, such as energy of electron - nucleui interaction. Some, however, are not, such as energy of electron-electron interaction. Still, several border cases were considered in theoretical physics. Specifically, functional for electron gas is known and was used. This gave raise to so known Local Density Approximation. It performs tolerable in many cases.
An interesting option is to add so-known exact exchange into the mix. Essencially, exact exchange is an attempt to enforce Pauli principle by hand, i.e. that two electrons with same spin cannot occupy the same spot. The problem is, that part of it is already included in the base LDA, so that member is usually considered with reduced weight, say, 0.25.
Another possible way is to try to include members dependent on electron density gradient, to aknowledge that electron density in a molecule varies from point to point. This is known as GGA approach.
B3lyp is a functional, that includes exact exchange and GGA corrections in addition to LDA electron-electron and electron-nuclei energy. The weights of the parts were fit to reproduce geometry of a test suite of small molecules. As such use of b3lyp for calculations with heavier atoms is questionable.
Density functionals perform poorly when dispersion interactions play significant role, thought correction schemes for this are also known.
Further details are not important for you at the moment. However, consider grabbing some book on DFT and quantum chemistry in general. If you ever end in 'real' chemistry, this would come handy, as computational chemistry papers are present in big numbers and often offer significant insight.


text:
How to calculate the magnitude of magnetic field of an orbital?
I read that according to QM  the electron in the ground state orbital produces no magnetic field, if that is true, can you explain why so and in what states there is a magnetic field and how that is produced? Why the difference?
Anyway, I am interested in learning how the classical model calculated it, since I read different values in different papers: for example hyperphysics says its value is .4 T, elesewhere I found .5 and .6, and If I use the classical formula I get .125 T
Does 0.4 Tesla refers to the exact value of the magnitude of the magnetic field od 1S state?

Orbitals don't "produce" magnetic field, only electrons. Where your original statement about ground states come from?

The hyperphysics page is referring to the 2p state, not the ground state.

@DavePhD,thanks,  do you know how they got .4 T and what would the value of ground state be? is it 1/8 T?

@user104372 hyperphysics means that given that the splitting between the two different 2p spin-orbit states is 0.000045 eV, and considering that this is due to the energy of the states being +/-
μB, where μ is the magnetic moment of the electron, B is 0.4T. For the ground state, 1s, there is no orbital momentum. The magnetic moment of the electron is only in the field of the proton, which is a much weaker field. The magnetic field would be equation 77 here: http://www.pha.jhu.edu/~rt19/hydro/node9.html

Hyperphysics means that the energy difference between the 2p J=3/2 state and the 2p J=1/2 state is as if the magnetic moment of the electron were aligned with and against a magnetic field of 0.4T.
The energy of the two states, relative to the ground state, is (according to NIST)
2p J=3/2 : 10.19885142904 eV
2p J=1/2 : 10.19880606470 eV
The difference is 0.00004538204 eV or $7.271004 \times 10^{-24}$ J
The electron magnetic moment is $−9.28476462 \times 10^{-24}$ J/T
So the energies are as if they are raised and lowered by being in a field of 0.3915556 T.
For the ground state, 1s, there is no orbital angular momentum.  The 1s state is only split by interaction with the proton.  This is sometimes referred to as hyperfine structure. The energy difference is $5.87433 \times 10^{-6}$ eV.

@tvb orbitals that are not s-orbitals have angular momentum and this creates a magnetic field.  A classical formula for the magnetic field is equation 42 here http://www.pha.jhu.edu/~rt19/hydro/node4.html#SECTION00031000000000000000 , but that is as if there are circular orbits, which is not true

Thanks, Dave, very helpful  answer and comment. Can you briefly say why s-orbital have no L while other do have, or should I ask a separate question?

@tvb "s" means l=0, "p" means l=1, "d" means l=2, etc.  Also, the angular momentum operators involve derivatives with respect to angular coordinates, while the wavefunctions for s orbitals are independent of angle, so such derivatives are zero.  For more see http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect8.pdf especially at the bottom of page 78.


text:
Would osmosis occur between two solutions of equal moles of water but unequal moles of a nonpolar molecule?
Earlier I learned that osmosis was, like diffusion, a result of the laws of entropy, kinetic energy, and random movement. However, if you have two solutions of equal moles of water, then net movement should be zero, right?
In the case above with equal moles of water, if you now have a semi-permeable membrane (allowing only water molecules to cross) and solutes dissolved in the water (at difference concentrations), then there would still be net movement of water molecules towards the concentrated solution. There was an explanation in my AP Bio textbook that the water molecules hydrating the solutes were unable to pass through the membrane, so net movement occurs because of a difference in free water concentration.
Now I am just wondering what would occur if you had a nonpolar molecule instead of a solute. Would the arrangement simply remain at this?
I just wanted to see if I truly understood this concept.

Nonpolar substance would not normally mix with water; but if you somehow form a lyophobic colloid; osmotic pressure would form but much lesser than lyophilic. https://books.google.co.in/books?id=2y7-BAAAQBAJ&lpg=PA15&ots=HO6bfxVbdm&dq=osmotic%20pressure%20in%20lyophilic%20colloids%20vs%20lyophobic%20colloids&pg=PA15#v=onepage&q=osmotic%20pressure%20in%20lyophilic%20colloids%20vs%20lyophobic%20colloids&f=false

Table 6 on https://books.google.co.in/books?id=2y7-BAAAQBAJ&lpg=PA15&ots=HO6bfxVbdm&dq=osmotic%20pressure%20in%20lyophilic%20colloids%20vs%20lyophobic%20colloids&pg=PA15#v=onepage&q=osmotic%20pressure%20in%20lyophilic%20colloids%20vs%20lyophobic%20colloids&f=false

Your example with the no-polar substance wouldn't qualify in a discussion of osmosis. There needs to be a solution on at least one side of the membrane. Since a typical non-polar substance would be insoluble in water, osmosis would not apply.
However, in your diagram, water would move from left to right due to a concept called pressure potential. You will eventually learn about this in AP Bio (I taught it for 24 years.) Pressure potential and osmotic potential work together, in a concept called water potential, to calculate which way water will move across membranes. You will probably perform a lab utilizing this concept this year. Pressure potential, in your example, is just a fancy term stating that because both the atmosphere and the non-polar substance on top of the left side are greater than just the atmosphere pushing down on the right, water will move to the right.
One of the things you should be careful about is that it is not the number of moles that are important, it is the concentration of substances. Get rid of your non-polar substance. Move the membrane over to the left so there is much more water on the right. Imagine the water levels are equal on both sides so we don't have any pressure potential to deal with. In this scenario, there would be no net movement of water even though there's far more total water on the right. This is because the concentration of water on both sides is the same, 100%

The floating nonpolar thing would work like just the wall of the container, or a piece of floater on it.


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Why is there a difference in available energy in saturated vs. unsaturated fatty acids?
On Wikipedia it states:

In cellular metabolism, unsaturated fat molecules contain somewhat less energy (i.e., fewer calories) than an equivalent amount of saturated fat.

Can someone explain why that is true in simple words? (that is, if possible, in a way understandable by someone who is not conversant with biology nor chemistry)

I think the 'energy' referred to here is Gibbs free energy. One way to think of things, admittedly an over-simplification,  is as follows: all of the ATP generated by oxidative phosphorylation is generated by electrons 'held' in C-C and C-H bonds. None comes from water or C-O bonds.  Count the total number of C-C plus C-H bonds in the saturated and unsaturated fatty acids. Saturated fatty acids have the potential to 'make' more ATP compared with their unsaturated equivalents, and so far as cellular metabolism is concerned  may be thought of as a better 'source' of free energy

The total energy stored in macromolecules is commonly quantified by the amount of $\ce{ATP}$ produced, as $\ce{ATP}$ is used throughout the body to couple reactions and make them thermodynamically favorable.
First, as the oxidation of a double bond in an fully saturated fatty acid requires a reducing agent, you here produce $\ce{FADH2}$ from $\ce{FAD+}$ which is equivalent to a $\ce{1.5 ATP}$ gain.[1]
Also, unsaturated fat has double bonds in cis positions that are unfavorable to beta oxidation.[2] An enzyme coupled reaction is needed for the double bond shift. Therefore, an enzyme coupled reaction that requires energy (in the reductant form $\ce{NADPH}$) is needed to make the molecule a trans fat with a shift in bond position in order to continue oxidizing the molecule completely. $\ce{NADPH}$, usually is defined as $\ce{2.5ATP}$ equivalents[1] and this extra energy expenditure in shifting the double bond is why less energy is ultimately produced.  Note, this $\ce{NADPH}$ is only used when unsaturated fats have an even numbered bond position.[2] This happens with polyunsaturated fats when the first bond is shifted.
For example:
18:2 linoleic acid will yield $1.5 + 1.5 + 2.5 = 5.5$ less $\ce{ATP}$ than 18:0 stearic acid.

References:
[1]: http://www.saylor.org/site/wp-content/uploads/2013/04/BIO101A-6.2.3-CellularRespiration.pdf
[2]: https://en.wikipedia.org/wiki/Beta_oxidation

Yes, it is a fact that saturated fats contain more gross energy due to higher H+ concentration. However, unsaturated fats are more digestible and have higher absorption rate, and therefore the metabolizable energy that unsaturated fats provide is higher than that provided by saturated fats.

