text:
Why should one perform an extraction with multiple small portions of solvent as opposed to one large portion?
I'm performing a liquid-liquid extraction on a 1 l water sample for GC analysis. The procedure calls for three extractions with 50 ml of dichloromethane. Would one extraction with 150 ml produce a different result?

Most definitely yes.  Qualitatively, you can perform a casual experiment to observe the difference the next time you wash your hands in a bathroom that offers paper towels for drying.  Compare how dry your hands get when you (a) pull three paper towels off all at once and dry your hands with them all at the same time; versus (b) pulling three paper towels, one at a time, tossing each paper towel after it feels like it's not getting your hands any drier.  (b) will get your hands dry much more effectively.

The important thing is the distribution ratio (for the extraction of a metal) and the partition coefficent for the extraction of a given chemical species. This can be used for a first approximation calculation.
The distribution ratio is the total metal concentration in the organic phase (grams of metal per litre) divided by the total metal concentration in the aqueous phase.
If we consider a batchwise extraction where a litre of water bearing x mg of a metal is extracted with an organic solvent. Assume that the distribution ratio is 2. We can do the calculation for the extraction.
The phase ratio (ratio of the volumes of the two phases) is 1, call this theta.
Theta D / (1 + (Theta D)) = the fraction extracted in each stage.
If D is 2, and theta is 1 then we will extract 66 % of the metal.
As an alternative, if we use one litre of water and extract this with 0.5 L of organic phase, then with D = 2 and Theta = 0.5 then we will extract 50 % of the metal.
If we repeat the extraction with another 0.5 L of solvent then we will recover 50 % of the remaining 50 % of the metal. Thus we will have recovered in total 75 % of the metal.
If instead we were to extract the 1 L of water four times with 0.25 L of water then in the first extraction we would leave 66 % of the metal in the lower phase. The easy way of working out the amount left in the aqueous later after n extraction stages with the same D and theta valve is to use
{1 - ( D Theta / {1 + (D Theta)})}^n = fraction left in raffinate. In solvent extraction raffinate is the term for the aqueous layer after an extraction has occured.
This gives me about 20 % left in the aqueous phase after four extractions with 0.25 L of solvent.
It is important to understand that the solubility of the substance is not what you should consider it is the distribution ratio or the partition coefficent which is needed.
If you were to take a dilute solution of DEHPA (150 mM) in aliphatic kerosene and to shake this with 0.1 nitric acid containing a small amount of Fe-55 with no stable iron as iron(III) then you might get a D value of 100 for arguments sake.
Now the organic phase can only extract 50 mM of iron per litre as to extract one iron atom three DEHPA molecules are needed. If we were to repeat the extraction experiment with a solution of 56 grams of iron per litre (roughly 1 moles per litre of iron) spiked with a little Fe-55 then we would measure a far lower distribution ratio.
This is due to loading of the organic phase.
Equally when you extract a large amount of an organic substance into an organic solvent the solvent properties of the organic layer at the end will be different to those of the pure solvent.
For example if I have 1 litre of water saturated with hexanol with 500 ml of water saturated hexanol floating on the top. Then if I was to add 500 ml of ether and then shake it. I would have a different Kd for hexanol to that which I would have if I was to shake a dilute solution of hexanol in water with ether.
I hold the view that the formula used above for the system with a D value of 2 will only hold perfectly true when the concentration of the metal is zero. This is why radioactive tracers are so useful for academic solvent extraction experiments. This is becuase they allow the measurement of distribution ratios with a zero metal concentration. The concentration in moles per litre of a 1 GBq per litre solution of Fe-55 will be very small.
The half life of Fe-55 is about 2.7 years, thus I have calculated that this very radioactive stock solution will contain 205 nanomoles of iron if the only iron atoms present are Fe-55. The best way to make the iron-55 would be to bombard managense with protons with a cyclotron and then to separate the iron from the manganese by solvent extraction or some other separation method.
I would in a lab expect to be able to run a solvent extraction experiment with only 700 Bq of Fe-55. If we assume that I was to shake 700 microlitres of aqueous phase with 700 microlitres of the organic phase. I would then be able to sample 200 microlitres of the two layers with a pipette.
If we assume that the highest distribution ratio I was to measure was 100 then I would expect that at the end of the shaking that the aqueous layer would contain 10 Bq per ml. If I was to sample 200 microlitres and to put this into a LSC vial and then to count it. Assuming that 100 % of radioactive decays will result in a count in the LSC machine then I would in one minute get 120 counts. If I was to count until I had 10000 counts (1 % error due to count number) I would have to wait about 83 minutes.
For a D value of 10 I would be able get 10000 counts in less than ten minutes for the least radioactive liquid phase. Thus it should be clear that with a radiotracer it is possible to measure the distribution ratio with less than 1 nanomole per litre of metal present.

By the way, you can use MathJaX for equations - it has mostly the same syntax as LaTeX, if you are familiar with that. see e.g. https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here


text:
How is the Nitrate Ion (NO3) formed?
I understand how the nitrite $\ce{NO2}$ ion can be formed and have a negative charge, but the nitrate $\ce{NO3}$ ion is confusing me. I made some simple drawings to try to explain what I don't understand:

Does the "electron from outside" that the oxygen receives, as indicated in the drawing, comes from the nitrogen atom? If so, then the N atom would be left with only 4 valence electrons instead of 5, thus being able to form another covalent bond with the extra oxygen, forming $\ce{NO3}$. Is this the correct line of thinking?

An organiker would look at it this way.

An inorganiker might look at it as literal N(+5) with two double-bonded oxygens.  It does redox as N(+5). Coordination structures are varied,

I was under the impression that organic chemists have a much greater tendency to expand octets than inorganic ones …

If you are thinking about it like literally adding O to the nitrite ion then yes, the two electrons the oxygen receives are from nitrogen. Imagine moving the nitrogen lone pair to the oxygen to give it a full octet and forming a single bond in the process.  You will notice that the Lewis structures will match the "organiker" structures of Uncle Al.

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text:
Which salt precipitates from a solution of ammonium oxalate and silver nitrate?
I am supposed to write the precipitation reaction for
$\ce{(NH4)2C2O4 + AgNO3 ->[\ce{H2O}]}$
Is my equation correct? I left out the Spector ions.
$$\ce{4(NH4) + 2 C2 +O4\rightarrow 4(NH4)2CO3}$$

How can your equation be correct if the number of each element on both sides are not the same?

Good morning! I took the liberty to add some tags and edited the title and the body of your question. I hope you are OK with the changes. I find it a tad more understandable this way, understood it and tried to give an answer that might help you to tackle questions like this.

Your current solution is wrong and you don't seem to be familiar with the concepts involved.
Let's walk through this together and change that!

You take solid ammonium oxalate $\ce{(NH4)2C2O4}$ and silver nitrate $\ce{AgNO3}$ and dissolve both salts in water.
The salts dissociate into their ions. Let's do that separately for both salts first.
$$\ce{ (NH4)2C2O4 ->[\ce{H2O}] 2 NH4+_{(aq)} + C2O4^{2-}_{(aq)}}$$
$$\ce{AgNO3 ->[\ce{H2O}] Ag+_{(aq)} +  NO3^{-}_{(aq)}}$$
Since you are dissolving both salts in the the same beaker with water, your solution now contains all the ions of both equations.
Forget about the spectator ion fanciness.
It is much more important to identify the chemical units (= the ions from which the salts are composed). Leave these ions intact and don't rip them apart into weird combinations of elements.
Precipitation will occur when the combination of these ions (chose a cation and an anion) yields a salt with a very low solubility in water. There are only two cations and anions here, so the number of possible combinations isn't too big. Look up the solubilities of the possible products to figure out which one will precipitate.
Do not forget to balance the reaction!

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text:
How much activation energy is required to combust propane?
According to WolframAlpha, the change in enthalphy in the combustion of propane is -2220 kJ/mol. How do I calculate the activation energy required start the reaction in the first place? Do you use the same principles as when calculating delta H or am I missing something mind bogglingly obvious?

have a look here http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Modeling_Reaction_Kinetics/Temperature_Dependence_of_Reaction_Rates/The_Arrhenius_Law/Arrhenius_Equation

http://www.engineeringtoolbox.com/fuels-ignition-temperatures-d_171.html
The autoignition temperature is an external variable.  It depends upon total pressure and oxygen partial pressure.  For paper it is famously 232.778 C (significant figures aside).


text:
Why don't molecules of ionic compounds exist?
My book says:

As single ions of a metal are not associated in the solid with single ions of a non metal, separate units of ionic compounds do not exist. It is, therefore, wrong to talk of a molecule of an ionic compound.

I know ionic solids exist in form of crystal lattice but why can't we isolate a single molecule of ionic compound?

Actually compounds considered ionic in solid state are molecular in gas phase.

Related: Will gaseous ionic compounds be free moving ions?

Ions are electrostatically bound. One need only separate them with integrated inverse squared distance work. If you mix multiple ionic compounds in solvent, a melt, gas phase, the original paired associations are spontaneously scrambled. Benzene plus xylene does not give you toluene.  "Separating" benzene is not electrostatic attraction.
Quaternized glycine, $\ce{^{+}N(CH3)3-CH2-C(=O)O^{-}}$, is betaine. It is ionic, but the ions are covalently bound into a molecule. Where are the discrete formula units in alumina? You can draw lines to connect closest atoms, but those are not bonds. (Note fractional atom counts for atoms embedded in planes, edges, and corners.)

How is this unit cell $\ce{Al2O3}$?  It seems to have six aluminum and six oxygen atoms.

This oppose the Mithoron comment above. In principle molecular ionic entities exist. The might require exotic conditions but it is not the point.

I agree with Oscar Lanzi in an old message dated November $21, 2017$. The drawn unit cell contains 6  $\ce{Al}$ atoms and $2+1+3$ = $6$ $\ce{O}$ atoms. This cell is $\ce{AlO}$ and not $\ce{Al2O3}$.

Molecules that have strongly ionic bonds actually do exist. In the solid phase where we see most ionic compounds electrostatic bonding with a high density of ions causes the molecules to lose their identity, forming the familiar ionic lattice structures. The vapor phase is a different story. Ionic vapors typically consist of small neutral clusters. Such clusters are, of course, strongly polar, the polarity appearing as dipoles or quadrupoles (two oppositely oriented dipoles attract and join together to form the quadrupole).
Sodium chloride offers an example. Ron's answer to a pervious question identifies sodium chloride vapor as neutral $\ce{NaCl}$ dipoles combined with dimeric $\ce{Na2Cl2}$ quadrupoles.

