Detecting impurities in polyethylene glycol 3350 I would like to test samples of polyethylene glycol 3350 ($\ce{H−(O−CH2−CH2)_n−OH}$ with an average molecular weight of 3350, so n is around 75) to determine if there are any impurities and if so the rough concentration of them. Some likely impurities that may be present are:

ethylene oxide
diethylene glycol
ethylene glycol

My goal is to test samples from different sources to find out which one has the lowest level of impurities.
If there happens to be an easy way for me to test this at home using supplies that I would be able to buy online for < $1000 that would be great. More likely, I am interested in finding out what equipment would be used to test this so that I can try to find a lab that would do this testing for me.
Level of detection:
In a sample of 10 grams, I would want to detect levels at least as low as the following (though detecting even lower levels would be preferable):

ethylene oxide: 0.07 mg
diethylene glycol: 1 mg
ethylene glycol: 4 mg

If there is a way of detecting arbitrary impurities rather than just these specific ones I'd be interested in knowing that too (mass spectrometer? $\mathrm{^{13}C \; NMR}$? these are things I've heard of but am not sure what would be applicable here).

Extraction and GC-MS or maybe LC-MS, if you really need to measure these specific impurities
I don't know a lot about this area, but you are looking for very low amounts of impurities.  For ethylene oxide, for example, your desired detection limit of 0.07 mg in a 10 g sample corresponds to 7 parts per million, or 7 ppm.  Your limits for the glycols correspond to 100 and 400 ppm.
I'm not an NMR expert, but I think NMR will not be remotely  sensitive enough for this purpose.   Even if pure, NMR on a 0.07 mg sample is challenging and far from routine.
The best route forward might be a selective extraction of these impurities.  You should chose a solvent that does not dissolve polymeric PEG at all, or as little as possible.  This can be a challenge, as PEG is soluble in water, alcohols, and even dichloromethane.  Ethylene oxide is a relatively nonpolar gas, so you may be able to dissolve it in hexanes or a similar nonpolar solvent.  Unfortunately the glycols are unlikely to be soluble in hexanes.
So for ethylene oxide, say you extract 10 g of sample with X mL of hexanes, and say X is ~100 mL.  The resulting solution would be 700 ug/L of ethylene oxide at your desired detection limit.  That should be detectable by GC-MS.
For the glycols, supposing it is somehow possible to find a solvent that selectively dissolves the impurities but not the polymer, you could then derivatize the solution of glycols with a reagent like TBDMS and analyze them by GC-MS.  Alternatively, you may be able to analyze them without derivatization by LC-MS, but sensitivity of electrospray-based LC-MS for aliphatic alcohols is limited.  GC-MS is probably the way to go...if you can find a solvent for the selective extraction.
If no selective solvent is available, you might be able to analyze the glycols by directly derivatizing the solid PEG sample with a reagent like TBDMS in a liquid solution of some (non-selective) solvent.  The derivatization will convert the glycols into trialkylsilyl ether derivatives.  These may be lower polarity than PEG or more volatile than PEG such that they could be selectively removed and analyzed after derivatization.
Do you really need these specific impurities?
Maybe thermogravimetric analysis (TGA) would work for you.  There are variants of TGA where a mass spectrometer samples the gas phase above a material while it is slowly heated and decomposed.  If you have a sample of PEG you know is "pure enough", you could use that as a reference material, and then spike it with known amounts of contaminants to verify that these contaminants result in a perturbation to the TGA signal, or to the TGA-MS signal.
You may or may not get a precise numerical estimate of the mass fraction of your target impurities, but you will probably get a sense of whether a given 10 g sample of PEG is pure "enough" as measured by similarity to known reference material.
Analytical chemistry is hard
No matter what, you'll need to do a lot of methods development work to verify the appropriateness and robustness of your method.  E.g. what's the best value of X to use for selective extraction?  Can you for sure detect the impurities at your target levels with your chosen procedure?  You'll probably need to spike reference material with known amounts of impurities to make sure.  And is the response of your analysis method or instrument linear with respect to impurity mass fraction?  If so, over what range?
These are things that analytical chemists spend a great deal of time fretting about.  It's not easy.
Good luck!


How do I build a home-made apparatus for identifying what minerals are present in water? While there are many readily available kits in the commercial market for testing the type of minerals present in water,

is there a way we can build one at home?

I am ready to spend some bucks on raw materials, if I can get the knowledge of how to build such apparatus

This probably not what you are looking for, but the cheapest and least time-consuming analysis is the one done by others for free.

Over here (Germany), local water suppliers are required by law (Trinkwasserverordnung - TrinkwV) to monitor the quality of potable water. This includes a large number of microbiological and chemical parameters. The details are given in the annexes 1-5 of the TrinkwV.

The results are published and are are available online at no cost.

Examples (mostly in German) are:

RheinEnergie (Cologne area)
RWW - Rheinisch-Westfälische Wasserwerksgesellschaft (one of the largest suppliers in Germany)


What happens to electrons in metal complexes after excitation by visible light? My book writes:

  When white light shines through a solution of a complex ion of a transition metal, photons of a particular frequency are absorbed and their energy promotes an electron from lower energy level to upper energy level. We perceive the complementary color to the light absorbed.

However, I have a question about this. Once the electron is promoted to a higher energy level, won't it fall back down and give off a photon?

Surely the electron cannot remain at the higher energy level forever, as that would mean that no further electronic transitions could taking place, and the solution would turn colourless.

Absorption of a photon typically results in a vibrationally excited higher electronic state of the same multiplicity.

$$\ce{S_0 ->[$h\nu_\mathrm{ex}$] S_1}$$

In most cases, the excited state deactivates through internal conversion in a radiationless process via vibrational energy exchange with solvent molecules. No light is emitted here, but the solvent is actually heated up, although it might not be measurable.

There are however cases, in which the initially formed excited singlet state species first relaxes to its lowest vibrational mode and then emits a photon to reach the ground state.

$$\ce{S_1 -> S_0 + $h\nu_\mathrm{f}$}$$

This deactivation via radiation is known as fluorescence. Fluorescence lifetimes are typically rather short (in the ns range) and the emitted photon has a lower energy than the absorbed one. This means that the emission spectrum is shifted to the red end of the uv spectrum ($\lambda_\mathrm{f} > \lambda_\mathrm{ex}$). The effect is known as Stokes shift.

A classical case for a fluorescent transition metal complex is $\ce{Ru(bpy)3^{2+}}$.

