Diffusion-controlled reactions, activation-controlled reactions and rate equation I am confused as to where the rate equation can be applied and where not to be applied.
https://en.wikipedia.org/wiki/Rate_equation

In diffusion-controlled reactions from what I understand reaction rate is analogous to substrate concentration and the rate determining step is the supply of substrate.
https://en.wikipedia.org/wiki/Diffusion-controlled_reaction

But in activation-controlled reactions does the same apply?
Does reaction rate also depend on substrate concentration?

To add a little to what @geo has written. In all types of bimolecular reactions in the liquid phase, the reactants diffuse together; they have to, there is no other way for them to meet. The reaction is diffusion controlled if the activation barrier is so small that the first time the reactants meet, reaction is effectively instantaneous. This means that the barrier is only a few times average thermal energy (3RT/2) and that reaction occurs before the species can diffuse apart again.

In an activation controlled reaction the species repeatedly meet by diffusion but only occasionally$^*$ do they acquire enough energy from the surrounding solvent molecules that, at that instant, have enough energy to surmount the activation barrier.  Once they have reached  the barrier top the energy fluctuations that caused the barrier to be surmounted in the first place must cease after the transition state is crossed and products are then formed. (Recall that molecules in solution are continually jostling about and, from the Boltzmann (exponential) distribution, there is always a small chance that a molecule has, by random fluctuations, and for a short time, many times the average energy ).
$^*$By 'occasionally' I mean perhaps 1 in a million or even tens of millions of encounters.


Can orbitals in a molecule be occupied via excitation like they can in atomic orbitals? I am facing a conundrum in the way I am viewing molecule orbitals due to the existence of excited states. What I wanted to know is, since atomic orbitals can exist in excited states (i.e. you get hydrogen accessing states outside the 1s orbital and such when a high-enough energy photon is used), is this the case with molecular orbitals, where 'technically' many more states exist than shown on a molecular orbital diagram? I mean, they tend to only show the valence and core electrons, but not past that, hence the question. For example, could methane, that only whose MO diagram only shows molecular orbitals composed of 1s, 2s, and 2p orbitals, have an electronic transition that occupies a 3s orbital? And if so, how would that occur? Would it occupy one particular atom or be 'spread out' like most electrons in MO theory?
Sorry, long question, but something that is bugging me for some strange reason.

Most of what applies for atomic orbitals in atoms can be transferred to molecular orbitals in molecules. This includes that electrons can be excited to higher lying orbitals, creating excited configurations.

Also, just as for the hydrogen atom, there are in principle infinitely many orbitals available. They are simply truncated since the higher lying ones are less important (and for numeric reasons).


Before the CIP system was created, how did chemists ensure that the D/L system matched the R/S assignments? The ᴅ/ʟ system I have referred to is the one chemists used before the Cahn-Ingold-Prelog (R/S) system where they arbitrarily assigned the (+) and (-) isomers of glyceraldehyde,  'ᴅ' and 'ʟ' respectively and related other carbohydrates via reactions with retention of configuration. They were eventually found to match the (R) and (S) assignments (for the lowest asymmetric carbon in Fischer projections) of those molecules today (particularly biomolecules), and that is where my question arises.

Even though the molecules were related to glyceraldehyde through reactions that proceeded with retention of configuration, it is also known that the resulting molecules may not always retain the absolute configuration of the original molecules (which depends on priorities of the new groups). And since it was before CIP's (R) and (S) system came into place, there was no way to know if the molecules had changed absolute configuration during the reactions. So how did the chemists manage to retain the configuration of all those molecules with respect to the glyceraldehyde isomers at all?

Chemists actually still use D- and L- prefixes today, especially for carbohydrates where using it saves a lot of characters: Compare D-ribose to (2R,3R,4R)-2,3,4,5-tetrahydroxypentanal. The nomenclature is also in common use for amino acids, since all proteinogenic amino acids (except for glycine) in animals are L-configured (Bacteria, smart as they are, have many exceptions), while cysteine has the opposite absolute configuration to all other amino acids due to sulfur’s high priority. But your question revolves around history.

When Emil Fischer first devised the D- and L- nomenclature in 1891, not much was known about stereochemistry at all. It was known that carbon took four bonds and what the elemental composition of monosaccharides was. And methods were known to find out what a compound was, but much less sophisticated than nowadays. You would essentially derive it back to known products via known reactions and measure melting point (solids) or refractive index (liquids) for additional confirmation. Optical activity had been known since 1850, so that was probably performed in Fischer’s lab, too. At the same time, Pasteur performed his experiments, separating a racemic mixture of tartrate crystals into the enantiomers.

The choice Fischer made, when he assigned the D-configuration to (+)-glycerinaldehyde (the plus sign meaning that linear polarised light is rotated clockwise), coming from Latin dexter meaning right, was arbitrary. Not even essentially arbitrary, it was completely arbitrary. He probably knew he had a $50~\%$ chance of getting it wrong but in no way was he going to get proven wrong anytime soon. He decided it fitted nicely: the glycerinaldehyde that polarised to the right (clockwise) shall be the one with the asymmetric hydroxy group pointing to the right. He used that as a basis for all further sugars, knowing that all reactions he performed would take place on the aldehyde and leave the hydroxy group unchanged.

There was a fair bit of understanding how reactions worked even then. The first step to most sugar syntheses starting from glycerinaldehyde is the addition of a cyano group to give a cyanhydrine: Attack to the aldehyde, forming a new carbon-carbon bond and nothing else.

$$\ce{HOH2C-CHOH-CHO + N#C- + H+ -> HOH2C-CHOH-CHOH-C#N}$$

In any case, once a reaction like that was performed, derivatisation back to known compounds was done, and they would have realised if something had gone wrong with the hydroxy group. So we can be pretty sure they knew what wouldn’t influence it.

It wasn’t until half a century later and the X-ray crystal structure of glycerinaldehyde that one actually found out what the configuration of its carbon atom was. When it was proven that he had backed the lucky horse and the hydroxy group of D-glycerinaldehyde indeed pointed rightwards according to his convention, Fischer probably smiled a big smile from up in heaven. But if he had been wrong, he would have shrugged and thought ‘better luck next time.’

Now how does all this connect to the absolute configuration? Not at all! The absolute configuration is assigned on a per asymmetric atom basis. You need to know what the atom’s surroundings are like and can then apply a descriptor directly to it. Most notably, it was not the molecule as a whole that served as the reference but only the neighbouring groups and their priorities (read: atomic charge values). Any similarities to other stereodescriptor schemes, in use or not, was entirely accidental and unintended.

You have to remember that the retention of configuration in atomic terms does not automatically mean that the absolute configuration is retained. Take for example the Mosher acid as depicted below (image was taken from Wikipedia, where a full list of authors is available):

(R) and (S) Mosher acid.

Assigning the (R) configuration to the structure on the left is trivial. Assume you were to turn the (R) Mosher acid into the corresponding acid chloride. Note that although the asymmetric carbon was not touched the acid chloride turns out to be (S)-configured! So absolute configuration is something far more ‘wobbly’ than stereocentres are; make a remote change and influence an absolute configuration although that part of the molecule wasn’t touched.

Incidentally, that is the beauty of the Fischer nomenclature, especially when you consider the age in which it was invented. You don’t need to know which oxygen is pointing where you don’t need to figure out a long chain of priorities. You know you started at D-glycerinaldehyde so your sugar is now D-ribose. And you can take a pen and draw the lowest hydroxy group pointing to the right.

Do note make the mistake and try to correlate Fischer nomenclature with the sign of optical activity or with absolute configuration; all attempts will fail. Each has their use, their place to shine and none treats all similar molecules in the same way. Use wisely.


What does the circle inscribed in a benzene ring represent? I have seen benzene being depicted in two different ways:

For the right-hand structure, does the inscribed circle represent the movement of double bonds or electrons?

The circle, if anything, represents the inability of our everyday physical intuition to cope with the quantum phenomena. See, you would often encounter those two pictures with "double bonds this way" and "double bonds that way", intended to give a vague impression that the molecule switches quickly between the two, but that's not true.

It does not switch; it just stays there, in some sense "halfway between" these structures. There is no movement of double bonds, and in fact there are no double bonds. (Also, there is no movement of electrons either. They just sit there on their molecular orbitals, which by itself is an approximation, though a pretty decent one). All $\ce{C-C}$ bonds in benzene are the same; they are neither double nor ordinary. Read the Wikipedia page on aromaticity, or just look at this picture from it:

That's what the circle stands for.


Why is carbon monoxide not a greenhouse gas? Carbon monoxide is considered to be a toxic gas and, given that it is usually found along with $\ce{CO2}$, why it is not considered a greenhouse gas?

$\ce{CO}$ is not considered a primary (or significant) greenhouse gas due to the weak (but non-zero) absorption of energy in the infrared. However, it does increase global warming by reacting with certain chemical species in the atmosphere which in turn lead to an increase in concentration of primary greenhouse gasses, most notably methane and ozone (see final paragraphs cited below). Thus, it is not completely accurate to dismiss it as a contributing factor.
From the National Oceanic and Atmospheric Administration's page on $\ce{CO}$ as a greenhouse gas:

Carbon monoxide (CO) is not considered a direct greenhouse gas, mostly because it does not absorb terrestrial thermal IR energy strongly enough. However, CO is able to modulate the production of methane and tropospheric ozone.

Carbon monoxide does contribute to global warming indirectly, as noted by the American Chemical Society's page entitled "Which Gases Are Greenhouse Gasses?":

Several other gases can have indirect effects on atmospheric warming. This occurs when chemical reactions in the atmosphere produce or destroy greenhouse gases, including tropospheric ozone. Indirect effects also occur when a gas influences atmospheric lifetimes of other gases or affects atmospheric processes like cloud formation that alter Earth’s radiative energy balance by increasing Earth’s albedo. Gases that can cause these indirect effects include carbon monoxide ($\ce{CO}$), oxides of nitrogen ($\ce{NO_{\rm x}}$) and non-methane volatile organic compounds (NMVOC).

The above is elaborated in the text Global Climate Change Linkages: Acid Rain, Air Quality, and Stratospheric Ozone (James C. White, Editor. Elsevier (1989), p. 106):

As has been noted, ozone also initiates reactions that generate the highly chemically active oxidants, including hydroxyl. The lifetime of radiatively active methane in the atmosphere is determined by hydroxyl abundance. Carbon monoxide and methane are both sinks for hydroxyl, $\ce{CO}$ being the more reactive of the two. Greater concentrations of carbon monoxide in the atmosphere lead to lesser concentrations of $\ce{HO\cdot}$, which in turn lead to higher concentrations of methane.
Carbon monoxide and ozone are linked in the non-urban atmosphere that is low in hydrocarbons through the interaction with $\ce{HO\cdot}$, as described above, and $\ce{NO2}$. Hydroxyl is destroyed by $\ce{CO}$, increasing the ambient concentration of $\ce{NO2}$. The resulting increased proportion of $\ce{NO2}$ to $\ce{NO}$ leads to a higher steady state concentration of ozone. Carbon monoxide, a non-radiatively active species, enhances the greenhouse warming by raising the levels of methane and ozone.