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text:
If absorbance is 0, what is the equilibrium constant?
If the absorbance of something is 0, what is the equilibrium constant? In a reaction, the absorbance is 0, so the concentration of the product is 0. The concentration of one of the reactants is also 0. What does this say about the equilibrium constant? You would have to divide by 0, suggesting that keq doesnt exist for this reaction.

Are we talking about uv-vis absorption?
Supposed that you are monitoring the right wavelength and that the extinction coefficients are not ridiculously small, you are outside the limits of experimental error.
If you still do not see any absorption, this means that the species isn't there (anymore).
If it isn't there anymore, it can't particiate in the equilibrium any longer - there is no measurable equilibrium anymore.
If you still need to measure something in the case of an irreversible and fast reaction, your experimental setup needs to change:
You might need to perform time-resolved  uv spectroscopy and use a stopped-flow setup for the controlled mixing of your reactants.
If you don't have anything but a uv-vis spectrometer, a cuvette and a microliter syringe, do it the old way using the standard addition procedure

Prepare the cuvette with a solution of one reactant with a decent concentration
Prepare a solution of the second reactant. This might be highly concentrated it you want to avoid dilution effects. If the reaction is fast, you can only measure the absorption of the first reactant and the product anyway
Measure the initial absorption of the first reactant
Add some volume of the second solution to your cuvette
Measure the uv absorption of your first reactant. Do you see any change?
Can you measure the absorption of the product too?

Repeat steps 4-6.


text:
Is photosynthesis endothermic and exothermic at the same time? ... somehow?
Photosynthesis is obviously an endothermic reaction, I mean, what would be the point otherwise, right? This is probably just sheer and utter stupidity on my behalf, but why does Wikipedia say the heat of formation of glucose is negative? As you can probably see I'm on a quest of understanding enthalpy as a concept and I'm still quite new, but some sort of ultranoob clarification would be very much welcome.

Consider that when $1$ molecule of glucose forms, this happens at the cost of $6$ molecules of  $\ce{CO2}$ and $6$ molecules of $\ce{H2O}$ :
$
\ce{6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)}
$
Considering that at $25\ \mathrm{^\circ C}$ and  $1\ \text{atm}$
$
\Delta H_{\text{f},\ce{CO2}}^\circ=-393.5\ \text{kJ/mol}
$
$
\Delta H_{\text{f},\ce{H2O}}^\circ=-285.8\ \text{kJ/mol}
$
$
\Delta H_{\text{f},\ce{C6H12O6}}^\circ=-1273.3\ \text{kJ/mol}
$
You get, by summing the enthalpies of formation for each of the involved species and considering that for $\ce{O2}$ that enthalpy is zero:
$
\Delta H_\text{r}^\circ=6\times393.5\ \text{kJ/mol}+6\times285.8\ \text{kJ/mol}-1273.3\ \text{kJ/mol}=2802.5\ \text{kJ/mol}
$
Note that, for $\ce{CO2}$ and $\ce{H2O}$, the change of sign is required, since you "lose" them, while you form glucose.
So, finally $\Delta H_\text{r}^\circ$ (heat to be absorbed to form one mole of glucose) is positive, indicating that photosynthesis is indeed endothermic.

If you're still there, a tiny by-the-side question: those 2802.5 kJ, they're stored in the bonds of the 1 mol of glucose, right? Is that number (2802.5kJ) synonymous to the amount of photon-energy that hit the plant?

@Uffe that's a good approximation, considering the complexity of such a process.

This is a great answer (+1).  I think people would be less confused by this topic in general if people included photons in their reaction.  The photon reactants would have an indicated wavelength (and thus energy), and would either be converted to nothing or to photons of lower wavelenths (thermal infrared).  If written this way with actually observed photon stoichiometries, it would also indicate in one reaction how inefficient photosynthesis is from an energy standpoint.

The heat of formation is the difference in energy between the current state (in this case, glucose) and the state of the elements at 25 °C (carbon, oxygen, hydrogen, etc).

Endothermic, exothermic. The difference is that an endothermic molecule is unstable. From the moment you remove the extra energy, it reverts back to its original structure. From original author, Park S. Nobel, 1974, it all depends on the manner you get the four protons in order to get the monomer. (five or six monomers to build a sugar). You get either four protons with two or four water molecules. Up to now, it is utilized two water molecules. So you say that the oxygen of photosynthesis is endothermic. And that can't be. (Park S. Nobel is the father of the expression Electron Chain Transport). He also proposed the possibility that the origin of the four protons is due to four molecules of water but for that one he did not propose a model, so the winning model is Two Water Molecules, only because the other model was not described.


text:
Carbon dioxide in lime ground
Is storing carbon dioxide in limestone a good idea? Wouldn't that lead to a reaction like
$\mathrm{C}\mathrm{O}_2 + \mathrm{H}_2\mathrm{O}\to {\mathrm{C}\mathrm{O}_3}^{-2} + 2\mathrm{H}^+$
And then the (unbalanced) reaction
$\mathrm{Ca}\mathrm{C}\mathrm{O}_3 + \mathrm{H}^+\to \mathrm{C}\mathrm{O}_2 + \mathrm{Ca}(\mathrm{O}\mathrm{H})_2$
So now the the ground turns into a water caustic water solution. Is this right?

Are talking about carbon sequestration http://water.epa.gov/type/groundwater/uic/wells_sequestration.cfm?

@GM yes, i do. And the risk of resulting errosion

Limestone is the natural sedimentary rock, e.g., the top of Mt. Everest that began as sea floor.  "Storing" $\ce{CO2}$ in limestone requires heating calcium carbonate white hot to dissociate it to lime, then scrubbing "other" $\ce{CO2}$ with the lime.  This is, to coin an adjective, smartless, since you also burned fuel to heat the limestone. Curing concrete imbibes and traps $\ce{CO2}$, but it was initially dissociated to release it.
Your second equation is terrible. You obtain a strong base by adding acid to a salt.  $\ce{CaCO3}$ plus carbonic acid, $\ce{CO2}$ in water, makes water-soluble calcium bicarbonate $\ce{Ca(HCO3)2}$.  That is how limestone caves are dissolved out, and why karst topography is stark (surface erosion) and unstable (sinkholes).
To vastly store carbon at a profit, plant forests, harvest, and build wood-frame buildings.  The photosynthesis bottleneck, carbon fixation, depends upon the slowest least efficient enzyme on Earth, RuBisCO.  A smart (at least, not corrupt) society then
1) Gene-gineers a better RuBisCO.  Ten times the productivity leaves it the worst enzyme on Earth.  Everybody would have a full belly.
2) Gene-gineers a 10x fast-growing forest with huge lignin content in its wood.  It won't rot, it won't feed termites, it will be immensely strong.  If wood-frame buildings become unfashionable, bury the wood in deserts to store carbon.
There is no problem so easily solved that social advocacy will not make it intractable, disastrous, and expensive.  The unabled maximize self-esteem to be credible - effectively selecting for the deluded, hence advocacy.  A society inundated with rules from religious and secular political classes ignoring productive ends will collapse.  USSR to Pol Pot, reality is what it appears to be - thermodynamics.  There is no cheating possible.

+1 I like you answer, but carbon sequestration doesn't  use calcium hydroxide as primary method to store carbon dioxide...

Don't underestimate the power of the farce, http://www.technologyreview.com/news/508051/a-cheap-trick-enables-energy-efficient-carbon-capture/ http://www.alrc.doe.gov/publications/proceedings/07/carbon-seq/data/papers/p1_002a.pdf  Do your chair parade before criticizing me for having done mine.

I've done mine sir! I've only point out that carbon sequestration usually doesn't use calcium hydroxide. From you answer seems that carbon sequestration uses only calcium hydroxide. I don't want to criticize nobody I do what I hope other users do with my answers: read and check if I have done the job in the best way (and inform me...) Best regards!

Carbon capture cuts a blanket then sews the piece back on to increase its area. Steric hindrance + ethanolamine capture gives facile CO2 release re energy efficiency. (Don't do it at all - iron fertilize the Southern Ocean with aluminum red mud). Advocate = liar.  http://fwrhn487tjy48uvw13uigsy9ht.wpengine.netdna-cdn.com/wp-content/uploads/2014/02/Chart.jpg  Global Warming, theory versus observations,
http://www.climate4you.com/images/MSU%20RSS%20GlobalMonthlyTempSince1979%20AndCO2.gif
http://fwrhn487tjy48uvw13uigsy9ht.wpengine.netdna-cdn.com/wp-content/uploads/2014/02/Chart.jpg

Please forgive my ignorance.  From what I understand, you are saying tress are the most viable means of CCS.  Why does it not mention planting trees in the CCS wiki page?https://en.wikipedia.org/wiki/Carbon_capture_and_storage

Sincerely I'm not sure that carbon sequestration injecting $CO_{2}$ in the ground is a good idea. The risk is that possible leaks cause asphyxiation to the living creatures (humans included) in the surrounding see this article, furthermore injecting gas can change the equilibria of the ecosystem or cause other hydrogeological instability.
I strongly recommend you to read this article, an excerpt of a natural $CO_{2}$ emission:

The year was 1986, the scene of the tragedy was Lake Nyos in Cameroon.
People were going about their daily lives when death struck; a large
amount of $CO_{2}$ exploded from Lake Nyos and more than 1,700 people died.
When medical examiners came upon the scene they found bodies sprawled
everywhere, many with clothes shredded and half-torn, a last attempt
to relieve the suffocation of asphyxiation.  Cattle had died, and
those who had survived had been a comatose state for hours.  Skin was
seared, bronchial tubes were burnt, and lungs filled with fluid.  It
was surely a quick way to die, but not something to be wished for.
The cause of the explosion was at first, but after multiple tests and
examinations it appears that Lake Nyos was connected to a volcanic
chamber that slowly released $CO_{2}$ into the lake.  After a large amount
accumulated, it erupted and rose to the surface.

However there are others research more optimistic this paper from 2009 can give you a good review however nobody can predict all the possible implications of these injections, and the topic is still debated the conclusion of the article clarify this:

“There is consensus that $CO_{2}$ storage in deep underground geologic has
great technological potential and  may be deployable on a widespread
basis. However, there are still substantial financial, institutional,
regulatory, and technical challenges that remain. To address these
challenges, multiple integrated $CO_{2}$  capture and storage system
projects are needed to prove out the technology. Also needed is an
array of  small, and intermediate and large-scale $CO_{2}$ injection field
tests in diverse geologies to adequately  characterize and validate
the U.S. geologic resource” (Position Paper of the Coal Utilization
Research Council  on the Status of Technology for Carbon Capture and
Storage (CCS), 2008).