In acetonitrile, it absorbs around $\lambda_\mathrm{ex} = 460~\mathrm{nm}$. The fluorescence is observed around $\lambda_\mathrm{em} = 620~\mathrm{nm}$.

Other bidentate ligands - more rigid than 2,2'-bipyridine (bpy) - often used are 1,10-phenanthroline (phen), dipyrido-[3,2-a:2',3'-c]phenazine (dppz), and benzo[i]dipyrido-[3,2-a:2',3'-c]phenazine (dppn).

In other cases, the excited singlet state undergoes a change in the spin multiplicity (Intersystem crossing) to a triplet state.

$$\ce{S_1 ->[ISC] T_1}$$

These excited triplet states are usually much longer-lived. When they eventually deactivate via emission of a photon, this process is known as phosphorescence.

$$\ce{T_1 -> S_0 + $h\nu_\mathrm{p}$}$$

Keeping all these processes in mind, one question arises - particular from younger students or pupils:

If our metal complex absorbs a photon and forms an excited state with such a short lifetime, why can we see a nevertheless observe permanent colour?

The answer is in the enormous numbers!

Continuous irradiation with white light means that we send a huge number of photons. Moreover, the solution of our transition metal complexes in a cuvette (or a test tube) also contains more than one molecule. In fact, much more!

As a result, we have a huge number of incidents

The complex absorbs a photon of a certain energy, which means that one colour is "swallowed". The complex forms a short-lived "excited state"
the solution reflects the remaining light
the excited state complex deactivates again

Since this happens all the time while we shine the light on our sample and for a lot of molecules, we observe a summary of the events: a coloured solution.


Mass of sucrose solutions What is the mass of 150 ml of 40% sucrose, and what is the mass of 150 ml of 20% sucrose? I don't know exactly how to figure it out without using a device to tell me the mass. All I know is the mass of 10 ml of 40% sucrose, which is about 15.7 grams.

If 10ml of 40% sucrose is 15.7 grams, then the mass of 150 ml of 40% sucrose is

$$ 15 \times 15.7 grams$$

You can not determine the mass of 150 ml of 20% sucrose from the information given.

You could calculate from this sucrose solution density table:
http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html

Note that the table shows the premise "I know is the mass of 10 ml of 40% sucrose, which is about 15.7 grams" to be incorrect.


Does a weak buffer solution still maintain a mostly steady pH if the acid or base load it has to buffer is small? I need to buffer the $\mathrm{pH}$ of a solution to $3.0$ using a citrate buffer.  I found a recipe to make a buffer solution that would use $82\%$ of $0.1\,\mathrm M$ citric acid (monohydrate) and $18\%$ $0.1\,\mathrm M$ trisodium citrate dihydrate to give a citrate buffer with a $\mathrm{pH}$ of $3.0$.  The purpose of buffering the $\mathrm{pH}$ to $3$ is that this solution contains an active ingredient that has a very short shelf-life at a more neutral $\mathrm{pH}$. The buffer solution will make up about two-thirds of the solution by weight, and isopropyl alcohol will make up about $30\%$ of the solution.  Assume all of the remaining ingredients are non-dissociable in the relevant $\mathrm{pH}$ range, except for the active ingredient which has the following properties:

Concentration of $0.03\%$ by weight
A single weakly basic moiety with a $\mathrm pK_\mathrm a$ of $11.45$
Molecular mass of approximately $420\,\mathrm{g\over mol}$

Since this will be applied to skin, I'd like to use the weakest buffer solution that would still maintain the $\mathrm{pH}$ at $3.0$.  My question is, if I use $0.01\,\mathrm M$ or $0.001\,\mathrm M$ solutions to make the buffer, would it still be able to buffer the $\mathrm{pH}$ at $3.0$?

First, a disclaimer: This answer addresses only the question of the chemical buffering capacity of the questioner's formulation, given the information provided. Neither I nor Stack Exchange / Stack Overflow can take any responsibility for the safety or efficacy of the proposed formulation, for the effectiveness of the buffer concentration indicated below for maintaining the formulation $\mathrm{pH}$, or for any other hazard, harm, or adverse event, without limitation, arising from the use of the information provided below.

To answer the general question of the title: Yes.

You haven't given enough information to answer your question in the particulars, though.

For your particular application:

Worst-case, buffer-wise, as long as the following assumptions are accurate, the maximum amount of buffering capacity you'd need can be estimated from the total number of moles of active ingredient:

None of the other ingredients have appreciable acid-base character (appears to be the case here)
Degradation of the other ingredients into acidic/alkaline species can be neglected
No acidity or alkalinity is likely to enter from the environment

At $0.03\%$ w/w, assuming the overall density of the formulation is approximately $1\,\mathrm{g\over cm^3}$ (probably reasonable, with $67\%$ w/w composed of aqueous buffer) a species with a molecular weight of $420\,\mathrm{g\over mol}$ will have a mole concentration of $0.714\,\mathrm{mM}$.  In general, I would probably want the buffering species to have a concentration about five to ten times my expected net acid/base load. So, I would choose no less than about $3.6\,\mathrm{mM}$ for the total concentration of all forms of citrate, perhaps going as high as $7.2\,\mathrm{mM}$ to be on the safe side.

Note: The $\mathrm pK_\mathrm a$ of the ionizable group on the active ingredient is sufficiently far from your target $\mathrm{pH}$ that it will be fully protonated in your formulation.  Thus, if it is possible to add this ingredient in its protonated form, the required buffer concentration is probably somewhat lower than the above.

General Principles

As long as the number of moles of acid or base to be neutralized is small compared to the total number of moles of the relevant buffering species in the solution, and as long as the target $\mathrm{pH}$ is within about one unit of the nearest $\mathrm pK_\mathrm a$, the $\mathrm{pH}$ should hold steady.

As the number of moles of acid/base to be buffered against approaches the number of moles of the buffering species, the target $\mathrm{pH}$ needs to be closer and closer to the $\mathrm pK_\mathrm a$ in order to hold the $\mathrm{pH}$ steady.

If the number of moles of acid/base to be buffered against exceeds the number of moles of buffering species, the buffer will in general fail to hold the target $\mathrm{pH}$. The main exceptions would be polyprotic species with multiple $\mathrm pK_\mathrm a$ values in close proximity (polyphosphates come to mind).