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NAMD free energy calculations (FEP, Alchemical)
I am trying to calculate the absolute free energy of a protein-ligand-complex using the free energy perturbation theory.
Following the alchemical route in this tutorial, I have calculated eight MD-simulations over 24 lambda values (500ps each).
1/2: Ligand in protein (Ligand restraint, forward/backward)
 Ligand ->(forward) vacuum ->(backward) protein

3/4: Ligand in water (Ligand restraint, forward/backward
 Ligand ->(forward) vacuum ->(backward) water

5/6: Ligand in protein (Ligand restraints scaled, forward/backward)
 Colvars 1 ->(forward) 0 ->(backward) 1

7/8: Ligand in water (Ligand restraints scaled, forward/backward)
 Colvars 1 ->(forward) 0 ->(backward) 1

The free energy contribution of 1/2 and 3/4 is calculated with the  ParseFEP-tool. All colvar contributions were printed to the output file for each lambda value.
$\Delta G_{1/2} = -152.65 ~~~~ \Delta G_{3/4} = 152.1$
Is it sufficient to integrate the curves for each colvar and sum them up afterwards?
$\Delta G_{5/6} = 12.3 ~~~~ \Delta G_{7/8} = 1.84$
Do I have to subtract the colvar contribution from binding and solvation energy to get the absolute binding free energy?
$\Delta G = (\Delta G_{1/2} - \Delta G_{5/6}) - (\Delta G_{3/4} - \Delta G_{7/8}) = -11.1\,$kcal/mol

Just a side note: Jonsca has been using the same avatar as you. I know caffeine is a common molecule, but you may want to consider making it more unique ;)

@SLx64 I am trying to do the same as you did, are you able to help me prepare the input files? I appreciate your thoughts


text:
Birch Reduction Of Anthracene

Which ring of anthracene is reduced during Birch Reduction? What will the final product look like? Why?

My teacher said that the more electron deficient ring should undergo the reduction but I couldn't figure it out still. Any ideas?

There are two ways of rationalising this. One way is to look up or qualitatively determine the frontier orbitals (HOMO and LUMO) of anthracene, the other is to think about the stability of various radical anions.
But first, a quick note on the Birch reduction and its mechanism. As you may know, the $\ce{Na/NH3}$ system effectively works as a single electron transfer reagent. The electron is added to the π system’s LUMO, and the resulting radical anion is trapped by a proton (donated from a proton source; e.g. added ethanol) to give a radical. This radical is still electron-deficient and thus susceptible to another single electron transfer reduction resulting in a second anion which again is captured by a proton resulting in the reaction product.
Frontier orbital analysis
A great resource for frontier orbitals is the WOrbit page of Hans-Ulrich Wagner from the LMU Munich. Among the systems whose orbitals you can look at is, luckily, anthracene. For posteriority, I have included the HOMO and LUMO pictures in the image below.

Figure 1: LUMO and HOMO of anthracene as calculated on the WOrbit page linked above.
The LUMO is of importance; after all, we are adding an electron. We see that the MO is most localised on carbons 9 and 10 of the central ring — thus, this is where the first electron will be added in a localised picture, and this is where the first proton will capture.

Scheme 1: First reduction sequence.
As I drew in scheme 1, this gives us the radical 3, of which I have only shown one resonance structure. Applying the same analysis for the second step will tell us that the second electron will be added exactly at that position where I have drawn the radical (data not shown). Thus, we find that the central ring is reduced as depicted in scheme 2.

Scheme 2: Second single electron transfer and subsequent reduction of radical 3 to 9,10-dihydroanthracene 5.
Stability of radical anions
If we consider the $+I$ effect on substituted carbons, we see that we only need to consider the three structures shown in scheme 3 below; reversing charge and radical in structure 2 would result in the negative charge being on the tertiary carbon which is much less favourable when compared to the secondary one.

Scheme 3: Possible radical anions of anthracene.
At first, it looks like 6 and 7 might be better stabilised structures: 6 effectively represents a methylnaphthalene radical and 7 a naphthylallyl radical. However, if we take a closer look at 2 we see that it can be rearranged in a manner shown in scheme 1 above. If you consider the above drawn structure 3 as a radical anion (i.e. before protonation), notice that the radical is now a diphenylmethyl radical. This diphenylmethyl radical wins in favourability against both naphthyl derivatives.
As a general rule, if you can preserve the aromaticity of a benzene ring the structure is ‘better’ than one where you can’t. And two benzene rings as are present in structures 3 and 4 (and 5) are better than the naphthyl rings of structures 6 and 7. Thus, it is most favourable to reduce the central ring.
Note, by the way, that depending on the conditions employed, 5 is not inert towards further Birch reduction. Mehda, Sen and Ramesh report a triple Birch reduction of anthracene leading to the very interesting tricyclic triene 8 in $80~\%$ yield (scheme 4).[1]

Scheme 4: Synthesis of 1,4,5,8,9,10-hexahydroanthracene.[1]

Reference:
[1]: G. Mehta, S. Sen, S. S. Ramesh, Eur. J. Org. Chem. 2007, 423. DOI: 10.1002/ejoc.200600610.


text:
Chemical to remove nail polish but retain ink
I have a 1954 team signed New York Yankees baseball with 5 Hall of Famers on it. When I was little and showed it to my friends, the signatures started to smudge a little so my Mother put clear nail polish on it to protect it. Is there a chemical I can use to eliminate the polish without hurting the signatures?

I have to admit, I was wondering what this question was doing on chem.SE. :D Next time, put the chemistry question in the title!

Depends on the type of ink but most likely not.

It would depend mainly on the chemical composition of the ink (permanent marker? standard ballpoint?)
If it was signed with standard ballpoint, you have a problem, because the main film forming / thickening agent used for ballpoint pen ink is the same used in the majority of nail polishes - nitrocellulose. This means that a solvent that removes nail polish (ethyl acetate, for example) will also remove the ink.

Nail polishes use solvents such as butyl acetate. If the solvent dissolves the ink, it would have destroyed the signatures when the polish was applied. Therefore you should try to use a similar solvent (e.g., ethyl acetate) to get rid of the polish. Use a cotton swab soaked in the solvent and try it first on the least valuable signature. If it works you can apply it to the rest of the baseball, but be warned that the signatures may appear more faded afterwards.

If butyl acetate destroys the ink, so will ethyl acetate. And probably so will most if not all other organic solvents.

@Jan But of course! I said that the signatures are still there, so it appears that the butyl acetate didn't affect the ink, and neither should ethyl acetate.

nononononononono

@Fl.pf. Actually yesyesyesyes. The ink in ball pens is insoluble in organic solvents once dried (>48h). See for yourself.


text:
Should I expect any surprise when trying to shoot green fireballs like this?
I wanted to use flash cotton (nitrocellulose) to ignite fireballs with an electric trigger. In order to make green fireballs, my attempt will be to put some drops of boric acid solution on each (nitrocellulose) projectile. Should I expect any... "surprise" there? (e.g. the compound exploding by itself in that moment).

Boric acid is toxic and causes damage to the unborn (teratogenic), so maybe not such a great idea to use on a party.

@ketbra according to wikipedia, boric acid is approximately twice as toxic as NaCl, so there are far worse things. https://en.wikipedia.org/wiki/Boric_acid. The teratogenicity is a concern, but I think the risks associated with nitrocellulose are at least as great.

@LevelRiverSt good to hear that you are trusting on wikipedia on this. I rather not make decisions on teratogenicity based on wikipedia, but I guess that's your choice to make, anyway. Why not just have a look at a trustworthy source instead (e.g. supplier information)? Cellulose nitrate is explosive, but not toxic at all. I just wanted to point out that boric acid is teratogenic as I would have wanted to know about it before working with it.

I would not expect boric acid to cause ignition of nitrocellulose. However, boric acid may react with un-nitrated hydroxyl groups on the cellulose backbone--likely a benign reaction. You might need more than a few drops of boric acid solution to see a green color - perhaps immersion in boric acid solution will be needed. Copper salts might give a good green color, so that is also worth trying. Either way, you will want to let the nitrocellulose dry thoroughly before trying to ignite it.
Also, be sure to carefully practice the words: "I AM OZ, THE GREAT AND POWERFUL" before you go public.

No. I will practice shouting Aveda Kedavrah! Seriously... i'm not a chemistry expert or even student, but I ask it as an addon to cosplay (death eater this time) and illusionism xD.
Thanks for your tip :D.

I once searched both expressions, and seem to be actually different ones. But it is still not clear to me since the sources claimed to be neither sure. Don't know where can you ask about this topic xDD.

Please be careful with boric acid, as it is toxic!

@ketbra: It's so toxic it will even poison a nuclear reactor :-)

I understand. Please tell me: using copper sulfur instead of boric acid would cause a reaction?

I doubt if there would be a reaction with copper sulfate.


text:
Conversion between anisotropic temperature factors and geometric ellipsoids
I am using the ANISOU of the PDB standard to represent ellipsoid shaped colloids.  I am writing a Monte Carlo simulation to transform the position and orientation of the ellipsoids and want to use a PDB file to represent the result.
I am not sure how to convert between a geometric representation of an ellipsoid to the anisotropic temperature factor tensor, $\mathbf{U}$.  I am able to define an ellipsoid aligned to the coordinate axes using the diagonal elements of $\mathbf{U}$: $\mathbf{U}(1,1)$, $\mathbf{U}(2,2)$, $\mathbf{U}(3,3)$, but I am not sure how to represent a rotated ellipsoid, i.e., I am not sure how to set $\mathbf{U}(1,2)$, $\mathbf{U}(1,3)$, and $\mathbf{U}(2,3)$.  I came across this resource which introduces the topic.  This second link mentions the similarity between tensor elements and geometric definitions and says the conversion is "rather complicated" but does not provide a link for further study.
Any guidance or direction to a good resource would be appreciated.