Back to the chemistry...
Your reaction (as stated Uncle Al) is wrong because:
$$\ce{CaOH + CO2 -> CaCO3 + H2O}$$
Has $K_{eq}$ of $4.60 \times 10^{19}$! So goes in the other side of your reaction. I think that you are being tricked by a typical reaction with an acid, like this:
$$\ce{CaCO3 + 2HCl -> CaCl2 + 2H2O + CO2 ^}$$
However if you substitute HCl with carbonic acid you have a series of complex equilibria:
$$\ce{CaCO3 + 2H2CO3 <=> Ca(HCO3)2 + H2O + CO2}$$
but:
$$\ce{H2O + CO2 <=>H2CO3}$$ However the ratio between carbonic acid and carabon dioxide is very low: $$\frac{[H_{2}CO_{3}]}{[CO_{2}]} ≈ 1.7×10^{−3}$$
In fact carbon sequestration normally take advantage of adsorption of gasses into the soils, but the carbonatation occur very slowly.
By the way with carbon sequestration you will have a soil with a lower pH, more acid... and maybe sparkling tap water too! ;-)


text:
Extraction of a pyridine tetrazole from the aqueous phase
I have synthesised a compound which is basically a pyridine ring connected to a tetrazole. The procedure suggested to precipitate it at pH 1, but it did not precipitate. It then suggested to extract it with ethyl acetate.
The problem is that it is not going to the organic layer but into the aqueous layer. I have tried both at pH 1 and 7 and it still does not move to the organic layer.
Any ideas how to solve this problem?

Concentrate under reduced pressure on the rotovap. Add a miscible co-solvent that does not solubilize the zwitterion at its isoelectric point but dissolves starting materials, like THF.
The pyridine is basic, the terazole is acidic. Catch it on an ion exchange resin column in water, then displace it with a polyvalent counterion.  Possibly down neutral alumina in water to slow it and concentrate it.  You can run a microcolumn in a disposable pipette with a tiny ball of glass wool inside at the taper.
Salt it out as a solid with saturated NaCl or potassium nitrate its isoelectric point. Precipitate it, perhaps as the tetraphenylborate or picrate (careful there).


text:
Is there an energy cost associated with flipping the spin of an electron?
THE STORY:
A common example used to illustrate the limitations of restricted Hartree-Fock (RHF) theory is the H$_2$ dissociation energy ($D_e$) curves.  RHF enforces electrons to be paired into spin orbitals, $\chi$, or two spatial orbitals $\phi$ with the same set of spatial coordinates $\mathbf{r}$ but a different spin function (i.e. $\alpha$ or $\beta$ for spin-up $\uparrow$ and spin-down $\downarrow$, respectively).  Unrestricted Hartree-Fock theory gives each spatial orbital its own set of coordinates, allowing for $\phi_1(\alpha$) to have a different energy than $\phi_1(\beta)$, giving rise to a proper description of the H$_2$ dissociation energy (see the graph below).  I didn't include units in this graph but the energy is in kcal mol$^{-1}$.  The red curve gives the appropriate (and approximate) $D_e$.  The RHF formalism (blue curve) gives a $D_e$ that is three-times greater than the 'right answer'.

A corresponding electron configuration picture is shown below.  Here we clearly see RHF forcing the electrons to be paired at infinite distances, giving rise to an H$^-$H$^+$ description of the system which is clearly wrong.  (Note that this is a 'state' that included into the wave function by RHF).  UHF allows the electrons to reside in their own spatial orbitals giving rise to two doublets at infinite separation.

Now here is where things get interesting.  In terms of the actual computations implemented to generate the curves in Figure 1, the electrons adopt opposite spins.  However, let us imagine a theoretical scenario where two lone hydrogen atoms are floating around in space at infinite distance.  The corresponding electrons (which we will assume is not interacting with each other) can adopt any spin they want so both electrons can have a spin-up configuration.  Now imagine these two hydrogen atoms approaching each other.  For a covalent bond to form, the electrons will be paired into a spin-orbital which means that one electron must flip its spin according to the Pauli-Exclusion principle.  (We should note at this point that orbitals are merely mathematical functions whereas spin is an actual physical property).  This now leads us to...
THE QUESTION:
Is there an energy cost associated with flipping the spin of an electron such as in the scenario I've described?
The obvious follow up question (if the answer is yes to the first piece), would the H$_2$ dissociation energy curves be different for two H atoms each with spin up electrons and two H atoms with electrons of opposite spin (and how different would they be)?

Yes.  (Cf: the 21 cm line in atomic hydrogen)

Your RHF and UHF description is not correct. In RHF at infinite distance there are still two linearcombinations of the orbitals (the same as in bonding distance) which are close to degenerate (the orbital with the electrons is lower in energy, since HF cannot describe empty orbitals correct). If one electron is at one H, then the other has a 50:50 probability to be at either H. In UHF the spin configuration stays formally the same (singlet). Since the restriction is lifted, it is no eigenfunction of the spinoperator anymore (spincontamination).

@UncleAl could you elaborate more, please. From my point of view, if two H with same spin electrons are put in close contact, then one electron would occupy the antibonding (excited) orbital. Relaxation then would release energy (negative singlet triplet gap). But I might be wrong - I am curious though. Or do electrons have to be of opposite spin to form a bond? If so, then spinflip would have to occour before - but how would the electron know?

Two magnets: ˄˅ is lower in energy than ˄˄.  Triplet oxygen has its unpaired electrons in orthogonal orbitals, plus other factors are at play.

@UncleAl Hmm, according to the first of Hund's rules an electron configuration of parallel spins is favored over one of paired spins. But I'm not entirely sure if this argument is admissible here.

@Martin I've attempted to clarify the RHF picture as a 'state' which is mixed into the description of the wave function.  Hopefully it doesn't sound misleading now.

This is still not correct. The RHF picture is qualitative wrong, because it fails to describe homolytical bond cleavage correctly. Since electrons are forced to occupy the same spacial orbital, they have a probability (it is not 1) that they might be at the same nucleus. HF can only describe one state, because it has only one determinant. For a dissociated molecule you need at least two determinants to come close to a qualitative picture. UHF does that by removing the restriction. It is still qualitatively wrong: Describing a mixture between singlet and triplet.

@Martin The electron configuration picture is merely illustrating the deficiency of RHF by showing the state that it is 'mixing' at infinite separation.  It is not there to show the ONLY state that RHF considers.  Is there a better way to refer to this piece of the linear combination of orbitals than 'state'?

Here we clearly see RHF forcing the electrons to be paired at infinite distances [...] is just very wrong. This is not the restriction of HF.

@Martin Please offer an example of what is correct.  I'm interested to know if the error of my words are truly wrong or if it is more of a semantically related thing.

Restricted HF makes the requirement that two electrons of opposite spin have to occupy one spatial orbital. This is causing the orbitals being spread over the whole molecule. Two hydrogen atoms at infinite distance are still a molecule in the HF formalism, but this approximation is not applicable any more. (In theory the amplitude of the wavefunction of the orbitals only vanishes at infinity, causing always some overlap.)

@Martin So, disregarding the figure for a moment... with RHF the electrons (if I understand you) are indeed paired (i.e. in the same spatial orbital).  RHF forces these electrons to be in the same spatial orbital (and therefore, RHF 'forces' the electrons to be paired at infinite separation - a result of using a  single reference which is a key characteristic of RHF...).  I'm failing to see the problem with this.

I have a hunch that this problem lies in a misunderstanding of what 'spin up / spin down' means. The up and down is the result of a measurement -- it is the spin projection along an axis. Spin half is an intrinsic property, but the 'direction' is the result of measurement. So I'm not sure you can say that two hydrogens are both spin up. Remember the Stern-Gerlach experiment! I know quantum measurement is tricky, but perhaps the interaction of two hydrogen atom spins results in a measurement of one spin up and one spin down? If they had a permanent 'spin up/down' we could distinguish them...eek

@jjgoings The up/down notation only applies if there is a point of reference right?  So if we have one electron by itself with no reference, one could not say whether it was spin up/down at all (I could stand on my head and give you a completely different answer for example).  As UncleAl pointed out, the nucleus is actually a point of reference (very interesting stuff).

Also, you can 'flip' spin in relativistic electronic structure theory. Only total angular momentum (j) is conserved, not orbital angular momentum (l) or spin (s). 'Spin-orbit coupling' is what you are looking for. This happens sometimes in ligand binding to transition metal surfaces. The binding event can change of the spin state of the metal.

@jjgoings So are you saying that, in the scenario I've prescribed, the point is... moot?

@LordStryker No, the nucleus in not the point of reference. Consider the silver atoms in the Stern Gerlach experiment. The reference is the measurement apparatus (the inhomogenous magnetic field).

@LordStryker Sorry, just saw your comment. I think it may be, but it is worth looking into. You could prepare a bunch of hydrogen atoms all in spin up with a magnetic field, and then see if they react. I just wonder if interacting spin 'counts' as a measurement. I find spin very tricky and so I am struggling to explain my (rather unformed) ideas!

@jjgoings Sorry, I was referring to the post (not the comment I wrote you... ack.)

If electrons would be paired, they would interact, i.e. correlate their movement. In the RHF picture this is not possible. One electron only sees the mean field of the other and vice versa. So they are completely uncorrelated - hence the error of HF - and not paired. Their probabilities only occupy the same space. You should be thinking wave not particle.

@Martin Okay now I see (finally) where you're getting at.  But, then I must ask... if the electrons aren't 'paired', RHF wouldn't care about the spin of the electrons but last I checked, it does...

@Martin  So they are completely uncorrelated - hence the error of HF  This is untrue.  HF has exchange correlation built-in.  Its a common thing for people to say that HF has no correlation but it does. (obviously for H2 there is no exchange correlation...).

This is true, I have to admit I was not thinking of exchange correlation for the Hydrogen molecule. HF Exchange correlation is also often referred to as the exact exchange. As for the wording 'pairing' - this might just be a philosophical question. Anyways, this whole rambelage does actually not help finding an answer. Sorry for that.

@Martin This discussion has been a pleasure and I thank you for participating.  We may be pushing a record for Chem:SE with 20+ comments :)

Am I missing something here? Can't this just be determined by ESR ?