In summary: $\ce{CO}$ is not a direct greenhouse gas due to low absorption in the infrared but is an indirect greenhouse gas because it increases the concentration of direct greenhouse gases (methane and ozone most prominently) by reacting with atmospheric $\ce{HO\cdot}$, which otherwise would decrease the concentration of those gasses. The toxicity part is not germane to the greenhouse gas issue.


What is the difference between an analytical standard and a certified reference material? I am analysing the analytical standard market, but every time I try to understand what an analytical standard is, I only get the definition of a certified reference material. Is there a difference?

An analytical standard is a compound of suitable purity and known concentration to be used as a calibration standard for an assay.

A certified reference material is a material which has been certified by some trusted organisation to be of a consistent quality and composition. Normally, but not always, the concentrations of the major components of a CRM have been carefully measured. For example, the US-NIST produce hundreds of reference materials for everything from limestone to baking chocolate to dried human lung tissue. Labs analysing these materials use the reference material as a quality control to check that they are consistent with other labs.


Does glucose react with Brady's reagent? A couple of reactions of glucose have been a source of great confusion to me. One of them is it's reaction with the Brady's reagent. It's pretty debatable whether the result is positive or negative. Could someone please shed some conclusive light onto this?

Furthermore, Glucose itself successfully produces an oxime with hydroxylamine. In contrast, it's pentaacetate fails to do so. Why does this occur? Why can't the pentaacetate, like glucose, "break out" of it's cyclic form to give a positive hydroxylamine test?

Edit:
I found out about the oxime test, I had overlooked the fact that glucose pentaacetate doesn't have any hemiacetal. The Brady's test though, is still open for debate.

From Wikipedia:

  The reaction of $\ce{NH2OH}$ with an aldehyde or ketone produces an oxime.
  $$\ce {R2C=O + NH2OH·HCl + NaOH → R2C=NOH + NaCl + H2O}$$

For the penta-acetate ester of glucose not to react with hydroxylamine would indicate that the ester is more stable, or there is steric hindrance to the reaction. Since hydroxylamine is heat sensitive, I am guessing that you did not heat the reaction. Given time, you should see the oxime form.

From Wikipedia:

  A positive test is signaled by a yellow, orange or red precipitate (known as a dinitrophenylhydrazone).
  $$\ce {RR'C=O + C6H3(NO2)2NHNH2 → C6H3(NO2)2NHNCRR' + H2O}$$

2,4-DNP also reacts with an aldehyde (or ketone), so both glucose and the penta-acetate ester of glucose should react. If you get a different color for the penta-acetate ester, it is due to the substituents on the glucose.


Is this a possible structural isomer for Butene(C4H8) Today in school, my teacher was explaining the different types of alkenes. Examples included ethene, propene, and butene.  She also showed us their structural formulas.
I tried to draw butene's formula and came up with this:

Is it correct?

Yes, the formula you drew is an example of isobutylene.  Isobutylene has the "official" IUPAC name of 2-methylpropene, and is an important industrial chemical.  Because it:

has a formula of $\ce{C4H8}$, and thus contains four carbon atoms;
contains a carbon-carbon double bond;
and doesn't contain any other functional groups

... it is indeed an isomer of butene (sometimes known as butylene).  You got this one right!$%edit$


Multi-step organic synthesis using benzene I have this project that I've been struggling with. I'm already given the reactant and product, so basically what I have to do is to figure out the steps needed to get the product.

I figured that some of the steps would include the Friedel-Crafts acylation and nitration, but after that, I'm not too sure what other steps I'd have to take. Any help will be very much appreciated!

@Waylander has provided a concise route to the nitro compound in question. I am also in accord with his comments regarding the OP's effort and his assessment that benzene, as a starting material, is not a choice by a practicing chemist. Now that he has opened the door to a solution, I offer a more traditional approach.
Classical nitration of ketoacid 7 should be reasonably selective in that the introduction of the nitro group in 8 is both para and meta directed. Keto acid 7 arises from indene 6 via exhaustive oxidation. To access the indene, I would start with benzaldehyde (2), whose preparations from benzene (1) are numerous. The Perkin condensation 2 $\rightarrow$ 3 is known. Hydrogenation of 3 forms carboxylic acid 4, which can be cyclized to indanone 5 with polyphosphoric acid. Friedel-Crafts techniques may also be employed. Reduction of this ketone followed by dehydration leads to the critical indene 6.


How are clathrate hydrates formed in the lab? How do chemists create clathrate hydrates in the lab (for example, those of $\ce{CO2}$, $\ce{O2}$, $\ce{H2}$ or $\ce{CH4}$)? I'm assuming that they just place a pressurized container of water and the gas in question into a freezer (or liquid nitrogen). However, I'm probably somewhat incorrect and don't have the resources currently to perform such an experiment.

I don't know about some of the clathrate hydrates you mentioned, but I've recently been reading some papers about both bromine and chlorine clathrate hydrates, and I imagine the procedure would be very similar for the systems you've mentioned.

I'll summarize from Spectroscopic Signatures of Halogens in Clathrate Hydrate Cages. 1. Bromine.

Basically, they're procedure is as follows. Get some really pure water and really pure bromine solution. Make a solution of bromine and water and place it in a quartz cell (cuvette). Place dry ice above the solution and the solution will begin to crystallize on the quartz cell wall. After it is clear that the solution has condensed/frozen onto the the side of the wall, place the cell into around $3^\circ C$ water. This is because at this temperature, both water and bromine will be liquid, but a bromine clathrate hydrate will be solid.

Thus, one can isolate the crystal by simply melting off the water and bromine which are not in a clathrate hydrate cage-type structure.

I believe this same sort of procedure could easily be applied to many of the other guest molecules you mention, however, because in general clathrate hydrates are going to have a higher melting point than ice. For instance, carbon dioxide clathrate melts at about $10^\circ C$ according to wikipedia.

I would suspect that methane clathrates and other clathrates with more weakly interacting guests would melt at a temperature very close to ice, so the procedure might be more complicated or sensitive. Indeed, that paper I cited above make some bromine-doped clathrates of dichloromethane, which is fairly insoluble in water, so a slightly different procedure is chosen.

One thing I find is that you can make these clathrate hydrates at around the melting point of ice, but then people frequently cool the clathrates down to very low temperatures to conserve the structures and keep the gases trapped in the cage, not necessarily because the this is needed in the formation of the clathrates.

After reading a bit more, I believe that a nearly identical procedure can be used for weakly interacting gases like methane, but the process is done under high pressure of the gas desired inside the cage which obviously shifts around the temperatures but doesn't change the idea presented above.


Energy level of molecular orbitals in heteronuclear diatomic compounds I have been studying the molecular orbital theorem. I read that the energy levels for the $\ce{p}$ orbitals in the first series of $\ce{p}$ block molecules are something to the tune of $$\pi_x=\pi_y<\sigma_z<\pi_x^*=\pi_y^*<\sigma_z^*$$

except for $\ce{O2 and F2}$, which have
$$\sigma_z<\pi_x=\pi_y<\pi_x^*=\pi_y^*<\sigma_z^*$$

But it got me thinking: what order would the orbitals take in a compound such as $\ce{CO, NO}$?

(Also, out of curiosity, what about $\ce{p-p}$ bonds in higher series of the $\ce{p}$ block? Does $\ce{Cl2}$ have the same order as $\ce{O2}$?)

My guess was that since in heteronuclear orbitals more electronegative elements have lower energy orbitals, compounded with the fact that the $\sigma$ orbitals it forms tend to be lower energy, it should follow the latter’s order.

This question requires idea of the Linear Combination of Atomic Orbitals (LCAO). It involves the following equation:
$\psi_n = \sum\limits_{i} c_{ni}\phi_{i} = c_{n1}\phi_{1} + c_{n2}\phi_{2} + ... c_{ni}\phi_{i}$
Here $\psi_n$ represents the wavefunction for the resulting molecular orbital, $\phi_{i}$ represents the atomic orbitals that contribute to the bond, and $c_{ni}$ represents the weight coefficient that can basically tell us how much of each atomic orbital is involved in the molecular orbital. If we look at the molecular orbital diagram for carbon monoxide, you are right, the bonding pi orbitals are lower in energy than the bonding sigma orbital:

The reason for this is due to the contributions from carbon. This comes from the relative energies of the atomic orbitals. From photoelectron spectroscopy, the energies of the carbon 2p orbitals and 2s can interact with the oxygen 2p orbitals because they are close enough in energy while the 2s oxygen orbital is a little too far to interact with the 2p orbitals of carbon. See image below

The important thing to note here is that the 2s orbital can interact as well. This means when we solve for the wave function using LCAO and the equation above, carbon can give a large contribution because two of its atomic orbitals interact, and it turned out it did. The same thing can be done for NO (which in fact is also similar to CO where the pi bonding orders are lower energy than the sigma bonding ones) and other molecules. However, it gets extremely complicated with larger and larger molecules.
I got this information from this public PowerPoint  https://www.chem.uci.edu/~lawm/10-9.pdf which come from lectures from UCIrvine which were released: https://www.youtube.com/watch?v=1CH6CeORL_g&list=PLaLOVNqqD-2GDVq2xFUuAltjIXPgQNNTL&index=7&ab_channel=UCIOpen


Solubility curve for uranium hexafluoride (in any solvent) I am doing a project and need a solubility curve or at least solubility data for uranium hexafluoride in any substance. It is impossible for water because $\ce{UF6}$ reacts violently with $\ce{H2O}$.

Is there any resource I could use for free to find this information?

So far I have found that it is soluble in:

Bromine Pentafluoride
Chlorotrifluoroethylene Liquid Polymer, Dichlorotetrafluoroethane, Perfluorotributylamine, Perfluoroisopropyl ether, 1,2-Dichlorohexafluoropropane, hydrogen fluoride and more.
chloroform,  carbon  tetrachloride  and  fluorocarbon  solvents;
liquid chlorine and bromine; dissolves in nitrobenzene to form a dark red solu-
tion that fumes in air

I think the resource given by Martin (2) is the best. Resource 1 gives a solubility curve, but resource 3 doesn't. However, I think there may be data floating around on compounds mentioned by 3.


How can I search molecules by constituent atoms Given a set of atoms, I would like to find (some / the most common) molecules that include only (some/all) atoms of the set. Is there any online search engine that can do this?

  E.g.
  Atoms={C,H}
  -->
  Molecules={CH$_4$, C$_2$H$_6$, ...}

Note:

I am looking for known molecules.
It is OK if I can only find chemical formulae, so without all possible configurations.

You can use ChemSpider's full search:

http://www.chemspider.com/FullSearch.aspx

and click on the 'Elements' section. Alternatively, there is a direct link here:

http://www.chemspider.com/ElementsSearch.aspx

This gives you two periodic tables that you can select elements from for your search. I think the upper one is 'allowed' and the lower one is 'forbidden', but I haven't tested this.

