Difference between backbonding and hyperconjugation and conjugation What is the difference between backbonding and hyperconjugation and conjugation?
I mean the basic differences as all are concerned with donation of electrons. What is the most striking effect of each that differentiate them from others.

OK, let's step through these terms and let's add "cross-conjugation" and "homoconjugation" for good measure to round out the series.
Backbonding:
This term is most often used to describe a type of bonding that occurs in inorganic compounds.  Metals bonding with carbon monoxide provide good examples of backbonding. The carbon atom in carbon monoxide has a lone pair of electrons that can be used to form a sigma bond with a metal.  Because carbon monoxide has low-lying $\pi^{\ast}$ orbitals, it can accept electrons back from the metal and further strengthen the bonding between the metal and the carbon monoxide ligand.  This process of "accepting electrons back from the metal" is termed backbonding. Here is a good picture of the $\ce{\sigma}$ and $\mathrm{d}-\pi^{\ast}$ (backbonding) bonding interactions.

It happens most often when 1) the metal is electron rich, 2) the ligand has low-lying $\pi^{\ast}$ orbitals and 3) the metal has available a $\mathrm{d}$ orbital with the proper symmetry to overlap constructively with the carbon monoxide $\pi^{\ast}$ orbital.  Here is a link to a great, one page article on backbonding if you'd like to read more.
Hyperconjugation:
This term is most commonly used in discussions about resonance.  Hyperconjugation is a stabilizing interaction that results from the interaction of the electrons in a $\sigma$-bond (usually $\ce{C-H}$ or $\ce{C-C}$) with an adjacent (empty or filled) $\mathrm{p}$-orbital. This interaction creates an extended molecular orbital that increases the stability of the system.  Most typically it involves the breaking of a carbon-hydrogen bond and then using what were the $\ce{C-H}$ bonding electrons to stabilize an adjacent charge. In resonance structures, we can't move nuclei, but we can move electrons. For example, below is a diagram showing resonance structures that involve hyperconjugation that serve to stabilize the n-propyl carbocation. Hyperconjugation has been used as part of the explanation for the carbocation order of stability being tertiary > secondary > primary.

Recently it has been suggested that hyperconjugation, rather than steric effects, is what stabilizes the staggered conformation of ethane (very interesting!).    Here is a link to a fairly extensive, but very well written article on hyperconjugation, it covers the effects of hyperconjugation on resonance structures and conformational stability.
Conjugation:
This term is used to describe a system of continuous, overlapping p-orbitals where electron delocalization can occur.  Examples of conjugated systems include the 1,3-butadiene, acrolein, benzene, beta-carotene, etc.

Cross-conjugation:
Sometimes you'll hear the term "cross-conjugation" used.  In this case, the continuity of the overlap between the p-orbitals does not follow a simple linear pattern as seen in the examples of conjugation pictured above.  Benzophenone and 3-methylene-1,4-pentadiene are examples of molecules with cross-conjugation.

Homoconjugation:
Sometimes $\mathrm{p}$-orbitals that are not directly conjugated (e.g., that are not directly connected, there is a "spacer" in between them) can still interact or overlap in a "through space" fashion.  The solvolysis of the exo- and endo-norbornenyl chlorides shown below provides a striking example. The exo-isomer generates the corresponding carbocation thousands of times faster than its endo-isomer. This difference is attributed to homoconjugation. The $\mathrm{p}$-orbitals on the remote norbornenyl double bond can interact through-space with the developing $\mathrm{p}$-orbital that is being formed on the carbon where the exo-chloride is leaving from, dramatically stabilizing the formation of the resultant carbocation. This interaction is geometrically impossible for the endo-isomer.  Again note, there is no direct conjugation between the remote double bond and the carbon atom where the carbocation will form, it is entirely a through space interaction that generates a pseudo-allyl system.


What is the density of carbohydrates/protein/fat? As in weight per volume? I mean.. you always see them specified by weight but they have to have a volume, right?

I suppose different fat(for example) molecules would have different densities but then how do they universally have the same energy density? Are they converted to like a standardized type of fat? Thats the fat I'm interested in.

I assume the "energy density" you are referring to are the widely known statistics that:

fat contains 9 "calories" per gram (actually kilocalories)
protein and carbohydrates both contain 4 "calories" per gram

How can all fats contain "9" kcal per gram but have different mass densities?  Here are a couple of reasons:

The 9 kcal/g figure is an approximation that assumes a typical dietary blend of fats.  The figure would if considering chemically pure fat molecules.
The mass densities of most common fats doesn't vary that much.  The densest "fat" molecules are probably the extremely short chain triglycerides like triformin and triacetin.  They have densities of approx. $1.3~ \frac{\mathrm g}{\mathrm{cm}^3}$.  These triglycerides have chain lengths of 1 or 2 carbons per fatty acid chain, a value far lower than common in natural fats.  Triolein is more representative of dietary fats and has a density of 0.95$~\frac{\mathrm g}{\mathrm{cm}^3}$.  So the mass density variation among different fat molecules is pretty small: a 30% increase is all that is really possible.


Kinetic isotope effect in nitration of benzene What is the effect of changing all the hydrogen atoms in benzene to deuterium? For example, which of the following reactions occurs at a faster rate?

Page 513 of Organic Reaction Mechanisms by Bansal references the work of Melander in Arkiv för Kemi, 2, 211 (1950). This states that the rates of nitration for benzene vs benzene-d6 are about the same, implying that the addition of the nitronium ion is rate controlling and not the weakening of the C-H / C-D bond. However in case of sulphonation H containing benzene reacts faster than D containing benzene.


Why are many chemical relationships logarithmic? I've noticed that in many relationships in chemistry, one variable is linearly related to the logarithm of another. Why is this the case?

For example, I carried out an experiment where the dependent variable, the amount of solute adsorbed to a surface, was almost perfectly linearly correlated (Pearson correlation coefficient = 0.93) with the pOH, which is a logarithmic quantity.

Edit: Would this perhaps be better in stats.stackexchange.com?

Some logarithmic behaviour is built into the way the world works and some is a consequence of how we perceive things which span very large ranges.

The simplest chemical and nuclear reactions (first order ones) are, for example, inherently logarithmic. Take any first order process (like radioactive decay). The process has a fixed probability of happening (a radioactive atom or an unstable molecule might have a 10% probability of decaying in a minute). What you observe in the bulk with processes like this is that the rate of the reaction decays exponentially. You start with a rate of say 12 units per minute. After a certain time half the initial material has gone and the rate is now 6 after the same time again the rate is now 3 and so on. This is exponential decay emerging from a very simple process with a fixed probability. A logarithm emerging from a probability, if you prefer. This sort of thing is very common in chemistry and physics.

The other cause of logarithms is human perception. Our perception of light and sound are both inherently logarithmic. This is partially because there is no sensible way to map the range of magnitudes of many natural phenomena to a linear scale. The intensity of sound and light vary over huge orders of magnitude. The Sun is about 15 trillion time brighter than the faintest visible star and it is fairly inconvenient to divide this scale into equal bands. So perception tends to divide the scale logarithmically with each perceived difference representing a fixed ratio not a linear difference (each star magnitude, for example is about 2.5 times brighter than the next lower one, but represents a just-about discernible difference).


What is burst nucleation? Can you tell me what is a burst nucleation process, and how can it be used for synthesising uniform nanoparticles?

Can you also provide (if possible) any links to papers or materials regarding mechanism of nanoparticles growth?

This is a part answer - an annotated short list of references regarding burst nucleation and mechanisms of nanoparticle growth.

Model of nanocrystal formation in solution by burst nucleation and diffusional growth (Robb and Privman, 2007), in this article the authors

  present a model with the assumptions of instantaneous thermalization below the critical nucleus size and irreversible diffusive growth above the critical size

A main feature of this article is that

  for the first time formulates LaMer's explanation of burst nucleation in a manner allowing quantitative calculations.

This is followed up by one of the authors the next year with the article Diffusional Nucleation of Nanocrystals and Their Self-Assembly into Uniform Colloids (Privman, 2008), where he

  review theoretical explanation of mechanisms of control of uniformity in growth of nanosize particles and colloids.

An explanatory diagram of LaMer's explanation (referred to in the 1st article) is shown below:

The image is from the article Controlled synthesis of colloidal silver nanoparticles in organic solutions: empirical rules for nucleation engineering (Sun, 2012) - unfortunately, the article is behind a paywall.

However, a sample chapter explaining the process in full entitled Growth of Nanocrystals in Solution (Viswanatha and Sarma), provides a greater detail about the processes involved with nanoparticle growth.


Copper Stoichometry Ratio Problem I was given two equations as part of a problem into Copper Wiring production for planes:

$$\ce{4CuFeS2 + 9 1/2 O2 + 4SiO2 \rightarrow Cu2S + Cu2O + 2SO2 + 4FeSiO3}$$
$$\ce{Cu2S + 2Cu2O \rightarrow 6Cu + SO2}$$

This question later became the source of a Chemistry exam question for the English IB equivalent.

I was given the following information.

  The previous equations are part of the manufacture of copper for wiring in planes.
  A passenger jet contains $4050\ \pu{ kg}$ of copper wiring.
  A rock sample contains $1.25$% $\ce{CuFeS2}$ by mass.
  Calculate the mass, in tonnes, of rock needed to produce enough copper wire for a passenger jet. ($\pu{1~tonne} = \pu{1000~kg}$)

Now, I tried to solve it as follows:

Adding both equations gave:

$$\ce{4CuFeS2 + 9 1/2 O2 + 4SiO2 + Cu2O \rightarrow 2SO2 + 4FeSiO3 + 6Cu + SO2}$$

Now, I can see that $4$ moles of $\ce{CuFeS2}$ from the ore would give us the required $\ce{6Cu}$ for the production.

However, the solution states that Cu and $\ce{CuFeS2}$ are in a $1:1$ ratio, which I couldn't explain why.

Using the equation I formed above, this idea didn't make sense.
Why is the ratio of Cu to $\ce{4CuFeS2}$ $1:1$?