I found the publication titled Atomic Displacement Parameter Nomenclature which seems to have the details.

If you would like to work out and post the details, please add as a separate answer and I will accept that.

When you need to deal with the anisotropic things, I think that you cannot transform or re-define an ellipsoid's matrix into U(m,n), where m is not equal to n. However, you can average the any anisotropic quantities that you want using the direct geometric transformation as long as you have the coordinate file (xyz) in your hand.


text:
Making big crystals of oxalic acid crystals from powder
How to transform oxalic acid small crystals into large crystals?

Like with most water-soluble compounds, the easiest way is to dissolve your oxalic acid in hot water and then cool down slowly. As the solubility decreases with cooling, much of your compound would precipitate, forming nice crystals (or not so nice, depending on the speed of cooling and the cleanliness of your glassware.)
Dissolving and letting the solvent to evaporate also works OK, just too slow (in case of water). As for melting, this is inherently bad idea for just about any organic compound. Even if you can keep your temperature precisely so as to avoid decomposition, you'll end up with an ugly lump of molten and recrystallized substance, probably containing some large crystals inside it, where they can't be seen, much like a beautiful sculpture hidden in a chunk of marble.

My best bet would be melt them and cool them down slowly or dissolve them and let the solvent evaporate.

I'm afraid your post is more of a comment than answer... Also why not recrystallize it normally?

This does provide an answer, even two answers; the problem is that one of them is wrong.


text:
Interpretation of miscibility curves
I stumbled across miscibility curves in the section Solubility of Liquids in Liquids (Organic Chemistry, by Wallwork and Grant) today. I'm not sure if my problem is with the way the content was laid out in the book or whether I'm not fully understanding the material, but I feel the book has dealt with the topic in a rather, obscure way.
I've taken snapshots of it and uploaded a portion of my text below.

What I want to know is if I interpreted the text (the bit about miscibility curves) correctly, which I seriously doubt.
Here's my interpretation:
This is the miscibility curve for phenol in water blah-blah-blah. All the points present in the area bound by the curve with the x-axis (concentration axis) are points of immiscibility. The curve itself represents the threshold of miscibility.
For example, at 20°C, phenol is soluble in water up till the concentration corresponding to point A on the curve. However, for phenol concentrations corresponding to points between A and B (lying on the line AB), the phenol-water mixture is immiscible and the phenol and water form separate layers. Although, at concentrations corresponding to point B (and beyond), the phenol is completely soluble. When for a constant temperature, the miscibility curve maps to two concentrations on the x-axis (eg; points A and B in the graph), which represent thresholds of miscibility at that temperature, the two mixtures (i.e- at A and B) are called conjugate pairs.
I labored for over two hours to come to the above conclusion, but when I saw the portion of the text which mentions that the liquids A and B exist together (at equilibrium) for the given temperature, I had to throw in the towel.
I have a very strong feeling that a large part of my 'interpretation' is horribly incorrect. If anyone's got the time, could you explain what the text actually means?

This is no different from what is observed in many binary systems. Here, the phenol and water exhibit a miscibility gap. As you noted, above point C, the liquid is one liquid with whatever composition it has. If one were to have a liquid at the composition corresponding to C, at a temperature above 66C. you would have one liquid. As you dropped the temperature to 66C, nothing would change.
Drop below 66C, and the liquid will phase separate in to two liquids of different compositions, given by the points on the miscibility curve. To figure out how much of each liquid you have, you do the appropriate weighted average as given by the book. That weighted average can be seen to be correct, since, say at 20C, sitting on point A you would be 100% liquid with the A composition, and sitting on point B you would be 100% liquid with the B composition. In between, you are balancing between the two liquids, which are in thermodynamic equilibrium.
If you have a book of phase diagrams handy, look at almost any phase diagram involving Pb, and you will likely find a liquid-liquid miscibility gap. For fun, look at Ag-Ni, two metals with fcc crystal structures, and you will see that the solids are immiscible, and there is even a liquid-liquid miscibility gap.

Your interpretation seems correct.  When you have a mixture with an overall average mass fraction of X, it separates into two fluid phases in equilibrium.  One of these fluid phases has mass fraction A, and the other fluid phase has mass fraction B.  You can determine the fraction of the total mass that is liquid of mass fraction A and liquid of mass fraction B as follows:  let f represent the fraction that is liquid of mass fraction B, and (1-f) the fraction that is liquid of mass fraction A.  Then $$m(1-f)x_A+mfx_B=mX$$where m is the total mass of both phases.  Cancelling out the m's yields:$$(1-f)x_A+fx_B=X$$Solving for f yields:$$f=\frac{X-x_A}{x_B-x_A}$$This equation is called the "lever rule," because it operates like as see-saw.  The fraction of the total amount of material that is in phase B is proportional to the difference between the average mass fraction and the mass fraction of your substance in phase A.  And, the mass fraction of material that is in phase A is proportional to the difference between the average mass fraction and the mass fraction of your substance in phase B.


text:
What aspects of a macromolecule/protein do the various contributions to its entropy relate to?
I've come across a few different contributions to entropy in macromolecules such as proteins: configurational, conformational and vibrational. The problem is that I can't seem to find a consistent definition and there may be some overlap.
In addition, I've only really understood entropy to be made up of rotational, vibrational and translational components and I'm struggling to reconcile this with the contributions to entropy for proteins described in the literature.

I've only really understood entropy to be made up of rotational, vibrational and translational components

There is also electronic entropy, but it's a fun exercise to show that (at least for the hydrogen atom) the corresponding partition function $q_{\text{elec}} = 1$ at room temperature, and for all intents and purposes there is only population of a single state. So, the other 3 kinds of entropy (translational, rotational, vibrational) are those directly from a statistical mechanical derivation that are relevant for most (bio)molecules.

I've come across a few different contributions to entropy in macromolecules such as proteins: configurational, conformational and vibrational.

Hopefully you agree that vibrational entropy in the context of proteins is the same as in the context of any other system in that vibrations can't be decomposed into other kinds of excitations (translational or rotational), regardless of whether or not the vibration is localized to a bond, highly delocalized, or quaternary structure subunits moving with a regular period.
Configurational and conformational entropy are being used slightly differently. From Wikipedia,

Conformational entropy – is the entropy associated with the physical arrangement of a polymer chain that assumes a compact or globular state in solution.
Configuration entropy - is the portion of a system's entropy that is related to the position of its constituent particles rather than to their velocity or momentum.

For proteins, I think the distinction between conformation and configuration is less clear than it is for more molecular systems. In my limited experience, a configuration of a protein is similar to a macrostate in statistical mechanics, which is distinguished from other possible configurations by some criteria related to the position of a protein's primary or secondary structure. More specifically, different techniques can be used for clustering the different protein conformations together into distinct configurations.
In these cases, the greater the number of conformations a protein can take, or the number of configurations it samples based on some metric, the greater the entropy. For something like an intrinsically disordered protein, its conformational and configurational entropy are likely to be higher than for a globular protein.


text:
What is the intuition and motivation behind the definition of the thermodynamic activity?
Stuck proving the relation $$\mu_i=\mu_i^\circ + RT\ln a_i,\tag{1}$$ I looked up the definiton of activity. It is
$$a_i=e^{\frac{\mu_i-\mu_i^\circ}{RT}},\tag{2}$$
but the the intuition/motivation for such a definition is unclear. Is it just because we wish for a relation like $(1)$ to exist by analogy with $(3)$? $$G_i=G_i^\circ+n_iRT\ln{\frac{p_i}{p^\circ}}\tag{3}$$
That is, if we define activity as given in $(2)$, we can take the natural logarithm of both sides, multiply with $RT$, add $\mu_i^\circ$ to both sides and arrive at $(1)$. And hence $(1)$ has a similar form to $(3)$ which holds for ideal gases.
Or am I mistaken and there are other reasons at play?

$\mu_i$ is chemical potential
The superscript $^\circ$ denotes the standard state
$a_i$ is activity
The subscript $_i$ denotes the $i$-th component of a mixture

So I'm taking Pchem right now so I'm no experts, but here is my two cents.
I'll be honest I essentially followed my books lead on the determination of $a_{i}$, and $\mu_{i}$. When looking at problems like this I try to go backwards. I see that $a_{i}$ is derived from $\mu_{i}$, and I try to find where $\mu_{i}$ comes from then see where that comes from until I get a definition or a well known relation like Gibbs free energy. I'm not savy enough to be like "Oh yea $\mu_{i}$, that's basically $\int dG(T, P_{2})$ I'll just show all that". But, I can slowly go backward. So, I followed the following path.
$$a_{i} \Rightarrow \mu_{i} \Rightarrow (\mu(T,P_{2})-\mu(T,P_{1}) ) \Rightarrow d \mu \Rightarrow d G$$
Then when you have this general path you reverse it to get the derivation.
Also remember that you have to keep track of all the assumptions you make. Its important to remember that. My answer is only applicable for a perfect gas at constant temperature, and a mixture of gas at low pressure.