Of course, as pointed out in the comments, the set {C, H} will give you most of the database! However, for rarer elements this can be a better filter.


Why BH3 its classified as a "Molecular Hydride" and AlH3 its a "Intermediary Hydride" First, I understand that, on $\ce{BH3}$, they eletronegativity difference between $\ce{B}$ and $\ce{H}$ is very small. Which mean that the $\ce{B-H}$ bond is less polarized.
But, I couldn't find a relation of this, to the fact that he is a Molecular hydride.
And, secondly, I couldn't understand why $\ce{AlH3}$ its "intermediary", and not molecular, as well...

Alane is not a molecule: https://en.wikipedia.org/wiki/Aluminium_hydride You can't isolate molecules easily because they polymerize.

Borane on the other hand is known to exist as a dimer: https://en.wikipedia.org/wiki/Diborane

The difference in bonding is likely due to "hypervalency" (or similar concept with different terminology) in aluminum.


Chemical consideration of biodegradeability We hear the term 'biodegradeability' being thrown around alot in science, but rarely ever is the science itself explained.

What are the chemical properties in a substance that make it biodegradeable?

Furthermore, what exactly are the chemical processes involved in biodegradation?

Biodegradability is not a chemical property, much like a price for which you can buy a compound is not its chemical property. Both are the properties caused by some external circumstances, which are not necessarily related to chemistry.

As for the processes, there is hardly anything they have in common. Life is pretty versatile and can consume a wide range of chemical compounds. Wide as it is, though, this range does not constitute all chemistry. There are a great deal of compounds that just never happened to appear in nature before humans made them, so no life form had the chance to develop the ability to consume them. Those are not biodegradable.


Does resonance apply to NO₂⁺? My textbook said $\ce{NO2+}$ and $\ce{NO2-}$ both have resonance structures, now I am a little bit confused since I cannot figure out the resonance structure for $\ce{NO2+}$.

What is it?

In the following below, please interpret formal charges as marked below the atom.
$$\ce{O=N=O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{O\equiv N-O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{O-N \equiv O} $$
$$\ \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \ \ +\ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\ \ +\ \ + $$

The triply charged species are valid resonance forms, but very likely are small contributors.


Is there any graph that depicts the deviation of real gases from Charle's law? I am looking for a graph that shows the basic deviation of a gas from the Charle's law. The graph should be for different values of constant pressures.This would help me evaluate how gases follow ideal behavior for low pressures and high temperatures.

Check out the van der Waals equation wikipedia page for a start ... it gives an explanation of one of the most common modifications of the ideal gas law that allows for modeling of deviations from ideality in the way you describe.  This approach is quite nice in my opinion, because it includes two coefficients (a and b) that account for specific deviations for a given gas ... a scales with the strength of intermolecular interactions, and b accounts for the actual volume of the gas molecules.

You should be able to use that information to do what you want to do regarding Charles Law.  The graph on the page shows isotherms, but you should be able to generate the isobar curves that you are after with a little bit of work.


How is an aqueous Ag+ the same as [Ag(H2O)2]+? $\ce{Ag^{+}_{(aq)}}$ can be written as $\ce{[Ag(H2O)2]+}$ but how would you know that it bonds with 2 water molecules?

I don't get when ligands randomly form 6, 4, 2 bonds.

There is no way to absolutely predict the co-ordination number in any complex entity, without knowing say for example, magnetic or geometric properties of the complex. All we can do is give a probable configuration based on intelligent guesses. The co-ordination number depends on the characetristics of the ligands and the central ion.

Silver ion, $\ce{Ag+}$ usually forms complexes with Co-ordination Number two. This is partly because the silver ion is small sized, making it difficult to accommodate higher co-ordination numbers around itself owing to increased ligand-ligand repulsion, and partly due to the low charge on silver ion which makes it difficult to hold more than two ligands together. Therefore, in most of its complexes, silver ion co-ordinates with two ligands to form a linear entity. But to confirm this, you need to study solvated silver ions spectroscopically, magnetically or geometrically (diffraction), without which the above arguments can only give you, at best, a good guess.

Other examples include $\ce{[Ag(NH3)_2]+}$, and the polymeric $\ce{[Ag(en)]ClO4}$, both of which have co-ordination number of the silver ion 2.


If copper has started to corrode, can it stop? I've recently obtained a couple of electronics units which have suffered corrosion on their copper inductor coils. They were operated in conditions of intense humidity/heat and were placed near the sea. This has resulted in the complete failure in one of these units, where the inductor has come apart due to being eaten away by corrosion, and the other coil shows noticeable amounts of corrosion (patina present).

If I were to operate the latter device in a more forgiving environment (ie: putting a filter in front of it in an air conditioned room) will the copper continue to corrode regardless or will the oxide layer stop spreading further?

The previous answers doesn't take into account the environment:

  They were operated in conditions of intense humidity/heat and were placed near the sea.

Near the sea means that there are certainly a lot of $\ce{Cl-}$ ions in your metal. Chloride ions are a very dangerous species for metals conservation they tends to depassivate the metal forming complex ions. In the case of copper occurs this series of reactions:
$$\ce{Cu^{+} + Cl^{-} <=> CuCl}$$

$$\ce{Cu^{+} + 2Cl^{-} <=> CuCl2- + Cl- <=> CuCl3^{2-} + Cl- <=>CuCl4^{3-} }$$

That lead to different copper corrosion product. (e.g. atacamite $\ce{Cu2(OH)3Cl}$, malachite $\ce{CuCO3.Cu(OH)2}$ (Did you see some blue-green pigment?), Cupric hydroxide $\ce{Cu(OH)2}$.

These are all dynamic equilibria so your copper with dry-wet, hot-cold cycle will continuously object of degradation until chloride ions are present.

Use distilled water with cotton to clean the salts deposit, dry quickly where you clean.

Reference:

The Open Corrosion Journal, 2009, 2, 71-76
Corrosion Science 46 (2004) 109–135


Is the rate of a zero order reaction always independent of the concentration of its reactants? Consider the following reaction :
$\ce{A + B}$ $\rightarrow$ $\ce{Products}$
where order of the reaction w.r.t. $\ce{A}$ is $-1$ and order of the reaction w.r.t. $\ce{B}$ is $+1$. So, overall order of the reaction $=(-1)+(1) = 0$. So, it is a zero order reaction. My teachers told me that in a zero order reaction, the rate of the reaction is independant of concentration of the reactants. Below I have derived the integrated rate law for the above reaction. You can clearly see that in the integrated rate law there are terms of concentration of $\ce{A}$ and
$\ce{B}$. Obviously, the integrated rate law is telling us that the rate of the reaction depends on the concentration of the reactants. Then why does everyone say that in a zero order reaction the rate of the reaction is independant of concentration of the reactants. Why?

When you look at the overall reaction, it is if you have an equal concentration of each reactant, then by the reaction progressing or adding equal amounts of both, the reaction rate won't change. For example, if you have 2M of both A and B, $[A]^{-1}[B]^1 = 1/2*2=1$, and increasing the concentration to 3M for both won't change the rate since $1/3*3=1$. The differential rate law applies to only one single species, so unless you have equimolar amounts of each reactant or make an approximation using an excess of one, you cannot apply the differential rate law.


Why does lavender oil dissolve plastic? A few weeks ago I decided to harvest the lavender in my garden and put it through a still to extract the essential oils. The still in question is essentially just an air cooled condenser.

I collected the first fraction in a clear plastic vial. Within a day the vial had become cloudy looking (like frosted glass - the lavender oil itself was not cloudy - just the container). After a week this had happened:

The whole top of the container has been eaten away and is now flexible where before it was rigid.

What's caused this, is it expected behaviour?

Is there anything I should consider before using the lavender oil in fragrances and soaps?

Based on appearance and extent of deformation, your bottle is likely made of PVC plastic which is not compatible with oils. Plastic bottles are made from blow molding. This process leaves residual stress in the polymer chains of the materials but  creates a smooth and transparent surface. Lavender oil can diffuse into PVC and make it softer. Once it is softer, the polymer chains can relax some of the residual stress which makes the surface rough and as a result appear cloudy due to increased scattering. Given Long enough more lavender oil can penetrate the plastic and soften it to the point that the plastic is allowed to flow which causes the observed deformation of the top.

Now I cannot say with certainty it was PVC that your bottle was made from, but if it was, then plasticizer would have seeped into the oil which isn't great for people. If you want to use your oil for air freshener or candles that is fine but I would not recommend it for skin contact applications.


Comparing size of ions lithium, gallium ions How is the size of $\ce{Li+}$ ($\pu{0.68}$) larger than $\ce{Ga^3+}$ ($0.62$)?

This is very difficult for me to understand  why they say that size of Li+ with 3 protons and just 2 electrons should be considerably more than that of $\ce{Ga^3+}$. I understand the poor shielding effect of $\ce{3d^{10}}$ electrons but it isn't obvious to me why it will make its size smaller than $\ce{Li+}$ which has just one shell with 2 electrons and more effective nuclear charge. Where is my logic wrong?

That is impressive.  What you may need to bear in mind is that as the nuclear charge increases, all the electron orbitals move in closer to the nucleus.  You may have been assuming, consciously or unconsciously, that the electron shells appear at a relatively fixed distance from the nucleus -- that is, that the n=1 shell is always at about the same distance from the nucleus, the n=2 shell is always at about the same additional distance, and so on.

Not the case!  For example, if you take a look at the quantum mechanical model of the hydrogen atom and solve it for a "hydrogen-like" atom with atomic number Z and one electron (so, He+, Li++, Be+3,....,U+92 ha ha) then you will find that the average orbital radius of the electron scales like 1/Z.  So if we ignored electron-electron repulsion entirely, the radius of each shell would decrease like 1/Z as Z increases.  That's quite a contraction.  It would mean that the n=1 shell in carbon, say, would be 1/6 the size of what it is in hydrogen.  In the case of Ga, Z=31, the n=1 shell would be 1/31 the size it is in H!  Each new shell would contract similarly.

So in the absence of electron-electron repulsion we would expect the size of atoms to shrink drastically as Z increases, going across the Periodic Table, and bumping up slightly at the beginning of each row with the addition of a new shell.  Of course, real atoms do have electron-electron repulsion, lots of it, so the effect is considerably muted.

Nevertheless, the size of atoms does shrink considerably going from left to right in the Periodic Table, and it's worth remembering this is because all of the electron shells shrink, not just the valence shell.


How do zeolites catalyse the cracking of hydrocarbons? Been learning about catalytic cracking of long chain hydrocarbons into shorter chain hydrocarbons, in which the catalyst zeolite is used so that the temperatures needed for the cracking are much lower (about $500\,^\circ\text{C}$ instead of $700\text{-}1000\,^\circ\text{C}$). The question is, how exactly does the zeolite help with the cracking process? Everywhere I look doesn't give a detailed answer, just that zeolite is porous, riddled with holes that can trap molecules. The only clue I have to the process I suppose is this video

https://www.youtube.com/watch?v=nTse5_wXup4

which shows an animation of some hydrocarbons whacking against a tunnel-like zeolite into smaller chain carbons. But to me the video can't be that accurate, because the whole thing seems out of proportion, the hydrocarbons are made from hydrogen and carbon which are pretty small atoms, whereas the zeolite is made from aluminium or silicon which are much larger. Even if this is the actual way it works, why not just have a plane surface to hit the hydrocarbons off to break them then? (Perhaps the rate of collision is much smaller? I'm just pulling at straws here.) Any light shed will be helpful, thanks.