@Lighthart What is ESR?  Care to share some thoughts?

ESR measures hyperfine structure based on electronic spin flips, so the energy value of the peaks in that spectrum should be directly related if not identical to the energy costs associated with spin flips.  It is possible I just don't understand the question, or that the questioner does not know about this experiment.

@LordStryker Is my answer actually helpful or do you have a lot open questions left?

@Philipp I'm still mulling your response over.  It goes without saying that yes, your post is indeed helpful.  If I have further questions I will be sure to let you know.

@LordStryker I use the opportunity to extend this comment section even further :) I updated my answer and expanded the explanation why a triplet-singlet transition is nearly impossible in the system you describe.

@Philipp The expansion on your response greatly strengthens the answer as a whole.  I appreciate you adding it.

I will try to describe what happens when two hydrogen atoms approach each other from infinity. At infinite separation the hydrogen atoms don't feel their mutual presence and each atom has one electron localized in its atomic 1s orbital. In the absence of magnetic fields it will not matter whether the spins of the electrons are parallel or antiparallel and you will basically have an ensemble of systems with an equal amount of triplet (parallel spins, $S=1$) and singlet (antiparallel spins, $S=0$) states, since $E_{\uparrow \uparrow} = E_{\uparrow \downarrow}$. But when hydrogen atoms are brought closer together the situation changes: For one thing each atom has a magnetic moment associated with its electronic spin and there will be a dipolar interaction between those magnetic moments. However this dipolar interaction will be very weak (in the order of $10^{-4} \, \mathrm{eV}$ at close interatomic distance) and won't have much influence. Another factor that comes into the game is spin-orbit coupling. Yet again, since $\ce{H}$ is such a light element spin-orbit coupling will be very, very weak and so it can be ignored too. But the predominant factor to be taken into acount is the exchange energy (its effect is in the order of $0.1$ to $1 \, \mathrm{eV}$). Here you can find a very nice treatment of the exchange energy for the hydrogen molecule which comes to the conclusion that

For the hydrogen molecule the exchange energy is negative [...] so the state when the resultant spin is $S=0$ has the lower energy than the state when $S=1$.

This source also provides the potential energy curves for the hydrogen singlet and triplet states.

These curves amount to the following electron configuration diagrams:

${}$

So, the situation for the approaching hydrogen atoms is like this: If both atoms start out with electrons of opposite spin (i.e. the system is in a singlet state) you have the common case that the hydrogen atoms will bond together thus lowering the total energy of the system. But if both atoms start out with parallel spins (i.e. the system is in a triplet state) the total energy will rise when they get closer to one another and the atoms will not bond together but keep their distance. This can be understood from the electron configuration diagram of the triplet hydrogen molecule: The antibonding $\sigma^{*}$ MO is more destabilized than the bonding $\sigma$ orbital is stabilized by the interaction (this is generally true for 2-orbital interactions) and the energy splitting becomes larger the closer the hydrogen atoms get. But because of the Pauli principle the two electrons of like spin in the system can't both occupy the $\sigma$ MO but one is forced to occupy the $\sigma^{*}$ MO and thus the system's total energy rises when the hydrogen atoms approach each other. And because there is virtually no spin-orbit coupling for hydrogen the total angular momentum (represented by $L$) and the total spin (represented by $S$) of the electrons are essentially conserved quantities and there is no mechanism to facilitate intersystem crossing, i.e. the process of changing from the triplet to the singlet state has such a low probability that it virtually doesn't happen. The situation changes of course if you add an external bias to the system, e.g. irradiation with high-energetic light. Then a triplet-singlet transition might be possible but as I read it that was not your question.
So, the way MO-diagrams are often pictured with electrons of like spin pairing up in a bonding MO does not reflect the "real" situation of two atoms approaching each other from an infinite distance. But it is not meant that way although it is often pictured as such.
Update
So, in summary I established that if you have two hydrogen atoms in free space without spin-orbit coupling and relativistic effects they are stuck with the spin state they initially start out with and can't change it during their time evolution, i.e. they have to stay either on the triplet or the singlet Born-Oppenheimer surface and no switching between those is allowed. But I want to delve a little deeper into the reasons for it. If the hydrogen atoms wanted to change from triplet to singlet state, the atoms would have to transfer $1 \hbar$ of spin angular momentum somewhere else in order to comply with the law of conservation of momentum. In the absence of any particles to collide with the only possible mechanism by which this can be achieved is to emit a photon whose energy is equal to the energy difference between the triplet and the singlet state, i.e. $\Delta E = \hbar \omega = E_{\uparrow \uparrow} - E_{\uparrow \downarrow}$, and which would "carry away" the $1 \hbar$ of spin. But in order to emit a photon there must be a coupling between the electrons and the electromagnetic field. The magnitude of this coupling or in other words the probability for a transition from an initial to a final state under the emission of a photon is described by the transition dipole moment
\begin{equation}
  \Theta_{\mathrm{i} \to \mathrm{f}} = \big\langle \Psi_{\mathrm{i}} \big| \sum_{n} q_{n} \mathbf{\hat{r}}_{n} \, \big| \Psi_{\mathrm{f}} \big\rangle
\end{equation}
where $q_{n}$ and $\mathbf{\hat{r}}_{n}$ are the charge and the position operator of the $n^{\mathrm{th}}$ particle in the system respectively, and the subscripts $\mathrm{i}$ and $\mathrm{f}$ signify the initial and final states of the transition. Under the conditions of the Born-Oppenheimer approximation the contribution of the nuclei to $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be zero for the situation at hand, so that we only need to be concerned with the electrons in the system. If there is a sizable spin-orbit coupling then the spin degrees of freedom and the spatial degrees of freedom are coupled via the total angular momentum operator $\mathbf{\hat{J}} = \mathbf{\hat{L}} + \mathbf{\hat{S}}$ because
\begin{align}
  \mathbf{\hat{J}}{}^{2} = \mathbf{\hat{L}}{}^{2} + \mathbf{\hat{S}}{}^{2} + \mathbf{\hat{L}} \cdot \mathbf{\hat{S}}
\end{align}
and $\mathbf{\hat{L}} = \mathbf{\hat{r}} \times \mathbf{\hat{p}}$ where $\mathbf{\hat{p}}$ is the momentum operator. However, if there is no spin-orbit coupling then $\mathbf{\hat{L}} \cdot \mathbf{\hat{S}} = 0$ and $\mathbf{\hat{J}}{}^{2} = \mathbf{\hat{L}}{}^{2} + \mathbf{\hat{S}}{}^{2}$, i.e.
the spin degrees of freedom and the spatial degrees of freedom are decoupled. That means you can make a product ansatz for the wave function where you seperate the spin-dependent part $| \psi(S) \rangle$ and the spatial part $| \psi(\mathbf{r}) \rangle$ such that $| \Psi \rangle = | \psi(\mathbf{r}) \rangle | \psi(S) \rangle$. So, for the transition dipole moment you get
\begin{equation}
  \Theta_{\mathrm{i} \to \mathrm{f}} = \big\langle \psi_{\mathrm{i}} (\mathbf{r}) \big| \sum_{n} q_{n} \mathbf{\hat{r}}_{n} \, \big| \psi_{\mathrm{f}}(\mathbf{r}) \big\rangle \underbrace{\big\langle \psi_{\mathrm{i}} (S) \big| \psi_{\mathrm{f}}(S) \big\rangle}_{= \, 0} = 0
\end{equation}
where $\langle \psi_{\mathrm{i}} (S) | \psi_{\mathrm{f}}(S) \rangle = 0$ because the initial state would be the triplet state ($S =1$) and the final state would be the singlet state ($S=0$) and spin wave functions for different $S$ are orthogonal.
So, no spin-orbit coupling means that the probability for a transition from a triplet to a singlet state under the emission of a photon is zero.
In real atoms and molecules there is of course some spin-orbit coupling, so $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be larger than zero. But the spin-orbit coupling in hydrogen is tiny and so $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be tiny too and thus the probability of a triplet-singlet transition will be tiny as well and can be neglected for all practical purposes.

Wow, I really have to think about this, thank you. Am I correct in assuming, that as long as there are other particles in the system, that can "carry away" the spin, transition from triplet to singlet is no problem?

Yup, collisions can provide a mechanism for spin-flips too. The efficiency of the process depends on the particle, of course.

This by far is the most comprehensive and insightful response I have ever received to this question.  It is also the most satisfying.  Though it leads to more questions (ones I cannot verbalize at the moment), the realization of knowledge generally does so and I applaud your efforts to provide such a rigorous answer.

@LordStryker Thanks for the kind words. Glad to hear it was helpful. It is a very interesting question and so I needed to provide a comprehensive and insightful answer to match it :) If the new questions ever reach a verbalizable state, just let me know. Helping is good for my karma :)

Nice answer. Just a remark: "In the absence of magnetic fields it will not matter whether the spins of the electrons are parallel or antiparallel and you will basically have an ensemble of systems with an equal amount of triplet ..."
You actually do not have an ensemble (since you only got 2 Atoms), but rather in the general case an undetermined superposition of both states unless you interact with a magnetic field (=measure). The probability of what you get then depends on how the state was prepared.

authors: Василий Свинко


text:
Why do organolithium and organomagnesium compounds react differently with CO2?
My textbook doesn't say anything about this but does say that on addition of organolithium compounds to carbon dioxide, they give ketones and not carboxylic acids like grignard reagents. This has been attributed to the greater nucleophilic character of Organolithium compounds compared to grignard reagents. I see that greater polarity of the C-Metal bond must relate to this according to many websites but I cannot understand why! Also, during formation of the RLi compounds using alkyl halides, what does the alkyl halide dissociate into? Please help!

Essentially, you are forming a carboxylate when you add the organolithium or Grignard compound to carbon dioxide, but the carboxylate may have covalent character in the metal-oxygen bond. Then this answer offers two hypotheses:
(1) Despite the difference in electronegativity, the lithium compound may be more covalent than the magnesium-halogen compound because the latter forms a three-center bond. This would tend to maintain more electrophilic character at the carboxylate carbon with lithium.
(2) The Grignard reagent imparts bulkier groups on the oxygen atoms, making the double addition required to form a ketone less favorable.

When Li reacts with an alkyl halide, a salt (lithium halide) is formed, whose stability is the primary thermodynamic driving force of the reaction: $\mathrm{2Li}+\mathrm{RCl} \rightarrow \mathrm{RLi}+\mathrm{LiCl}$.
Due to the larger difference between the electonegativities of C and Li, the C-Li bond in organolithium compounds has more ionic character and C is more negatively polarized. The higher polarization makes R in RLi a harder nucleophile compared to RMgX, which explains why lithium organyls and Grignard reagents react differently when adding to certain carbonyl compounds.