From deferential of Gibbs free energy the chemical potential of a pure gas is,
$$dG = -SdT +VdP = d\mu$$
at constant temperature,
$$d \mu = V_{m}dP = \frac {RT}{P}dP$$
To get $\mu$ we integrate with constant temperature essentially  expanding the gas on an isotherm for $P_{1}$ to $P_{2}$.
$$\int _{1}^{2} d \mu = RT \int _{P_{1}}^{P_{2}} \frac{1}{P}dP$$
$$\mu(T,P_{2})-\mu(T,P_{1})=RT \ln \Big{(}\frac {P_{2}}{P_{1}}\Big{)}$$
Let $\mu(T,P_{1})=\mu^{\circ} $,  $P_{1}=P^{\circ}=1bar$, $P_{2}=P$, and $\mu(T,P_{2})=\mu_{i} $,
$$\mu = \mu_{i}+ RT \ln \Big{(}\frac {P}{P^{\circ}}\Big{)}$$
Side note your equation 3 comes from this equation. $\mu_{i}=G_{i}$.
For a mix of gas at low pressure, the activity is equal to the ratio of the partial pressure of the gas over the standard pressure,
$$a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)}$$
$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i}$$
I think that's how you prove that relationship. I hope that helped.
$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i} \Rightarrow a_{i}= e^{\frac{\mu_{i}-\mu^{\circ}}{RT}} $$
This relation is trivial and just some algebra. When you are trying prove something its important to start as far back as possible.

Citations:
Chemical Activity
Levine, Ira N. Physical Chemistry. 6th ed. pg 175-176 . New York: McGraw-Hill, 1978. Print.
Notes from my professor (sorry).

OP asked for the motivation/intuition behind the definition, perhaps you should try and address that? :)

Is that paragraph a good comment on motivation, or should I actually show the intoution for each step of the derivation, and do the derivation backwards?

I have to agree with @getafix. Note that IUPAC defines activity as $$a_i=e^{\frac{\mu_i-\mu_i^\circ}{RT}}.\tag{1}$$ It is not possible to say that (1) holds because the following $\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i}$ is true. How do you then know that $$a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)}?$$ To keep your argument from being circular, you must then define activity in an alternative manner. How would you go about it? In essence you are still saying that hey, since we have a pretty relationship for ideal gases, let us not mess that up!

I'm confused by your question then? In my opinion the relationship between your equation 1 and 2 is trivial, and my teacher wouldn't except that as an answer. Your equation 1 has to come from somewhere I showed one way of finding that for a pure gas at near standard pressure.  $a_{i}$ is equal to a lot of things if you have it in terms of activity coefficient and molar fraction $\gamma-{i} x_{i}$ you can prove it that way, but I did in terms of partial pressure so I found a relationship between partial pressure and activity. Correct me if I'm wrong but that's the way I play the derivation game

My point was that your argument is circular in a mathematical sense. You cannot say activity is equal to anything before defining it first. Right now you have the mathematical relationship $$\text{definition of activity} \implies a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)} \implies  a_i=e^{\frac{\mu_i-\mu_i^\circ}{RT}}.$$ Do you see the problem?

The activity is as you have defined it and can easily be expressed in mathematically equivalent ways. It arises from the need to use equations developed for ideal gases to the behaviour of real gases (or solutions, or electrolytes etc.)
The variation of the free energy G of an ideal gas at pressure p (bar) from its standard state $G^0$ is expressed as
$$G=G^o +RT\ln \left(\frac{p}{p_0}\right) $$
where $p_0$ is 1 bar pressure and is added only so that the log is dimensionless. (Many books do not include this but this means that the log has dimensions, which it cannot have. You can also use chemical potential as this is just free energy/mole. You can use atm. instead of bar if you want). For clarity I remove the $p_0$ in the following equations thus
$$G=G^o +RT\ln \left(p\right) $$
When we want to measure free energy of a real gas then it would be useful to still to be able to use this equation. However, real gases deviate in their behaviour from an ideal gas, the more so as the pressure is increased.
To cope with this we define a new quantity, the activity a. Some authors use fugacity $f$ which is equal to activity for a gas. The activity is used to define a new standard state which is that at which the activity would be equal to $1$ bar assuming that the gas behaves ideally to this pressure.  Thus an ideal gas always has an activity equal to the pressure but a real gas deviates from this and has, perhaps, an activity that is smaller than the pressure. This may be due to attractive intermolecular forces.
Although a real gas may never reach the standard state of unit activity (fugacity) it does not prevent us from using it since measurements are always relative to the standard value. The standard states for liquids or solids is always taken to be 1 bar, whereas the standard state for a gas is not that of unit pressure but  of unit activity.
It can be shown that as the activity and hence free energy at low pressure has  ideal behaviour then$^*$
$$G=G^o+RT\ln(a)   \tag 1$$
If we compare this to the equation for an ideal gas then we can write it as
$$G=G^o+RT\ln (\gamma p) $$
where $\gamma = a/p $ is called the activity coefficient.
Similarly to low vs. high pressure gases, activity is used for higher concentrations in solution instead of concentration.

$^*$ As activity and pressure are equal at low pressure we can arrive at eqn (1) by subtracting the free energy at some pressure (represented by activity a ) from that at low pressure in which case its activity is equal to the pressure itself thus
$G-G_{low} = +RT\ln(a) - RT\ln(p_{low})$
and now do the same with the standard where $a=1$
$G^o - G_{low}= +RT\ln(1)- RT\ln(p_{low}) = - RT\ln(p_{low})$
then by subtracting, eqn. (1) follows.

I suggested a few edits but the sentence the activity a [---] that defines a new standard state as that at which the activity would be equal to 1 bar if the gas behaved ideally up to 1 bar and the footnote are a little confusing.

have tried to make things clearer.

There are still a few problems, some of which were reintroduced after your edit. $(1)$ It is more correct to denote Gibbs energy per mole with a subscript $m$. $(2)$ For a gas, the standard state is the gas itself under standard pressure and ideal behavior. Thus, activity as such is not part of the definition of a standard state. $(3)$ Activity does not have dimensions, hence it does not make sense to say 'activity is equal to 1 bar'. $(4)$ Fugacity is not equal to activity, note that fugacity has the same dimensionality as pressure.

ok here goes! (1) don't care whether m is subscript you can define it how you want. The symbol $\mu$ is usually used for chemical potential /mole. (2) yes for an ideal gas, but we do not deal with ideal gases but with real gases when we use activity/fugacity. (3) for a gas activity = fugacity, $a=f$ , and $a/p=1$ as $p \rightarrow 0$. Often in books we see $\ln(f)$ but this really means $ln(f/1 atm)$ , yes its confusing ; same applies to $ln(p)$. The fugacity is equal to the vapour pressure when the vapour is an ideal gas, and in general it may be regarded as a corrected pressure.

$(1)$ The subscript $m$ is for the molar Gibbs energy, that is $G_m$. No, the symbol $\mu$ denotes chemical potential itself. It is just that since $$\mu = \frac{\partial G}{\partial n}$$ and $G = n\cdot G_m$, then $$\mu = \frac{\partial nGm}{\partial n}=G_m.$$ In other words, chemical potential is molar Gibbs free energy. $(2)$ See here for proper definition of a standard state. $(3)$ The limit you gave may equal $1$ but that hasn't got to do with anything :). And still, activity $\neq$ fugacity. The correct relationship is $$a=\frac{f}{p^\circ}.$$

Sorry careless! I meant free energy /mole for $\mu$ not chemical potential/mole: .  In your (3) this is just what I put in my (3) but its in the limit  $p\rightarrow 0$ . Activity is dimensionless only if mole fractions are used as these are dimensionless. Usually we make the activity coefficient dimensionless and this is what is usually tabulated. Activity in gases is taken to be equal to the fugacity with dimensions of pressure. In solution we use activity in concentration units which is always divided by 1 unit so use $\ln(a/1 unit)$ and  activity coefficients are always dimensionless.

As wikipedia reports on the first lines:
"
In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution.
"
In other words, you should and have to find the same equations of an ideal solution. It is just that X and Y might have different activities/concentrations from Xideal and Yideal.
The motivation is very simple: keep ideal equations, models but characterise
the chemicals "just" by the activity. Or mathematically: fit on the activity as
free parameter the ideal equations.
I hope it fits also the intuition part.


text:
Difuran Synthesis
We've been asked for the mechanism for the synthesis of the following difuran from the starting materials shown below:

The ytterbium complex will act as a Lewis acid and activate the oxygen of the benzaldehyde. However, I don't know what the next step in the mechanism would be.

Electrophilic aromatic substitution?

This is actually a series of two electrophilic aromatic substitutions. The ytterbium stabilizes the benzyl cation enough for it to attack position 5 of the 2-methylfuran. After the new ytterbium complex dissociates away, the resulting cation can then attack another 2-methylfuran, and the final product is formed.


text:
Why is there such a violent reaction when NaCl is formed?
In school I saw how $\ce{NaCl}$ was formed and I noticed that there was a violent reaction. Can someone tell me why there is such a violent reaction? The $\ce{NaCl}$ was formed by reacting sodium metal and chlorine gas.

It will depend on what the reaction was - what other reagents and conditions were involved? NaCl can be formed many ways.

Solid sodium and chlorine gas.

Sodium's single unpaired electron is partially occupying a much higher energy orbital than the unfilled orbital on the chlorine which will be occupied when the two chemicals are mixed. The excess electronic energy generated will rapidly become thermal energy giving a "violent" but rarely explosive reaction (as there is no gas generation only thermal expansion).