The video you reference to does not tell the whole story.  Yes, zeolites are minerals with open pores, similar to a filter, occurring in nature and are produced commercially, too. Equally to their structure, their composition is interesting -- as both are used to tailor the reactivity and selectivity of this material.  ZSM-5 is one example of designed remarkably high Brønsted acidity.  Hydrocarbons in contact with these sites split into fragments and rearrange because of the intermediate generation of carbocations.  It is advantageous, as these reactions are heterogeneously catalysed by these "solid state acids".


Dipole moment in NF3 versus NH3 The usual explanation for the molecular dipole moment of $\ce{NF3}$ being smaller than that of $\ce{NH3}$, despite the $\ce{N-F}$ dipole being stronger than the $\ce{N-H}$ dipole, is that influence of the lone electron pair on nitrogen is to oppose the net $\ce{N-F3}$ dipole, while enhancing that of $\ce{N-H3}$. See, for example, this good example.

What bothers me about this explanation is that when the $\ce{N-F}$ dipole is calculated (experimentally, I think), the effect of the electron pair on $\ce{N}$ is already factored in. Likewise for the calculation for $\ce{N-H}$.

Putting the question in terms of the language of the linked article:

  [T]he lone pair's negative concentration will augment the polarity contribution from the polar bonds while for the $\ce{NF3}$, the lone pair's dipole would subtract from the dipole contributed by the NF bonds.

If the lone pair on $\ce{N}$ subtracts from the $\ce{N-F}$ bond in the molecule it must have had the same effect in the original diatomic dipole calculation. Likewise for $\ce{N-H}$.

If this is true, assuming identical $\ce{N-F}$ and $\ce{N-H}$ bond angles in $\ce{NF3}$ and $\ce{NH3}$, it seems that the vector calculation should still give a stronger dipole moment for $\ce{NF3}$. [Please regard this as a question!]

One thing that would lessen the molecular $\ce{NF3}$ dipole moment is a smaller $\ce{NF}$ bond angle (compared to $\ce{NH}$), which increases the lateral component of the dipole at the expense of the axial, making the (axial) resultant smaller. I think fluorine is larger than hydrogen and maybe this would tend to push the fluorine atoms out laterally, decreasing the resultant axial dipole. However I do not see this as an explanation in texts so I am guessing it is wrong.

My confusion is that the vector algebra doesn't seem to support the argument but there is probably something I am not taking into consideration.

You should probably be cautious in general applying the concept of bond dipoles to predict molecular dipole moments, and it does seem ad hoc if not downright flimflam to invoke a dipole to the nitrogen lone pair to predict dipole moments in the examples you present (particularly in the case of nitrogen trifluoride).

Now if you read the wikipedia page on nitrogen trifluoride, it explains that

  $\ce{NF3}$ is slightly soluble in water without undergoing chemical reaction. It is nonbasic with a low dipole moment of $\pu{0.2340 D}$. By contrast, ammonia is basic and highly polar ($\pu{1.47 D}$).[9] This difference arises from the fluorine atoms acting as electron withdrawing groups, attracting essentially all of the lone pair electrons on the nitrogen atom. $\ce{NF3}$ is a potent yet sluggish oxidizer.

The low basicity of $\ce{NF3}$ supports the argument that electron density associated with unpaired electrons on N is withdrawn by the fluorine atoms. So it would appear that in the case of $\ce{NF3}$ there might not be a significant dipole moment between the $\ce{N}$ nucleus and an electron pair (or the charge opposite the three fluorines). Instead the electron density (relative to ammonia as a reference) is polarized toward the fluorines. This makes the application of a method of linear summation of bond dipoles somewhat sketchy.

Continuing along similar sketchy lines, the website to which you provided a link argues that the difference in molecular dipoles cannot be explained simply in terms of the order of electronegativity ($\ce{F > N > H}$). But note that electronegativities are computed from dissociation energies, electron affinities, etc, and not from partial charges or dipole moments, and the relation between charges or dipole moments and energies is bound to be highly non-linear, making quantitative prediction of dipole moments based on electronegativities difficult.

What seems necessary is to perform some ab initio electronic structure calculations.


Most stable conformation of aldehydes and ketones Why is the most favorable conformation of ketones and aldehydes the one where the alkyl group is anti to the other alkyl group (or hydrogen atom in an aldehyde) as attached? Why isn't more stable the conformation where the bond between the alpha and beta carbon atoms can participate in hyperconjugation with the carbonyl pi antibonding MO?

By contrast, the slightly more favorable conformation for butene is the right one:

I found in Ian Fleming's Molecular orbitals and Organic Chemical Reactions, page 93-94 the answer (the aldehyde is propanal):

  The polarisation of the H-C bonds in the methyl group leaves a weak
  positive charge on the outside and this is attracted to the partial
  negative charge on the oxygen atom, lowering the energy of this
  conformation

(the difference in energy of the two eclipsed conformations of propanal is 8 kJ/mol).


Synthesis of trans alkene with organoborane Trans alkenes can be synthesized with organoboranes. The following sequence is represented in Smith's Organic Synthesis

How does the mechanism look like after the first step (hydroboration)?

I think this might be a alkyl shift that results in an $\mathrm{S}_{N}2$ type substitution at an $sp^{2}$ center. Cool. I've only ever seen $\mathrm{S}_{N}2$ type substitution at an $sp^{2}$ center for vinyl iodides. Unfortunately, I don't have my teaching materials on me at work, so I can't find the reference for this for you.

EDIT:

Note the "inversion" at the $sp^{2}$ center.

Greg Fu had this great example of a substitution on a vinyl iodide, like I mentioned above. I can't seem to find this reference, but I just emailed him, so I'll update when he writes me back.


Why don't trigonal S and P compounds undergo inversion at room temperature? Most molecules containing nitrogen atoms in trigonal pyramid configuration undergo a relatively fast process of inversion at room temperature. On the other hand, the free energy barrier for phosphines, sulfoniums and sulfoxides are high enough that they are optically stable: the rate of racemization is slow at room temperature.

I wonder what effect is responsible for this difference in behaviour (the larger energy barrier). I expect that it has to do with the size of the central atom (N being one row higher than P and S), but is that all there is to it? And how does size impact the free energy barrier for the inversion: energetically or entropically (or both, of course)?

Azane all by itself has an inversion barrier of about $6\ \mathrm{kcal/mol}$, which is  low compared to phosphine's approximately $30\ \mathrm{kcal/mol}$ barrier. Pyramidal nitrogen inversion requires a planar transition state. One can use a classical theory to model the thermal rate and we could write $$k \propto \mathrm e^{-E_\mathrm a/(RT)} $$ and using models like these we would predict that ammonia at $300\ \mathrm K$ would have a rate on the order of $10^8\ \mathrm{s^{-1}}$, instead what is observed is something on the order of $10^{10}\ \mathrm{s^{-1}}$

Assuming the potential energy surface has two minima equal in energy and a higher energy TS, two routes are possible: a thermal process going over the maxima or quantum-mechanical tunneling. Determining a tunneling frequency requires writing down vibrational wave functions, finding their linear combination and overlap. If we were to do all this for a particular vibration, we'd see that the tunneling frequency decreases exponentially with a dependence on increasing $\mu$, thickness of the barrier/shape, and the barrier height. As an example, $\ce{ND3}$ has approximately an order of magnitude slower rate of inversion than $\ce{NH3}$ at a sufficiently low temperature – this has been attributed to a decrease in its tunneling frequency.

But why the difference in the energy barriers and corresponding rates depending on the atomic position in the periodic table? Perhaps the easiest/quickest explanation rests on considering what the TS requires for interconversion. Ideally the classical conversion requires passing through a trigonal planar structure with bond angles on the order of 120 degrees. If you consider the series of azane, phosphane, arsane then you'll see bond angles of 108, 94, 92 degrees respectively. In the absence of other effects, the barrier goes from smaller to larger. Indeed, asane-like molecules were the first of this series resolved, followed by phosphane-like. To the best of my knowledge the fastest have yet to be resolved, because rates on the order of $10^{-5}\ \mathrm{s^{-1}}$ are required at RT for a decent shot at isolating them. Naturally, this isn't the whole story, as I've previously alluded to above. Let me know if you would like to know more.


Are all fatty acids monocarboxylic acids? The answer to this question might solve a multiple choice in my previous chemistry exam, which caused an argument with my teacher.

We know that fatty acids form fats and oil due to their carboxylic groups by forming triesters with glycerol. But is it possible to have a fatty acid with two or more $\ce{-COOH}$ groups to exist naturally? If so, is the property "fats and oils are neutral" just a general property, or does it have an exception?

You are asking two separate questions.

Not all fatty acids are typically found in fats and oils (common name for storage lipids, which is also somewhat a common name...)  Identifying fatty acids that have an additional charged tail does not inherently mean you'll find them in your fats and oils - fatty acids do have other roles. There may be exceptions, but they're not going to be common.

From what information you have given us, the larger question your teacher is driving you towards is to think about what the physical nature of fats and oils must be given their biophysical properties and how that correlates with their biological role, and secondly what the physical nature of fatty acids would be that drives those properties.

Considered macroscopically, your storage lipids will remain neutral, and pack together. If you added charge to your storage lipids in any significant amounts, you would turn them into something that behaves like a membrane lipid.


Why was Lawrencium's symbol changed? The initial proposal by Lawrence Berkeley National Laboratory scientists for naming element 103 was "Lawrencium" with the symbol "Lw", However, while IUPAC accepted the name "Lawrencium" for the element, it changed the symbol to "Lr".

Was there a stated (or unstated) rationale for this alteration? Was it in anyway connected with the competing claims from the Joint Institute for Nuclear Research in Dubna regarding who discovered the element first?

Ghiorso, Sikkeland, Larsh, and Latimer of the Lawrence Berkeley Laboratory reported the discovery of element 103 in 1961. It is correct that the discoverers suggested the name lawrencium, symbol Lw, for the new element in honor of E. O. Lawrence, inventor of the cyclotron and founder of the Lawrence Berkeley Laboratory. The original proposal can be found in Ghiorso, A.; Sikkeland, T., Larsh, A. E.; Latimer, R. M. New Element, Lawrencium, Atomic Number 103. Phys. Rev. Lett. 1961, 6 (9), 473–475:

  In honor of the late Ernest O. Lawrence, we respectfully suggest that the new element be named lawrencium with the symbol Lw.