Metal Evaporation in GFAAS? Our lab is doing aqua regia digestion of an ore matrix for the determination of Platinum Group Elements (PGE). We're getting white smoke when we run it through a Graphite Furnace assisted Atomic Absorbance Spectroscopy (GFAAS). Smoke forms above 1000 C, so we think it's some metal evaporating. Any thoughts or ideas? Thanks in advance!

No, you do not evaporate the metals in question.  The use of aqua regia in the analysis aims to dissolve them, to form salts, which in turn are significantly more volatile than the metals or will decompose in the gaseous state.  (Just have a look alone at the melting points of the pure metals, which still are considerably lower than the boiling points, too.)

Without further detail about the colour of the "smoke", it remains a speculation that the sample placed on the L'vov plate still might have been a bit wet (from the acidic digestion).  If so, it were not good for the instrument, as it may lead to corrosion to metal parts of the housing around the graphite tube.


Increasing available area of a solid will increase effective concentration? If I say that making a solid reactant into a powder will increase the yield of the product, because it increases the available area of the solid, hence increasing its effective concentration. Is this claim valid?

The concentration of solid in itself is always the same, like m3wolf said in the comments (unless the solid is not pure, which is not relevant anyway, because grinding wouldn't change that either). So no, you can't change the effective concentration this way.

If anything, you will have higher surface area and hence higher overall rate of reaction (or dissolution, if the reaction happens in the solution and not on the surface, and thus has to be preceded by dissolution). In this manner, you will get the same yield, but in less time.

Secondary effects can eventually change your yield in an unpredictable way, as likely down as up.


Gibbs Free Energy and Maximum Work I am a high school student and my professor mentioned that the Gibbs free energy is the maximum amount of work (or useful work) that a system can do, whereas entropy is a measure of the non-available enthalpy.

I am not able to understand this and it would be helpful if you could explain this or link to some resources that could explain this at a high school level (I know some multivariable calculus, but not that comfortable with it) as I have difficulty finding them.

Also, we used the fact that Gibbs free energy is the maximum amount of work to derive the equation $∆G = -nFE$ and then proceeded to derive the Nernst equation.

In order for a process to happen, it has to increase the combined entropy of the system in which it happens and of the surrounding (2nd law of thermodynamics). As we will see in a bit, the more work the process does on the surroundings, the less the entropy of the surroundings increases, putting a limit on how much work the process can do.

What happens to the energy leaving a system?

When a process brings the system from a higher to a lower energy, the energy leaves the system in the form of work done on the surroundings or heat transferred to the surroundings (1st law of thermodynamics). For a given set of initial and final states, the amounts of work and heat can vary, but the sum has to be the same. For example, if our system is a charged battery, we can either run a motor with it, lifting up a rock (work done on the surrounding). Or we can short the battery, creating thermal energy that is transferred to the surrounding in the form of heat. We can even have a situation where heat and work have opposite signs (e.g. do more work and have heat transferred into the system to get the same energy balance).

How is heat and work related to entropy changes?

The change in entropy inside the system depends solely on the initial and final state. For the surrounding (with nothing else going on in the surrounding), the change in entropy depends on the heat transferred. The more heat is transferred to the surrounding, the more the entropy of the surrounding increases. To maximize the work, you want to minimize the heat transferred to the surroundings (or even transfer heat to the system), but only to the point the that total entropy still increases (otherwise the process could not run). So if you match the entropy change in the system with the opposite entropy change in the surrounding (via heat exchange), you get the maximum work out of your process.

Can I see a picture?

Sure. The picture shows that the sum of heat and work is equal to the enthalpy change of the system. Values shown with a downward arrow are negative and with an upward arrow positive. For heat and work, negative values mean energy is transferred out of the system. For example, work done by the system on the surrounding is a downward arrow, and exothermic reactions have a downward arrow for $\Delta H$.

The picture also indirectly shows the entropy (via the arrows labeled heat and $-T \Delta S$. If the arrow for $-T \Delta S$ is downwards, entropy in the system increases (scenarios labeled "more dispersed"). If the arrow for heat is downwards, entropy in the surrounding increases. To calculate the actual entropy, you would have to divide by the temperature, but the way it is shown everything can go into a single picture.

I have three scenarios, depending on whether the process is exothermic or endothermic and whether the entropy in the system increases or decreases. One scenario (endothermic reaction with a decrease of the system's entropy) is not shown (you would have to do work on the system to drive such a process, and we are trying to make the process do work for us instead).

In scenario 1, the process is exothermic. Some of that available energy is transferred in the form of heat (to make sure that the overall entropy does not decrease) and some of it is transferred in the form of work done by the system.

In scenario 2, the process is also exothermic, but different from scenario 1 the entropy of the system increases in this scenario. The work done by the system is (surprisingly, perhaps) larger than the enthalpy change, and the extra energy comes from heat transfer from the surroundings.

In scenario 3, the process is endothermic. You might think that this process can't do any work because - where would the energy come from? It comes from the surrounding, partially to provide the energy for the endothermic process and partially to be transferred back into the surrounding in the form of work. This can happen because just as in scenario 2, the entropy of the system increases.

  [...]the Gibbs free energy is the maximum amount of work (or useful work) that a system can do, whereas entropy is a measure of the non-available enthalpy.

This sort of describes scenario 1, which I pasted again below with two boxes for emphasis.

The pink box represents the 1st law of thermodynamics. There is some enthalpy available from the process, and this will leave the system either as work or heat or some combination, as long as it adds up to the enthalpy change in the system. The blue box represents the 2nd law of thermodynamics. The entropy increase of the surrounding (through transfer of heat to it) has to be larger or equal to the entropy loss in the system. If you combine the two laws, you get the maximum work the system can do. In this scenario, the change in entropy of the system times the temperature represents that portion of the enthalpy not available for work (labeled $-T \Delta S$ in the diagram). You can't, however, use that label in the other scenarios where the entropy of the system increases. In those scenarios, the entropy change in the system results in more work available than the enthalpy change would suggest.

This is all rather complex and might make you dizzy. The Gibbs energy helps us to summarize all this in a single quantity. It combines entropy and enthalpy changes in the system in a neat way that tells us about the maximum work. In the absence of work, it also tells whether the process will happen in that direction or not.

Disclaimer for those that have a physical chemistry background

This is for a closed system at constant pressure and temperature (i.e. P-V work is done against a constant pressure, and initial and final state have the same temperature). Whenever it says work, it refers to non-P-V work only. Labels $\Delta H$, $\Delta G$, $\Delta S$ should read $\Delta H_\text{sys}$, $\Delta G_\text{sys}$, $\Delta S_\text{sys}$.