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


text:
Is the valence band neutral?
While studying about band theory of semiconductors, I observed that when the electrons were excited from the valence band to the conduction band, they left behind holes in the valence band. From my existing knowledge, I believe that the valence electrons alone occupy the valence band which tells me the valence band is negative. For the holes to exist, the valence band has to be neutral. So, why is the valence band neutral?
If my reason for the valence band to be negative was wrong, I would like to know the reason.

I think if you look at the overall structure is neutral. I don't think that has much sense take valence band it self...

Using an organic example, as you move from ethylene to butadiene to hexatriene, you have 1, 2 and 3 bonding and antibonding pi molecular orbitals respectively.  If you continue this chain extension and wind up with a very large polyene, the energy difference between the all of the bonding (or antibonding) orbitals becomes small and instead of having discreet molecular orbitals, you can imagine them "fuzzing" out into a "band" of molecular orbitals.  All of the bonding orbitals would become what is termed the valence band, while the antibonding orbitals would comprise the conduction band.  Just as ethylene is neutral, so is the large polyene.  If you excite one of ethylene's pi electrons to the antibonding molecular orbital you have created an excited state which is still neutral.  Scientists working with the solid state would use different terms to describe this process.  They would say that the antibonding orbital contains 2 holes and the boding orbital 2 electrons. When the electron is promoted to the excited state a hole is also promoted (holes have lower energy the further they are from the nucleus) from the antibonding orbital to the boding orbital.  Just like the ground state, the excited state molecule is still neutral, the electron and hole are just in different orbitals from where they started.


text:
Conceptualizing Multiple Equilibria
Take the reaction:
$$ (1) \; \ce{A + B <=>X + Y}, \; K_1=\frac{[X][Y]}{[A][B]} $$
Now assume that X isomerizes to Z according to:
$$ (2) \; \ce{X <=> Z}, \; K_2=\frac{[Z]}{[X]} $$
We can then add the equations to obtain:
$$ (3) \; \ce{A + B <=> Z + Y}, \; K_3=K_1K_2=\frac{[Z][Y]}{[A][B]} $$
It's pretty easy to show that $ K_3=K_1K_2 $ when considering the ratio of products and reactants, but this makes less sense when considering the actual numeric value of $K_3$. The fact that $ K_3=K_1K_2 $ can be explained by the statement of multiple equilibria: all equilibria must be satisfied simultaneously. This is what I'm having trouble conceptualizing. Does the equilibrium constant in $(1)$ already take into account $(2)$, and vice versa? Equation $(3)$ will reach a state of maximum entropy in which $K_3$ applies. This must also occur when both $K_2$ and $K_3$ are satisfied? I can do the math all day, but I'm not quite getting the link between the three equilibrium constants.

When there is a situation of multiple equilibria, all of them must be satisfied. If this is the case, the whole system will be at equilibrium.
For your example and as you report, the maths let you write that
$
K_{3}=K_{1}\times K_{2} \quad (1)
$
This final equation, beyond being the mere mathematical combination of two initial equations, is obtained by considering that Gibbs free energy is a function of state: iIt describes "quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state" (Wikipedia). One may get to equilibrium via an unique straight step, or through several steps. But the final destination will be the same.
This definition allows us to write:
$
\Delta G_{3}^{o}=\Delta G_{1}^{o} + \Delta G_{2}^{o}  \quad (2)
$
Equation (2) can be converted into equation (1), by remembering that $
\Delta G^{o}=-RT\text{ln} K$.
We can first of all say that $K_{1}$ and $K_{2}$ relate to the two "pieces" of the reaction path one has to cover to reach the overall equilibrium. Knowing them is essential, if one wants to obtain the quantities for all the chemical species, at equilibrium.
$K_{3}$, on the other hand, relates to the "shortcut" (the unique "ideal" step connecting the initial and final states of a system) and is linked to $K_{1}$ and $K_{2}$, through equation (1).
Using the examples you provide and assuming initially 1 mole for $A$ and 1 mole for $B$, we write:
$
K_{1}=\frac{(y-z)y}{(1-y)^2}
$
$
K_{2}=\frac{z}{y-z}
$
We could more generally write:
$
K_{1}=f(y,z)
$
$
K_{2}=g(y,z)
$
The two constants are different functions: but they both depend on the same parameters, $y$ and $z$, which are temperature and pressure (for gases) dependent. If temperature changes, both constants will accordingly change.
To determine the equilibrium quantities, at given temperature and pressure, we'll need the values for $K_{1}$ and $K_{2}$, since knowing only $K_{3}$ won't be enough.
Beyond all this, kinetics will affect the time needed to reach system equilibrium: but that's another story.

Ok, I can accept that $ (1) $ is true due to $ (2) $. Is there any way to explain why $ K_1 $ and $ K_2 $ are also both satisfied? That is, why are ALL equilibria satisfied? You could prove that both $ K_1 $ and $ K_2 $ have to be true in order to reach $ K_3 $ due to $ (2) $ being true. I feel like this is more along the lines of working backwards from the solution, however. Is there any way to explain this in terms of energy/entropy?

@David Wouldn't be correct to say that since all equilibria have to satisfied, each $\Delta G$ (related to a single equilibrium) has to be $0$ ? (short reminder: $\Delta G  \neq  \Delta G^{0}$). $\Delta G=0$ is the condition that always applies once a chemical equilibrium is established: $G=H-TS$, so in the definition of $G$ you find enthalpy and entropy....

The only thing I can think of is the following: $ ∆G_3 = ∆G_1 + ∆G_2 $. Since $ ∆G_3 = 0 $, $ ∆G_1 + ∆G_2 = 0 $ and $ ∆G_1 = -∆G_2 $. This is clearly true for $ ∆G_1 = ∆G_2 = 0 $, but I'm not sure how to prove this is always the case. If $ ∆G_i = 0 $, then $ K_i $ must apply.

@David one further point: if $\Delta G_{3}=0$, then $\Delta G_{1}=-\Delta G_{2}$. One option is that both are equal to $0$ and that means that the two reactions are at equilibrium. Another option is that $\Delta G_{1}=-\Delta G_{2} \neq 0$: so if $\Delta G_{1} < 0$, then $\Delta G_{2} > 0$. That can't work: reaction $1$ would be spontaneous, while reaction $2$ would not. In these conditions, the overall reaction $3$ would never reach equilibrium, since reaction $2$, essential in order to complete reaction $3$, would never occur.

You say reaction $ (2) $ would never occur for $ ∆G > 0 $, but I feel like it would since it is coupled to reaction $ (1) $. This is essential to the idea of spontaneous processes forcing non-spontaneous ones to occur. i.e. The coupling of ATP hydrolysis to non-spontaneous enzymatic reactions makes the overall process spontaneous. Am I wrong, or is there another explanation? Does the coupling I've explained react differently when the coupled reaction is at equilibrium, as opposed to being simply positive or negative?


text:
Yama Cold Brew Tower: what are these lab parts / components?
When viewing this on a shopping web site, one of the reviewers indicates "This is about 50-60 [dollars] worth of pyrex glass and another 20 [dollars] worth of wood."  The company charges $499 for this setup, so I'd like to find out if there is any truth to this statement by identifying the parts and seeing how much they cost.  Image of the setup below:

It might be 20 dollars worth of wood, but the glass is substantially more; I would guess about 250 dollars. I think the comment was to indicate that it was cheaply made and not on the actually cost. BTW, what is the purpose of it?

A well-jointed 3 litre round-bottomed flask alone can set you back at least 70 dollars. Is that bottom bit joined, or is it just a tube that drips into the flask openly?

It's a cold brew coffee maker.  The person indicated it was pyrex glass.

I think that you can make your own cheaper Yama cold brew tower however what they are selling to you is the design! What you see in the picture is not chemistry glassware but if you want to make your own with chemistry glassware (this will cost more money because good chemistry glassware is purer and must have many expansive features that normal glassware don't have) I've written the analogous components in the figure. You need an open-top cylindrical dropping funnel with a stopcock valve.

See here and here for reference. You can use a normal funnel and attach to it a stopcock valve if you want. Then there is something similar to a Buchner funnel, I imagine that there is a filter in the bottom, may be included into the funnel, in this case, a 90-150 micron filter should be sufficient. See here for more info. Then there is a glass coil spiral, up to my knowledge this has no practical use is only beautiful and geeky but will cost a lot. Then you have a normal carafe.
(probably a piece of filter paper is placed over the grounded cafe to allow the diffusion of water in the whole volume of the cylinder and not only in the centre).

With some filter paper and a couple of half cut plastic bottles you can do your Yama tower with less of 5 $: you can make a little hole in the screw cap of your first bottle to make drip the first funnel, and use a piece of filter paper in the second.  If you want to use glassware don't use chemistry glassware you don't need it really, try to find normal glassware to adapt to your project and you can make your own cheaper and nice Yama tower.

Thanks for such intricate details.  I saw a plastic cold brew maker for $35, but prefer the 'Breaking Bad' style look.  If not so expensive, I would consider it.

@Javy yes maybe is better... Right I think I should make a link with http://chemistry.stackexchange.com/questions/7439/chemistry-behind-gales-coffee-maker-in-breaking-bad/7491#7491

No need for Pyrex this all occurs at room temp, replace the valve on any water dispenser with adjustable valve, set to drip. For the funnel, empty wine or liquor bottle, cut off bottom, you all set. For bottom flask, glass milk jug. Easily under $50 and two hours.


text:
Chemistry Spreadsheet Software?
Are there any chemistry spreadsheet-like software that:

Generates 2D structure for the entered compounds (based on name or SMILES)
Easy to search and create customized reports (such as similarity search)

that preferably runs on linux and is not too expensive?
I'm aware of ICM Chemist and JChem for Office (Windows), and I'm wondering if there are any other alternatives available.

Spreadsheets are for calculations (you can describe functional relationship between cells etc). What kind of calculation / spreadsheet functions do you want to use?

I can't comment on spreadsheet-like tools.
But I'm frequently using Open Babel, which is free (as in free speech and free beer).
If you're not so much into using command-line tools or using programmes via their bindings to programming languages, a GUI for Open Babel is available too, but I've never used it.