At the XXII Conference of the IUPAC in London 1963, the name was accepted but the symbol was changed to Lr. According to Comptes rendus / XXII Conference, London 5th to 9th July, 1963, the reason was simply the difficulty with the letter “w”:

  Following the observation by H. Remy that “w” is an uncommon letter in many languages and difficult to pronounce it was proposed to alter the spelling to Laurentium. The Chairman emphasized that the Commission had no right to modify the spelling of the name, but that it could be pronounced according to the custom of the language. To make the name more acceptable it was decided to recommend the chemical symbol Lr and not Lw for element 103.


Calculating rate constant and order of a multi-step reaction If I’m given a multi-step reaction, say:

$$\ce{2 NO + O2 -> 2 NO2} \label{rxn:R1}\tag{R1}$$

with the following mechanism:

$$
\begin{align}
\ce{NO + O2 &<=>[$K$] NO3} &\quad &\text{(fast)} \label{rxn:R1.1}\tag{R1.1}\\
\ce{NO3 + NO &->[$K_1$] NO2 + NO2} &\quad &\text{(slow)} \, , \label{rxn:R1.2}\tag{R1.2}
\end{align}
$$

then will the rate constant of the whole reaction be equal to the product of the equilibrium constant of the fast reaction and the rate constant of the slower reaction?

Because this is how they made us solve it at my institute:

$$K_\mathrm{eq} = \frac{[\ce{NO3}]}{[\ce{NO}][\ce{O2}]}$$

$$
\begin{align}
\text{Rate of slow step \eqref{rxn:R1.2}} &= K[\ce{NO3}][\ce{NO}] \\
&= KK_\mathrm{eq}[\ce{NO}][\ce{O2}][\ce{NO}] \\
\Rightarrow\text{Rate of reaction} &= K'[\ce{NO}]^2[\ce{O2}]
\end{align}
$$

where $K' = KK_\mathrm{eq}$, and the order is $2 + 1 = 3$.

So, is $K'$ the rate constant for the overall reaction? And is the reaction a 3rd order reaction?

The method you've indicated in your solution is correct. Regarding:

  Then will the rate constant of the whole reaction be equal to the product of the equilibrium constant of the fast reaction and the rate constant of the slower reaction?

In your question, the final $K'=K_1\cdot K_{eq}$, but this is not always true.

Remember that you've to adjust the final rate law expression in such a way such that it is free from all intermediates. In doing so, you might need to multiply the $K$ of another reaction, as you did above. But you may also need to divide or multiply the square/cube/etc. of another reaction.

Hence, $K_{final}$ is not always equal to $K_1\cdot K_2\cdot...\cdot K_n$. So, your assumption is incorrect.


Determination of solubility equilibrium using galvanic cell reactions This is the problem:

We can determine the solubility equilibrium for silver bromide using cell:

Ag (s) | AgNO3 (aq) || KBr (aq) | AgBr (s) | Ag (s)

We know that:

$\text{AgBr} + \text{e}^- \rightarrow \text{Ag} + \text{Br}^- \ \ \ \text{E}^0=0.095 \text{V}$

$\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}\ \ \ \text{E}^0=0.799 \text{V}$

First of all isn't that cell diagram incorrect? As far as I understand it is the silver reducing from silver nitrate to metallic silver thus being cathode. And vice versa for the silver bromide that is being produced more as silver is oxidating. Shouldn't the cell diagram be like this instead:

KBr (aq) | AgBr (s) | Ag (s) || Ag (s) | AgNO3 (aq)

Anyways, if we consider the Nernst equation:

$$E = E_0-\frac{RT}{nF}\ln{k}$$

and assume that the galvanic cell is in equilibrium when there is no voltage (i.e. $E=0$). We can then write:

$$\ln{k}=\frac{E^0nF}{RT} \\ \leftrightarrow k=e^{\frac{E^0nF}{RT}}$$

Then plug in the numbers:

$$\ln{k}=\frac{(-0.095+0.799)\ \text{V}\cdot 1 \ \text{mol}\cdot 96485.31 \ \frac{\text{C}}{\text{mol}}}{8.31451\ \frac{\text{J}}{\text{mol}\cdot \text{K}}\cdot 293.15 \ \text{K}} \\ \leftrightarrow
k= 1.26754\cdot 10^{12}$$

But the right answer is:

$$1.26 \cdot 10^{-12}\ \text{mol}^2/\text{dm}^6$$

Any ideas what am I doing wrong?

The leftmost and rightmost parts of the galvanic cell notation are supposed to be electrodes, not electrolytes (no KBr (aq) and the likes of it there). What is more, the double vertical line represents the interface where the junction potential has been eliminated ((c) Atkins), i.e. salt bridge, which in this case is the interface between electrolytes. The left hand side of the cell is anode, where the oxidation occurs, and the silver ions are reduced at the cathode, which is written on the right hand side. It seems to me that the correct notation is:

Ag (s) | AgBr (s) | KBr (aq) || AgNO3 (aq) | Ag (s)

In turn, the cell reaction is $\ce{Ag+ + Br- -> AgBr}$, and the EMF of this reaction is +0.704 V. It is not surprising that the equilibrium constant of this reaction is much larger than unity, otherwise silver bromide wouldn't be so stable and so insoluble. What you're missing is that the solubility product, which you want to calculate, is an equilibrium constant of the inverse process - not the formation, but the dissolution of $\ce{AgBr}$. So take the inverse of the equilibrium constant which you obtained from the galvanic cell EMF, and you'll get $7.9\cdot10^{-13}$, which is pretty close to the reference values.


What is the oxidation state of bromine in BrO3? I came across this compound $\ce{BrO3}$ in my textbook, and was trying to find the oxidation state of bromine.

This is how I proceeded:

Since this is not a superoxide or a peroxide, or any other case where the oxidation state of oxygen is not −2 according to this, so the oxidation state of oxygen should be −2.
For oxidation state of Br as $x$, I can have $x + 3(-2) = 0$, which gives $x = +6$.

But, according to Wikipedia, the +6 oxidation state of Br is not listed.

I was confused because my book (NCERT Part I for class XII) says that "The bromine oxides Br2O, BrO2, BrO3 are the least stable halogen oxides (middle row anomaly) and exist only at low temperatures."

Please tell me where I am going wrong.

P. S. This is not an ion $(\ce{BrO3-})$ but a neutral compound.

Bretherick’s Handbook of Reactive Chemical Hazards lists bromine trioxide $\ce{BrO3}$ structure as the one composed of cationic and anionic parts [1, p. 116]:

The solid produced at $\pu{—5 °C}$ by interaction of bromine and ozone is only stable at $\pu{—80 °C}$ or in presence of ozone, and decomposition may be violently explosive in presence of trace impurities [...]. The structure may be the dimeric bromyl perbromate, analogous to $\ce{Cl2O6}$ [2].

so that formula written as $\ce{BrO3}$ is fine being a formula unit, but more correctly bromine trioxide should be written as $\ce{[BrO2]+[BrO4]-}$, implying that bromine is in two oxidation states: $+5$ and $+7$.
References

Bretherick, L.; Urben, P. G.; Pitt, M. J. Bretherick’s Handbook of Reactive Chemical Hazards, 7th ed.; Elsevier: Amsterdam, The Netherlands ; Boston, MA, 2007. ISBN 978-0-12-373945-2.
Pflugmacher, A.; Rabben, H.-J.; Dahmen, H. Zur Kenntnis der Bromoxyde. Über das Bromtrioxyd. Zeitschrift für anorganische und allgemeine Chemie 1955, 279 (5–6), 313–320. https://doi.org/10.1002/zaac.19552790507.


CH4 + O2 => CO2 + 2H2, what's wrong with it? It says in my textbook that:

$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$$

which seems like a displacement reaction to me. Shouldn't the reaction be:

$$\ce{CH4 + O2 -> CO2 + 2 H2}$$

since $\ce{4H}$ are displaced by $\ce{O2}$ (becoming $\ce{2 H2}$), which makes the reaction balanced?

Why is the reaction $\ce{CH4 + O2 -> CO2 + H2O}$, which is further balanced as my textbook says?

The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Then not only methane is burnt, but also the arised hydrogen.

So in "first" step:

$\ce{CH_4 + O_2 -> CO_2 + 2H_2}$

but then the hydrogen will be also combusted in reaction :

$\ce{2H_2 + O_2 -> 2H_2O}$

so the overall reaction will be:

$\ce{CH_4 + 2O_2 -> CO_2 + 2H_2O}$

As was pointed out in the others answers the real mechanism of a methane combustion at a low oxygen pressure is more complicated:

49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida, http://enu.kz/repository/2011/AIAA-2011-94.pdf

another:

Intermediate Species Proﬁles in Low-Pressure Methane/Oxygen Flames Inhibited by 2-H Heptaﬂuoropropane: Comparison of Experimental Data with Kinetic Modeling, http://www.nist.gov/el/fire_research/upload/Williams-Intermediate-Species-profiles-in-Low-pressure-Methaneoxygen-Flames-inhibited-by-2-heptafluoropropane.pdf


Which one of the following is optically active? To be optically active the presence of chiral centres are very important but if there exists a plane centre or axis of symmetry then the molecules are optically inactive.  How to determine the optical activity when more than one ring are present?  Is it sufficient to check those two mentioned condition?

How to determine the optical activity when more than one ring are present? Is it sufficient to check those two mentioned condition?

As Ivan Neretin accurately noted above:

the rules for determining optical activity of a molecule are the same for all molecules, whether it has one ring, several rings, or no rings at all.
axes of symmetry are irrelevant in determining the optical activity of a molecule. We only have to consider center and plane of symmetry.

That said, the given molecules have several chiral centers, which is a great sign that the molecule may be optically active.

If you will try to find the plane of symmetry, you won't find any. While the rings may appear planar on the paper, they (even the fused ones) actually exist as conformations in 3D.

Moreover, the center of symmetry is a point that requires identical groups equidistant from itself; a quick glance over both the molecules shows there are no such centers.


Grouping of symmetry classes in chemistry character tables In the C3v point group, it lists three sigma Vs together in one column (class), and it doesn't list the  cVs separately. Consequently there are three irreducible representations.

In the C2v point group, however, it lists 2 separate cVs in separate columns. Consequently there are four irreducible representations.

My question is, why are the cVs grouped together in C3v, but not grouped together in C2v?

The $\sigma_\mathrm v$ mirror planes in the $C_\mathrm{3v}$ point group are themselves related by symmetry: note that they can be interchanged via rotation by 120 degrees about the preexisting $C_3$ axis. To be technically precise, they belong to the same conjugacy class, in the sense that applying $C_3$, then one mirror plane, and then the inverse of $C_3$ is the same as applying a different mirror plane:

$$C_3\sigma_{\mathrm v, i}C_3^{-1} = \sigma_{\mathrm v, j}$$

As a result of this, they have to be treated together and not separately, in the sense that the characters in the character table represent how each irrep transforms under all the three mirror planes collectively. Unfortunately, this is hard to explain properly without delving into the actual matrices. But it should be possible to see that the two mirror planes in $C_\mathrm{2v}$ are not related to each other in the same way.