No serious mining operation would just dump a third of their desired product.
 The question does say that the reactions are "part of the manufacture of copper." This does not preclude other reactions. With the given reactions alone we are not able to handle the surplus of $\ce{Cu2S}$.
$$\ce{\color{\red}{4CuFeS2} + 9 1/2 O2 + 4 SiO2 \rightarrow \color{\red}{Cu2S + Cu2O} + 2SO2 + 4FeSiO3}  \tag{1}$$
$$\ce{\color{\red}{2} Cu2S + 2 Cu2O \rightarrow 6Cu + SO2 +\color{\red}{Cu2S}}\tag{2}$$
However we can infer from the first reaction that the sulfide can react with oxygen to form the needed oxide to balance reduction and oxidation.
$$\ce{ Cu2S + 1 1/2O2 ->  Cu2O + SO2}\tag{3}$$
The second equation can now become:
$$\ce{\color{\red}{2} Cu2S + 2 Cu2O +\color{\red}{1 1/2 O2} \rightarrow 8 Cu + 2 SO2 \tag{4}}$$
Thus in the first step $\ce{Cu2O \text{and} Cu2S}$ are produced according to quation 1.  In the second step they are reduced with aditional oxygen according to equation 4. To pove they balance first double equation 1.
$$\ce{8CuFeS2 + 19 O2 + 8 SiO2 \rightarrow \color{\red}{2Cu2S + 2Cu2O} + 4 SO2 + 8 FeSiO3}  \tag{5.1}$$
Add $\ce{1 1/2 O2}$ to each side
$$\ce{8CuFeS2 + \color{\red}{20 1/2 O2} + 8 SiO2 \rightarrow \color{\red}{2Cu2S + 2Cu2O + 1 1/2 O2 }+ 4 SO2  + 8 FeSiO3}  \tag{5.2}$$
Substitute the red products with equation 4 reactants.
$$\ce{8CuFeS2 + \color{\red}{20 1/2 O2} + 8 SiO2 \rightarrow \color{\red}{8 Cu + 2 SO2}+ 4 SO2  + 8 FeSiO3}  \tag{5.3}$$
This Gives a final balanced equation of
$$\ce{8CuFeS2 + 20 1/2 O2 + 8 SiO2 \rightarrow 8 Cu + 6 SO2  + 8 FeSiO3}  \tag{5.4}$$
As you can see on a molar eatio $\ce{CuFeS2\text{:}Cu}$ is 1:1 after accounting for the additional oxygen needed.


Solubility of solids in liquids How does crystallisation happen and why?

I have been told that when a solid is dissolved in a liquid solvent, an equilibrium is set up between processes of dissolution and crystallisation. But if this is so then why does all solute dissolves (I am talking about an unsaturated solution), isn't it supposed to recrystallise simultaneously too?
I am certainly missing a lot and most importantly what crystallisation is and how and why it happens?

I don't want to know anything about the separation technique using the principle of  crystallisation. Just want to know the principle itself.

The principle of solubilization — in the ideal case — is closely related to that of melting. Think of solubilization as a two step process:

melt the solute;
mix the solute and the solvent.

Of course, step 1 involves raising the temperature to the melting point, or at least adding energy (in the form of heat) at the melting temperature in order to break attractive bonds between solute molecules. We will consider only the case where we are already at the melting point (an isothermal process), so you only need to add heat in order to turn the solid into liquid.

Step 2, on the other hand, is spontaneous if the solute and solvent mix ideally. You don't have to add heat (energy) for this step to proceed, assuming both substances are liquids and mix ideally.

Now what happens when you solubilize something is that you need to add less heat than you would have needed in step 1 to melt all of the solute, since some of the required (free) energy comes from the process of mixing.

By a similar line of argument, based on the properties of ideal solutions, you can relate (approximately) the solubility to the melting point and the heat of fusion as follows:

$$\log(x_2) = \frac{\Delta H^\circ_\mathrm{fus}}{R}\left(\frac{1}{T_\mathrm{fus}}-\frac{1}{T}\right)$$

Since $\Delta H^\circ_\mathrm{fus}$ is generally positive (heat is required to melt something) then increasing $T$ tends to increase the solubility of solids.

Crystallization is just the reverse of solubilization! And yes, there is always an equilibrium between soluble/insoluble solute, it's just that sometimes the soluble or insoluble part is negligible, depending on the values of $\Delta H^\circ _\mathrm{fus}$, $T_\mathrm{fus}$, and $T$.


Is an imine considered a secondary amine? I am looking at the difference between primary, secondary, and tertiary amines.

What about if the nitrogen is double bonded to a carbon?

Is this carbon counted twice, and is the molecule above a considered tertiary amine?

No. Imines cannot be considered an example of secondary amines any more than pyridine can be considered a tertiary amine.
The easiest way to think about this is with oxidation state: imines are at a higher oxidation level than amines, just as aldehydes are at a higher oxidation level than alcohols.
Aside from this 'intuitive' explanation, the IUPAC gold book (a source of chemical nomenclature) defines both amines and imines, clearly differentiating the two:

amines. Compounds formally derived from ammonia by replacing one, two or three hydrogen atoms by hydrocarbyl groups, and having the general structures $\ce{RNH2}$ (primary amines), $\ce{R2NH}$ (secondary amines), $\ce{R3N}$ (tertiary amines).
imines. Compounds having the structure $\ce{RN=CR2}$ (R = H, hydrocarbyl). Thus analogues of aldehydes or ketones, having NR doubly bonded to carbon; aldimines have the structure RCH=NR, ketimines have the structure $\ce{R'2C=NR}$ (R' ≠ H). Imines include azomethines and Schiff bases. Imine is used as a suffix in systematic nomenclature to denote the C=NH group excluding the carbon atom.


How are half lifes calculated for radioactive substances that decay over many years How are the half lives (or whole lives) calculated for slowly decaying radioactive substances such as tellurium? I must be having a fundamental misunderstanding of how we are able to predict the amount of time it will take for an isotope or radioactive material to decay fully including for ones that can remain unstable for practically eons.

Suppose you have a sample of $\ce{N}$ atoms of a radioactive material. As it is radioactive, you may state that a small fraction k is decomposed by second. This fraction does not depend on the total number of atoms. If this sample is placed in a Geiger counter, each decomposition of its atoms produces a ray that crosses the inner volume of the Geiger counter, and it will be detected and counted. In a given time, the number of rays is equal to the number of disintegrations. If dN/dt is the number of atoms disintegrated per second (or per minute), the factor k can be calculated by dividing the number of rays detected every second in the counter, by the total number of atoms N of the radioactive material. k = $\ce{\frac {dN}{dt}·\frac{1}{N}}$
Once k is known, the half-life T can be derived from the integrated law, if $\ce{N_o}$ is the original number to atoms at t = $\ce{0}$  $$\ce{N/N_o = e^{-kt}}$$ The half-life is the time T after which $\ce{N = N_o/2}$, which gives $$\ce{1/2 = e^{-kT}}$$ with the following result : $$\ce{T = \frac{ln 2}{k}} = 0.693/k $$
So the period may be obtained from the knowledge of k. It is not necessary to wait up to the radioactive material has decreased to one half.
For example, if $0.0238$ g, or $\ce{6.02·10^{19}}$ atoms U-$238$ is inserted in a Geiger counter, $292$ rays will be detected per second. From this measurements, the constant k can be calculated : $$k\ce{ = \frac{292 s^{-1}}{6.02·10^{19}} = 4.87·10^{-18} s^{-1}}$$ and the half-life is $$\ce{T = \frac{0.693}{4.87·10^{-18}} = 1.423·10^{17} s = 4.51· 10^{9} years}$$


Terminals Of An Electrochemical Cell Does the positivity and negativity of terminals of a battery depend on the terminal where the electrons enter the external circuit?

The positive terminal of a battery is where 'conventional current' comes out of the battery and flows through an external circuit, returning to the battery via the negative terminal. Since electrons are (by convention) negatively charged, the negative terminal of a battery is where electrons are 'released' by the battery.

NOTE: Of course, if it is a secondary battery which is 'recharging', electrons will 'enter' the battery at the terminal marked negative.


Sublimation of Iodine from Iodine Povidone Iodine povidone is a complex of iodine and the polymer povidone. I am wondering if one were to heat it enough, the iodine would sublimate and could be collected. Normally iodine boils at 184.3 °C, according to wikipedia.

A video of how the procedure might work, except iodine crystals would replaced with boiled down iodine povidone: Sublimation of Iodine Time lapse

This can be used to purify iodine, so from my understanding, you would be able to vaporize the iodine and leave the polymer behind.

Does the complex with povidone change how this would work?

I apologize if my question is ignorant, as I do not have a great understanding of how complexes work.

A procedure is available on Erowid for this process.

  Make or buy a coldfinger (see below), boil your fluid to absolute dryness, without burning the dry restant, then hang your coldfinger in an appropriate Erlenmeyer, and heat gently the now dry restant. Make your coldfinger fat-free with acetone first (on the outside where the I2 will condense). The I2 will sublimate from the solid form into the gas form, without getting a fluid first (that's sublimation), and the gaseous I2 will resublime from the gaseous form into the solid form against the cold wall from the coldfinger. A little bit will of course also resublime against the upper part of the Erlenmeyer, but that you can also scrape off later, when all your 10% I2 hangs on the cold glass walls.


What is the nature of the Fe–O2 binding in oxymyoglobin and oxyhemoglobin? Deoxymyoglobin ($\ce{Mb}$) is known to have iron in the +2 oxidation state; I believe this was deduced from its magnetic moment, which corresponds to four unpaired electrons in high-spin $\mathrm{d^6}$ Fe(II).

However, upon binding of $\ce{O2}$ (which is itself paramagnetic) to Mb, the resultant complex oxymyoglobin ($\ce{MbO2}$) is found to be diamagnetic.

This observation has been rationalised in several ways over the years. As far as I know, the two main camps are:

The oxygen ligand binds to low-spin Fe(II) in the form of singlet oxygen, via $\ce{O2}\rightarrow\ce{Fe}$ σ donation and $\ce{O2}\leftarrow\ce{Fe}$ π backdonation, which mirrors the binding mode of the isoelectronic $\ce{NO-}$. Neither metal nor ligand possesses unpaired electrons;
Fe(II) is oxidised to Fe(III), with reduction of $\ce{O2}$ to $\ce{O2-}$; thus both the metal and the ligand possess one unpaired electron. These two electrons interact via antiferromagnetic coupling to produce a singlet state.

The former viewpoint was championed by Linus Pauling, and the latter by Joseph Weiss, leading to a rather heated debate in Nature.1–3

These descriptions of myoglobin of course also apply to hemoglobin, where each monomer is structurally similar to myoglobin.

What experimental evidence supports either side, and is there any general consensus as to which is correct?

References

1. Weiss, J. J. Nature 1964, 202 (4927), 83–84. DOI: 10.1038/202083b0.

2. Pauling, L. Nature 1964, 203 (4941), 182–183. DOI: 10.1038/203182b0.

3. Weiss, J. J. Nature 1964, 203 (4941), 183. DOI: 10.1038/203183a0.

TL;DR: It is not really "one or the other" - it is more of "somewhere in between", like a resonance hybrid. If I had to choose, I'd personally lean towards Weiss's $\ce{Fe(III)-O2-}$ model. However, there will always be people who disagree, and there is still no general consensus amongst the scientific community.

There is an excellent - and recent - perspective article by Bren et al.,1 describing the research carried out on the structure of hemoglobin ever since Pauling and Coryell's initial report2 on the diamagnetism of oxyhemoglobin.

Model systems

A common technique in bioinorganic chemistry is to synthesise "inorganic versions" of enzyme active sites; for example, much work has been done on dinitrogen complexes in the hope of understanding the enzyme nitrogenase. Myoglobin and hemoglobin are not enzymes in the strict sense of the word,3 but the same technique applies: if stable $\ce{Fe(II)-O2}$ complexes could be synthesised, they might prove to be more amenable to experimental investigation.