Well, you know that hydrogen and oxygen burn in a violent reaction. Compare the following enthalpies of formation (amount of heat given off when one mole substance is formed from component elements in their standard states) for sodium chloride and water:
\begin{array}{|c||c|}
\hline
\mathrm{Substance} & \Delta H^{\varnothing} \left(\mathrm{kJ\,mol}^{-1}\right)\\ \hline\hline
\ce{NaCl} & -411.12 \\
\ce{H2O} & −291.83 \\ \hline
\end{array}
The formation of sodium chloride releases even more energy, so the reaction is more violent.
This is partially because sodium chloride is an ionic compound, so the transfer of an electron from electropositive sodium to electronegative chlorine is not only favored, by releases a lot of energy. More so than what we get out of making a few bonds between hydrogen and oxygen.


text:
Is the following rate law possible for the given reaction?

A researcher is investigating the following overall reaction:
$$\ce{2C + D -> E}$$
The researcher claims the rate law for the reaction is written as follows: $\text{Rate} = k[\ce{C}][\ce{D}]$.

Is the rate law equation possible for the given reaction?
If so, suggest a mechanism that would match the rate law. If not, explain why not.

The answer is yes. I don't know how to arrive at this answer. From what I  have learned, the following must be true for a) to be true:

The rate law for the slow step agrees with the experimental rate law
The elementary steps of add up to the overall reaction.

I would assume the following elementary steps from the rate the researcher has confirmed:
$$
\begin{align}
\ce{C + D &-> CD}\\
\ce{C + CD &-> E}
\end{align}
$$
But since $\ce{CD}$ has a larger size than $\ce{D},$ wouldn't it be less likely to have the proper alignment in a collision to enter the active state? Thus, wouldn't it be a slower reaction than $\ce{C + D}?$
Please inform me where my logic is wrong and why the answer is yes.

Steric factors are not the only thing that controls the rate of each elementary step. Perhaps most crucially you have to look at the activation energy as well. That second step may have a smaller steric factor than the first step but that doesn't necessarily mean it's slower.

The rate law does not usually (i.e. almost never) indicate what the actual mechanism is. The only way to determine mechanism is by experiment. The classic example is $\ce{H2 +Br2 = 2HBr}$ which occurs via  five recactions involving $\ce{H\cdot , Br\cdot}$ radicals with rate  $\ce{d[HBr]/dt= a[H2][Br2]^{1/2}/(1+b[HBr]/c[Br_2])}$ where a, b, c are  rate constants (or combinations thereof) of the mechanism.

To comment on your particular reaction (a) if the rate quoted is true then this C+D represents the rate limiting step, i.e. slowest and this means that CD reacts with C more rapidly than D reacts with C, i.e. CD+C has a lower activation barrier. (b) Alignment has a very small effect compared to activation energy $\Delta E$ , look up Arrhenius equation $k=A.\exp(-\Delta E/(RT))$ where $A$ is rate constant at infinite temperature (or zero activation energy)

We know norhing about molecular mechanism we are just provided that c reacting d will result in cd and c reacting with cd will result in e.

You should use steady state approximation which leads to the mentioned answer:
$$\frac{\mathrm{d}[\ce{CD}]}{\mathrm{d}t} = 0$$
$$
\begin{align}
\ce{C + D &->[$k_1$] CD}\\
\ce{C + CD &->[$k_2$] E}
\end{align}
$$
$$\ce{->[QSSA]}\quad k_1[\ce{C}][\ce{D}] = k_2[\ce{C}][\ce{CD}] \quad\to\quad k_1[\ce{D}] = k_2[\ce{CD}]$$
$$\frac{\mathrm{d}[\ce{E}]}{\mathrm{d}t} = k_2[\ce{C}][\ce{CD}] = k_1[\ce{C}][\ce{D}]$$

QSSA: quasi-steady-state assumption

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


text:
How do you calculate the equilibrium constant with the absorbance of a substance and the absorbance constant?
When given the equation: $$\ce{Fe^3+_{(aq)} + SCN^-_{(aq)} <=> FeSCN^2+_{(aq)}}$$
How do you calculate the equilibrium constant when given the slope of the absorbance vs concentration graph ($\pu{4317 M-1}$) and the absorbance of $\ce{FeSCN^{2+}}$ (0.276)
The following information is also given: $2.000\ \mathrm{mL}$ of a $0.00200\  \mathrm{M}$ solution of $\mathrm{KSCN}$ with $5.00\ \mathrm{mL}$ of $0.00200\ \mathrm{M}$ solution of $\ce{Fe(NO3)3}$ and $3.00\ \mathrm{mL}$ of water is combined.
So from this, I assumed that the concentrations of $\ce{SCN-}$ and $\ce{Fe^{3+}}$ are both $0.002\ \mathrm{M}$.
I found the concentration of $\ce{FeSCN^2+_{(aq)}}$ to be $\pu{6.39e-5 M}$ using this equation:
$$\mathrm{Absorbance} = \mathrm{slope}\cdot \mathrm{conc.}$$
$$\mathrm{conc.} = \frac{\mathrm{Absorbance}}{\mathrm{slope}}$$
$$\mathrm{conc.} = \frac{0.276}{\pu{4317 M-1}}$$
$$\mathrm{conc.} = \pu{6.39e-5 M}$$
Using this value, I used the equation for the K constant of an equilibrium:
$$\mathrm{K} = \frac{[\ce{FeSCN^2+}]}{[\ce{Fe^3+}][\ce{SCN^-}]}$$
$$\mathrm{K} = \frac{\pu{6.39e-5}}{0.002^2}$$
$$\mathrm{K} = 15.975$$
This did not turn out to be the correct answer, but I'm not sure why. I suspect the concentrations for the two reactions are not correct since the volumes are also given.

The thing you did wrong is to assume that the concentration of the substance you have at the beginning is the same in the 'endmix'. Because you see, when you add all these things together the volume is bigger thus changing the concentration of the substances you added previously. Also besides that you should then correct in the denominator for concentratioms Fe3+ and SCN- that have reacted by substracting with concentration of formed FeSCN2.

I'm following the outline from the comment by user21398.
Initial concentrations after mixing
The volume of the mixture is $V_\text{mix} = \pu{10 mL}$. Using the dilution law, I get the following concentrations:
$$\ce{[Fe^3+]_\text{initial}} = \pu{2.00 mM} \cdot  \frac{\pu{5 mL}}{\pu{10 mL}} = \pu{1.00 mM}$$
$$\ce{[SCN-]_\text{initial}} = \pu{2.00 mM} \cdot  \frac{\pu{2 mL}}{\pu{10 mL}} = \pu{0.400 mM}$$
Equilibrium concentrations
The equilibrium concentration of the complex is already calculated, $\ce{[FeSCN^2+]_\text{equil}}=\pu{6.39e−5 M}.$
To get the equilibrium concentrations of the reactants, we have to consider that some reacted:
$$\ce{[Fe^3+]_\text{equil}} = \ce{[Fe^3+]_\text{initial}} - \ce{[FeSCN^2+]_\text{equil}} $$ $$ = \pu{1.00e-3 M} - \pu{6.39e−5 M} = \pu{0.94e-3 M}$$
And similar for thiocyanide:
$$\ce{[SCN-]_\text{equil}} =  \ce{[SCN-]_\text{initial}} - \ce{[FeSCN^2+]_\text{equil}} $$ $$ =\pu{0.400e-3 M} - \pu{6.39e−5 M} = \pu{0.336e-3 M}$$
Equilibrium constant
This is copied from the question, using the values calculated in this answer above and dropping the unit M because the standard state is 1 M:
$$\mathrm{K} = \frac{[\ce{FeSCN^2+}]}{[\ce{Fe^3+}][\ce{SCN^-}]}$$
$$\mathrm{K} = \frac{\pu{6.39e-5}}{\pu{} \pu{0.94e-3}\cdot \pu{0.336e-3}}$$
$$\mathrm{K} = 202$$


text:
How to relate a reaction barrier to the time the reaction needs to proceed?
As I am writing this I am at a conference and one of the participants just asked a question where he linked reaction barriers to durations for the reaction to complete. To paraphrase:

From our experience a reaction with an activation barrier of 15 kcal/mol should occur instantaneous at room temperature. An activation barrier of 20 kcal/mol takes about one to two minutes and an activation barrier of 25 kcal/mol needs about 10 hours.

I would like to rationalise this statement as it seems quite hand-weaving to me. How can I judge from (a possibly also computed) activation barrier how long a reaction will need to complete? For the the sake of the argument, let's only consider reactions that proceed in one step; generalisations may also be implied, but may be too complex.
In the comments (and the already existing answer) the Eyring equation is mentioned. If the connection between the energy values and the duration can be made with that, an illustrative example would be nice.

We do have the Eyring equation $k = \frac{k_{B}T}{h}e^{-\frac{\Delta G^{\ddagger}}{RT}}$. Have you tried plugging in some values?

@Zhe I'm aware of the equation and no, I have not tried plugging in some values. It was a statement I heard at the conference today and this was my way of making a mental note, so that I won't forget to investigate. I hope that this will provide some more insight further down the line. From this statement it seems that this is well established, although I never heard it, maybe established by experienced, and I wonder where did it come from and under what circumstances it will break down. I think the Eyring equation might give a first clue, but certainly won't be enough to justify it alone.

I plugged in some numbers for $T = 298,\mathrm{K}$. For $15,\mathrm{kcal/mol}$, the half-life of a first order reaction is 11 ms. For $20,\mathrm{kcal/mol}$, 0.8 minutes. For $25,\mathrm{kcal/mol}$, 64 hours. The top value is a bit off, but the others line up fairly well with "experience."