What does the format nF10.8 mean? I was trying to assign weights to each state in a CASSCF PES scan using nroot=2 in Gaussian. I specified 0.500000 0.500000 at the end of .com file. But it caused the program to assign the weight for the first calculation only. All the rest points of the scan had weights 0 and 0, which I don't want to have. I referred to the Gaussian manual which states

  StateAverage: Used to specify a state-averaged CASSCF calculation. All states up to NRoot are averaged. This option requires the weighting of the various states to be input in format $n$F10.8 (no trailing blank line).

But I don't understand what is this nF10.8 format means and if it could help me solve the problem to assign the weight in each step of the scan. I tried writing 0.500000 0.500000 one after the another for say 50 points and I could see the required weight has been assigned. Can someone clarify the meaning of the nF10.8 format? I searched on internet but I couldn't find any helpful material.

It is a Fortran format specifier.

It means

$n$: repeat the input $n$ times, once for each weight
F: input is a floating-point number
10.8: the number is of total width 10 (including the decimal point), and has 8 digits to the right of the decimal point

This means the number cannot have more than one digit to the left of the decimal point, which makes sense since weights should all be $< 1.0$.


Reason for negative Joule Thomson coefficient of Helium and Hydrogen at NTP conditions Recently, while reading my textbook I came to know that Helium, Hydrogen and Neon are the only gases which have negative Joule Thomson coefficient at NTP conditions, i.e heating effect is observed when these gases expand from a region of high pressure to a region of low pressure. I was wandering about possible explanations for this strange behaviour (since almost all gases display cooling effect during free expansion at NTP conditions) but nothing came to my mind. Can anyone give me a reason why these gases behave differently at room temperature and 1 atm pressure?

One thing I noticed is that all the three gases in question have compressibility factor(Z) always greater than 1, i.e. negligible intermolecular forces of attractions, but how can this explain their abnormal behaviour?

Note: You can skip section I, and go straight to section II and/or the end of section III (specifically the conclusions subsection), if you are already familiar with the basic mathematical machinery/definitions

I. Preliminaries

Feel free to skip all of this if you are familiar with it

First, we have an adiabatic expansion, i.e $\mathrm{q} = 0 $.

Second, our process is carried out as follows: a gas at pressure $p_i$ flows into a region of lower pressure $p_f$ without significant change in kinetic energy. This is called the Joule–Thomson expansion.

Obviously, this process is inherently irreversible. Additionally, one can show that this expansion isenthalpic (i.e $H_i = H_f$). Unlike a free expansion, work is done, causing a change in internal energy; the exact specifics of this change will depend on the initial and final conditions and also on the nature of the working fluid.

What we measure experimentally is a change in temperature with respect to pressure at constant $H$, and we call it $\mu_{\mathrm{JT}}$ (Joule-Thomson Coefficient).

$$ \mu_{\mathrm{JT}} = \left(\frac{\partial T}{\partial P} \right)_H = \frac{V(T \alpha-1)}{C_p} \tag{1}$$

where $\alpha$ is the coefficient of thermal expansion $$\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p$$

All real gases have an inversion point at which the value of  $\mu _{\mathrm  {JT}}$ changes sign. The Joule–Thomson inversion temperature, depends on the pressure of the gas before expansion.

$$\begin{array}{cc}
\hline
\text{Temperature} & \text{sign of}\ \partial P & \text{sign of}\ \mu _{\mathrm  {JT}} & \text{thus sign of}\ \partial T \\ \hline
\text{below  inversion temp.} & - & + & - \\
\text{above inversion } & - & - & + \\
 \hline
\end{array}$$

The final pressure is always lower than the initial pressure and thus the change is always negative. Depending on whether you are above or below the inversion temperature, your change in temperature can either be positive (warming) or negative (cooling). I would like to stress that the inversion temperature depends both on the temperature and pressure.

II. Qualitative Analysis

The arguments given below might seem a little "hand-wavy"/crude, but bear with me.

$$ H = U + pV$$
$$ \Delta H = \Delta U + \Delta (pV)$$

During a Joule-Thomson expansion (isenthalpic, adiabatic), the change in $pV$ represents the work done by the fluid.

If $PV$ increases,at constant enthalpy, then U must decrease as a result of the fluid doing work on its surroundings; the working material is using its own internal energy to do work. This produces a decrease in temperature and results in a positive Joule-Thomson coefficient.

Decrease in$pV$ means that work is done on the fluid and the internal energy increases. This can thus lead to an increase in temperature and the Joule-Thomson coefficient will be negative.

For an ideal gas, $pV$ does not change during a Joule-Thomson expansion.

As a result, there is no change in internal energy; also, there no change in temperature; and the joule-thomson coefficient is zero.

In real gases, $pV$ does change. The deviation from ideality (in general) is contained in $Z$, the compressibility factor. Thus, it would make sense if we could make some connection to it.

The ratio of the value of "real" $pV$ to that expected for an ideal gas at the same temperature is called the compressibility factor,

$$Z := \frac{pV}{nRT}$$

For a gas, this is typically less than unity at low temperature (attractive forces dominant) and greater than unity at high temperature.

At low pressure, the value of $Z$ always moves towards unity as a gas expands (tend towards ideality).

In general at low temperature, $Z$ and $PV$ will increase as the gas expands, resulting in a positive Joule-Thomson coefficient.

At high temperature, $Z$ and $pV$ decrease as the gas expands; if the decrease is large enough, the Joule-Thomson coefficient will be negative.

It follows, that for the $Z>1$, Joule-Thomson coefficient will be negative and the gas gets warmer. This is the case for the gases in question at the given conditions.

A graph from wikipedia showing how the Joule-Thomson Coefficient changes sign (for nitrogen):

Note: The question originally asked about gases, but I'll make a small point about condensed phases and supercritical fluids (since you can see them in the graph). At high pressures, these phases can't respond to an increase in pressure by reducing their volume, and $pV$ increases as pressure increases. The molecules are being forced together, and such the Joule-Thomson coefficient is negative.

Of course intermolecular forces, attractive and repulsive, and the limits in which they are dominant are key in explaining the changes of $Z$, and consequently of $\mu_{\mathrm{JT}}$. This point is further developed in the next section.

III. A slightly more quantitative, but ultimately Qualitative Analysis

To see how intermolecular interactions come into play, let's look at two model systems: an ideal gas (no intermolecular interactions) and a Van der Waals gas (both attractive and repulsive interactions).

Using equation (1) and the definition of $\alpha$ we can write

$$ \mu_{\mathrm{JT}} = \frac{V(T \alpha-1)}{C_p} = \frac{T \left(\left(\frac{\partial V}{\partial T}\right)_p -V\right)}{C_p} \tag{2}$$

For a perfect gast $\left(\frac{\partial V}{\partial T}\right)_p = \frac{V}{T}$ and so, $ \mu_{\mathrm{JT}} = 0 $

Since the Van der Waals gas equation is cubic in $V$, I will use the first few terms of the virial expansion and solve for $V$ to obtain $\left(\frac{\partial V}{\partial T}\right)_p$. I am omitting explicit computations, you may do them as an exercise.

$$Z = 1 + \frac{1}{RT} \left(b- \frac{a}{RT} \right)p + .... \tag{3}$$

Since $Z = \frac{p\tilde{V}}{RT}$ where $\tilde{V}$ is the molar volume of the gas; I am doing this for convenience's sake, I don't want to lug around a $n$ all over the place.

$$\tilde{V} = \frac{RT}{p} +  \left(b- \frac{a}{RT} \right) + .... \tag{4}$$

After differentiating we obtain

$$\left(\frac{\partial V}{\partial T}\right)_p = \frac{R}{p} +  \left(- \frac{a}{RT^2} \right) + .... \tag{5} $$

Substituting (4) and (5) into (2), and dropping all of the higher order terms.

$$ \mu_{\mathrm{JT}} \approx \frac{\frac{2a}{RT}-b}{C_p}$$

Now, consider the low temperature limit of this expression. The $\frac{2a}{RT}$ becomes large, and $b$ is negligibly small compared to it so

$$ \mu_{\mathrm{JT}} \approx \frac{2a}{RTC_p}$$

All the quantities on the right hand side of this equation are positive. so $\mu$ is positive. The attractive forces are dominant, since only the paramter $a$ appears in our expression.

Similarly, the high temperature limit is

$$ \mu_{\mathrm{JT}} \approx \frac{-b}{C_p}$$

Here, $\mu$ is clearly negative. The repulsive forces are dominant, since only the parameter $b$ appears in our expression.

Conclusions

The existence of an inversion temperature becomes obvious in a gas with both attractive and repulsive forces, since $ \mu_{\mathrm{JT}}$ changes sign from positive to negative.
The connection to attractive and repulsive forces also becomes more explicit.
In the low temperature limit, attractive forces are dominant and the gas needs to break free from these interactions as it expands. This requires energy, which has to be supplied by $U$ the internal energy of the gas. Thus the gas cools.
At high temperatures, the repulsions are dominant. As the gas expands, the over potential is lowered, and energy is released. Thus, the gas warms up.
Obviously, the exact magnitude of the high and low temperature limit depends on the nature of working fluid, since the parameters $a$ and $b$ are different for each gas. The gases you are talking about, have barely an attractive interactions.

Caveats

Our analysis based on the Van der Waals gas seems to suggest that the inversion temperature and $\mu_\mathrm{JT}$ is a function determined by temperature alone. This has experimentally proven to be false. This discrepancy arises because we dropped the higher order terms in our expansion which contained the pressure dependance.
This model is ultimately qualitative. It will give you the wrong numbers when you try to compute something, but it explains the general trends well.


What is the partial molar volume of epsomite (magnesium sulfate heptahydrate)? I am trying to calculate the expected volume of an aqueous solution of epsomite and NaCl (308 deg K but willing to go with 298 K as close enough, and 1 atmosphere pressure).  I realize the two may interact, but as a first approximation I'm looking for the partial molar volumes of both. I found NaCl's in a paper (Archer (1992) J Phys Chem Ref Data 21(4): 793-829), but I can't find the values for epsomite.

I have a known quantity of salts and am trying to figure out how much solution I'll end up with after adding enough water to get a particular overall specific gravity. For the curious, I've got 600 pounds of epsomite, 50 pounds of NaCl, and I'm shooting for a specific gravity of 1.25.  Knowing the amount of water to add is straightforward and, unfortunately, not sufficient for this particular situation. I really need to calculate the total final volume after the salts have dissolved.

Thank you.

As you know, partial molar volume of given substance is the change in solution volume after one mole of this substance is added. When you add $\ce{MgSO4.7H2O}$ to water it dissolves to magnesium cation and sulfate anion but the hydration water simply mixes with the rest of water. So, I suppose that partial molar volume of the magnesium sulfate heptahydrate is the sum of partial molar volume of magnesium sulfate and seven times molar volume of water. If $\widehat{V}_i$ denotes partial molar volume of substance $i$, and $V_i$ denotes molar volume, then: $$\widehat{V}_{\ce{MgSO4.7H2O}}=\widehat{V}_{\ce{MgSO4}}+7V_{\ce{H2O}}.$$
Nevertheless, partial molar volume is function of composition and commonly quoted values are for very low concentration (adding substance to large quantity of water). If you want to obtain solution with higher concentration, finding its exact volume requires knowledge about $\widehat{V}_i$ as a function of composition and such function has to be integrated.