The main problem was the fact that dioxygen readily oxidised simple Fe(II) complexes to Fe(III). Initially, Co(II) complexes (which only have one unpaired electron) were used: EPR studies showed that the unpaired electron was localised on oxygen,4,5 and the length of the O–O bond resembled that of superoxide:6

$$\begin{array}{ccccc}
\text{Species} & \ce{O2^2-} & \ce{O2^-} & \ce{O2} & \ce{Co(II)-O2}\text{ complex} \\ \hline
\text{Bond order} & 1 & 1.5 & 2 & ? \\
\text{Bond length / Å} & 1.49 & 1.28 & 1.21 & 1.26
\end{array}$$

Of course, Co(II) complexes do not necessarily have the same electronic structure as their Fe(II) counterparts, so such data can only be of limited use. The key to overcoming the early difficulties (of Fe(II) oxidation) was the fact that such oxidations tend to proceed via dimerisation; therefore, if a sterically hindered Fe(II) complex could be made, the oxidation could be prevented. Based on this, Collman and coworkers synthesised so-called "picket fence" complexes,7 which reversibly bound dioxygen, the concept of which is shown below:

Access of other iron centres, as well as the imidazole ligand, to the top face of the metal ion was blocked by the bulky substituents on the porphyrin ring (in one example, $\ce{-C6H4NHC(O)^tBu}$ was chosen). The reported data from different "picket fences" strongly favoured the iron(III)–superoxide model: in particular, the O–O stretching frequencies ranged from $1163\text{–}\pu{1150 cm−1}$ (cf. $1200\text{–}\pu{1050 cm−1}$ in superoxide complexes, $\pu{1556 cm-1}$ in $\ce{O2}(^3\Sigma_\mathrm{g}^-)$, and $\pu{1484 cm-1}$ in $\ce{O2}(^1\Delta_\mathrm{g})$).8

A controversy over magnetic data

It was realised that a key difference between the two schemes was in their predicted magnetic susceptibilities. For the $\ce{Fe(II)-O2}$ binding scheme, one would expect the compound to be diamagnetic at all accessible temperatures (to obtain a magnetic state would require electronic promotion). On the other hand, the $\ce{Fe(III)-O2-}$ binding scheme would be expected to have a temperature-dependent magnetic susceptibility, as antiferromagnetic coupling could be overcome with higher temperatures. In other words, there is a thermally accessible triplet state.

In 1977, Cerdonio et al. published a paper claiming to have observed a temperature-dependent magnetic susceptibility for $\ce{HbO2}$.9 However, their findings were refuted in a later 1984 paper by Savicki et al., which found that $\ce{HbO2}$ was in fact diamagnetic at all temperatures.10 The incorrect magnetic data was attributed to residual deoxyhemoglobin, which is paramagnetic, and in a private communication with Savicki, Cerdonio agreed that his earlier findings were in error.

Spectroscopic data on the proteins

Using infrared spectroscopy, the O–O stretching frequencies in both oxymyoglobin11 and oxyhemoglobin12 were identified as $\pu{1103 cm−1}$ and $\pu{1107 cm-1}$ respectively, again consistent with superoxide.

Furthermore, $\ce{^57Fe}$ Mössbauer spectra of $\ce{HbO2}$ showed large quadrupole splittings, which is consistent with iron being in the +3 oxidation state.13

Computational studies

Early quantum mechanical calculations by Karplus14 showed that the properties of $\ce{HbO2}$ could be explained by a mixture of two $\ce{Fe(II)-O2}$ resonance forms: one with both bonding partners in the singlet state, and one with both in a triplet state. The latter was actually put forth as a separate - third - model, most notably by Goddard in 1975.15

Of course, these are fairly primitive by today's standards. In 2008, Shaik et al. revisited the topic,16 and provided an overview of the numerous computational studies performed up until then, which had failed to reach a conclusion. Instead of trying to discern the appropriate bonding model from indirect parameters, such as the net charge on oxygen or the form of the occupied MOs, the authors opted to transform a CASSCF wave function into a generalised valence bond wave function. They concluded that the Weiss $\ce{Fe(III)-O2-}$ model was dominant for a gas-phase oxygen–heme complex, and that the interaction with the protein further reinforces the $\ce{Fe(III)-O2-}$ character of the wave function.

More recent experimental work

In an entirely separate endeavour, Gray and coworkers found that polyhalogenated iron porphyrins, which possessed unusually high Fe(III)/Fe(II) reduction potentials, did not bind to oxygen.17 This has been taken as evidence for the necessity of electron transfer from Fe(II) to dioxygen in $\ce{Hb-O2}$ binding, and hence the identification of the bonding as being $\ce{Fe(III)-O2-}$ charge-transfer in nature.

X-ray absorption spectra by Wilson et al. further complicate the matter, though. Firstly, their investigation of picket fence porphyrins led to the conclusion that these generally showed characteristics of both models, but were more in line with an Fe(II) description. The effective nuclear charge on iron was found to be closer to that of Fe(II) than to that of Fe(III), and the spectra lacked a low-energy transition characteristic of Fe(III) complexes (arising from the hole in the $\mathrm{t_{2g}}$ orbitals on iron).18

A further study on the protein itself revealed that crystalline $\ce{HbO2}$ mainly has $\ce{Fe(II)-O2}$ character, whereas solution $\ce{HbO2}$ has $\ce{Fe(III)-O2-}$ character.19 They therefore concluded that the bonding was multiconfigurational in nature, and that "secondary changes in the protein environment, crystal packing, and so forth, can tip the electronic structure in either direction".

Notes and references

1. Bren, K. L.; Eisenberg, R.; Gray, H. B. Discovery of the magnetic behavior of hemoglobin: A beginning of bioinorganic chemistry. Proc. Natl. Acad. Sci. 2015, 112 (43), 13123–13127. DOI: 10.1073/pnas.1515704112.

2. Pauling, L.; Coryell, C. D. The Magnetic Properties and Structure of Hemoglobin, Oxyhemoglobin and Carbonmonoxyhemoglobin. Proc. Natl. Acad. Sci. 1936, 22 (4), 210–216. DOI: 10.1073/pnas.22.4.210.

3. Hemoglobin is famously known as an "honorary enzyme"; for an overview of its history, see e.g. Brunori, M. Hemoglobin is an honorary enzyme. Trends Biochem. Sci. 1999, 24 (4), 158–161. DOI: 10.1016/S0968-0004(99)01380-8.

4. Hoffman, B. M.; Diemente, D. L.; Basolo, F. Electron paramagnetic resonance studies of some cobalt(II) Schiff base compounds and their monomeric oxygen adducts. J. Am. Chem. Soc. 1970, 92 (1), 61–65. DOI: 10.1021/ja00704a010.

5. Getz, D.; Melamud, E.; Silver, B. L.; Dori, Z. Electronic structure of dioxygen in cobalt(II) oxygen carriers, singlet oxygen of dioxygen(1-). J. Am. Chem. Soc. 1975, 97 (13), 3846–3847. DOI: 10.1021/ja00846a064.

6. Gall, R. S.; Rogers, J. F.; Schaefer, W. P.; Christoph, G. G. The structure of a monomeric oxygen carrying cobalt complex: dioxygen-N,N'-(1,1,2,2-tetramethyl)ethylenebis(3-tert-butylsalicylideniminato)(1-benzylimidazole)cobalt(II). J. Am. Chem. Soc. 1976, 98 (17), 5135–5144. DOI: 10.1021/ja00433a015.

7. Collman, J. P.; Gagne, R. R.; Reed, C. A.; Halbert, T. R.; Lang, G.; Robinson, W. T. Picket fence porphyrins. Synthetic models for oxygen binding hemoproteins. J. Am. Chem. Soc. 1975, 97 (6), 1427–1439. DOI: 10.1021/ja00839a026.

8. Collman, J. P.; Brauman, J. I.; Halbert, T. R.; Suslick, K. S. Nature of O2 and CO binding to metalloporphyrins and heme proteins. Proc. Natl. Acad. Sci. 1976, 73 (10), 3333–3337. DOI: 10.1073/pnas.73.10.3333.

9. Cerdonio, M.; Congiu-Castellano, A.; Mogno, F.; Pispisa, B.; Romani, G. L.; Vitale, S. Magnetic properties of oxyhemoglobin. Proc. Natl. Acad. Sci. 1977, 74 (2), 398–400. DOI: 10.1073/pnas.74.2.398.

10. Savicki, J. P.; Lang, G.; Ikeda-Saito, M. Magnetic susceptibility of oxy- and carbonmonoxyhemoglobins. Proc. Natl. Acad. Sci. 1984, 81 (17), 5417–5419. DOI: 10.1073/pnas.81.17.5417.

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Procedure to assign the irreducible representation of a derivative operator How can we assign an irreducible representation to a derivative operator like
$\displaystyle\frac{\mathrm d}{\mathrm dx}$ or the whole gradient operator $\nabla$ in a given point group?

Since the Hamilton operator can be assigned to the totally symmetric representation, it seems like there is a way to to do so for derivatives or operators in general, but I have no clue how to do so. I am well aware how to determine the representation for normal functions, but I have never seen it done for operators.

The transformation of a derivative operator under a linear transformation $\hat R $ is given by
$$
\frac{\partial}{\partial x_i} \rightarrow \sum_j R_{ji} \frac{\partial}{\partial x_j} = \sum R^T_{ij}\frac{\partial}{\partial  x_j}
$$
where $R_{nm}$ are the matrix elements of the matrix representation of the transformation operator in regard to cartesian coordiantes. The matrix is defined as the coordinate transformation matrix that takes the initial coordiantes to the transformed coordinates.
$$
\hat R : x \rightarrow  Rx
$$
The character of an representation with the derivative operators as basis is therefore the same as the character of the corresponding coordinate representation, since the trace of a matrix and a matrix transpose are identical. The derivative operators form thus a basis for the same irreducible representation as their coordinates and belong to the same irreducible representation.


Are Isopropyl Alcohol (2-Propanol) and 1-Propanol tautomers? They both are structural isomers, but are they tautomers too?

Tautomers need to be able to interconvert readily and quickly. This is done via proton transfer. For example in keto-enol tautomerism (https://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism#/media/File:Keto-Enol-Tautomerie.png) the proton on the alpha carbon donates its electrons to form a pi bond between the 2 carbons. As a result the pi bond between the carbonyl carbon and oxygen breaks and the now negatively charged oxygen picks up the proton which was originally lost.

In propan-1-ol and propan-2-ol, the oxygens are on different atoms and there is no way they can interconvert, so they are not tautomers.


Why radioactive elements emit alpha beta and gamma rays I am confused about this that why radioactive elements emits alpha beta and gamma rays WHILE other elements can't do so.

The stability of nuclei is really a sophisticated topic in theoretical quantum mechanics. But there is a simple way to think about what is happening that doesn't get too intense with the quantum mechanical theory.