How to calculate the concentration after a certain timespan from the specific reaction constant? A container at $30~^\circ\pu{C}$ contains $\ce{HNC_{(g)}}$ reacting to form $\ce{HCN_{(g)}}$. Knowing that the specific reaction constant is $4.4\cdot10^{-4}~\pu{s-1}$ and that you initially have $\pu{1 g}$ of $\ce{HNC}$: how many moles of each gas are in the container after one hour and a half?

I'm pretty sure that the reaction is reversible here. However, I have no idea how to calculate how many moles of each gas will there be after $\pu{1.5 hrs}$ (even if it was not reversible).
I have many doubts:

Did they skip the concentration because there's only one reactant?
Does $K$ take into account the $30~^\circ\pu{C}$?
What's the equation for "change of concentration over time" here?
How can I figure out how many moles of $\ce{HNC}$ and $\ce{HCN}$ the container will have after 90 minutes, ignoring the fact that it's a reversible reaction?

To answer this, you should know what is the concentration of HNC after 90 minutes. First, you should know the initial concentration of HNC.

After that, look at the rate constant. It tell you if the reaction is a first order. So, you can apply the data in integrated first-order rate law equation:

Don't forget if you must change 90 minutes to seconds :)

Then, now you find the concentration of HNC after 90 minutes. To figure out the moles of HCN after reaction, we know that the initial moles of HNC is the same as the moles of HNC after reaction plus the moles of HNC that changed to HCN.

Then after submit the initial moles of HNC and moles of HNC after reaction, you get the moles of HNC that changed to HCN. This is the same of moles of HCN that
present in the container.


How can the gauche-effect be explained? Generally, anti-conformations are more stable. But in the case of 2-fluoroethanol, the gauche confirmation is more stable.

It's given in March's advanced organic chemistry book that intramolecular hydrogen bonding is not the correct reason for the stability. Wikipedia explains it using hyperconjugation, but I am not able to understand it. What is the actual reason for the gauche effect?

The gauche effect is commonly explained with LCAO-based bond orbitals. LCAO is short for linear combination of atomic orbitals and implies that we can take two atomic orbitals $\phi_1, \phi_2$ and create two molecular orbitals $\psi_1, \psi_2$ from them by linear combination: $\psi_1 = a_1 \phi_1 + b_1 \phi_2$ and $\psi_2 = a_2 \phi_1 + b_2 \phi_2$.

Considering two neighbouring $\ce{sp^3}$ hybridised carbon atoms, we can create a $\unicode[Times]{x3c3}$-orbital between them by overlapping a hybrid orbital of each; and if $a_1 = a_2$ and $b_1 = - b_2$ (or vice-versa) we can create an antibonding $\unicode[Times]{x3c3}^*$-orbital; both are shown in figure 1 (the bonding in the bottom row, of course). Note that the lobes on each side are the same size.

Figure 1: $\unicode[Times]{x3c3}$ and $\unicode[Times]{x3c3}^*$ orbitals of $\ce{C-C}$, $\ce{C-H}$ and $\ce{C-F}$ bonds (schematically).

If we consider a $\ce{C-H}$ bond rather than a $\ce{C-C}$ bond, we need to take electronegativity into consideration. Compare with the middle column of figure 1: The more electronegative element, carbon, contributes more to the $\unicode[Times]{x3c3}$ orbital, while the less electronegative element hydrogen contributes more to the $\unicode[Times]{x3c3}^*$ orbital. (Note that hydrogen uses a 1s-orbital for binding.)

Similarly for the $\ce{C-F}$ bond, although now fluorine is more electronegative than carbon. And note how fluorine uses a p-orbital for bonding: No energy or anything else would be gained by hybridising fluorine’s orbitals. Neither the size difference between hydrogen’s and carbon’s nor that between fluorine’s and carbon’s orbitals is to any scale and both are grossly exaggerated.

The gauche effect, as a special form of hyperconjugation, relies us on looking at how these molecular orbitals can (or cannot) interact with each other. If we look at the relevant part of the molecule (compare figure 2), we can see that if hydrogen and fluorine are anti to each other, we get a $\unicode[Times]{x3c3}_{\ce{C-H}}$ bond and a $\unicode[Times]{x3c3}^*_{\ce{C-F}}$ bond that have favourable geometries and favourable lobe sizes for overlap.

Figure 2: Potentially overlapping orbitals in the gauche and anti cases of 2-fluoroethanol.

On the other hand, as also depicted in figure 2, the overlap would be bad if oxygen and fluorine be anti due to the small contribution of the less electronegative carbon to the $\unicode[Times]{x3c3}_{\ce{C-O}}$ orbital. (For simplicity, I have rendered oxygen using a plain p-orbital. Oxygen usually has orbitals somewhere between p and $\ce{sp^2}$.)

All this so far tells us why the interaction is geometrically favourable, but we yet did not deduce why it is energetically favourable. And here is where we are, more or less, linear combining the two molecular orbitals we created by linear combination earlier. The combining we are doing is a very general thing, observed not only in the gauche effect here or in hydrogen peroxide, but also is the basis for the anomeric effect in carbohydrates and hyperconjugation. The principle is: If we combine an occupied orbital with an unoccupied one, the occupied orbital gets lowered in energy while the unoccupied gets raised. Because only the occupied orbital contains electrons and their energy is lowered, the entire system gains stability. (You can read the two bolded half-sentences as one.) Compare figure 3.

Figure 3: Stabilisation by combining occupied and unoccupied orbitals

What we have done, is lower the energy level of an already lowered (because bonding is favourable) occupied orbital; the entire system gains. We often state this by saying things such as:

  Electron density is transferred into the antibonding orbital.

But it really boils down to stabilising occupied orbitals. However, that transfer of electron density helps us understand part of the consequences associated with this favourable mixing. If both the bonding and the antibonding orbitals of a certain bond are occupied, you can think of that bond as non-existant; as if both sides had an electron pair. If, on the other hand, neither orbital is occupied, the bond does also not exist, but due to lack of electrons. We can use that analogy to create a mesomeric description of the gauche effect with the two resonance structures depicted in scheme 1.

Scheme 1: A possible depiction of resonance stabilisation corresponding to the gauche effect.