Interpreting equilibrium equations correctly I've watched over this Khan video titled "Reactions in Equilibrium" and am still confused on the interpretation of equilibrium. Reactions in Equilirbium

$\ce{N2(g) + 3H2(g) <=>2NH3(g)}$

Whenever I see an equilibrium like the one above, the amounts of each molecule are only what happens at the start of the reaction, correct? In the video, it was made out as if the starting coefficients next to each molecule were what was needed to reach equilibrium. However, each reaction at a different temperature has a different equilibrium constant. With that said, the amount of each molecule will change which also changes the concentration so the amount of $\ce{NH3}$ would not necessarily be 2. He does this in other videos too. He starts off with something such as:

$\ce{aA + bB <=> cC + dD}$

and immediately plugs those into an equilibrium equation. I guess what he is getting at is that we figure out the final concentrations necessary to reach equilibrium but I'm not positive.

Towards the end of the video, arbitrary (I guess) concentration amounts of each molecule were plugged into the formula to output the matching equilibrium constant. He raised each concentration to the power of the starting coefficient for that molecule. If you start out with $\ce{3 H2}$ and you need a wildly different concentration to reach equilibrium, why are we raising the concentration to the power of the starting amount of $\ce{H2}$?

Finally, let's say the concentrations I have for a particular reaction were not the concentrations at equilibrium. If I still plugged them into the equation, that would tell me not only if I were at equilibrium (if I looked up the constant) but also if the forward or backward reactions were favored right?

However, each reaction at a different temperature has a different equilibrium constant. With that said, the amount of each molecule will change which also changes the concentration so the amount of NH3 would not necessarily be 2.

This is true. I have not watched the video, but if he is suggesting that the stoichiometric coefficients define the ratios of the equilibrium concentrations of the compounds, then he is wrong. Stoichiometric coefficients do play an important role in equilibrium expressions, but not that one.

  He raised each concentration to the power of the starting coefficient for that molecule. If you start out with $\ce{3H2}$ and you need a wildly different concentration to reach equilibrium, why are we raising the concentration to the power of the starting amount of $\ce{H2}$?

This is how the equilibrium constant (and more broadly the reaction quotient) is defined.

Take your generic reaction:

$$\ce{aA + bB <=> cC + dD}$$

The reaction quotient $Q$ is defined as the product of the concentrations of the products divided by the product of the concentrations of the reactants. Since $\ce{a}$ equivalents of $\ce{A}$ are in the reactants, $[\ce{A}]$ is multiplied $\ce{a}$ times for $[\ce{A}]^{\ce{a}}$.

$$Q_c = \frac{\prod\lbrace[\text{Products}]_i\rbrace}{\prod\lbrace[\text{Reactants}]_i\rbrace} = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}} $$

The pressure version of $Q$ replaces concentrations with partial pressures $P_i$.
$$Q_p = \frac{\prod\lbrace (P_{\text{products}})_i\rbrace}{\prod\lbrace( P_{\text{reactants}})_i\rbrace} = \frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$

The equilibrium constant ($K_c$ for concentrations and $K_p$ for pressures) is defined as the reaction quotient at equilibrium concentrations/pressures. At equilibrium:

$$Q_c = K_c = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}}$$

$$Q_p = K_p =\frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$

The values of $K_c$ and $K_p$ are determined by experimentally measuring the concentrations or pressures of $\ce{A}$, $\ce{B}$, $\ce{C}$, and $\ce{D}$ and plugging them into the expressions above.

For the specific reaction of nitrogen and hydrogen to form ammonia, the equilibrium constant (at a given temperature) is determined by measuring the equilibrium concentrations/pressures and plugging them into the expressions below.

$$\ce{3H2(g) + 2N2(g) <=> 2NH3 (g)}$$

$$K_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2}$$

$$K_p = \frac{P_{\ce{NH3}}^2}{P_{\ce{H2}}^3 P_{\ce{N2}}^2}$$

Once we know the value of $K_c$ or $K_p$, we can then use it and some partial concentration/pressure data to predict the equilibrium concentrations/pressures.

At 300 K, the $K_c$ of this process (the Haber Reaction) is $4.34\times10^{-3} \text{ M}^{-3}$. If we started with $[\ce{H2}]=3.0 \text{ M}$, $[\ce{N2}]=2.0 \text{ M}$, and $[\ce{NH3}]=2.0 \text{ M}$, we can calculate the equilibrium concentrations of all three species when the dust has settled. First we calculate $Q_c$ and compare to $K_c$. If $Q_c$ is greater than $K_c$, there is too much product, and the reaction will shift toward reactants. If $Q_c$ is less than $K_c$, there is too much reactant, and the reaction will shift toward products.

$$Q_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2} = \frac{(2.0 \text{ M})^2}{(3.0 \text{ M})^3 (2.0 \text{ M})^2}= \frac{1}{27 \text{ M}^3}=3.70\times10^{-2} \text{ M}^{-3}$$

$$ 3.70\times10^{-2} > 4.34\times10^{-3} \space \therefore \space Q_c > K_c  $$

At this combination, the reaction shifts toward reactants. $[\ce{H2}]$ and $[\ce{N2}]$ increase, and $[\ce{NH3}]$ decreases, all by a factor of $x$ multiplied by the stoichiometric coefficients of each species. The new concentrations are $[\ce{H2}]=3.0+3x \text{ M}$, $[\ce{N2}]=2.0+2x \text{ M}$, and $[\ce{NH3}]=2.0-2x \text{ M}$. The equilibrium expression is:

$$K_c 4.34\times 10^{-3} = \frac{(2-2x)^2}{(3+3x)^3 (2+2x)^2}$$

Since I chose the starting concentrations to be those values implied by the stoichiometry, I ended up (coincidentally) with a reaction near equilibrium. Thus, I cannot make assumptions about the value of $x$ relative to the concentrations, so no simplification is possible. To avoid mistakes, I will let WolframAlpha handle the algebra and solve for $x=3.18\times 10^{-1}$. The equilibrium concentrations are thus:

$$[\ce{H2}]_{eq} = 3+3(3.18\times 10^{-1})=3.95 \text{ M}$$
$$[\ce{N2}]_{eq}= 2+2(3.18\times 10^{-1})=2.64 \text{ M}$$
$$[\ce{NH3}]_{eq} = 2-2(3.18\times 10^{-1})=1.36 \text{ M}$$

Note that if we had started at different initial concentrations, we would end with different equilibrium concentrations. The maths are easier if you start with zero of one or two species.

I could also go on about the other way to determine the equilibrium constant (i.e. from the relationship $\Delta G^o=-RT\ln{K}$), but I have gone on long enough.


What do short-range and long-range corrections mean in DFT methods? Currently I am looking for the most accurate calculation method for a simple non-conjugate molecule consist of C, N, O and H. Normally I would try each method for a similar and known molecule then compare them with the experimental results but I don't have anything to compare this time.

So I need to predict a method for ground state and excited state calculations. While I was searching I kept coming across the terms "long-range" and "short-range". In the original publication of CAM-B3LYP for example, the authors claim that CAM-B3LYP provides additional long-range corrections, the weak part of B3LYP. What does it suppose to mean to me? I was unable to find any definitions of these terms.

Any additional tips about choosing a method are appreciated, also.

There are many good review articles, but I really like the freely available "The ABC of DFT" by Kieron Burke (and friends) and a recent review article that tries to keep to understandable language.

The problem is treating the Coulomb interactions. Conventional DFT is not "asymptotically correct." That is, at long range, the 1/r behavior is not met, and dispersion interactions (e.g., van der Waals forces) are not properly handled either.

For example, in Neon atom (from ABC of DFT by Burke):

The problem of course is that most DFT methods start from the local density approximation (LDA) which only considers the electron density at a particular (local) point. LDA indicates that the long-range behavior falls off exponentially instead of 1/r:

$$v_{xc}^{\mathrm{LDA}}(\mathbf{r}) = \frac{\delta E^{\mathrm{LDA}}}{\delta\rho(\mathbf{r})} = \epsilon_{xc}(\rho(\mathbf{r})) + \rho(\mathbf{r})\frac{\partial \epsilon_{xc}(\rho(\mathbf{r}))}{\partial\rho(\mathbf{r})}\ $$

So some recent efforts have used "range separated" functionals, e.g.,

$$ \frac{1}{r} = \frac{1-g(r)}{r} + \frac{g(r)}{r}$$

The first (short-range) term can be handled by the generalized gradient approximation (GGA) used by basically all modern functionals. The second term is usually handled by Hartree-Fock (i.e., exact exchange), and there's some function g(r) which smoothly scales between the two.

I think modern range-separated hybrid functionals are quite good, but choosing a functional depends a bit on the property that you want to predict. Thermodynamics? Excitation energies? Oxidation/reduction potentials from orbital eigenvalues? etc.


Metal carbonyl bonding I have been going through this topic in my coordination chemistry chapter. I read that in metal carbonyls, -CO group acts as a ligand and donates its lone pair of electrons into a vacant orbital of the metal to form a sigma bond. Also, there's a pi bond formation by the donation of electrons from the filled d orbital of the metal into pi* MO of the -CO molecule, and this pi bond strengthens the metal to ligand bonding.

But filling of electrons in the antibonding MO means decrease of bond order and correspondingly decrease of bond strength and bond lengths. It sounds contradictory to what's written in my book.

Where am I making a mistake?

Carbon monoxide bonds to transition metals using "synergistic π* back-bonding." The bonding has three components, giving rise to a partial triple bond. A sigma bond arises from overlap of the nonbonding (or weakly anti-bonding) sp-hybridized electron pair on carbon with a blend of d-, s-, and p-orbitals on the metal. A pair of π bonds arises from overlap of filled d-orbitals on the metal with a pair of π-antibonding orbitals projecting from the carbon atom of the $\ce{CO}$. The latter kind of binding requires that the metal have d-electrons, and that the metal is in a relatively low oxidation state ($<+2$) which makes the back donation process favorable.

As electrons from the metal fill the π-antibonding orbital of $\ce{CO}$, they weaken the carbon-oxygen bond compared with free carbon monoxide, while the metal-carbon bond is strengthened. Because of the multiple bond character of the $\ce{M-CO}$ linkage, the distance between the metal and carbon atom is relatively short, often $< 1.8~\mathrm{Å}$, about $0.2\ \mathrm{Å}$ shorter than a metal-alkyl bond.