Nuclei are made from two particles: protons and neutrons. But protons are positively charged and repel each other. The electromagnetic force is very strong and therefore this force is very large. So the first mystery is why all nuclei don't just fly apart.

The reason they don't is that there are two very short-range but very strong forces that bind the nucleus together: the strong and weak nuclear forces. Without getting into mind-bending topics in theoretical physics we can understand something about their net effect like this.

The interaction of the electromagnetic force and the two nuclear forces has some structure (it's quantum stuff, just accept it). Some combinations of protons and neutrons are more stable than others. Each combination has an energy level and some combinations have lower energy than others. Nuclei with even numbers of protons and neutrons are more stable than odd-odd combinations and nuclei with wildly unbalanced neutron to proton ratios are less stable. Neutrons act a little like a glue, helping protons stick together (this is an oversimplification as too many neutrons is also a source of instability: this is a consequence of a complicated interplay of several forces). But bigger nuclei are less stable and need a higher ratio of neutrons. And some large nuclei are just too large for the forces to keep them together so beyond a certain point all nuclei are unstable.

Some nuclei can be transformed into a more stable (lower energy) nucleus by various forms of radioactive decay. Nuclei with too many neutrons can emit a beta particle (this decay mode converts a neutron into a proton); elements with too many protons can emit a positron (converting a proton into a neutron). Bigger nuclei can become more stable by kicking out an alpha particle (which makes the nucleus significantly smaller, moving it towards the stable zone). Gamma radiation is associated with some of these modes: the high energy photons "mop up" the excess energy (I'm simplifying a lot).

This picture (more detail on wikipedia) sums up what we know about nuclear stability:

To sum up: nuclei are held together by the interaction of several forces but the forces not equally effective for all possible combinations of protons and neutrons. Some nuclei have so much energy that they are unstable and can decay into more stable nuclei in various ways depending on the mix on neutrons and protons they have. In some cases the forces hiding the nuclei together are strong enough that they never decay.


What are the effects of storing drinking water in a copper container? I recently learned that Ayurveda recommends storing drinking water in copper containers for positive health benefits.

Upon my simple experimentation, I've noticed the following effects:

Water Type : activated carbon filtered drinking water

Duration of storage: 8 hours

The taste of the water changed, it tasted more metallic. I think this is because of the increase of dissolved copper in the water.

If I'm drinking 4-5 liters of water per day, would that be still under the safe limit of 2 mg/day? ref 2 Note: I don't have any apparatus to measure the copper content in the water.

I've researched a bit and found that it does have some positive effects on killing some microbes such as bacteria ref 1. Note: I'm not able to deduce any conclusion about the increase in copper content from this paper even though the author said it is "within the permissible limits."

I would like to know more about the studies related to this claim and I'm really interested to know about the side effects.

As you state, copper storage containers have long been used to prevent growth of pathogens in potable water.

As for the danger of copper toxicity, it depends on the pH and water temperature. Between pH 8 and 11, about 1 ppm of copper dissolves, so if you're consuming ~5 l/day or about 5,000 g/day of water, and if the water has had sufficient time to become saturated, then you'd get ~5 mg/day of copper, exceeding suggested safe limits for long-term exposure.

That said, I think that, if the water pH is between 8 and 11, and if it is refrigerated (e.g. 4 C), and if it has been standing quiescently for no more than eight hours, it would not have time to be saturated with copper, so it probably would be safe to drink if that were the only source of copper in your diet.

Given all those mealy-mouthed caveats, then, you might at least check the pH of the tap water, and use a different container if it's acid. You would not want to be like the drinkers from Yudono spring, unless you're really into Heavy Metal.


Is the hydrogen atom attached to the benzylic carbon in benzaldehyde acidic? If benzaldehyde loses hydrogen atom at the benzylic carbon, the negative charge will disperse into the benzene ring via resonance. So shouldn't the conjugate base formed be stable, implying that the hydrogen was acidic?

Yes and no.

What you discuss is only possible if the $\ce{C-H}$ bond is perpendicular to the ring plane. However, since the carbonyl group is planar and better conjugation can be achieved with the $\ce{C=O}$ bond participating in resonance with the benzene ring the predominant configuration of unsubstituted benzaldehyde should be planar, i.e. the $\ce{C-H}$ bond being in the ring plane.

If you introduce steric hindrance, e.g. 2,6-di-tert-butylbenzaldehyde, the chance of the carbonyl group turning out-of-plane is much greater resulting in a better overlap of the $\ce{C-H}$ bond with the ring system and thus a lower $\mathrm{p}K_\mathrm{a}$.


Can we force a reversible reaction in equilibrium to achieve 100% completion? Can we force a reversible reaction in equilibrium to become a "complete" one? If so, will the equilibrium constant change?

It's said about reversible reactions that they never go to completion & I know that's because in reversible reactions, the mixture obtained in equilibrium is more stable than the extremes (i.e. either reactant or product) But if we force the reaction to proceed in a certain direction, say in the direction of products, will it ever go to 100% completion or near 100%?

My second question is regarding the equilibrium constant. I have read about equilibrium constant that its value doesn't change if we alter the initial concentration of reactants, which I understand why, but consider this situation:
 if we force equilibrium to proceed in the direction of products by altering $p,T,V$ (only and not adding or removing any substance), the equilibrium will get disturbed and obtain a new equilibrium. Wouldn't the equilibrium constant for this new equilibrium be different from the earlier one?

The two questions are sort of related, so i'll answer to the best of my ability.

A reaction in dynamic equilibrium can get close to completion depending on its equilibrium constant, which is a ratio between products and reactants at a certain temperature. No, the equilibrium constant won't change if you add more reactants in an attempt to push the reaction to completion. We refer to Le Chatelier's principle which states that if one were to add another parameter, say a reactant, the reaction will act to stabilize the stress that has been placed on it.

Take this reaction $\ce{PCl3 + Cl2 <=> PCl5}$. If I were to add more $\ce{PCl3}$, the change we would witness is a shift in equilibrium, that is to say, the system will react to consume more reactants and produce more products. What we see essentially is the system reacting to stabilize the changes we made.

This principle isn't just limited to adding reactants or products, however. We see the system tries to stabilize changes made to pressure, temperature, and volume. If we increase pressure, how will the system act to stabilize it? Pressure is a result of the gas particles hitting the walls of the container. Thus we would see the system reduced the number of gas particles in order to reach equilibrium again. Essentially, it will shift to the side with the least amount of gas particles. Volume and pressure are obviously related in this principle. An increase in volume is a decrease in pressure and a decrease in volume is an increase in pressure.

Take this reaction for example $\ce{CaCO3(s)<=>CaO(s) + CO2(g)}$. If we increase the volume (decrease the pressure), we would see the system shift to the right because it has the most moles of gas and we need more gas particles to fill the volume and stabilize the pressure.

Lastly, when changing heat we view it as a product or a reactant for exothermic and endothermic reactions respectively. An exothermic reaction has heat written as a product $\ce{A <=> B} + \text{heat}$ and an endothermic reaction has heat written as reactant. Thus, when we change the heat of an endothermic or exothermic reaction, we are changing its reactants or products and we treat it in the same way we treat a change in concentration.

The point is that the equilibrium constant remains constant, otherwise it would be called the equilibrium variable. We do, however, see a change in equilibrium position.


When you report Einstein units do you capitalize it? When you report Einstein units do you capitalize it? Since we're using it in the context of a unit (and not a name) is capitalization necessary?

Example:

The measured value was reported as 10 Einstein.

or

The measured value was reported as 10 einstein.

The non-SI unit einstein is not accepted for use with the SI and should be avoided.

Anyway, unit names are normally printed in upright type, and they are treated like ordinary nouns. Names of units are spelled with a lower-case initial in English, except in the beginning of a sentence when a capital initial is used. This also applies to unit names that are derived from a proper name of a person such as kelvin, newton, joule, pascal, watt, hertz, coulomb, volt, farad, ohm, siemens, weber, tesla, henry, and thus also einstein.

For SI units, it is only the unit name degree Celsius that contains a capital letter. In keeping with the rule, the unit degree begins with a lower-case d and the modifier Celsius begins with an upper-case C because it is a proper name.

However, in quantity names containing a person’s name, the person’s name is spelled with a capital initial, for example, the Einstein transition probability for spontaneous emission $A_{jk}$.

These writing rules are in accordance with following references:

ISO/IEC: ISO/IEC 80000 Quantities and units (14 parts)
BIPM: The International System of Units (SI)
IUPAC: Quantities, Units, and Symbols in Physical Chemistry (Green Book)
NIST: Guide for the Use of the International System of Units (SI)
ACS: The ACS Style Guide


Femtosecond spectroscopy Hi I am new to laser pump-probe spectroscopy and want to understand the concepts better. At the moment I am struggling to understand the following: When one takes absorption measurements with lasers in a pump-probe manner, the spectra that are collected need to be normalized. What does the Normalization mean and how does one do it? Also, why are relative changes important as opposed to absolute changes (in absorption) because I think they should give the same information?

I'm not sure exactly what you mean by normalising as it can mean more than one thing. At each probe wavelength ideally you measure two signals simultaneously from the same probe pulse which is split into two parts before entering the sample. One part probes the sample at the same place that it is excited and the other probes unexcited sample. You then record the transmission of these two signals and calculate optical density change or work with transmission depending upon choice.  It is clearly necessary to normalise to the pump laser intensity since a more intense pulse will produce a bigger effect. Second the probe pulses have to be monitored and adjusted as necessary (without exciting sample) so that these signals are kept equal or at least in a constant ratio so they can be normalised later.
Pump-probe is just like any other transient absorption measurement (used to be called Flash-Photolysis) and a difference signal is measured. This produces a transient spectrum which might be positive or negative. If you have the exact optical density the you can construct the spectrum of a transient species knowing the ground state spectrum. If there is no ground state absorption where you have a transient spectrum then this must be positive and is the transient spectrum. If it is negative it might be due to transient gain, i.e. the probe causing stimulated emission from the sample. The schematic shows one way the experiment can be set up. The absorption change at t = 0 is exaggerated for clarity; a 1% change is a huge signal.


What is the difference between moisture regain and water holding capacity? Water retention is how much water the fibre can retain. Water holding is function of hydrophilicity and construction of the fibre (fabric). What is the difference between those two? Because to retain is also to hold, in my knowledge.

According to the FabricLink Textile Dictionary:

  Moisture Regain - The amount of water a completely dry fiber will absorb from the air at a standard condition of 70 degrees F and a relative humidity of 65%. Expressed as a % of the dry fiber weight.

Water Holding Capacity is the amount of water a completely dry fiber will retain after soaking in liquid water and allowed to "drip dry".


Acidic nature comparison; benzoic acid and phenylacetic acid The $\ce{pK_a}$ values of benzoic acid and phenylacetic acid are around 4.2 and 4.31 respectively.
In benzoic acid, you have the resonance being the dominating effect, destabilizing a conjugate base anion. In phenylacetic acid, resonance doesn't work anymore, and only the -I inductive effect remains, which should stabilise the anion.
Why, then, is benzoic acid the stronger acid?