Many chemists out there will vigorously refute this depiction as invalid. Indeed, the resonance structure on the right-hand side of the mesomery arrow shows charge separation and a non-optimal number of bonds. We can therefore assume that it only contributes ever so slightly to the complete picture — which is in line with the electron density only minutely being transferred and also the new bonding orbital (figure 3) being a lot closer in shape and size to the $\unicode[Times]{x3c3}_{\ce{C-H}}$ orbital rather than $\unicode[Times]{x3c3}^*_{\ce{C-F}}$ orbital. But the depiction certainly helps us understand why oxygen cannot do the job: Imagine $\ce{HO+}$ instead of $\ce{H+}$. Looks ugly and unstable from a chemical point of view.


Difference between Reaction Rate and Rate Law? Lets say you have the reaction

$$\ce{$a$A + $b$B -> $c$C + $d$D}$$

The rate of the reaction is given by the change in initial concentration over the change in time.

$$-\frac{\Delta \left[\ce A \right]}{a\Delta t} = -\frac{\Delta \left[\ce B \right]}{b\Delta t} = \frac{\Delta \left[ \ce C \right]}{c\Delta t} = \frac{\Delta \left[\ce D \right]}{d\Delta t}$$

However, there is also the rate expression, which is of the form

$$\text{reaction rate} = k \left[\ce  A \right]^{m}\left[\ce  B \right]^{n}$$

What is the difference between these two, and what purpose does each one serve?

Further, what does the $k$ represent?

Let's use the following gaseous chemical reaction as an example:
$$\ce{H2 + I2 -> 2HI}$$
When we study reaction kinetics, we oftentimes can't observe the changes in concentration of each reactant and product.  Therefore the rate expression helps us to relate the rates of appearance/disappearance of the species present:
$$-\frac{1}{1}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\;
= \; -\frac{1}{1}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\;
=\; \frac{1}{2}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{rate expression}\label{rateex}$$
Note that each rate is modified by the inverse of the stoichiometric coefficient.  I explicitly put $\frac{1}{1}$ in the expression just to emphasize this point.  We can see that multiplying the coefficients of the reaction will not change the rate expression in the following example: let's multiply the coefficients in the chemical reaction by 2 and then write the corresponding rate expression.
$$\ce{2H2 + 2I2 -> 4HI}$$
$$-\frac{1}{2}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\;
= \; -\frac{1}{2}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\;
=\; \frac{1}{4}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{new expression}\label{newex}$$
The $\ref{rateex}$ and $\ref{newex}$ are numerically equivalent.

Full disclosure here: the term rate expression is ambiguous and is also used as a synonym for the rate law. I've also seen the rate expression referred to as the rate relationship.

The rate law or differential rate law relates the rate of a reaction to the concentration (or pressure) of the reactants.  The rate of a reaction is proportional to the concentration (or pressure) of the reactants modified by some experimentally determined number called the reaction order.
$$\begin{align}
-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}
&\propto\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\\
-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}
&= k\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\tag{rate law}
\end{align}$$
Here, $m$ and $n$ are the reaction orders for $\ce{H2}$ and $\ce{I2}$, respectively, and the sum, $m + n$ yields the overall order of the reaction.  The proportionality $a\propto b$ can be replaced with an equality $a = k b$.  In kinetics, this constant of proportionality is referred to as the rate constant.

In summary: the rate expression tells us how the appearance/disappearance rates of the products and reactants relate to one another; the rate law tells us how the rate is related to the concentrations (or pressures) of reactants and the rate constant is a constant of proportionality that comes out of the rate law.

This answer is a summary of information found at the boundless website and from Tro's most recent general chemistry textbook.


Is pressure equalization "diffusion"? There is another question on the site which is:

  Two containers, A and B, connected by a closed valve were evacuated and the Chlorine gas (Cl2) with a pressure of 200kPa was put into A and Oxygen (O2) with a pressure of 150kPa into B. a)If Container A has a volume of 10L and container B has a volume of 5L, find the total pressure within the system if the valve were opened and the gases allowed to mix b)Which of the gases would diffuse through the valve at a faster rate?

It has been a long time since I took chemistry and blundering around with Google didn't really seem to answer my question about a technical point about question (b) for this problem.

Obviously when you first open the valve between containers A and B the pressure is going to equalize. Since A has a higher pressure, gas will flow from A into B. Is this diffusion?

In a weird way the pressure equalization does fit the basic notion of diffusion - molecules mover from an area of higher concentration to an area of lower concentration. Thus if both containers had oxygen, then container A would be more concentrated (e.g mol $\rm L^{-1}$) since it has a higher pressure. Thus the molecules will move from A to B because there is a lower "concentration." However this doesn't really seem to fit my notion of either  diffusion or effusion. So is there a better proper term for this or is the phrase "pressure equalization" about as good as can be done? Or am I just wrong?

Obviously after the pressures equalize, then the oxygen from the oxygen/chlorine mixture will diffuse into the chlorine faster since oxygen molecules have a lower mass.

The gist is that the question would be simple to answer if containers A and B had the same pressure. But the difference in pressure seems to make the question turn on a very exacting definition of diffusion.

Its not diffusion, it's effusion. The difference is that with diffusion the particles/molecules are migrating through a permeable medium where as for effusion the particles/molecules are migrating through a pore/hole.

For example, a scent travelling across a room is diffuision, a scent from the room traveling outside is effusion. Also fun fact: the helium escaping through a balloon skin is actually effusion not diffusion.

helpful source

Side note: I am starting to flashback to all the times my thermo professors said diffusion when they meant effusion.


Why do carbonates, oxides, and pure metals precipitate before metal sulfides? Water tends to leach metals near the surface and deposit them deeper through precipitation in two layers. The upper layer is usually made of metal carbonates, oxides, and occasionally pure or native metals, while the second and deeper layer is metal sulfides. The following excerpt is summarized above.
Source: "Secondary Enrichment of Ore Deposits" by Samuel Franklin Emmons, which was published in the American Institue of Mining Engineers, Transaction 30, (1901), p. 177-201.
The paper does not go into detailed chemistry for the most part but mentions that the chemistry varies from one morphology of soil/rock to another. The author postulates a thermodynamic reason along the lines of reduction potential that facilitates this stratification:

The actual changes observed by me in a great body of pyrite ($FeS_2$) carrying galena (PbS or Lead(II)Sulfide) in a limestone ($CaCO_3$)country-rock, which had undergone partial decomposition from the periphery inwards, are as follows: the original fresh pyrite or marcasite crystals are first disintegrated and slightly pitted on the surface, then changed to  melanterite or hydrated ferrous sulfate and the galena becomes anglesite. In the outer or more fully oxidized zone, the iron-vitriol has changed in part to yellow, basic sulfate; in part to limonite with a separation of native sulfur.
The theoretical changes that are assumed to take place by the action of waters carrying oxygen or oxidizing agents are: first, an alteration of the iron sulfide ($FeS$) to ferrous sulfate (Iron(II)Sulfate or $FeSO_4$) with the formation of sulfuretted hydrogen (Hydrogen Sulfide? $H_2S$) and sulfur which may have oxidized to sulfuric ($H_2SO_4$)or sulfurous acid ($H_2SO_3$). By further oxidation the ferrous sulfate will become, in part at least, ferric sulfate (Iron(III)Sulfate or $Fe_2(SO_4)_3$), and this in its turn will react upon the remaining ferrous sulfate, or upon the sulfides, and form more ferrous sulfate or sulfates of the other metals which are present. By this cycle of reactions, a supply of both ferric and ferrous sulfates would seem to be provided in the oxidized zone, but the extending downwards of the ferric salts would decrease as the supply of oxygen in the waters became less abundant.
It may be assumed that the sulfates of the metals thus formed would be transported for greater or lower distances, generally in proportion to their solubility, the iron sulfates being the most soluble; next, those of copper and zinc; silver sulfate is less soluble and also more readily decomposed, while lead sulfate is extremely insoluble.

Why Do Carbonates, Oxides, and Pure Metals Precipitate Before Metal Sulfides?

They don't. I cannot get to the particular paper you read, but I can provide both a logical and a quantitative argument that the metal sulfides precipitate first. The thermodynamics of this situation do not involve reduction potentials so much as simple solubilities. Metal sulfides are much less soluble in water than metal carbonates.

From your description, I imagine a system like the following image. There are two strata of sedimentation: an upper layer of metal oxides and a lower layer of metal sulfides.

There are three reasonable explanations why the sulfides are on the bottom.

The sulfides did not precipitate from the water; they were always there. The carbonates and oxides deposited on top of them.
The carbonates and oxides were deposited first, and the sulfides were deposited second, settling through the carbonates and oxides to form a deeper layer.
The sulfides were deposited first, and the carbonates and oxides were deposited second.

Explanation 1 is very reasonable, but counterproductive. We will ignore it.

Explanation 2 is the one supposed by the OP in the question, and perhaps it was also provided in the paper. Let us examine this idea a little. In order for this to happen, the metal sulfides would have to be denser than the metal carbonates and oxides to settle through. All it takes the counterexample of iron to see that this is not always the case. Iron (II) sulfide is denser than iron (II) carbonate but less dense than iron (II) oxide.
 - Iron (II) oxide, $\ce{FeO}$, 5.745 g/mL
 - Iron (II) carbonate, $\ce{FeCO3}$, 3.961 g/mL
 - Iron (II) sulfide, $\ce{FeS}$, 4.84 g/mL

Additionally, if the sulfides were settling through, the layers would not be so well stratified - the boundary between them would be indistinct.

  Metal sulfides are much less soluble in water than metal carbonates.

That leaves with Explanation 3, which is the correct one. Metal sulfides are almost universally less soluble that metal carbonates. We'll get to where the oxides come from later. Both metal carbonates and metal sulfides are nearly insoluble in water, so the minute amounts present are hard to quantify in the "grams of solute per liter of solution" kind of way. The saturated solution may only have few parts per billion of the compound. Instead, we talk in terms of the solubility equilibrium, in particular, the solubility product (constant), $k_{sp}$, the equilibrium constant associated with the combination dissolution and dissociation.

We can determine the concentration from the $k_{sp}$. Consider barium carbonate $k_{sp}=5.1\times10^{-9} \text{ M}^3$. The solubility equilibrium is:

$$\ce{BaCO3(s) <=>Ba^{2+}(aq) + CO3^{2-}(aq)}$$

And the equilibrium constant is (the concentration of solid barium hydroxide is constant and is thus part of $k_{sp}=k_{obs}$.
$$k_{sp} = [\ce{Ba^{2+}}][\ce{CO3^{2-}}]=5.1 \times 10^{-9} \text{ M}^3$$

Since $[\ce{CO3^{2-}}]=[\ce{Ba^{2+}}]$ (mass balance):
$$[\ce{Ba^{2+}}][\ce{CO3^{2-}}]=[\ce{Ba^{2+}}]^2$$
$$[\ce{Ba^{2+}}]^2=5.1\times 10^{-9} \text{ M}^3$$
$$[\ce{Ba^{2+}}]= 7.14\times 10^{-5} \text{ M}=>0.0141 \text{ grams per liter}$$

Now, let's look at carbonates and sulfides. All $k_{sp}$ came from here.

Cadmium:

Carbonate $\ce{CdCO3}$, $k_{sp}=5.2\times 10^{-12} \text{ M}^2$, $[\ce{Cd^{2+}}]=2.28\times 10^{-6} \text{ M}$
Sulfide $\ce{CdS}$, $k_{sp}=8\times 10^{-27} \text{ M}^2$, $[\ce{Cd^{2+}}]=8.94\times 10^{-14} \text{ M}$

Cadmium sulfide is less soluble than cadmium carbonate by a factor $10^8$ (100 million). The same behavior exists for cobalt, copper, iron, lead, manganese, nickel, silver, and zinc (all metals for which data for both the carbonate and sulfide are listed).

  Why might this paper suppose that the sulfides are deposited second when the sulfides are much less soluble than the carbonates?

The authors of the paper probably did not know that (and perhaps few others did either). That paper was published in 1901. Svante Arrhenius presented his thesis on the behavior of salts in water in 1884. It was not well thought of initially. He received a Nobel Prize for the work in 1903 (after the paper was published), but there is little reason to believe the authors (mining engineers) would have been taught controversial and cutting edge physical chemistry in whatever minimal introductory chemistry courses they had taken (if any). I may mention recent Nobel work in chemistry in my intro courses, but I certainly don't teach recent Nobel work in chemistry in into courses.

  Where do the metal oxides come from?