What do the stereochemical labels D and − mean in tartaric acid? Why is the isomer of tartaric acid above labelled as ᴅ and (−)?

it rotates plane-polarized light clockwise
it rotates plane-polarized light anticlockwise
it is an equimolar mixture of the enantiomorphs
its configuration is related to that of ᴅ-glyceraldehyde

If I am not wrong, the ᴅ in ᴅ-glyceraldehyde refers to (+)-glyceraldehyde, and likewise ʟ-glyceraldehyde is (−)-glyceraldehyde. The (+) and (−) labels indicate that these isomers are dextrorotatory and levorotatory, respectively.
So in ᴅ-(−)-tartaric acid, what does the ᴅ mean, if not dextro, and what does the (−) mean, if not levo (the minus sign indicating anticlockwise rotation of plane-polarized light)? How can one molecule be both?

As far as I remember, small d always means the same as (+), not small cap ᴅ. So, a dextrorotatory compound is prefixed with either (+) or d while a levorotatory compound is prefixed with either (−) or l.

Small caps ᴅ and ʟ prefixes have different meaning: if an enantiomer can in principle be synthesized from naturally occurring (+)-glyceraldehyde without affecting its configuration, it is considered the ᴅ-form. Otherwise, it is considered the the ʟ-form. And, say, the ᴅ-form of a molecule is not necessarily a dextrorotatory one.

So, the answers to your question are options 2 (the (−) sign indicates that it rotates plane-polarized light anticlockwise), and option 4 (the ᴅ label indicates that it is related to ᴅ-glyceraldehyde).


What chemical process occurs when dead leaves make a mark on concrete? In autumn, leaves that fall on concrete make marks that remind me of fossils.

Is this caused by a chemical reaction that occurs between the concrete and the leaf? Or would a leaf leave a stain on any surface as it decays? What circumstances can intensify the process?

The shape of the leaf is consistent with a type of maple tree leaf, and the brown coloring is characteristic of the water soluble tannins that would be extracted by water. For comparison, here are two photographs, one of tanic acid in water and one of a dried mixture of tannins. The photos were taken from this Wikipedia article.  Although the tannic acid in the photograph below was extracted from different parts of a redwood tree, it is a ubiquitous tannin found in most hardwood tree leaves and the color is characteristic of many such tannins.


Why is heat capacity defined as a partial derivative? I understand that the heat capacity is the gradient of the slope of q against T, which implies that C = $\mathrm{ \frac{dq}{dT}}$, so how do we arrive at C = $\mathrm{\frac{∂q}{∂T}}$?
At constant volume, $\mathrm{C_V}$:

$$\mathrm{dU = dq + dw}$$

$\mathrm{dw = p(dV)}$, but $\mathrm{dv = 0}$, so $\mathrm{dU = dq}$

How do we make the jump to $\mathrm{C_V = \frac{∂U}{∂T}}$?

3. At constant pressure, $\mathrm{C_p}$:

$$\mathrm{H = U + pV}$$
so $$\mathrm{dH = dU + pdV = (dq + dw) - dw = dq}$$
hence $\mathrm{dH = dq}$

How do we make the jump to $\mathrm{C_p = \frac{∂H}{∂T}}$?

I think you are starting on the wrong foot.

$U$ and $H$ are state functions. They both depend on several variables. $U$ is a function of entropy $S$ and volume $V$ and composition. For a pure fluid there is no dependence on composition, so we can just say $U = f(S,V)$. Similarly $H$ is a function of entropy and pressure $p$ i.e., $H=f(S,p)$. You go from U to H by using a Legendre Transform (but ignore Legendre Transforms for now... I will give you a good reference to read for thermodynamics at the end, it will make your life better).

Important

Reality has a way of changing things. For instance experiments are done almost always at a constant $T$, and either constant $p$ or $V$. Therefore, we like thinking of things in terms of these variables. Lets face it... entropy is tricky.

Assumption

Okay, lets now say that $U=f(T,V)$ and $H=f(T,p)$ this is alot easier to handle. In thermodynamics we are always after changes in things. Therefore, the differential forms of $U$ and $H$ are desirable.

Also, this is not a haphazard switch, entropy and temperature are intricately related to each other $T \equiv \left( \frac{\partial U}{\partial S} \right)_{x's}$ where $x's$ are all other independent variables being held constant.

The differentials for both $U$ and $H$ are as follows (with respect to our new independent variables).

$$U(T,V)$$
$$dU = \left( \frac{\partial U}{\partial T} \right)_V dT + \left( \frac{\partial U}{\partial V} \right)_T dV$$
and
$$H(T,p)$$
$$dH = \left( \frac{\partial H}{\partial T} \right)_p dT + \left( \frac{\partial H}{\partial p} \right)_T dp$$

Endgame

It is here that we now define what $C_V$ and $C_p$ are

$$C_V \equiv \left( \frac{\partial U}{\partial T} \right)_V$$

$$C_p \equiv \left( \frac{\partial H}{\partial T} \right)_p$$

These are fundamental definitions and can't be tampered with. In introductory thermodynamics teachers play games and try to frankly dumb things down, and this leads to, in my opinion, improper definitions and understanding. This link provides a definition (without much context) that ties $Q$ into the mix.

If we insert these definitions into our differential equations we have

$$dU = C_V dT + \left( \frac{\partial U}{\partial V} \right)_T dV$$
and
$$dH = C_p dT + \left( \frac{\partial H}{\partial p} \right)_T dp$$

I leave it as an exercise, if you wish, to find the definitions for $\left( \frac{\partial U}{\partial V} \right)_T$ and $\left( \frac{\partial H}{\partial P} \right)_T$

Final Resource

Herbert B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd Edition

The man called Callen

The Textbook


Distillation: Simple vs Fractional Here is the excerpt of my lab manual:

"Simple distillation is more useful for purifying a liquid that contains either a non-volatile impurity or small amounts of higher or lower boiling impurities. Fractional distillation allows for several condensation-condensation cycles in a single operation. It can be used to separate liquids with comparable volatilities and to purify liquids that contain relatively large amounts of volatile impurities."

So, why does simple distillation put emphasis on small amounts and fractional distillation, on large volumes?

So, why does simple distillation put emphasis on small amounts and
  fractional distillation, on large volumes?

It's not small and large volumes of material being distilled, but rather small and large volumes of impurities present in the material to be distilled.  Both simple and fractional distillations can be used to distill large quantities of material.

The efficiency of a distillation column is measured in "height-equivalent theoretical plates" (HETP), the more plates the more efficient the column, the more efficient the column the more likely that it will be able to separate two closely boiling materials.  If you have a small amount of a lower boiling impurity in your solution and you want to separate it from a large amount of the desired material, then you can probably get away with using a simple distillation column and discarding some of the forerun (the first drops of liquid that distill over).  On the other hand, if there is a large amount of impurity and it boils close to the desired product, then you will need a column with a higher efficiency (fractional distillation column) to effect a good separation between the two fractions (or "cuts" as they're sometimes called) and get pure product.

To sum up, perhaps the original question would be better written as

  So, why does simple distillation put emphasis on small amounts and
  fractional distillation, on large amounts of impurities?

and the answer would be that even an inefficient (fewer HETP) simple distillation column has enough power (enough theoretical plates) to separate a small amount of impurity from a large amount of product, especially if you are willing to discard the initial fore-run from the distillation.  If there is a larger amount of impurity, then we may need a more powerful (higher efficiency column) to produce the desired separation.

The higher the efficiency of a distillation column, the greater its ability to separate two things.  If we have to separate two materials that are both present in large quantity, or distill at boiling points close together, then we need a more efficient column than if they boiled further apart or one was present in only a small amount and we were willing to throw away the forerun (e.g. throw away some of the desired product in that early fraction that contained the impurity).  When we call a column a fractionation column, we're really saying that it has greater separating efficiency than a simple distillation column.  The more efficient the column, the better the job it will do on difficult separations.


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

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

Also, be sure to carefully practice the words: "I AM OZ, THE GREAT AND POWERFUL" before you go public.


Living-cationic vs cationic polymerization To my understanding, cationic polymerization is a type of living polymerization but my teacher's notes and Wikipedia seem to suggest they're different. I can't see the difference.

The same goes for anionic polymerization. From what I've studied so far, the difference must be subtle, if there is one.

Is there such a difference?

This difference between living and non-living polymerizations is small in words but large in effects. In principle cationic and anionic polymerizations can be living. In practice, it is not as easy as it looks on paper. See below.

A living polymerization is any chain (or addition) polymerization that is prevented from terminating, i.e. the ends of the chains are still reactive.  In a living polymerization, chain initiation has one initiation and propagation continues as long as there is monomer. Termination (almost) never happens.  If your polymerization stops because your monomer runs out, adding more monomer restarts the process and the existing chains get longer. The number of chains formed matches very closely with the number of equivalents of initiator, and the length of the chains depends on the ratio of monomer to initiator.

Most chain polymerizations (radical, anionic, cationic, transition-metal insertion, and ring-opening) do not behave this way. There is a mechanism for termination that becomes increasingly kinetically favored as the concentration of free monomer decreases. Sadly, the wikipedia article for chain termination is pretty sad. Most chain polymerizations propagate and terminate quickly, with termination usually initiating the next chain. Thus, in a non-living polymerization, new chains are constantly being initiated and old chains are constantly terminating until their is no more monomer.  In these cases, adding additional monomer later would not accomplish anything because the chain ends are "dead". Experimental evidence for this phenomenon includes the formations of more chains than there were equivalents of initiator. The length of the chain is more dependent on the relative rates of propagation and termination than on the initial ratio of monomer to initiator.

For radical polymerization, termination mechanisms include fragmentation by β-hydrogen abstraction or by recombination. Both of these mechanisms are inherent in the radical nature of the polymerization, and so are not preventable by exclusion of other species. Radical polymerizations are made living by the addition of species that "protect" the radical at the end of the chain to prevent it from terminating. Examples include ATRP and RAFT. The wikipedia articles for both are pretty good.

For cationic polymerization, termination mechanisms include β-hydrogen abstraction, chain transfer, and attack by nucleophiles like water. In principle, a cationic polymerization could be a living polymerization if great pains were taking to exclude all species which might induce termination. In practice this is doable, but difficult, as termination by chain transfer is always possible. See the wikipedia article on living cationic polymerization. The article is not great, but suggests that cationic polymerizations done in nonpolar solutions (not in bulk) with nonnucleophilic Brønsted or Lewis acids at low temperatures can become living polymerization.

For anionic polymerization, termination mechanisms include protonation by protic molecules, chain transfer, and reaction with electrophiles, which can include O2 in some reactions. Like cationic polymerizations, anionic polymerizations can be made living by rigorous exclusion of terminating species. Again, this is difficult, due to the possibility of chain transfer termination. However, with the correct choice of conditions, it is doable, as suggested by the wikipedia page for living anionic polymerizations. Another way to accomplish some anionic polymerizations in a living fashion is group transfer polymerization, even though it does not  follow a strictly anionic mechanism.

I know that Contemporary Polymer Chemistry by Allcock, Lampe, and Mark, Prentice Hall 2003 covers this topic well, but the book is a little intense to read. In particular, the book derives the kinetic expressions for various polymerization types in the living and non-living cases. I expect other textbooks focusing of the chemistry of polymers might also cover this topic in one fashion or another.