Generates 2D structure for the entered compounds (based on name or SMILES)

Names might be tough in general, but SMILES (Simplified Molecular Input Line Entry System) can be processed with Open Babel when the identifier is read from a file. Output to large number of chemical formats and image formats is possible.
obabel -ismi stuff.smi -omol -O stuff.mol
would convert a SMILES string to a mol file.
Converting a collection of SMILES to MOL and writing them to a SDF file is not a big thing:
#!/usr/bin/env python

import os
import pybel

# dictionary of SMILES taken from chemspider.com
insecticides = {
    'parathion' : 'CCOP(=S)(OCC)OC1=CC=C(C=C1)[N+](=O)[O-]',
    'disulfaton' : 'CCOP(=S)(OCC)SCCSCC',
    'permethrin' : 'C1(C(C1C(=O)OCC2=CC(=CC=C2)OC3=CC=CC=C3)C=C(Cl)Cl)C',
    'furathiocarb' : 'CCCCOC(=O)N(C)SN(C)C(=O)OC1=CC=CC2=C1OC(C2)(C)C',
    'methomyl' : 'CC(=NOC(=O)NC)SC'
}

# write dict items to multi-component SDF file
outdir = '/home/klaus/workspace/chemistry/stackexchange/similarity/'
outname = 'insecticides.sdf'
outpath = os.path.join(outdir, outname)
outfmt = 'sdf'
datafile = pybel.Outputfile(outfmt, outpath, overwrite = True)

for name, smiles in insecticides.iteritems():
    mol = pybel.readstring('smi', smiles)
    mol.title = name
    datafile.write(mol)

datafile.close()

EDIT

Easy to search and create customized reports (such as similarity search)

For the substructure search, I used the Drug-likeness NCI subset from ligand.info. The unpacked SDF file (445.2 MB) has 193,323 entries on 14,757,262 lines (wc -l).
From the SDF file, I first generated a binary fastsearch file using
obabel data.sdf -ofs -O data.fs
Searches are based on SMILES and SMARTS patterns.

I thought that 2-aminobenzothiazole might be an interesting fragment to search for and with a bit of fiddling, I wrote the suitable SMARTS pattern: c12ccccc1nc(N)s2
If in doubt about the correct representation, it is always a good idea to convert the pattern to an image file here: obabel -:"c12ccccc1nc(N)s2" -opng -O pattern.png.
The search was performed using:
obabel -ifs data.fs -s"c12ccccc1nc(N)s2" -osvg -O results.svg -xd -xc5 -xA -xC
and generated a SVG table with 233 entries.

Oooh, you've got my attention now. Can't wait to see the substructure analysis thing that you've got up your sleeve ;)

Thanks. I'm a fan of OpenBabel (and RDkit), but for this case I need to store (and easily-visualize) many types of chemical data, including structures. I'm kind of thinking to role out my own solution (RDKit + SQL database + web interface) but want to make sure I'm not reinventing the wheel.

You forgot a link. Check the SMARTS pattern where?

@LDC3 No, I didn't ;) But I added the obabel command to generate a PNG from the pattern.

I use ACD/Labs Chemsketch in Windows. There is a version that runs on Linux.

Linux users can download a compatible version of ACD/ChemSketch
  Freeware that can be run with Code Weavers' CrossOver Office.

http://www.acdlabs.com/resources/freeware/chemsketch/

Isn't Chemsketch mainly a drawing program rather than a spreadsheet/database program?

After looking at JChem and ICM Chemist, I agree ChemSketch does not provide the same functionality that the others do. I do know that it can create structures from SMILES and can name the structures. I'm not certain how well it  translates to common names. It will use 2-methyl-2-propanol instead of t-butanol.


text:
Simple enantiomer in powder form?
I want to do some experiments on enantiomers. As it involves some calculations of potential energy surfaces, I would like to do this experiment on a simple enantiomer (For example $\ce{CHBrClF}$). But this is in liquid form I think. I need an enantiomer sample in the powder form. Can anyone suggest a simple sample? Either one I can buy or can easily make in a chemical lab with the help of a chemist.

Chirality was discovered by observing that tartaric acid crystallised from wine had crystals which are also chiral. It is widely available: how about that?

Enantiomorphs are non-superposable mirror images (all signs of one coordinate axis reversed).  What is the maximum enantiomeric divergence possible?  If you have a  quantitatively negligible chiral divergence, there is nearly no quantitative difference to be realized in calculation or measurement.  BTW, the maximum energy difference between enantiomers cannot be more than a part-per-trillion difference/average.  If it were larger, there would be easily detectable enthalpy of fusion anomalies for calorimetry secondary standard benzil.
Chirality is geometry.  Geometric chirality is easily observed (no ${S_n}$ symmetries[A]) and easily calculated[B].  alpha-Quartz and gamma-glycine single crystals have $\text{CHI} = 1 \text{, COR} = 1 \text{, DSI} = 0$, maximum mathematical chirality[C], but it cannot be measured.  Solution optical rotations ignore atomic mass distribution[D].  Silver thiogallate, $\ce{AgGaS2}$ in achiral space group ${I}$-${42d}$, rotates 522°/mm along [100] at 497.4 nm[E].  ${P3_121}$ quartz and ${P3_221}$ berlinite are both  levorotatory[F].  Achiral $\ce{PhCOCH3}$ impurity changes resolved $\ce{PhCH(OH)CH3}$ optical rotation[G].  CIP notation flips with composition at constant geometry (look at natural protein cysteine versus alanine, both L-amino acids but opposite CIP notations).
How do you obtain $100\%$ optically pure maximally divergent enantiomers to your specs?  No problem!  Benzil, $\ce{(C6H5CO)2}$, is achiral when molten, gas phase, or dissolved. Solid state crystal lattice forces twist and stack benzil molecules into homochiral helices, space groups ${P3_121}$ (right-handed)or ${P3_221}$ (left-handed)[H].

You obtain (racemic) benzil, modestly resolved (1 $\mathrm{S}$)-(-)-alpha pinene and (1 $\mathrm{R}$)-(+)-alpha pinene (both enantiomers), and mixed xylenes.  $20$ ml scintillation vial, foil-lined cap.  $100$ mg of benzil or so, $10$ ml resolved, alpha-pinene, heat to $100$ C. Drip in mixed xylenes until the molten benzil is just soluble, add another drop or two, very slowly cool undisturbed.  Only one space group crystallizes out.
Enantiomorphic space groups[I] are mathematically perfectly divergent geometric chiralities.  Here is the calculation of benzil crystal lattice's atomic mass distribution from Petitjean's software, CHI = $1$ (rapidly asymptotic) COR = $1$ DSI = $0$

If you want a discrete molecule as opposed to a crystal lattice, resolved camphor comes in around $\text{CHI} = 0.5$.  Resolved ${D_3}$-trishomocubane is a much better geometric example and is easily synthesized in large quantities starting with benzoquionone and cyclopentadiene (Diels-Alder).  If you have a methyl group sticking up from its ${D_3}$-axis, calculated $\text{CHI} = 0.996$.
Look at that methyl-bearing carbon!  It has no ${S_n}$ symmetries, for R must reflect or invert as S.  This chiral center's sense of chirality cannot be named.  Try doing it with CIP notation.

Uncle Al has been very naughty with geometric chirality, "[6.6]Chiralane: a remarkably symmetric chiral molecule," Symmetry: Culture and Science 19(4) 307-316 (2008).  Note that the core carbon atom has four rigorously identical substituents yet it is a maximally geometrically chiral center.  Chemical chiral nomenclature is fundamentally defective, for said chiral center (and the four to which it is attached) cannot be assigned a direction of handedness by any means, even in theory.
[A] http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
  pp. 11-13.
[B] J. Math. Phys. 40, 4587 (1999)
http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM
[C] http://www.mazepath.com/uncleal/qzdense2.png for alpha-quartz
http://www.mazepath.com/uncleal/glydense2.png for gamma-glycine
[D] http://www.mazepath.com/uncleal/norbors.gif
http://www.mazepath.com/uncleal/norone.png
[E] Appl. Cryst. 33, 126 (2000)
[F] J. Appl. Crystallogr. 19, 108 (1986)
[G] J. Org Chem. 38(10), 1870 (1973)
[H] Acta Cryst. B43 398 (1987)
[I] http://elib.mi.sanu.ac.rs/files/journals/publ/69/7.pdf Section 2: eleven pairs of enantiomorphic space groups;
http://www.math.ru.nl/~souvi/papers/acta03.pdf  Section 3ff.

If you value ease of acquisition more than structural simplicity, then chiral protein aminoacids are very cheap, easily obtained crystalline white powders, L-alanine being the simplest among them. Its structure is quite small, but its zwitterionic character could possibly introduce additional complexity depending on how you intend to use it. L-Malic acid is a slightly larger option that's also cheap and has no zwitterionic character. There are probably other examples of relatively small, chiral, biologically important compounds which are isolated as solids in large scale and have good enantiomeric purity.
For maximum structural simplicity (such as a molecule with a single carbon atom that is also chiral), you will need some rather polar substituents in the structure to make sure that such a small molecule is a solid at room temperature, or some very heavy atoms such as iodine and bromine.
Edit: After a bit more searching, lactic acid may be the best combination of simplicity and low cost for an enantiomeric molecule which is a solid at ambient conditions. An enantiomerically pure sample is a white crystalline powder that melts at around 50°C.

http://en.wikipedia.org/wiki/Mosher's_acid  Note that only one atom of 16 non-hydrogens is chiral.  For lactic acid, one heavy atom of 6 is chiral.  For D_3-trishomocubane, 8/11 skeletal atoms are chiral for having no S_n axes.  The molecule is absolutely rigid.  Its 8 chiral centers are  wonderfully equivalent by symmetry: apex (2) and tertiary (6).  One can argue the methylenes are also chiral, since their other two substituents cannot be split by a mirror plane.

Sucrose, common table sugar, is a pure enantiomer that exists as a powder.  Many drugs are manufactured as a single enantiomer, Lipitor would be an example.  Any chemical supply company will sell numerous pure enantiomers; d- and l-carvone would be a common example.

Sucrose and Lipitor are very complex molecules!!! I am looking for a simple molecules like CHClBrF.

Is an amino acid like d-alanine too complex for your purpose?


text:
An issue with Bouger-Lambert Law
I'm certain that I misunderstand this law. It states that:
$$I = I_o e^{-kb} $$
and after rearranging:
$$ A = ln{I_o}-lnI $$
where $A$ stands for absorbance.
1) Does is mean that when the intensity of the exiting light is very low than the absorbance is extremely high?
2) But from the graphs in my textbook, I've seen absorbance which was up to 1 or 1,2. Does it mean that most objects give little "resistance" to light?
3) My textbook says that molar absorption coefficient depicts the probability that photon will excite the energy states (not their words exactly :) ). Than why is f(c) = epsilon a constant function? Shouldn't more concentration increase the probability of change in energy states in some for example molecule?