Metal oxides form either by decomposition of the carbonates or by oxidation of pure metals (which probably occur from the decomposition of the sulfides):

Decomposition of sulfides:

$$\ce{FeS(s) <=> Fe(s) + S(S)}$$

Oxidation of metal:

$$\ce{6Fe(s) + 3O2(aq) -> 3Fe2O3(s)}$$
$$\ce{2Fe(s) + 3H2O(l) -> Fe2O3(s) + 3H2} \text{ (acid is helpful here)}$$

Decomposition of the carbonate. Since the carbon dioxide would be carried away by the water, this equilibrium is continually pushed toward the oxide.
$$\ce{FeCO3(s) <=> FeO(s) + CO2(aq)}$$


Equation for the equilibrium constant for the hydration of carbonyls using UV spectroscopy I was looking through the literature when I found a paper by Greenzaid et al. [1] which states that the equilibrium constant $K_\mathrm{hyd}$ is:

$$K_\mathrm{hyd} = [\text{hydrate}]/[\text{carbonyl}] = (\varepsilon_0^\mathrm{w} - \varepsilon^\mathrm{w})/\varepsilon^\mathrm{w}, \label{eqn:1}\tag{1}$$
where $\varepsilon^\mathrm{w}$ is the molar absorption coefficient measured under conditions of hydration, and $\varepsilon_0^\mathrm{w}$  is the molar absorption coefficient of the carbonyl compound in water in the absence of hydration.

(In water solvent, $\varepsilon_0^\mathrm{w}$ usually taken as the molar extinction coefficient in cyclohexane.)
How did the researchers get to the above equation in terms of the molar absorption coefficients?
As $[\text{hydrate}]$ is equal to $[\text{carbonyl}]_0 - [\text{carbonyl}],$ the equation \eqref{eqn:1} turns to:
$$K_\mathrm{hyd} = \frac{[\text{carbonyl}]_0 - [\text{carbonyl}]}{[\text{carbonyl}]}$$
However, the Beer–Lambert law states that
$$A = c\varepsilon l,$$
where $A$ is the optical density, $c$ is the concentration, $\varepsilon$ is the extinction coefficient and $l$ is the path length.
Denoting the corresponding parameters for carbonyl without hydration as $A_0$ and $\varepsilon_0$, and the parameters for carbonyl with hydration as $A$ and $\varepsilon$, the following can be written:
$$
\begin{align}
K_\mathrm{hyd} &= \frac{[\text{carbonyl}]_0 - [\text{carbonyl}]}{[\text{carbonyl}]} \\
  &= \frac{\frac{A_0}{\varepsilon_0l} - \frac{A}{\varepsilon l}}{\frac{A}{\varepsilon l}} \\
  &= \frac{\frac{A_0}{\varepsilon_0} - \frac{A}{\varepsilon}}{\frac{A}{\varepsilon}} \\
\therefore  &= \frac{\varepsilon}{A}\left(\frac{A_0}{\varepsilon_0} - \frac{A}{\varepsilon}\right) \\
  &= \frac{\varepsilon}{A}\frac{A_0\varepsilon - A\varepsilon_0}{\varepsilon\varepsilon_0} \\
  &= \frac{A_0\varepsilon - A\varepsilon_0}{A\varepsilon_0} \\
\end{align}
$$
But we require
$$K_\mathrm{hyd} = \frac{\varepsilon_0 - \varepsilon}{\varepsilon}$$
Shouldn't this mean that $K_\mathrm{hyd}$ is actually proportional to a fraction containing the $1/\varepsilon$ values?
References

Greenzaid, P.; Rappoport, Z.; Samuel, D. Limitations of Ultra-Violet Spectroscopy for the Study of the Reversible Hydration of Carbonyl Compounds. Trans. Faraday Soc. 1967, 63 (0), 2131–2139. DOI: 10.1039/TF9676302131.

The law is not as you state although it often gets reported as this. The '$A$' you quote is the optical density. The Beer-Lambert law is $I_{tr}=I_0e^{-\epsilon_\lambda [C]\ell}$ where $I_{tr}$ is the intensity of transmitted light for a molecule at concentration $[C]$ at wavelength $\lambda$ and cell path length $\ell$, and $\epsilon_\lambda$ is the extinction coefficient at wavelength $\lambda$.

From the definition you can see where the expression you ask about comes from even with their different notation; it looks as if $w=-{\epsilon_\lambda [C]\ell}$ where ${\epsilon_\lambda [C]\ell}$ is the optical density.


Is chlorophenylmethane the correct IUPAC name rather than chloromethylbenzene? I thought that the correct IUPAC name for the molecule below (benzyl chloride) would be chloromethylbenzene because the ring always gets priority over the chain unless a primary functional group is present. However, it seems the correct name is chlorophenylmethane. Can someone explain why that is?

The preferred IUPAC name is, indeed, (chloromethyl)benzene, built from substitutive prefix chloromethyl. The parentheses around "chloromethyl" are required because it is a compound substituent (a simple substituent attached to another simple substituent, specifically, methyl and chloro in this case). The Blue Book [1, p. 66] has plenty of examples similar to yours:

  $$
\begin{array}{lll}
\ce{C6H5-CH2-NCS} &\quad &\text{benzyl isothiocyanate} \\
  & &\text{(isothiocyanatomethyl)benzene (PIN)} \\
  & & \\
\ce{C6H5-NC} &\quad &\text{phenyl isocyanide} \\
  & &\text{isocyanobenzene (PIN)}
\end{array}
$$

Note that common names for this molecule also include: benzyl chloride and α-chlorotoluene.

References

Favre, H. A.; Powell, W. H. IUPAC “Blue Book” Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, England, 2013. ISBN 978-0-85404-182-4.


The partition coefficient for a chemical species across a membrane What factors influence the partition factors across a membrane for a chemical species? Wikipedia defines the partition coefficient as:

  In the physical sciences, a partition-coefficient (P) or distribution-coefficient (D) is the ratio of concentrations of a compound in a mixture of two immiscible phases at equilibrium.

If we let these "two immiscible phases" be either side of the membrane, In this situation the equilibrium concentrations would be determined by the solution either side of the membrane? If there is water either side of the membrane would they have a P of 1?

I'm not a chemist (or biologist for that matter) so apologies for any misuse of terminology or ambiguity.

Partitioning across a membrane is better known as Osmotic pressure. Think of a tube with a "bag" made out of the membrane attached to it. The more concentrated solution goes inside the bag, the less concentrated solution suspends the bag. The liquid from the suspending liquid then diffuses into the bag causes the fluid level to rise against gravity. The height of the rise yields the Osmotic pressure. So it is the solvent that diffuses not the solute.

If the solute diffuses through the membrane then it is just a diffusion problem such that the rate of diffusion is related to pore size. Bigger pores give faster diffusion.

To a chemist a "partition coefficient" would also really seem to apply best to a system at equilibrium. If you're considering drugs in the body then an equilibrium probably isn't obtained. The blood sweeps the drug away to react somewhere else in the body. So the drug is continually transported into the blood.