The acidity of benzoic acid $(\ce{C6H5-COOH})$ and its substituted derivatives, $\ce{R-C6H4-COOH}$ $(e.g., \ \ce{4-NO2-C6H4-COOH})$ have been studied extensively using the Hammett Equation. Similarly, acetic acid $(\ce{CH3-COOH})$ and its substituted derivatives, $\ce{R-CH2-COOH}$ $(e.g., \ \ce{Cl-CH2-COOH})$ have been studied extensively as well using similar Hammett Equation. You may read some brief discussion of Hammett Equation here.
Actually, you are comparing benzoic acid and a $\ce{R-CH2-COOH}$ where $\ce{R = Ph}$. Their is no relationship between these two acids except for both having $\ce{Ph}$ functionality. Benzoic acid has $\ce{Ph}$ directly connected to $\ce{COOH}$ group while acetic acid derivative has it connected through $\ce{-CH2\!-}$ group. Because of that, acidity of $\ce{Ph-CH2-COOH}$ is depend solely on inductive effect. It isworth noting that I cannot find any comparative studies of these two type of carboxylic aid in one single research.
The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20$ $(K_\mathrm{a} = 6.3 \times 10^{-5})$ while the $\mathrm{p}K_\mathrm{a}$ of acetic acid is $4.75$ $(K_\mathrm{a} = 1.77 \times 10^{-5})$. Thus, benzoic acid is about 3.6 times acidic than acetic acid. That is because when you consider $\ce{R-COOH}$ acid, it is more acidic if $\ce{R}$ group is electron withdrawing group (EWG) such as $\ce{Ph}$ group, compared to $\ce{R}$ group being electron donating group such as $\ce{CH3}$ group (EDG) like in the case of acetic acid.
Now, let's increase the electron withdrawing ability of $\ce{R-CH2-COOH}$ acid compared to acetic acid (where $\ce{R = H}$): The $\mathrm{p}K_\mathrm{a}$ value of 2-chloroacetic acid (where $\ce{R = Cl}$) is $2.87$ $(K_\mathrm{a} = 1.36 \times 10^{-3})$, which is about 76.8 times acidic than acetic acid. Thus, we can conclude that $\ce{ClCH2}$ group has larger inductive effect than that of $\ce{Ph}$ group. Thus, if we change $\ce{Cl}$ in $\ce{ClCH2}$ group to $\ce{Ph}$ group, then resulting carboxylic acid should be less acidic than that of 2-chloroacetic acid.
As given in OP's question, this carboxylic acid, 2-phenylacetic acid $(\ce{Ph-CH2-COOH})$ has a $\mathrm{p}K_\mathrm{a}$ value of $4.31$. It is more acidic than acetic acid, but not quite acidic as benzoic acid because of the extra $\ce{CH2}$ group. Remember, inductive effect is weaken by additional $\ce{C-C}$ bonds (c.f., ortho-, meta- and para-inductive effects).
Source of  $\mathrm{p}K_\mathrm{a}$ values


Mathematical basis of why enthalpy of mixing is 0 for ideal gas For this case, let's say there are two identical chambers connected by a small tube, constant pressure and temperature. There's a closed valve that separates the two chambers. Within the chambers there are two ideal gases (doesn't matter which, I think; they're ideal gases). I release the valve and the contents of the two chambers mix.

I understand that heat of mixing for ideal gases is 0 conceptually, because the molecules don't have interaction forces between them so introducing the gases to each other doesn't do much with respect to this. But what is the mathematical basis for heat of mixing being 0 for an ideal gas?

quick edit: above might not be clear. I mean that dH=TdS+VdP. In isothermal, isobaric conditions, dP could be zero, but dS isn't necessarily, that'd imply adiabatic conditions. So how is dH=0 for mixing an ideal gas?

Does the above line of thought apply for mixing? If so, do we take the terms before or after the volume increases? (in this case, final volume = 2*initial volume) I feel this is a frame of reference issue, one that comes up a lot for me in thermo.

A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}}H = \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S$ later): for isothermal and isobaric conditions it is

\begin{equation}
  G = \sum_{i} n_{i} \mu_{i} + \mathrm{constant} = n_{\mathrm{tot}} \sum_{i} \underbrace{x_{i}}_{= \, \frac{n_{i}}{n_{\mathrm{tot}}}} \mu_{i} +  \mathrm{constant} \hspace{4em} (1)
\end{equation}

with $n_{i}$, $\mu_{i}$, and $x_{i} = \frac{n_{i}}{n_{\mathrm{tot}}}$ being the amount, the chemical potential, and the mole fraction of the $i$th component in the system, respectively, and $n_{\mathrm{tot}} = \sum_{i} n_{i}$ being the total amount of gas in the system.
So, now we need an expression for the chemical potential of an ideal gas.
It can be shown that this is given by

\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{i}}{p^{0}}\Bigr) \hspace{18em} (2)
\end{equation}

where $\mu_{i}^{0}$ is the chemical potential of the $i$th component in a given standard state, $R$ is the universal gas constant, $T$ is the temperatur, $p_{i}$ is the partial pressure of the $i$th component, and $p^{0}$ is the standard pressure (i.e. the pressure connected to the standard state).
For ideal gases we can apply Raoult's law stating that

\begin{equation}
  p_{i} = x_{i} p_{\mathrm{tot}}
\end{equation}

where $p_{\mathrm{tot}}$ is the total pressure of the system.
Using this with equation (2) we get

\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\Bigl(\frac{x_{i} p_{\mathrm{tot}}}{p^{0}}\Bigr) = \mu_{i}^{0} + R T \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \hspace{6em} (3)
\end{equation}

Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.

\begin{equation}
  p_{\mathrm{tot}} = \sum_{i} p_{i}
\end{equation}

Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process.
At the initial state the chambers are seperated (whereby chamber $i$ contains only one sort of gas labeled by $i$) and so these subsystems are completely independent from each other.
Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the ambient/total pressure $p_{\mathrm{tot}}$ in the system, since $x_{i} = 1$ (each chamber/independent subsystem holds just one kind of gas).
According to equation (1) the Gibbs free energy of the $i$th chamber is then given by

\begin{align}
  G_{i} &= n_{i} \mu_{i} = n_{i} \biggl( \mu_{i}^{0} + R T \underbrace{\ln 1}_{= \, 0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
        &= n_{i} \biggl( \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}

and the Gibbs free energy of the total system at the initial state is

\begin{align}
  G_{\mathrm{ini}} &= \sum_{i} G_{i} \\
                 &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} x_{i} \underbrace{R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr)}_{= \, \mathrm{constant}} \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}

At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before.
So now, the complete system of connected chambers must be treated via equation (1)

\begin{align}
  G_{\mathrm{final}} &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} R T x_{i} \ln x_{i} + \sum_{i} x_{i} \underbrace{R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr)}_{= \, \mathrm{constant}} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}

Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:

\begin{align}
  \Delta_{\mathrm{mix}}G &= G_{\mathrm{final}} - G_{\mathrm{ini}} \\
                       &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)  - n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)  \\
                       &= n_{\mathrm{tot}} R T \sum_{i} x_{i} \ln x_{i} \hspace{22em} (4)
\end{align}

Differentiating equation (4) with respect to the temperature we get the entropy of mixing since $\frac{\partial G}{\partial T} \big|_{p, \{ n_{i} \}} = - S$

\begin{equation}
  \Delta_{\mathrm{mix}} S = - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \bigg|_{p, \{ n_{i} \}} = - n_{\mathrm{tot}} R \sum_{i} x_{i} \ln x_{i}
\end{equation}

Finally, you get the desired result by using the famous equation $G = H - TS$

\begin{align}
  \Delta_{\mathrm{mix}}H &= \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S \\
                        &= n_{\mathrm{tot}} R T \sum_{i} x_{i} \ln x_{i} - T n_{\mathrm{tot}} R \sum_{i} x_{i} \ln x_{i} \\
                        &= 0
\end{align}


How to determine if a compound is a strong, weak, or non-electrolyte? I'm stuck on questions asking whether a given chemical formula is a strong or weak electrolyte in an aqueous solution. I did some research online, but could not get a specific solution that I could readily comprehend.

My conception right now is that ionic compounds are strong electrolytes? But I have no idea when it comes to organic compounds or covalent bonds.

Can somebody explain?

It's best to start with basic definitions: electrolytes are compounds that allow a solvent to conduct an electric current. The mechanism is generally formation of ionic species, that is, solvated atoms or molecules carrying a net charge. As charge carriers, ions are the most important contributors to the electrical conductance of solutions.

Salts represent a reliable source of ions in solution. They tend to dissociate to some extent into their component ions in solution. It is usually ok to assume that in the presence of solvent the ions in a salt can exist either as part of the solid lattice or as free ions in solution. Only the free ions contribute significantly to conductance. Therefore solubility and conductance are directly related. A salt that dissociates to a large extent is referred to as a strong electrolyte. One that does not dissociate much is called a weak electrolyte. Generally speaking the degree of dissociation depends on the lattice free energy and the free energy of solvation of the salt in the particular solvent under given conditions of temperature and pressure. Degrees of dissociation for different salts or conditions are generally tabulated as solubilities and may be described qualitatively or quantitatively (say as solubility product $K_{sp}$). Qualitative trends are usually memorized by chemists and are taught at an early level to students. While detailed explanations can usually be found for differences in solubility, they are not usually the subject of introductory courses. Memorization (and mnemonics) tend to be the rule.

Some compounds that would not generally be regarded as salts (although are sometimes prepared in salt form) contain ionizable groups that can dissociate to form oppositely charged ions. Good examples are acidic groups, such as carboxylic groups that generate an conjugate anionic (negatively charged) molecule and hydronium (in water) upon dissociation, and basic groups, such as amines, that can accept a proton from a donor compound (such as water) to form a cation (positively charged ion). Generally speaking molecular organic compounds are regarded as weak electrolytes, since the degree of dissociation of the ionizable groups may be incomplete to a large extent. This is not universally true. Some groups readily dissociate, for instance sulfates. The degree of ionization depends inasmuch on the nature of the group in question as on the structure of the compound and the solvent conditions, including in particular identity of the solvent and temperature.

Compounds that are insoluble can obviously not contribute to the conductance and are therefore not electrolytes. Some compounds readily solubilize in the solvent (for instance nonionic polar compounds in water) but do not dissociate into ionic forms. Such compounds are not electrolytes.


Magnesium Bicarbonate Water Reaction With MSM - Methylsulfonylmethane The following is a recipe for a mineral water containing magnesium and bicarbonate.

Magnesium hydroxide powder is dissolved in a plain bottle of carbonated mineral water resulting in magnesium bicarbonate. This reaction causes the $\ce{CO2}$ to be consumed and the bottle caves in on itself.

But when I added the supplement MSM (methylsulfonylmethane) and the bottle is shaken, the water inside fizzes and pressure builds up again.

So my question: what causes the bottle to re-pressurize and fizz?