1) Yes that is correct.  Generally, when the intensity of the light transmitted or reflected is low, then the absorbance is high.
2) It is hard to say just what "most" means, but if we are speaking of visible light (remember absorbance is wavelength dependent), there are a LOT of colorless materials that do not absorb in the visible region.
3) Yes, more molecules will increase the probability of light being absorbed by the solution, but as you noted, the absorption coefficient is a "molar" absorption coefficient.  Therefor when using the molar absorption coefficient, the absorbance is adjusted for the number of molecules in the solution.

Thanks for your answer! 2) I know that transmittance is expressed as I/Io, but in everyday language what is absorbed and what is transmitted should sum up to 100%, but this isn't true. So how should I think of these two terms and their relationship (logs are hard to imagine)? 3) so it is used only in the formula with concentration, right?

If the glass cell is clean and polished, then there should be little light scattered or reflected and what is absorbed and transmitted should be close to 100%. 3) Yes

Reflectance equations given angle and refractive indices across dielectric interfaces, http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/freseq.html  Each impedance mis-matched interface reflects a few % incident light.  http://www.thorlabs.com/images/TabImages/UVFSLong_780.gif  8% reflection losses.


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Determine standard enthalpy of formation of salt given other reactions
I'm a little confused by this question, perhaps my understanding of enthalpy isn't as good as it should be.

For the compound $\ce{KNO3}$, which of the following reactions is used to determine the standard enthalpy of formation, $\Delta H^\circ_\mathrm f$?
$$\ce{K(s) + N (g) + O3(g) -> KNO3 (s)\\
2K + N2 + 2O3 -> 2KNO3\\
2K + N2 +3O2          -> 2KNO3\\
K  + 1/2 N2 + 3/2 O2  -> KNO3}$$

I know that the $\Delta H^\circ_\mathrm f(\ce{KNO3})= \pu{−494.6 kJ/mol}$ from reading a table, I'm just confused about how do I get from that number to determine the formation of a compound.

The textbook is asking which of the 4 equations is used to determine the heat of formation (only one is correct). Since the standard state of oxygen is $O_2$, we know the first 2 equations are not used to calculate the heat of formation.

Since the standard enthalpy of formation refers to the formation of 1 mole of product, equation 4 should be correct, as it gives 1 mole $\ce{KNO3}$ as the product. Equation 3 is equation 4 multiplied by 2, which would give $2\Delta H$.

The standard enthalpy of formation $\Delta H_\mathrm f$ refers to the formation of 1 mole of product, with the educts and product being in their respective standard states. The fourth equation is therefore correct, as it gives 1 mole $\ce{KNO3}$ as the product. Equation 3 is equation 4 multiplied by 2, which would give $2\Delta H_\mathrm f$. The first two equations can be ruled out because they do not contain the educts nitrogen and/or oxygen in their standard states, which are $\ce{N2}$ and $\ce{O2}$, respectively.


text:
Thin Layer Chromatography & Stationary Phase

There are two types of stationary phases used in thin layer chromatography. One stationary phase is made of aluminium oxide. Which of the following molecules would be expected to have the smallest $R_f$ using aluminium oxide as the stationary phase on the glass slide?
(a) $\ce{CH3CH2CH2CH2CH2COOH}$
(b) $\ce{CH3CH2CH2CH2CH2OH}$
(c) $\ce{CH3CH2CH2CH2NH2}$
(d) $\ce{CH3CH2CH2CH2CH3}$

$\ce{AL2O3}$ is a non-polar stationary phase, therefore if the molecule is also non-polar it would display high adsorption  to the stationary phase, and therefore a smaller $R_f$ value.
The answer is (a).
Isn't (a) polar?
Wouldn't this increase the $R_f$ value, because it doesn't adsorb to the non-polar stationary phase? Can someone please explain, correct me?

A is the most polar of the group, and D is non-polar. Since the answer is A, then your statement "Al2O3 is a non-polar stationary phase" is in question. "In extreme situations, very polar substances on Aluminum Oxide do not migrate very far from the starting point" from http://infohost.nmt.edu/~jaltig/TLC.pdf

so AL2O3 is polar then?

Go to Figure 7.5 here: http://courses.chem.psu.edu/chem36/Experiments/PDF's_for_techniques/TLC.pdf

Aluminum Oxide is more polar than Silica Gel and polar compounds do not migrate far from the starting point.
Thin Layer Chromatographic Analyses
7. Thin-Layer Chromatography

"There are two types of stationary phases used in thin layer chromatography."  Sigma-Aldrich sells silica gel (unmodified, modified/bonded, chiral and high purity), aluminum oxide, cellulose (fibrous, microcrystalline), and polyamide 6 stationary phase TLC plates.  That barely scratches the surface.
Alumina is highly polar and highly basic unless modified.  Yellowed tetramethylenediamine run down a Dispopipette of dry Activity 0 basic alumina has everything bothersome ripped out of it within an inch or so.
The carboxylic acid will ionize and salt bond, given an Rf of about zero.


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Why does C=C bond replace C-O bond when two graphite platelets are rubbed against each other edge on edge?
When two small 2D carbon platelets are pushed into proximity, why do they release the edge attached molecule (maybe O) and then bond C to C?
I think the answer I seek would address the relative difference in bond strengths, and what forces are involved to remove attached molecules, probably Oxygen.   .....
BACKGROUND:   I experiment with electrostatic properties of conductive carbon films for oscillation and physical displacement applications. I use a mechanical 'tribological' method of creating 2D Carbon sheets.  In this case the method is: smearing powered graphite onto a smooth substrate, followed by agitation and polishing to a smooth highly conductive film.  The result is assumed to be hexagonal graphene film of several layers.


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How to model iso-surface of molecular orbitals of singlet and triplet excited states
I've modelled the iso-surface of the highest occupied molecular orbital of a molecule with DFT in Gaussian.If I were to model the iso-surface of the molecular orbital of the singlet and triplet excited states, how would I do that?Using tdDFT, I've already calculated the excitation energy of both the triplet and singlet manifold.

Are you looking for a solution using Gaussian?

Yes, I hope to find results with Gaussian.

On a research paper, I've read that the Highest Energy Singly Occupied MO (HE-SOMO) is equivalent to the first excited singlet state.  I was wondering why that is the case.

It is maybe the perfect case when you strictly want to distinguish orbitals and states, and clarify for yourself and for us what exactly you want to visualize.


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Why do only certain acids disassociate in water?
I understand that most acids are weak acids, i.e. they do not fully disassociate in solution. However, certain acids, such as $\ce{H2SO4}$, do. I have been unable to find out what the defining characteristic is that differentiates strong acids from weak ones.
For example, HCl is strong, yet both HF and HBr are weak (even though they're based on halogens as well). Shouldn't HF be even stronger?
What characteristic makes only a select few acids fully disassociate (and hence belong to the strong acid group)?

We had questions on hydrogen halides (HX) and the relevance of resonance structures for the basicity of actate here. You might find the answers given there helpful.

@Just_a_fool  HBr is not a weak acid, in fact it is stronger than HCl

There are many factors that determine the strength of an acid:

electronegativity of the non-metal atom: an electronegative atom can attract more easily  the electrons of the hydrogen an so generate the proton.
atomic radius greater the radius of the conjugate base greater the acidity, this because the electrons are dispersed around the atom and there is a less attraction between the positive proton and the negative anion.
stability of the conjugate base if the conjugate base is stable due to some particular electronic configuration (e.g. stability due the resonance structure) the species are more prone to be dissociated.


text:
Reaction Pathways: What to do when IRC geometries are relaxed (optimized) and the energy changes significantly
The Story:
Reaction pathways can be difficult things to deal with using electronic structure methods.  Intrinsic reaction coordinates (IRCs) can be determined a number of ways (a discussion of these methods lie way outside the scope of this post).  Suffice to say, the reaction pathway is an 'energy' pathway of reaction, connecting reactants and products, usually through some intermediate and/or transition state (TS).
For the purposes of this question, imagine implementing some sort of IRC technique which begins with a previously resolved transition state (Figure 1).  The computation follows the reaction coordinate in both the forward and reverse directions from the TS energetically downhill until it reaches a (for the sake of simplicity) a minimum energy point with features which satisfy the method used.  In general, the structure and energy of this final IRC geometry should be very representative of the structure which is connected to the TS via the pathway determined.  If you were to optimize this final geometry, you should get good agreement between the optimized structure and the IRC configuration.

Now consider this scenario.  An optimization was performed on the IRC resolved structure B and the energy relaxation was significant and the structure significantly changed, giving rise to a new (and lower) energy.  One could no longer (within the formalism of IRC methods) say that the optimized structure B lies on the reaction pathway, and therefore is not a representative configuration at point B on the reaction pathway associated with the TS.
The Question:
Does anyone know of an example in the literature where this type of behavior occurs?  How does one 'deal' with a situation like this?
Other Information:
Let us assume the following:

That this scenario occurs with a few IRC procedures (not just one).
That the IRC computations were given a ridiculously large amount of
cycles and steps
That various stepsizes were sampled in parallel IRC jobs with no
significant difference in the results

(I will add more as I think of them)
Potentially Helpful Imagery
Imagine two somewhat complex molecules coming together to form a single covalent bond.  You run an IRC from the TS (which describes the bond formation) in the reverse direction and the intermolecular separation begins to increase.  This separation continues to increase until the IRC job terminates properly.  Now imagine running an optimization on these two (slightly complex) molecules which are separated by maybe 4.0 Å.  Obviously these molecules are going to interact because they are not infinitely separated.  You observe the (slightly complex) molecules moving toward each other during the optimization procedure (rotating, twisting, gyrating along the way) until you now have a minimum energy structure which is much lower in energy than the resulting IRC computation.  These two molecules interact via intermolecular interactions (no covalent bonds are formed between the molecules).  This optimized structure is analogous to a reactant complex (as opposed to separated reactants) and has an energy a few kcal mol$^{-1}$ lower than the corresponding IRC geometry.  How on earth does this fit in to the reaction pathway?