Because of the exponential relationship between activation energy and rate constant this becomes very small for relatively small changes in activation energy and can easily become years. Just as well as most organic material, i.e. us are exothermic wrt combustion with oxygen. Incidentally, the Eyring eqn as quoted above is only for first order reactions, need to multiply by ratio of partition functions for other reactions, but it is a good 'rule of thumb'.

This earlier answer may provide some useful background.

If I understand your last statement correctly, what you would like to have is the reaction time $t$ as a function of the reaction barrier $\Delta G$. However, $t$ also depends on the conversion $c$ (for (pseudo)first order reactions, as is the assumption in the Eyring equation, conversion can never be 100%) and temperature $T$.
Although it has been mentioned, just for the sake of completion, here is the Eyring equation giving us the rate constant $k$:
$$k = \frac{k_B T}{h}e^{-\frac{\Delta G^\ddagger}{RT}}$$
We know that the half life $\lambda$ is:
$$\lambda = \frac{\ln(2)}{k}$$
The conversion $c(t)$ is related to this:
$$c = 1 - \frac{1}{2^{\frac{t}{\lambda}}}$$
If we solve this for $t$, we get:
$$t = \frac{\ln(\frac{1}{1-c})}{\ln(2)}\lambda = \frac{\ln(\frac{1}{1-c})}{k}$$
Where we can insert the Eyring equation for $k$, to get this final result:
$$t(\Delta G, c, T) = \frac{h \cdot \ln(\frac{1}{1-c})}{k_BT} \cdot e^{\frac{\Delta G^{\ddagger}}{RT}}$$
Here is a plot of the reaction time for some typical conversion rates at room temperature:

And another logarithmic plot, which makes it easier to get the involved time scales, from 10$^{-9}$ h (3.6 $\mu$s) to 100 h:

As you can see, reactions around 20 kcal/mol lie around the "typical" regime, from seconds to several hours, while reactions with $\Delta G^{\ddagger}$ < 15 kcal/mol proceed within milliseconds and reactions with $\Delta G^{\ddagger}$ > 25 kcal/mol may take days or weeks to complete.

Here are some back of the envelope numbers from using the Eyring equation:
$$k = \frac{k_{B}T}{h}e^{-\frac{\Delta G^{\ddagger}}{RT}}$$
Let's just assume we're at $298\ \mathrm{K}$ for the reaction, and the reaction is relatively simple:
$$\ce{A->B}$$
I constructed the following table by plugging in values. $t_{1/2} = \frac{\ln 2}{k}$
\begin{array}{|c|c|}
\hline \Delta G^{\ddagger}\ (\mathrm{kcal\,mol}^{-1})& k\ (\mathrm{s}^{-1}) & t_{1/2} \\ \hline\hline
15 & 63.4 & 10.9\ \mathrm{ms} \\
20 & 0.0138 & 50.2\ \mathrm{s} \\
25 & 2.98\cdot 10^{-6} & 64.6\ \mathrm{h} \\\hline
\end{array}
The values for 15 and 20 $\mathrm{kcal\,mol}^{-1}$ seem pretty consistent with your rule. The top value is a bit off, but we're working with a very small numbers at this point, and there may be other sources of error that we're not accounting for in the model.

Thanks for doing this. You derived the value for the half-time from the first order rate law like $$\begin{align}c&=c_0\cdot e^{-kt}&\Leftrightarrow\ \ln c &=\ln c_0 -kt &\Rightarrow\ \ln\left(\frac{c_{1/2}}{c_0}\right) &= -kt_{1/2} &\Rightarrow\ t_{1/2} &=k^{-1}\ln2\end{align}$$ I am probably going to ask some follow up questions to this, especially why the last value is so far off.

The activation barrier you got, is it $\Delta G^‡$ or $E_a$ (the Arrhenius activation energy)? Depending on this you might either use the Eyring equation or the Arrhenius equation.
It is actually quite common to use the Eyring equation to calculate $\Delta H^‡$ and $\Delta S^‡$ and with that $\Delta G^‡$ from experimental rate constants, so doing it the other way round and using the equation for prediction is completly fine. But I would not say something like

An activation barrier of 20 kcal/mol takes about one to two minutes

but rather use half-lifes or 95% conversion or something similar. Just be careful if the reaction is not 1st order, since only there the half-life is independend on the concentration.
As an example see this paper from Joe Fox. At page two, bottom of the left column they predict one reaction to be 29-times faster than another one, purely based on calculated $\Delta G^‡$ values, using the eyring equation to calculate those rate constants. On page 21 of the supporting information you can see that they used the eyring equation to calculate $\Delta G^‡$ from experimental data.

As an extension to this, a similar approach should be possible to also judge the temperature of a reaction and linking it to the reaction duration and reaction barrier.

Yes it is, if you assume that your $\Delta G^‡$ is constant over that that temperature range.

This is not really what I was looking for. I know the Eyring equation and all its faults. Hence your answer is more or less at the same level of handwavyness as the original statement. I was looking for a more rigorous approach, a judgement of the degree of handwavyness,  or at least a few calculated examples like in Zhe's comment to the question. The statement also was given as quoted, so there has to be some justification why it was said in that way.

Well, your question was quite different before the edit and imo my answer wasn't that bad to the questions present in the original post. I find it very questionable to alter the question in such an extreme way that the answers below seem to be quite offtopic.
And as I said, people usually use the eyring equation in case of calculated activation energies to determine (relative) rates. If you want "authoritative uses", K.N. Houk is using this regularly in his papers, for example: http://pubs.acs.org/doi/abs/10.1021/ja505569a

I amended my question, I did not really change it. Your answer is as off-topic (if you really se it that way now) as it was before. It generally just boils down to: Use the Eyring equation, others do it, too. Which is not really a justification and exactly the same what Zhe said. When I amended my question I was careful enough not to invalidate what you have wrote (that would have been if I would have deleted the first paragraph). You can still [edit] your answer and obtain the bounty - I just found it not very helpful, sorry.

For example: I don't know what the kind of reaction energy the guy on the conference was going on about, probably Gibbs, what would be the difference in those approaches? You have not even stated the Eyring equation once and how do I get from the rate to a duration. That might be trivial for you, but the connection is not really that obvious to me. I don't understand what you mean with "but rather use half-lifes or 95% conversion or something similar." Links to papers are fine, but if they are behind a paywall it would be better to include the relevant portion in your own words or as quotes.

Before it was "how can I predict a rate from (calculated) energies?", now it's "if I use the eyring equation to predict rates, what are the problems, how to fix those and what are the alternatives ", those are quite different questions. My whole argument that you can use the eyring equation to do so seems quite stupid now that the eyring equation is mentioned in the question, and it also completly changes the "level" of discussion. Before it seems to me that we are talking on basically an undergrad level, leading to this simple answer. Now it's a much more complex and elaborate topic.

I'm sorry about this. I was looking for an rationalisation of the statement 10 kcal/mol - instantaneous, 15 kcal/mole - one,two minutes, 20 kcal/mol - 10 hours. I'm not particularly looking for a connection with the Eyring equation, but since that's the only thing that came up in the comments, I thought to include it in the question, too. I simply want this question to be of a general reference type, therefore the bounty, and only because it already had a high score. For me it's just a nifty little factoid...

Again I apologise for my messup. The reason for this is probably because I didn't fully understand your answer. I have rolled the question back and only added that I'm looking for an illustrative example. Also I removed the second part as I intend to ask that separately. I hope you consider adding an example to make it easier for me to understand. (Feel free to flag these comments for removal once you read them.)


text:
Why do Asn-Ser and Gln-Thr have different H bonding patterns?
I noticed that Asn-Ser H-bonding is different from the H-bonding between Gln-Thr.
However both have the same functional groups. I wasn't able to find any information about this. Maybe it's due to steric effects, but because these striking similarities between Asn and Gln and on the other side Ser and Thr I don't think that is an explanation (but maybe it is, I'm just guessing).

excuse me for the drawing, but I couldn't find an image showing this clear

I found these H-bonding patterns in Fundamentals of Organic Chemistry, McMurry, seventh edition.

update

Are al these H-bonds possible?

Your threonine in your picture is missing the extra methyl group. Right now it looks the same as the serine.

You are right! I forgot that one @BenNorris

Can you clarify what do you mean when you say that the bonding is different? Energy-wise, its distance...?

I mean the H-bonding, Asn-Ser the H-bonding exists between a H from the NH2 and the O from the alcohol. But Glu-Thr H-bonding exists between the carbonyl O and the H from the alcohol @Variax

I don't have the textbook you've gotten this from, but unless there's a statement to the contrary, I suspect these are intended as examples of hydrogen bonding, rather than an exhaustive list of the only hydrogen bonding pattern which can exist.
If you start to look at real structures, you will see both "directions" of hydrogen bonding (hydroxyl donor/carbonyl acceptor and nitrogen donor/hydroxyl acceptor) in real protein structures.
Take, for example, the structure 1GKT. S297 is very clearly in a hydrogen bonding orientation with N303, with the hydrogen of the hydroxyl pointing toward the carboxylate oxygen of the asparagine. Also, Q22 is very likely in a sidechain-sidechain hydrogen bond with T29, where the hydrogen on the glutamate nitrogen is pointing toward the oxygen of the threonine.
If you spend some time looking at various experimental structures from the PDB website, you should be able to find examples of a wide range of hydrogen bonding patterns - basically, any acceptor and any donor can be paired, depending on the geometry of a particular protein.