And is it still safe to drink, maintaining the original minerals, magnesium bicarbonate and msm?

On the reactions happening

It helps to have clear view of what actually happens at what point of the experiment. Before you add anything to your mineral water, $\ce{CO2}$ and water are in the following equilibrium (I am explicitly ignoring all other ions present, even if they are not spectator ions because they too can take part in acid-base equilibria):

$$\ce{CO2 (g) + H2O <=> CO2 (aq) + H2O <=> H2CO3 <=>\\ <=> HCO3- + H+ <=> CO3^2- + 2 H+}\tag{1}$$
$$\ce{CO2 (g) + 3 H2O <=> CO2 (aq) + H2O <=> H2CO3 + 2 H2O <=> \\ <=> HCO3- + H3O+ <=> CO3^2- + 2 H3O+}\tag{1'}$$

(Equation (1') is merely an equvalen version of equation (1) with extra water molecules to show that naked $\ce{H+}$ does not exist in solution.)

Under equilibrium conditions, most of the $\ce{CO2}$ is either dissolved ($\ce{(aq)}$) or present as minute gas bubbles ($\ce{(g)}$) and the amount of carbonic acid is rather low. Adding $\ce{Mg(OH)2}$, an Arrhenius and Brønsted base, shifts these equilibria rightwards by consuming protons:

$$\ce{Mg(OH)2 (s) + 2 H+ <=> Mg^2+ (aq) + 2 OH- (aq) + 2 H+ <=>> Mg^2+ (aq) + 2 H2O}\tag{2}$$

Since reaction (2) will actively push equilibrium (1) rightwards, gaseous $\ce{CO2}$ is dissolved and then reacted, lowering the bottle’s overall pressure. Now you are adding methylsulphonylmethane. This compound can act as an Arrhenius and Brønsted acid by the following equation:

$$\ce{H3C-SO2-CH3 <=> H3C-SO2-CH2- + H+}\tag{3}$$

Hence, it is able to shift equilibrium (1) back leftwards, counteracting the force of reaction (2). The acidity of dimethyl sulphone (as it is also called) is due to the strong $-I$ effect of the sulphone group, which has a formal $\ce{S^2+}$ atom in its centre. The resulting anion can be understood in a similar way as phosphorus ylides can.

Since equation (3) pushes the equilibrium (1) leftwards, it again becomes more favourable to release $\ce{CO2}$ into the surroundings, i.e. form gaseous $\ce{CO2}$, i.e. increase the bottle’s pressure.

On the safety aspects

You should never drink anything that has come into contact with laboratory grade chemicals. These are not tested to be safe for consumption, they are tested and controlled only to be safe for use in a lab (and hence in chemical reactions). There is at least one urban legend of a fatal accident happening due to mixed-up chemical bottles.

That taken care of, it seems that dimethyl sulphone be a relatively safe additive. According to Wikipedia, its $\mathrm{LD_{50}}$ value is larger than $17.5~\mathrm{g / kg}\text{ body mass}$, making it a safe chemical. It is sold as a dietary supplement, showing that the risk seems to be very low. As such, it will probably be safe to drink the soda after addition of both compounds, provided the pH value is where it should be (i.e. dimethyl sulphone has been added sufficiently to counteract the pH increase induced by $\ce{Mg(OH)2}$.)


How to create the equivalent of luminol for a compound? How would you go about creating a compound, like luminol, that would visibly react with monomorine I (5-methyl-3-butyloctahydroindolizine) on a household surface (like a kitchen floor)?

It is ok if the compund reacts with other similar compounds as well, as long as they aren't common to a household.

Background:

Luminol is the chemical used by forensic investigators (like those on CSI) to detect trace amounts of blood at crime scenes. The luminol is sprayed over an area and when it comes into contact with blood (specifically, the iron), it exhibits a blue glow.
Monomorine I is an ant trail pheromone (a trail guiding other ants to food).

Objective:

I'd like to be able to apply a compound on a surface and see the trails left by ants.

Probably not easily, especially if the concentrations are really low. Luminol + $\ce{H2O2}$ only works with blood because it's a catalytic reaction, i.e. it doesn't take much iron to create a lot of chemiluminescence. I'm not an organic chemist, but I'm not aware of any sort of similar reaction specific for this class of compounds. I did see this reaction: http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1521-4168(19980501)21:5%3C315::AID-JHRC315%3E3.0.CO;2-V/abstract Unfortunately, it's not open access and I can't share my institutional access, but it's basically a photo-generated ruthenium complex that reacts with tertiary amines and indoles which might be applicable here, though I'm guessing is not sufficiently sensitive. It's not terribly selective so you might run into interference problems and in that paper they use it for HPLC  detection so it may not be bright enough to see with the naked eye. (I have seen similar chemiluminescent compounds before and they're pretty dim, even with the lights out)

There are ways to produce visible changes in response to small concentrations of an analyte: LSPR, ELISA, etc. but they're a lot of work to implement for new analytes and probably too expensive for looking at a large area. Chemical imaging based on mid-IR or Raman spectroscopy may be applicable, though as far as I know, they're not geared for measuring large areas.

It seems that directly visualizing pheromone traces is a non-trivial task. Groups seem to do things like use image analysis (http://www.ploscompbiol.org/article/info:doi/10.1371/journal.pcbi.1002592) to estimate the levels of pheromones. I would like to have seen a comparison of their model with actual pheromone measurements and the way I would propose to do it is to take surface samples after the experiment. If a big sheet of paper was used as the surface, you could cut it up in a grid and extract the pheromone for analysis by LC-MS or whatever (or maybe even used directly as a paperspray emitter for MS). An ambient-ionization mass spec technique like LTP or DART would also probably be a good way to construct a surface map of pheromone concentrations.

Briefly: There are certainly ways of measuring trace amounts of things like pheromones, but a simple CSI-esque naked eye visualization probably isn't easily achievable.


Which steps are most critical in driving glycolysis foward? I'm assuming that the non-reversible steps (1,3,10 on the chart below) in glycolysis are the most critical in driving it forward, but I have a feeling it might be something else.  If so, why are they the most critical?

I've been using this chart as a reference.

The second half of my question is how mass-action ratio (Q) affects these reactions.

Those steps you mentioned are not merely critical steps, in glycolysis but are however the control points of the process and I would prefer to call them rate controlling steps. There is nevertheless one critical step which is believed to be the first committed step in glycolysis (not really sure if there are more).

There are three irreversible steps in glycolysis, and the differences
between glycolysis and gluconeogenesis are found in these three
reactions. The first of the glycolytic reactions is the production of
pyruvate (and ATP) from phosphoenolpyruvate. The second is the
production of fructose-1,6-bisphosphate from fructose-6-phosphate, and
the third is the production of glucose-6-phosphate from glucose.
Because the first of these reactions is exergonic, the reverse
reaction is endergonic. Reversing the second and third reactions would
require the production of ATP from ADP, which is also an endergonic
reaction.

Having said that, there are factors whih affect the process of glycolysis:
Enzymes
Control of hexokinase
The enzyme that catalyses the first reaction is hexokinase. The substrate of hexokinase is not necessarily glucose; rather, it can be any one of a number of hexoses, such as glucose, fructose, and mannose. Glucose-6-phosphate inhibits the activity of hexokinase; this is a control point in the pathway.  Hexokinase is inhibited by high levels of its product, glucose-6-phosphate. When glycolysis is inhibited through phosphofructokinase, glucose-6-phosphate builds up, shutting down hexokinase. (However, the liver contains a second enzyme that phosphorylates glucose-glucokinase).
Control of phosphofructokinase
The phosphorylation of fructose-6-phosphate is highly exergonic and irreversible, and phosphofructokinase, the enzyme that catalyses it, is the key regulatory enzyme in glycolysis. Phosphofructokinase is a tetramer that is subject to allosteric feedback regulation. fructose-2,6-bisphosphate an important allosteric activator of phosphofructokinase (PFK), the key enzyme of glycolysis; (it is also an inhibitor of fructose bisphosphate phosphatase (FBPase), which plays a role in gluconeogenesis.)
Control of Pyruvate Kinase
The final step of glycolysis is also a major control point in glucose metabolism. Pyruvate kinase (PK) is allosterically affected by several compounds. ATP and alanine both inhibit it. (It would be wasteful to breakdown more glucose is there is already vast amounts of ATP). Alanine is essentially pyruvate with an amino group. (converted to pyruvate via a transaminase enzyme). Therefore, a high level of alanine indicates that a high level of pyruvate is already present, so the enzyme that would make more pyruvate can be shut down. Fructose-1,6-bisphosphate allosterically activates pyruvate kinase so that the incoming products of the first reactions of glycolysis can be processed.
Substrate
The reaction in which fructose-6-phosphate is phosphorylated to give fructose- 1,6-bisphosphate is the one in which the sugar is committed to glycolysis. Glucose-6-phosphate and fructose-6-phosphate can play roles in other pathways, but fructose-1,6-bisphosphate does not. After fructose-1,6-bisphosphate is formed from the original sugar, no other pathways are available, and the molecule must undergo the rest of the reactions of glycolysis.
It is frequently observed that control is exercised near the start and end of a pathway, as well as at points involving key intermediates such as fructose-1,6-bisphosphate. This is the summarised version of control factors of glycolysis:

Remarks
To sum up, the reason I explained your mentioned reactions not to be critical is because (with reference to reaction 1 and 3):

Even if hexokinase is inhibited, glycolysis can still take place via catalysis by glucokinase in liver.