I am not entirely familiar with the techniques that you are using but my hunch is that this is a matter of kinetics =/= thermodynamics. When running this IRC simulation, are you forcing the reaction to take place? In your last paragraph it sounds like you're taking a TS, separating the two molecules and then doing an energy minimization. The resulting complex now has a lower energy state than the covalently bonded version of the two molecules. Is this correct?

The cartoon I've included is not quantitative (my apologies).  An IRC simply follows the imaginary mode of vibration in the TS downhill along the corresponding coordinate (its not forced).  The resulting reactant complex may or may not be lower in energy than the covalently bound molecule (the products).

I guess this is a known problem with Angstroms (http://tex.stackexchange.com/questions/24255/angstrom-not-working), so I pasted it in.  Stellar question!

@LordStryker but the chart does qualitatively show that the associated complex has a lower free energy as compared to the covalently bound complex? I could certainly see that being a plausible result...

@CTKlein These are ELECTRONIC energies, not free energies.  The point of the graph is to illustrate that some geometry on the reaction pathway (in this case, not a TS) qualitatively changes (in structure and in energy) when the resulting IRC geometry is relaxed via an optimization procedure.

@LordStryker I'm curious: does the optimized TS have one (and only one) imaginary mode? And when the IRC completes are the vibrations for the products all real?

@jjgoings Yes, the TS can only have one imaginary mode of vibration (by definition).  WRT your second question... not necessarily.  The vibrations for a structure only mean something if the geometry is optimized (without constraint unless its symmetry related).  An IRC =/= an optimization.  However, one would EXPECT only a small change between an IRC geometry and an optimized geometry (common practice to do a check).  I've yet to see what happens if and when there is a LARGE change between the two.

An IRC calculation is a constraint optimisation, hence it freezes the reaction (transition) mode and optimizes the rest of the molecules. It should not be used to detect minima. However, if one would do that the program should terminate with the message that it reached a minimum (G09: PES minimum detected on this side of the pathway.). That structure should indeed be a stationary point on the PES. It is not necessarily a local minimum (only along the specified motion). Run frequencies on that point and check if it is a minimum. Could you edit-in your example?

@Martin Correct.  I should have put IRC=/= full optimization.  The problem could be (with respect to the IRC) that the coordinate that an IRC follows becomes flat on the PES and therefore the computation thinks its done, giving rise to a geometry that satisfies the IRC conditions but is not necessarily a minimum on the PES?

About your bounty, please tell us exactly how your IRC did terminate properly. (Does the program say it found a stationary point.) There are certainly many possibilities, that this could be achieved. What cut-off values are you using. Also how did you optimize your IRC geometry? Have you done a frequency calculation at the final point? Finally, you are describing a bond break - is it homolytic or heterolytic. Did you take that into account choosing the level of theory?

@Martin In this particular case, no, the program did not find a stationary point. As such, a harmonic vibrational frequency computation would yield meaningless results since a minimum is not guaranteed. Optimized geometries were performed at the same level of theory used in the IRC computation.  The bond breaking is heterolytic (can use restricted methods). Two levels of theory were used for all pathways (MP2 and B3LYP) with the same basis set.  IRCs used tight thresholds/cut-offs in general.

I do not really understand then. If the IRC did not find a stationary point, then you could be almost anywhere on the potential energy hypersurface. Have you tried giving it more steps until it finds a minimum along the transitional mode? I have found a paper about minimum energy paths, but I am now at home and do not have access, I am also not done reading it. Maybe tomorrow I can post the link here. This is a remarkable question - however we might not have the right audience here. Have you thought about asking the question to the ccl mailing list?

@Martin I have not gone to CCL.  The more specific I'm supposed to get, the more I want to delete this post.  I was hoping to find some sort of sufficient answer by being as vague as possible.

I have not had the time to read it, but it might help you anyway: http://www.mathematik.uni-leipzig.de/~quapp/JTCCpreprint.pdf DOI: 10.1142/S0219633603000604

IRC is a gradient following algorithm that tries to follow the imaginary mode. Check the algorithm (several one implemented, so really depends on which you are talking about) and you will see that unlike in optimisation algorithms, there is a big emphasis on following the real gradient (and not just jump to the endpoint as fast as possible).
It works well around the TS where the gradient is large and has a well define direction. However getting closer to the optimised geometry, the requirements of following the exact gradient and having an effective optimization algorithm are contradicting. In analytical sense it may be not obvious, but when you have numeric gradients, small second derivatives (these are almost always very soft modes), the direction of the gradient may have big errors and you should keep a small stepsize to check it thoroughly.
In summary, the situation you wrote down is more or less common at some extent in almost all the IRC calculations due to the numerical nature of the problem. Harder imaginary frequencies may help.

The IRC algorithm usually follows the gradient from TS to product state B and by definition has to go always downhill in energy. In best case, it treats the Potential Energy Surface (PES) around the given point as quadratic and just integrates Newtonian equation of motion and does it until it reaches local minimum.
This procedure is different from geometry optimization, which can use powerful methods to find energy minimum.
Static picture, T = 0K shows that the reaction will end in the "upper" B state. If you then ask for energy minimum, reorientation of any kind (intermolecular or conformational) takes place and you find "lower" B. Do not be decepted by your plot, the optimized B lies on different point of reaction coordinate (however you define it) than the IRC-type B.
Dynamic picture, T > 0K shows that IRC is not the only way to proceed from TS to product, although it is still important representative. Upon running molecular dynamics (MD) from TS, you will end up in the equilibrium state, probably well represented by "lower" B.
It would be quite difficult to find exactly this type of behavior in literature, because such technical topics are quite often swept under the carpet and only indirect hints remain. Similar topics are discussed in Levine: Molecular Reaction Dynamics. As a simple example take the E2 elimination from cyclohexane, where the leaving groups must be axial, so the reactive species is different from energy minimum.
And how to deal with it? It is up to you, where you set zero to your relative energy scale. It will depend on which energies you want to compare, if you propose some series of reactions.

It would be quite difficult to find exactly this type of behavior in literature, because such technical topics are quite often swept under the carpet and only indirect hints remain.  This is what I was afraid of.  Wordsmithing is as much an art as it is a skill when it comes to publishing.

On the other hand, it really IS a just a technical question, where to set zero and how to treat the pre-reaction complex. As an illustration how flexibly is the IRC used you can look at the X-axis labeling in the papers. This ranges from no labels to weird labels, while only occasionally you can see some units and values.

Well if someone is proposing a reaction mechanism and an optimized geometry is drastically different from the one on the IRC, how does one say that 'this structure' is representative of the one on the actual pathway?  Doesn't this sort of 'sink' the whole "We've mapped out the reaction pathway..." conclusion?

It is possible to map the PES around the reactant state and find transition states for the conformational (or similar) change. You often find, that the energies involved in this process are order of magnitude smaller than for the main reaction and treat it using steady-state approximation.

@LordStryker "Although the reaction coordinate' does not correspond to any actual trajectory on a given potential energy surface, it gives a path about which many actual trajectories, especially those of low energy, are likely to travel." IRC should be used to map/ approximate the immediate surrounding of a TS, therefore ensuring that your geometry optimization will end up in the most likely/ closest local minimum. And that is also enough, because all of the local minima/ TS are on the same PES.


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How to Find Each Element in Nature
I tried to google this but 'element' as a programming term complicated the search.
I'm looking for a site or article that covers how one would find each element in nature, in its most abundant form. For example, what's the best natural source/form of Sulfur? Beryllium? Tugsten? etc.
Any help is appreciated.

http://en.wikipedia.org/wiki/Timeline_of_chemical_element_discoveries

The Wikipedia page for each element is probably the easiest place to start with. If you want some significantly more in-depth data, the USGS publishes free-access mineral commodity summaries on many (but not all) elements, including information about their origin, abundance and applications. Other good sources are books, especially descriptive inorganic chemistry books, such as a few of John Emsley's publications including Nature's Building Blocks and The Elements, as well as many other college-level inorganic chemistry texts.

The book Chemistry of the Elements by N. N. Greenwood is an absolutely huge handy tome full of information, which is organised by element or clusters of similar elements.
You'll find this as a handy overall chemistry reference, but most common ores of the elements (where applicable) and geological extraction information are also given. An essential to have on any chemistry shelf.

.. or any good inorganic chemistry text book (but this is indeed a good book).

From experience, not all inorganic books include the information like this one does. This book is essentially an inorganic textbook and data book at the same time.

You should try this site. You will need to click on each element to get the information you need.

http://elements.vanderkrogt.net/index.php is in my opinion the best site about how have been finden and discovered the elements.

http://www.geokem.com/global-element-dist1.html#ge is more specific for your needs,more concise, less history more about actual source of the element.


text:
What is the energy change of ethene reacting with hydrogen gas?
$\ce{CH2CH2(g) + H2(g) -> CH3CH3(g)}$
Using this table, source: http://www.kentchemistry.com/images/links/Kinetics/TB08_003.GIF

C-C enthalpy = 348 kJ/mol
C-H enthalpy = 413 kJ/mol
H-H enthalpy = 436 kJ/mol

The way I see it, H-H has to break, C-C has to break, and then two C-H bonds have to form.
(Units omitted) Therefore, should the answer be: $413 \cdot 2 - 348 - 436 = 42 \, \text{kJ/mol}$
So then it has to be an endothermic reaction.  Is this correct, or am I wrong?

Your reasoning is correct, but your numbers are wrong.
You are not breaking a C-C bond, but changing a C=C bond to C-C. So you will also need the energy for C=C which is 614 kJ/mol.


text:
How to estimate the dielectric constant of a liquid?
Are there any approximate relationships between the (static) dielectric constant of a liquid and properties of the individual molecules?
For example, suppose I know the polarizability and dipole moment of a molecule, as well as it's density in the liquid phase. Can I estimate the dielectric constant of the liquid from these quantities (or other similar ones)?

Yes, such relations exist.
Notably there are two general theories for polar and non-polar fluids that connect the polarizability $P$ with the dielectric constant $\epsilon$ as reviewed in this article by Harvey and Lennon, International Journal of Thermophysics, Vol. 26, No. 1, January 2005.
The Clausius-Mosotti equation for non-polar liquids is:
$$P = \frac{\epsilon-1}{\epsilon +2}$$
Whereas an equation developed by Kirkwood for polar liquids is:
$$ P = \frac{(\epsilon-1)(2\epsilon+1)}{9 \epsilon}$$
The article then continues to explain that you can estimate the polarizability from theoretical relations involving the molar density $\rho$ and the dipole moment $\mu$, but in your case the two equations above are sufficient since you assume that you know $P$.