One caveat I should mention here is that because of the way X-ray crystallography works, it's often difficult to tell which way the sidechains of the ASN and GLN residues are oriented. With electron density, the NH2 group and the =O group look very similar. Often it's a bit of guesswork on the crystallographer's part to tell which orientation they have, often relying on the layout of neighboring residues to determine the most likely hydrogen bonding network.
However, 1GKT was not determined by X-ray crystallography, but by neutron diffraction. In neutron diffraction it's very clear where the protons are located. (I picked a neutron diffraction structure as the example for this very purpose.) I also double checked that there wasn't any reason for flipping those particular ASN/GLN residues, both by manual visualization and with the MolProbity web server.

You only took two examples into consideration. What about serine and glutamine or threonine and asparagine? Also, serine can hydrogen bond to another serine or a threonine …
What I was trying to point out in this first paragraph is that these examples are really only single examples. Maybe they are trying to show how the same group (hydroxy group and amide group) can be both acceptor and donor depending on the structure. Maybe McMurry just picked four for fun. In any case, you can hydrogen bond from any $\ce{O-H}$ or $\ce{N-H}$ to any lone pair of $\ce{O}$ or $\ce{N}$ in a protein as long as that lone pair is not participating in resonance — this last caveat is really only to capture amides, $\ce{RC(=O)-NH2}$, wherein the nitrogen cannot be a hydrogen bond acceptor due to the resonance structure $\ce{RC(-O^-)=NH2+}$.
And furthermore, unless you have unambiguous side chains — arginine which can only donate or glutamate/aspartate which can only accept at neutral $\mathrm{pH}$ — the orientation of hydrogen bonds is typically left to the guesswork of the crystallograph for X-ray diffraction analysis. But the other answer already goes into detail there.

Thankyou! I added another picture to my question, so every H-bond shown here is possible? @Jan

@RickBeeloo 1 and 2 are; 3 is not. Read the caveat sentence on amides and maybe read on about amide resonance structures.

Thankyou found the information here: http://chemistry.stackexchange.com/questions/42085/can-an-amide-nitrogen-be-a-hydrogen-bond-acceptor


text:
Openbabel: Generate 3D depiction?
The process of creating 2D depictions is explained in detail here. This way, it is possible to create a 2D image via running e.g.
obabel  -O 2d_depiction.png
However, how do I generate a ball and stick 3d depiction (as image) instead?

Try: obabel -L pov

By the way, there is a specific Open Babel forum to ask such questions..

Open Babel does not generate "ball and stick 3D depictions" directly, because it's not a molecular visualization program.
That's why we started Avogadro.
There are a few formats supported in Open Babel to write out 3D depictions, e.g.:

POV-Ray
STL - for 3D printing
Point Cloud - outlining the Van der Waals surface

But you can make a pseudo-3D "ball and stick" depiction in the SVG format:

obabel [file] -O file.svg -xS

Personally, I'd recommend using a visualization software to make sure I get the view angle set the way I want. The only reason we added the -xS option was to generate previews for 100,000 molecules in PQR search results.

Nice answer, thanks : )
Yeah, I know that automatic 3D structure depiction will most often lead to a bad view angle. But I just need them for quick evaluation of MD results and I am planning to integrate the visualization into the IPython notebook.

If you're integrating into iPython, you might also want to look at 3DMol.js which will give you interactive JavaScript views. e.g. py3Dmol

3DMol.js how could I overlook that, and I looked for molecular graphics packages a lot. If I could I would up vote two times for that comment : )

I just tried it and managed to get py3Dmol running in like 2 min.  $-$ interactive molecule visualization in the IPython notebook made easy : )


text:
Why hot water is more effective than cold water in cleaning
Today while washing curtains, my mother dipped them in hot water(containing detergent) for an hour and then washed them.When i asked her that why she done that,her answer was simple(it will make the cleaning easier and effective). It seems to me that there is a scientific phenomena is lying behind it,but i don't know what it is.
The question i m asking is why hot water is more effective than cold water in cleaning (cloths,curtains,bedsheets etc).
I shall be thankful to have your answers.

Many physical and chemical reactions are affected by temperature. Usually higher temperature causes the reaction to occur faster, and lower temperature will slow it down.
Water solubility (the ability of substances to go into solution in water, i.e. bond with water molecules) is a reaction that is affected by temperature for many substances. Changes in temperature may increase or decrease the solubility of a certain substance in water, depending on the substance properties.
Since high temperature will cause water molecules to move around relatively faster, it will also cause the hydrogen bonds between them to form and break at a higher rate (that is why hot water vaporizes more quickly than cold water). In addition, the high temperature increases the water's energy, and this energy may be used to break bonds between dirt and curtains for which colder water will not have enough energy.
This means that the detergent/water solution will be able to dissolve (form bonds with) the dirt faster and better, because the bonds between the detergent and water and the bonds between the detergent and dirt will form faster and will be strong enough to prevent the dirt from reforming the bonds with the curtains. As mentioned earlier, these bonds will also break faster, but the presence of a detergent helps to stabilize the solution and keep the dirt dissolved in the water. Once the dirt is dissolved in the water, the curtains are clean, and the dirty water can be easily discarded.
In summary, hot water helps the cleaning process (i.e. moving the dirt from the curtains to the water) because it can dissolve the dirt faster and break strong bonds (up to a limit, of course) of dirt with the curtains.
It is worth mentioning that between the detergent and the hot water, the detergent contributes more to the dissolution of dirt in the water than does hot water alone (that is the same reason it is better to wash your hands with soap and water than with water alone). However, for optimal results, using both detergent and hot water will clean the curtains better (and faster).
Here are some images I found online that may help clarify this point:

Detergent breaks surface tension by disrupting the bond between water molecules. This is because detergent has a polar and non-polar heads on its molecules.
The hydrogen and oxygen molecules in water form hydrogen bonds, which form a type of skin on the surface of the water. This skin is strong enough to support the weight of very light objects. That is why ants seem to walk on the surface of a body of water and why a needle carefully placed on top of the water will float.
Detergent molecules have a polar head on one edge that attracts water and a non-polar head on the other edge that repels water. These opposing heads interfere with the hydrogen bonds in the water molecules and weaken the strength of the skin on the surface, breaking the tension.

I found this table which shows the value of surface tension at different temperatures. Its clear from this table that hot water is more effective when used with detergent.
Source(data table): https://srd.nist.gov/JPCRD/jpcrd231.pdf

Most water soluble substances are more soluble at higher temperatures. Hot water will be more effective for getting rid of these types of substances.

Because the bicarbonates of mg and ca decomposes respectively into their products on adding hot water. Then water can produce effective lather easily


text:
Why the reduction potential of the reaction AgCl (aq) /Ag + Cl- is +0.22V
The point is $\ce{AgCl_{(aq)}}$ is the same as $\ce{Ag+}$ + $\ce{Cl-}$ (again, in water)
So my reduction reaction would be
$$\ce{Ag+ + Cl- + e- -> Ag + Cl-}$$ then the equation is simplified by removing $\ce{Cl-}$ from both sides (Spectator Ion).
Now there is $\ce{Ag+ + e- -> Ag}$ which has a potential of 0.80V, so why the original equation is 0.22V.
PS: I know it takes more energy to break the bonds and thus Ag is more stable with Cl and less likely to reduce. But the subscribed (aq) doesn't mean the bonds have already been broken by water?

I don't know what your source is, but it should definitely be an $\ce{AgCl(s)}$ in your first equation. Wikipedia gets it right and if you find any textbook that doesn't explicitly state the phase of $\ce{AgCl}$, you can be pretty darn sure they meant solid because talking about aqueous $\ce{AgCl}$ makes little to no sense precisely because it is so insoluble in water.
So, essentially you've already answered your own question; precipitation of $\ce{Ag(I)}$ as $\ce{AgCl}$ stabilises it towards reduction and lowers the standard reduction potential relative to $\ce{Ag+/Ag}$.


text:
Dissociation of diatomic molecules
At high enough temperature, diatomic molecules in a dilute gas should dissociate. Where do their electrons go? Will they dissociate into neutral atoms or into ions? Does it depend on the substance? What for $\ce{H2}$, $\ce{O2}$, $\ce{HCl}$, $\ce{HF}$? Which quantum mechanical technique can decide the question - Hartree-Fock?

If you are talking about the ground state then adding energy causes the molecule to climb the ladder of vibrational and rotational levels (for example as in a Morse potential) until the bond is so long that any extra energy will dissociate the molecule and two neutral atoms are produced. Why bother with theory when the best and only real test would be to do the experiment as has been done for numerous molecules.

Where do their electrons go?

IIRC, in the gas phase, the energy required for heterolysis is usually much greater than that required for homolysis. Consequently, usually, each of the atoms of a diatomic molecule retains one of the originally bonded electrons on dissociation.

Which quantum mechanical technique can decide the question -
  Hartree-Fock?

In principle, one could scan the potential energy surface1) by elongating the bond using some quantum chemical method to look on what actually  happens. Beaware though, that the most used variant of the Hartree-Fock (HF) method, the so-called restricted Hartree-Fock (RHF) method is know to be ill-suited for such calculations: it is known to fail completely when the homolytical dissociation of a closed-shell molecule into two open-shell fragments is considered.
Such behaviour is usually refered to as the Hartree-Fock "dissociation catastrophe", although, strictly speaking, only the restricted variant of the Hartree-Fock method is a subject to it. The unrestricted Hartree-Fock (UHF) method does not suffer from the "dissociation catastrophe" and tends to give at least qualitatively correct dissociation limit. So, use UHF at least. Even better, use some correlated method.