There are other feeder pathways for glycolysis even if the glucose sugar is not present, the saccharides (maltose, lactose, trehalose, and sucrose, mannose and galactose) can still be used to produce fructose 6-phosphate.

e.g $$\ce{Fructose + ATP  (Mg2+) → fructose 6-phosphate + ADP}$$

The last reaction would be bypassed by transaminase (is not important since the reaction would have been completed):

So these are major controlling steps, meaning glycolysis can still take place  (if committed step is reached) though at reduced rates until optimum conditions are restored. The second part of your question can’t not answer it because it is a different concept. Hope this helps
References

Harpers illustrated Biochemistry
Lehninger Principles of Biochemistry
Biochemistry (Campbell and Farrell)


What noble metal is most oxidation resistant? To clarify, by noble metals in this case I mean all platinum metals + gold, no copper, silver or rhenium. These noble metals are described as having low reactivity, but which one is least reactive or most "noble" of them all?
I am chemistry noob and I have feeling this might greatly depend on what type of element/molecule is trying to react with given noble metal so to make this simple I will narrow this question down to oxidation by diatomic oxygen in air at room and elevated temperatures.
I will now writte my current knowledge and thoughts on why some noble metals might be more noble than other. Its just window into my head so you can see what I think so you are able to correct what is wrong, approve what is right and expand on what is missing.
My current understanding of what makes noble metal noble, is that its somewhere in middle of periodic table, their electron shells are half full. Elements on the left side are strongly reactive because their shells are only sparsely filled, so they give away electrons easily, they have low ionisation energy.
Elements on the right side, excluding noble gasses, are strongly reactive because they have almost full electron shell and they have tendency to tear out and "steal" electrons from less electronegative elements. That is why noble metals are in middle,they don't want to give away or get inside electrons very much.
That would be my first trait of noble metal, the more in middle it is, the more noble it is. Since periodic table is 18 numbers wide, 18 is even number so there is no integer exactly in middle, the middle is between 9 and 10.
That means elements in group 9 and 10 should be more noble than those in 8 and 11 group because they are closer to middle, so rhodium, palladium, iridium, and platinum should be more noble than gold, ruthenium, and osmium.
Another thing that I believe is that the heavier elements don't hold their electrons as strongly as lighter elements because the inner electron shells shield the valence electron shell from attractive force of nucleus. If this is true, then noble metals in period 5 should be more noble than those in period 6, that is ruthenium, rhodium, and palladium more noble than osmium, iridium, platinum, and gold.
Third factor in my opinion is first ionisation energy, the higher it is, the more it should be resistant to reacting with oxygen. From highest to lowest first ionisation energy: Gold = 9.2255 eV, iridium = 8.967 eV, platinum = 8.9587 eV, osmium = 8.4382 eV, palladium = 8.3369 eV, rhodium = 7.4589 eV, ruthenium = 7.3605 eV. If this is true, gold should be most noble, and ruthenium least noble.
I read one study about oxidation resistance at high temperatures, the study said rhodium was most resistant, even more than iridium or platinum. I also heard Platinum doesn't oxidise in air at any temperature bellow melting point, that seems contradictory to the paper I read.Furthermore, spark plugs are mode from iridium and platinum but not rhodium.
If we had all these noble metals side by side in identical shape, lets say cube with highly polished sides, and we heated them in air atmosphere to 100, 500, and 1000 degrees Celsius, which one would oxidise least?

Ruthenium, Rhodium, Palladium, Osmium, Iridium, Platinum, Gold

  ... I will narrow this question down to oxidation by diatomic oxygen in air at room and elevated temperatures.

According to this criterion, gold is the most oxidation resistant of the above metals.

First of all, gold oxide cannot be obtained by reaction of gold and oxygen ($\ce{O2}$) at standard pressure. Gold oxide ($\ce{Au2O3}$) decomposes to gold and oxygen above $\pu{160°C}$.

Ruthenium reacts with oxygen (air) above $\pu{700°C}$ to give $\ce{RuO4}$. Ruthenium powder burns in air.
Rhodium reacts with oxygen (air) above $\pu{700°C}$ to give $\ce{Rh2O3}$.
Palladium powder burns in air.
Osmium powder reacts with oxygen (air) at room temperature to give $\ce{OsO4}$.
Iridium reacts with oxygen at high temperatures to give $\ce{IrO2}$ or $\ce{IrO3}$. Iridium powder burns in air.
Platinum powder burns in air.


How do you calculate the equilibrium constant with the absorbance of a substance and the absorbance constant? When given the equation: $$\ce{Fe^3+_{(aq)} + SCN^-_{(aq)} <=> FeSCN^2+_{(aq)}}$$

How do you calculate the equilibrium constant when given the slope of the absorbance vs concentration graph ($\pu{4317 M-1}$) and the absorbance of $\ce{FeSCN^{2+}}$ (0.276)

The following information is also given: $2.000\ \mathrm{mL}$ of a $0.00200\  \mathrm{M}$ solution of $\mathrm{KSCN}$ with $5.00\ \mathrm{mL}$ of $0.00200\ \mathrm{M}$ solution of $\ce{Fe(NO3)3}$ and $3.00\ \mathrm{mL}$ of water is combined.

So from this, I assumed that the concentrations of $\ce{SCN-}$ and $\ce{Fe^{3+}}$ are both $0.002\ \mathrm{M}$.

I found the concentration of $\ce{FeSCN^2+_{(aq)}}$ to be $\pu{6.39e-5 M}$ using this equation:

$$\mathrm{Absorbance} = \mathrm{slope}\cdot \mathrm{conc.}$$

$$\mathrm{conc.} = \frac{\mathrm{Absorbance}}{\mathrm{slope}}$$

$$\mathrm{conc.} = \frac{0.276}{\pu{4317 M-1}}$$

$$\mathrm{conc.} = \pu{6.39e-5 M}$$

Using this value, I used the equation for the K constant of an equilibrium:

$$\mathrm{K} = \frac{[\ce{FeSCN^2+}]}{[\ce{Fe^3+}][\ce{SCN^-}]}$$

$$\mathrm{K} = \frac{\pu{6.39e-5}}{0.002^2}$$

$$\mathrm{K} = 15.975$$

This did not turn out to be the correct answer, but I'm not sure why. I suspect the concentrations for the two reactions are not correct since the volumes are also given.

I'm following the outline from the comment by user21398.

Initial concentrations after mixing

The volume of the mixture is $V_\text{mix} = \pu{10 mL}$. Using the dilution law, I get the following concentrations:

$$\ce{[Fe^3+]_\text{initial}} = \pu{2.00 mM} \cdot  \frac{\pu{5 mL}}{\pu{10 mL}} = \pu{1.00 mM}$$

$$\ce{[SCN-]_\text{initial}} = \pu{2.00 mM} \cdot  \frac{\pu{2 mL}}{\pu{10 mL}} = \pu{0.400 mM}$$

Equilibrium concentrations

The equilibrium concentration of the complex is already calculated, $\ce{[FeSCN^2+]_\text{equil}}=\pu{6.39e−5 M}.$

To get the equilibrium concentrations of the reactants, we have to consider that some reacted:

$$\ce{[Fe^3+]_\text{equil}} = \ce{[Fe^3+]_\text{initial}} - \ce{[FeSCN^2+]_\text{equil}} $$ $$ = \pu{1.00e-3 M} - \pu{6.39e−5 M} = \pu{0.94e-3 M}$$

And similar for thiocyanide:

$$\ce{[SCN-]_\text{equil}} =  \ce{[SCN-]_\text{initial}} - \ce{[FeSCN^2+]_\text{equil}} $$ $$ =\pu{0.400e-3 M} - \pu{6.39e−5 M} = \pu{0.336e-3 M}$$

Equilibrium constant

This is copied from the question, using the values calculated in this answer above and dropping the unit M because the standard state is 1 M:

$$\mathrm{K} = \frac{[\ce{FeSCN^2+}]}{[\ce{Fe^3+}][\ce{SCN^-}]}$$

$$\mathrm{K} = \frac{\pu{6.39e-5}}{\pu{} \pu{0.94e-3}\cdot \pu{0.336e-3}}$$

$$\mathrm{K} = 202$$


NMR of phenylethane amine I did an undergraduate experiment and could use some help in interpreting the $\ce{^1H}$ NMR spectra.

This was the task:
About $\pu{5-10mg}$ (1-2 drops) distilled amine is weighed in a NMR tube. Put the top on the tube immediately to prevent reaction between the amine and $\ce{CO2}$  from the air. Weigh exactly 1.1 equivalent of (S)-alpha-acetoxyphenylacetic acid. Dissolve the acid in about $\pu{0.4 ml }\ce{CDCl3}$ and transfer the solution to the NMR tube.

This here is the $\ce{^1H}$ NMR:

Could I get affirmation of the choices drawn in red and what does the shift at $\pu{4.2ppm}$ correspond to? Could it be because $\ce{CO2}$ reacted with the amine?

(ChemdDraw was used to get the molecular figure and chemical shifts on the figure.)

I think some of your assignments are not correct. The doublet at $\pu{1.3 ppm}$ is correctly assigned. But close inspect of that signal suggests $\pu{12 Hz}$ coupling constant. Therefore, the signal given by the single proton causing that splitting should be a visual quartet with $\pu{12 Hz}$ coupling constant. But, your assignment for that proton is broad singlet at $\pu{4.75 ppm}$ with 2.7 integration (keep in mind that ChemDraw suggestion is not always true). I think that signal at $\pu{4.75 ppm}$ belongs to $\ce{NH3+}$. The small signal at $\pu{4.1 ppm}$, which is seemingly a quartet with large coupling constant is your $\ce{H-C(Ph)(CH3)NH3+}$ resonance. My best bet for the broad singlet next to it at $\pu{4.2 ppm}$ is crystalline water from  (S)-alpha-acetoxyphenylacetic acid ((+)-O-Acetyl-L-mandelic acid), which is crystalline solid.


Can an elimination reaction be visualised as the combination of the loss of a leaving group and proton transfer? Specifically in the conversion of a haloalkane to an alkene, i believe this is the reaction mechanism.

Extrapolating on this, do all elimination reactions function as such, or is the term elimination reaction a reference to the ends rather than the means?

Elimination refers to the ends rather than the means. On top of that, it does not necessarily refer to the loss of a proton + a leaving group; it can in general refer to the loss of any small molecule. So, for example, a loss of CO2 is also considered an elimination, even though there is no proton there.

As far as proton + leaving group eliminations go (elimination of HX), the order in which the leaving group leaves and the proton falls off determines what type of elimination it is.

The leaving group can leave first - that is called an E1 elimination.

The proton can fall off first - that is called an E1cb elimination.

And if both of them happen at the same time, it is an E2 elimination.


What's the difference between povidone K29/32 and povidone K30? I'm specifically looking at the two as inactive ingredients in the drug lamotrigine. How do the povidone K values differ chemically and/or in the context of lamotrigine?

They're obviously designed to be completely inert, but is there any chance they're acting differently in the body?

Firstly, Povidone (polyvinylpyrrolidone, PVP)  is  a synthetic polymer used in the pharmaceutical industry as vehicle for dispersing and suspending drugs.

Polyvinylpyrrolidone

There are quite a number of commercial grades of povidone (and also its highly cross-linked counterpart called crospovidone). The K-value of different types of povidone is characterised by their viscosity in solution  and also the mean molecular weight:

The method of obtaining these values is stated in various pharmacopoeias e.g British Pharmacopoeia

These different chemical characteristics offer a variety of uses in pharmaceutical formulations, including as a binder for tablets and capsules, a film former for ophthalmic solutions, to aid in flavoring liquids and chewable tablets, and as an adhesive for transdermal systems.

These also affect the formulation processes during tablet manufacturing for example, the low viscosity and high adhesive properties of K29/32 make it highly efficient as a binder for wet granulation. On the other hand, K30 has the ability to complexate with the active ingredient providing taste masking and solubilization.

  The grade of povidone that is selected depends mainly on its molecular
  weight, as this dictates the viscosity, binding effect, the
  complexation capacity and how readily it is eliminated by the body.

Povidone is relatively inert and won't necessarily affect processes in the body instead, on the formulation providing different release characteristics of drugs for example modifying its release properties etc

References

British Pharmacopeia ed
Povidone K29/32
Kollidon® 30 – The Original from BASF


How to convert ethanol to 1,1-dichloroethane? How to convert ethanol to 1,1-dichloroethane?