1) Simply a curve in this case, since there is only single degree of freedom (bond length).

If you are thinking about the ground state then thermally adding energy causes the molecule to climb the ladder of vibrational and rotational levels (for example as in a Morse potential) until the bond is so long that any extra energy will dissociate the molecule and two neutral atoms are produced.
In the case of ionic molecules such as alkyl halides, for example NaI,  when the molecule dissociates Na and I radicals are formed. The ionic state is higher than this by the difference in ionisation potential of the metal and the electron affinity of the halogen. The figure below shows the potential for NaI in the gas phase.

Because the covalent state has symmetry in common with the ground state the two interact and this causes the splitting in the potentials when they are close in energy and this is at around 6.9 angstrom. The molecule changes its character when moving across the interaction region, from largely ionic to being covalent, then the atoms separate as the bond stretched further.
Incidentally, this molecule was used in some famous experiments by Ahmed Zewail that showed for the first time the direct observation of product, Na and I, being formed in real time and in a stepwise manner as the wavepacket in the covalent state reached to crossing point and then repeatedly rebounds in the upper potential well. Zewail was awarded the Nobel Prize in Chemistry for this work. There are many papers in the scientific literature but the Scientific American article, 'The Birth of Molecules' gives a good introduction (Sci. Am. v262, Dec 1990, p108). A paper describing many experiments is 'Femtosecond Real-Time Probing of Reactions IV' J. Chemical. Physics. 91, 1989, p 7415.

Thanks. For more details see
Zewail, Ahmed H. "The birth of molecules." Scientific American 263.6 (1990): 40-46.
https://www.its.caltech.edu/~femto/SciAm.pdf
Jouvet, C., et al. "Experimental femtosecond photoionization of NaI." J. Phys Chem A 101.14 (1997): 2555-2560.
http://pubs.acs.org/doi/abs/10.1021/jp9639049
Engel, Volker, et al. "Molecular state evolution after excitation with an ultra-short laser pulse: A quantum analysis of NaI and NaBr dissociation." Chem Phys Lett 152.1 (1988): 1-7.
http://www.sciencedirect.com/science/article/pii/0009261488873196


text:
IR spectra and hydrogen bonds
Normally we see IR signals that correspond to the vibration of covalent bonds. Can a (strong) hydrogen bond (or a vibration thereof, to be precise) correspond to an IR signal as well? How about other non-covalent forces? What are examples were hydrogen bonds can be seen in the IR spectrum?

Hydrogen certainly shows up nicely in Raman spectroscopy. And, given that the simple harmonic approximation gives maximum energy for minimum mass ($\omega = \sqrt{k/m}$ it yields a fairly large frequency shift. As an example, for hydrogen in silicon, the H-Si peak is at higher energy than the Si-Si peak (again, for Raman, not straight IR).

@JonCuster Hey Jon,
Very kindly, could you specify "hydrogen bonds show up...", or, "hydrogen bonding shows up..." instead of just "hydrogen". Also, Raman is based on scattering rather than absorption and is thus fundamentally different than infrared spectroscopy, finally, we are talking about hydrogen bonds so I do not understand why you talk about silicon - hydrides. Sorry.

Your question implies that hydrogen bonds give rise to discrete peaks in the vibrational spectrum. It's better to say that hydrogen bonding affects the position of already existing peaks, in particular those vibrational modes coming from the $\ce{X-H}$. A hydrogen bonding network may also broaden low-frequency peaks coming from librations and hindered rotations.
For example, take the $\ce{O-H}$ stretching region in water.

The vapor spectrum consists of water molecules effectively in the gas phase that aren't interacting with each other, or are doing so rarely enough that the molecular configurations are "countable", shown by the discrete peaks between ~3600-3900 wavenumbers. A lack of water molecules interacting means hydrogen bonds aren't forming.
Moving to a bulk phase, where the water molecules are now interacting with each other all the time, hydrogen bonds are being formed and broken much more often. This results in a continuous peak, as more configurations are being sampled. In the case of water, charge transfer into the $\ce{O-H}$ bond causes a red-shift (lowering) of its frequency, moving the center of the distribution to 3400 wavenumbers. The distribution begins to narrow again in the solid phase, as the number of configurations is still large but dynamical switching between them slows.
It is also possible to experimentally measure spectra of small clusters through more advanced techniques such as mass spectrometry-detected rare gas tagging. Note the appearance of peaks around 3400 wavenumbers as the size of the cluster grows. (The presence of a proton doesn't affect these spectra in the hydrogen bonding region.)

A similar effect can be seen in molecular dynamics simulations. See how the peaks broaden and shift to lower energies when going from 1 water to the dimer, then 3 in a line/triangle, and finally 4 in a square.

First image taken from http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html. Experimental cluster spectra taken from http://pubs.acs.org/doi/abs/10.1021/acs.jpca.5b04355. BOMD plot my own creation (unnormalized dipole ACF spectra from HF/STO-3G, 0.36 fs time step, 10000 steps).


text:
Applying Henderson–Hasselbalch equation to amino acids: which pKa to use to calculate Z-/HZ ratio?
I was wondering which $\mathrm{p}K_\mathrm{a}$ to use when calculating the ratio of $\ce{HZ}$ to $\ce{Z-}$ of amino acids, the Henderson–Hasselbalch formula used:
$$\mathrm{pH} = \mathrm{p}K_\text{a} +\log\frac{[\ce{Z-}]}{[\ce{ZH}]}$$
Because most amino acids have two $\mathrm{p}K_\mathrm{a}$'s, one for the amino group and one for the carboxyl group (and sometimes a third $\mathrm{p}K_\mathrm{a}$ for the side chain). For example, leucine:
$$\begin{array}{cc}
\hline
\text{Residue} & \mathrm{p}K_\mathrm{a} \\
\hline
\ce{NH3} & 9.74 \\
\ce{COOH} & 2.33 \\
\hline
\end{array}$$
Let's say I want to know the ratio of $\ce{HZ}$ to $\ce{Z-}$ for a $\mathrm{pH}$ of $1.00$ and $11.00$, which $\mathrm{p}K_\mathrm{a}$ should be used in the Henderson–Hasselbalch equation and why?

Both? I don't see why you can't use both at the same time. When $\mathrm{pH} = \mathrm{p}K_{\mathrm{a}}$, the corresponding dissociation is at 50%. When they're not that close, then one form will greatly predominate over the other...

how would you calculate the Hz/Z- ratio based on the two Pka's given at Ph  = 1 e.g.? Offcourse it's possible to use both the Pka's in two different calculations but how would you combine these to get the Hz/Z- ratio? @Zhe

Check out this page

By a rule of thumb, pK farther than 3 units from pH can be omitted, as the component is ( at 99.9% level ) exclusively present in just of the A/B forms.

There are two approaches.

The two groups are treated as independent. You would calculate the ratio of protonated and deprotonated group one using the pKa for that group. Then, you would repeat that for the second group. Independent treatment would be a good approach, for example, to estimate the average charge of the hundreds of side chains in a typical protein that have a pKa near physiological pH.
The two pKa values are obtained experimentally from a titration.$$\ce{AH2 <=> AH- + H+ <=> A^2- + 2 H+}$$ Here, the $\ce{AH-}$ species could be ambiguous if the pKa values are close to each other (group 1 deprotonated and group 2 protonated or vice versa). Or, for a case like hydronium, water and hydroxide where the same atom can accept two protons, $\ce{AH-}$ would represent a single defined species. You can get the ratios of the concentrations from the pH and the respective pKa ($\ce{AH2}$ vs $\ce{AH-}$ via $K_1$, and $\ce{AH2}$ vs. $\ce{AH-}$ via $K_2$), and turn that into fractions of the total for each species. You can use that for a weighted average to calculate the average charge, for example.

Let's say I want to know the ratio of HZ to Z− for a pH of 1.00 and 11.00, which pKa should be used in the Henderson–Hasselbalch equation and why?

This depends what HZ and Z- refer to. With two ionizable groups, there are really four species, $\ce{HZYH, HZY^-, ^-ZYH and ^-ZY-}$. If the pKa values are far apart, one of the mixed species would be very unlikely in aqueous solution (in your example, protonated carboxylate with deprotonated amino group). At pH 1.00, the amine would be 100% protonated, and the carboxylic acid would be mostly protonated (comparing its pKa of 2.33 to a pH of 1.00). At pH 11, the carboxylic acid would be 100% deprotonated, and the amine would be mostly deprotonated (comparing its pKa of 9.74 to a pH of 11.00).

If you want to calculate with $\mathrm{pH} = 1$ that means the solution is predominantly acidic hence use $\mathrm{p}K_\mathrm{a} = 2.33,$ but if you want to calculate with $\mathrm{pH} = 11,$ the solution is predominantly basic so you solve with $\mathrm{p}K_\mathrm{a} = 9.74.$

Unfortunately, this doesn't seem to add much to the already existing answer. Also, please note that the official language of Chemistry.SE is English.

@andselisk Sometimes, the perfect answer is not the one, where you cannot add anything, but the one where you cannot omit anything. Karsten's answer is very well elaborated, but less experienced chemist can get lost. Oo65's answer is rather laconic, and that is its added value.

@Poutnik I agree, and that's why I didn't delete or converted this answer to a comment. The thing is, this would make much more sense if this were the first answer and Karsten's were the second. Currently it just looks redundant despite its simplicity.