My approach: Convert ethanol to ethene followed by $\ce{HCl}$ addition. But that's no use. Well if it is possible to create gem-diol from here, then maybe it can be done. But not getting a suitable reaction. Can't seem to figure out anything. Please help!

convert ethanol to 1,1-dichloroethane?

A standard route for producing gem-dichloro compounds involves treatment of a ketone or aldehyde with $\ce{PCl5}$ (see the second to last paragraph in the "Chlorination of organic compounds" section here and page 18 here).

(image source)

Therefore, you could carefully oxidize your ethanol to acetaldehyde and treat it with $\ce{PCl5}$.


Why is an alcohol proton so shielded? Considering how electronegative oxygen is, why is the chemical shift of alcohol protons (1H-NMR) not so high? The chemical shifts of the protons on 3-propanol are about 1.20 for the carbon-1 hydrogens, about 2.2 for the alcohol proton, and about 4.01 for the carbon-2 hydrogens. Why on earth would the electrons on the carbon-2 hydrogens feel more pull from the oxygen than those on the alcohol proton?

Oh, I was just reading about this. This excerpt points out the factor contributing to the chemical shift observed in alcohols

  The extent of hydrogen bonding varies with concentration, temperature, and solvent, and changes in the degree of hydrogen bonding can cause substantial shift changes. This is very evident in the nmr spectrum of ethanol taken at different concentrations in CC14 (Figure 9-29). The hydroxyl resonance will be seen to move upfield at the lower concentrations. This is the result of decreasing association by hydrogen bonding through equilibria...

Source

So basically the reason (or a reason) for alcohols moving upfield due to hydrogen bonding, which results in decreasing association.

I looked at your link and there are a few notable things. They mention *variable, highlighting that it is not necessarily constant based on different conditions, which connects to hydrogen bonding. For carbon-2, we have the normal ~1.7 for methine protons + 2.5 from oxygen of alcohol, coming to be a pretty close value. So no, it's not "why on earth would the electrons on the carbon-2" but what on earth is going on with that alcohol?


What are identical molecules? I don't know what identical molecules are but these are by thoughts.

Since from the word identical the molecules should have the same number and kind of atoms attached in the same way. Since in stereo-chemistry the isomerism is due to spacial orientation all stereo-chemical isomers are identical.

Is my thinking correct? What factors determine whether two molecules are identical or not?

Identical compounds have the same chemical and physical properties (and other properties too).

Of the two types of stereoisomers, diastereomers (non-mirror image stereoisomers) are clearly nonidentical. Consider the isomers of tartaric acid. Meso-tartaric acid and (+)-tartaric acid have different melting points (165 $^\circ$C and 171 $^\circ$C, respectively). They also have different acidities.

The (+) and (-) isomers (enantiomers) are more challenging, but they are distinguishable by one physical property - the way they rotate plane-polarized light. If the two molecules were identical, they would not be isomers - we would not be able to distinguish them. If they are distinguishable, they are not identical.


What's Standardization, and why do it? Surprisingly, there are very few resources online (outside of obscure lab pages) that explain exactly what standardization is (not even Wikipedia!). The context is the phrase standardize the titrant for a titration. Does this mean you titrate it with a known solution?

Imagine that you need to titrate the sample S with a solution of the reagent R.

Unfortunately, R isn't available as a stock solution and maybe not even in solid form. All you have is it's precursor P and a recipe for the reaction $\ce{P + X + Y -> R}$. You never know how good that works out.

At this point, titration of S with R doesn't make any sense - unless you can determine the concentration of R!

Here, a stock solution with an exact concentration of the helper H, maybe in a sealed vial from a commerical source comes into play.

The titration of R with H to give the exact concentration of the solution of R is the standardization of the titrant.


Why is the equilibrium constant of water with its vapour equivalent to its vapour pressure? Can someone conceptually explain why the equilibrium constant is equivalent to the vapour pressure of water for the above equilibrium reaction?

For example if $K = x$, then it is said that vapour pressure $= x$ (in bars). However I seem to cannot understand why.

Lets start by writing the definition of the equilibrium constant, for some general reaction:

$$\sum_i^{n_\text{reac}} a_iA_i = \sum_j^{n_\text{prod}}b_jB_j$$

The above is a general reaction, where $a_i$ and $b_j$ is the stoichiometry and $A_i$ and $B_i$ is the molecules. For this reaction we can write the equilibrium constant:

$$K=\frac{\prod_j^{n_\text{prod}}\frac{B_j^{b_j}}{U_j^{\circ}}}{\prod_i^{n_\text{reac}}\frac{A_i^{a_i}}{U_i^{\circ}}}$$

Here $\prod$ is the symbol that denotes we take the product over a range. The $U_i^\circ$ are a standard unit, to ensure that our equilibrium constant is well defined.

Now if we look at your reaction:

$$\mathrm{H_2O\left( l \right)} \rightleftharpoons \mathrm{H_2O\left( g \right)}$$

We can from the above equations identify that; $a_1 = 1$, $b_1=1$, $A_1=\left[ \mathrm{H_2O} \right]$ and $B_1=p_\mathrm{H_2O}$.

We can thus write our equilibrium constant as:

$$K=\frac{\frac{p_\mathrm{H_2O}}{1\text{ bar}}}{\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}}$$

As you might note, I have also inserted the $U$s. For pressure $U=1\text{ bar}$ and for concentrations $U=1\text{ M}$. Now we need one last ingredient. You might have learned that the activity of liquids is $1$, i.e. the means that we have to set $\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}=1$, this is due to the definition of activity. If interested I would suggest to take a look at this. As our final result we can see that:

$$K=\frac{p_\mathrm{H_2O}}{1\text{ bar}}$$

I.e. we have that the equilibrium constant equals the partial pressure of water in standard units.


Oxidation state of the sulfur atoms in the thiosulfate Ion What are the oxidation states of the two inequivalent sulfur atoms of the $\ce{S2O3^{2-}}$ molecule? For all I know is that Lewis Structure does matter in most of the cases.

According to Oxidation state of sulfur in thiosulfate and implications for anaerobic energy metabolism

  according to the currently held view, the two sulfur atoms of thiosulfate exist in the oxidation state of sulfate (+6) and sulfide (−2) and do not change their respective oxidation states upon disproportionation. Our results based on XANES spectroscopy indicate that the two different sulfur atoms in thiosulfate have charge densities corresponding to +5 and −1 oxidation states

and in Using X-ray absorption to probe sulfur oxidation states in complex molecules

  the oxidation states were established for the outer and inner sulfur atoms of thiosulfate as −1 and +5, respectively. The revised values give a far better understanding of the role of thiosulfate as an energy substrate for bacteria because intra-molecular electron transfer, which generates metabolic energy with the production of sulfide and sulfate, cannot be envisaged with the previous view that the oxidation states were −2 and +6 for the outer and inner sulfur atoms of thiosulfate

Although both of the above articles are by the same author.

On the other hand, Determination of Changes in Sulfur Oxidation States
in Prostate Cancer Cells

  there is a difference between formal and apparent
  oxidation state received from XANES measurements, for
  example the apparent oxidation states of sulfur in thiosulfates,
  obtained from XANES data, are -1 and +5,
  while the formal ones are -2 and +6. Such a difference
  could occur due to the calibration curve which has been done
  on the basis of chemical compounds in which there are
  ionic bonds which isn't true in the case of thiosulfates.

However, it must be kept in mind that according to the current IUPAC definition, the oxidation state is not an actual property of a substance but instead a charge "an atom might be imagined to have", specifically:

  [Oxidation State] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion; (3) hydrogen has an oxidation state of 1 and oxygen has an oxidation state of -2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of -1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of -1 in peroxides, e.g. H2O2); (4) the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.

The current IUPAC definition of oxidation state doesn't specify how to apportion electrons of bonds between two atoms of the same element.  This causes various oxidation states to meet the IUPAC definition depending upon how one chooses to apportion the electrons of the S-S bond between the two S atoms.  Also, experimentally, the S-S is somewhere between a single bond and a double bond based on its length, so the number of electrons to be apportioned is also in question.  One could say the oxidation states are of the S atoms are -2, +6; -1, +5 or 0, +4 without contradicting the IUPAC definition.

In "Expanded Definition of the Oxidation State" there is a proposed addition to the definition:

  Bonds between like atoms (having the same formal charge) are cleaved homolytically.

If an S=S double bond of thiosulfate is cleaved homolytically, 0 and +4 oxidation states result.

If an S-S single bond of thiosulfate is cleaved homolytically, -1 and +5 oxidation states result.

However, the article concludes 0 and +4 oxidation states even with the single bond, because when there is a single S-S bond the two S-S atoms do not have the same formal charge. They give both electrons from the S-S single bond to the central S. Below is Fig. 4 from the article:

The most recent (June 2014) information (and probably most authoritative) I can find is Toward a comprehensive definition of oxidation state (IUPAC Technical Report) Pure and Applied Chemistry. Volume 86, Pages 1017–1081

In this article, another definition of Oxidation State is proposed requiring

  Bonds between two atoms of the same element are divided equally.

with no exceptions for formal charges.  This clearly rules out the possibility of -2, +6 oxidation states for the S atoms in thiosulfate, but leaves ambiguity between-1, +5
 if the S-S bond is single and 0, +4 if the S=S bond is double.

Specifically, concerning thiosulfate the article states:

  Structural properties of sodium thiosulfate suggest that all its terminal atoms carry some of the anion
  charge. In other words, all form ionocovalent bonds to a cation. The calculated Mulliken charges (−1.4 for
  O and −1.1 for terminal S) confirm this but, being different, they suggest unequal SS and SO bond orders. The
  SS bond distance of 2.025 Å (202.5 pm) is shorter than the single bond of 2.055 Å (205.5 pm) in crystalline
  $\ce{S8}$ or 2.056 Å (205.6 pm) in the $\ce{H_2S_2}$ gas, but substantially longer than the double bond of 1.883 Å
  (188.3 pm) in $\ce{S2O}$ or 1.889 Å (188.9 pm) in $\ce{S2}$. Given this uncertain situation, the approach chosen
  here is to draw in Fig. 24 two limiting Lewis formulas.

Then they proceed to draw two Lewis formulas, similar to above, one on the left with S-S single bond and -1 and +5 oxidation states for the sulfurs and one on the right with S=S double bond and 0 and +4 oxidation states for the sulfurs, and go on to say:

  The result in Fig. 24 means that the individual OS of the two sulfur atoms in thiosulfate are ambiguous,
  and no formula in-between these two limits yields reasonable values. The formula on left provides an
  unusual yet not entirely illogical OS of −1 for the terminal sulfur, reminiscent of the oxygen OS in peroxides.
  The formula on right, while having an SS bond order that is too high, suggests OS values that at times
  are used...

The article also comments that:

  A nominal OS is one that has been nominally adjusted against the generic OS definition in order to suit a specific purpose. That purpose may be...polarizing
  a homonuclear bond in thiosulfate to OS [of S] = +6 and −2 in order to emphasize the uniqueness of the central atom and similarity of all four ligands.