The other side of this is something like urea in urine and blood. The kidneys actually concentrate urea from the blood plasma into the urine. However this biological process consumes energy. So this sort of process is different than a simple equilibrium.


Predicting the products of the radical chlorination of propane Question:

When propane reacts with chlorine, two different monochloropropanes are formed.
a) Name and show the full structural formulae of the two products.
b) What kind of isomerism is shown?

Based on my understanding, the answer for part a is $\ce{2C8H18 + Cl2 -> 2C8H17Cl +H2}$.
Name: chloropropane, formula: $\ce{CH3CH2CH2CH2CH2CH2CH2CH(Cl)}$.
For part b, is it structural isomerism? Are my answers correct?

Let me give you two helpful hints.

  Propane

If you use an internet search tool of your choice, you should quickly find out that propane is $\ce{C3H8}$, which has the structure $\ce{CH3CH2CH3}$. The formula and structure you are using is octane. The product you drew, 1-chlorooctane $\ce{CH3CH2CH2CH2CH2CH2CH2CH2Cl}$, is one of the monochlorination products of octane, so you should feel good about being able to correctly apply the reaction type and produce a correct product.

  What is the question asking?

A careful read of the intro part of the question When propane reacts with chlorine, two different monochloropropanes are formed indicates that we should be expecting two different organic products with chlorine atoms attached to the carbon skeleton. In other words this reaction should give two products with the formula $\ce{C3H7Cl}$.

However, part a) is not worded well to indicate that you mostly likely are to draw the structures of the two monochloropropane products. You are correct in noting that a radical halogenation reaction will have two products, though the inorganic byproduct is usually considered to be $\ce{HCl}$ and not $\ce{H2}$ or $\ce{H2O}$.

$$\ce{C3H8 + Cl2 -> C3H7Cl + HCl}$$

So let me ask your question in a clearer way:

  Propane reacts with chlorine to produce two different monochlorination products with the formula $\ce{C3H7Cl}$. (a) Draw the structures of  both monochlorination products and provide their names. (b) Identify the type of isomeric relationship that exists between the two structures that you drew in (a).


How was it concluded that the H3O+ rather than H+ is the "acid" ion? I suspect that initially, scientists believed that the acid ion was $\ce{H^+}$ since $\ce{H2}$ is released through electrolysis, right? But what experiment was done to change the standpoint to assume that it is instead the $\ce{H3O^+}$ ion?

Or perhaps there's both $\ce{H^+}$ and $\ce{H3O^+}$ ions?

Actually, the initial theories before Lewis suggested that $\ce{H+}$ is the cause of acidity. However, it soon turned up that an ion as small as the nucleus of hydrogen (you may simply call it a proton) can't be created in low energy reactions due to its high polarising power. So, $\ce{H+}$ is though the cause of acidic nature in aqueous solutions, $\ce{H+}$ never exists as $\ce{H+}$ but as $\ce{H3O+}$, $\ce{H9O4+}$, etc. It is because the proton is heavily hydrated. Though we are not very sure of what is the actual hydrated form, we usually refer it to as $\ce{H3O+}$ or $\ce{H+ (aq)}$.


Balancing a REDOX reaction using the oxidation number method, in acidic solution I am trying to balance the REDOX equation shown below, but I got stuck after balancing the major atoms. Where can I go from here? I know I need to balance the H and O atoms, and make sure the charges are balanced.

Second Attempt (Answer Only)

Is this correct?

Whenever you balance a redox reaction in acidic or basic solution, you have to account for interactions with the solution itself. The easiest way of doing this is by the half-reaction method.

This is a key detail that I find a lot of students miss - you cannot balance a reaction like this without considering the solution, because the solution itself is participating. Your clue that this is happening is the phrase "in acidic (or basic) solution" - any time you see that in a redox problem, you should be thinking "half-reaction method." Another clue is that H and O appear on only one side of the equation. If you are familiar with the algebraic method of balancing equations, another clue is that the system of equations would be unsolvable.

At any rate, once you know that you need to use the half-reaction method, the next step is to use it!

This tutorial has a good description of the method, your book probably does as well, and I'll give a brief description here for good measure.

These are the major steps:

Split the reaction into two half-reactions. Then, for each reaction
Balance everything except for H and O
Balance O by adding $\ce{H2O}$ (water is available in solution, so this is legal)
Balance H by adding $\ce{H^+}$ (it's an acidic solution - so this is also legal)
Balance charge by adding electrons ($\ce{e^-}$)
Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel - this is necessary because electrons are not created or destroyed in chemical reactions.
Add the equations together, and if the solution is basic, add $\ce{OH^-}$ to both sides to remove all protons.

I'll use an example from my copy of Brown, LeMay, Bursten and Murphy's "Chemistry: The Central Science" to demonstrate (all phases are aqueous):

Balance the following reaction in acidic solution:

$\ce{MnO4^- + C2O4^2- -> Mn^2+ + CO2}$

Break the reaction into two half-reactions:

$\ce{MnO4^- -> Mn^2+}$

$\ce{C2O4^2- ->CO2}$

It's not always easy to see how to do this. The quick way is to make sure that the elements that are not O and H are in their own equations. The proper way is to figure out what is being reduced and what is being oxidized, then split them so that the oxidation happens in one reaction and the reduction happens in the other. In this case, Mn is reduced and C is oxidized.

Balance everything except for H and O:

$\ce{MnO4^- -> Mn^2+}$

$\ce{C2O4^2- -> 2CO2}$

This step is usually pretty easy.

Balance O by adding $\ce{H2O}$:

$\ce{MnO4^- -> Mn^2+ + 4H2O }$

$\ce{C2O4^2- -> 2CO2}$

We start with O because we are using water to balance it. Each time we add a water, we also add two H, so it doesn't make sense to start with H.

Balance H by adding $\ce{H^+}$:

$\ce{8H^+ + MnO4^- -> Mn^2+ + 4H2O }$

$\ce{C2O4^2- -> 2CO2}$

Remember that we can do this because it's in acidic solution. You are probably wondering what we do for basic solution. The answer is - use protons! There is a trick we can do at the end to "fix" it for basic solutions, and this is much easier than iterating between $\ce{OH^-}$ and $\ce{H2O}$

Balance charge by adding electrons:

$\ce{5e^- + 8H^+ + MnO4^- -> Mn^2+ + 4H2O }$

$\ce{C2O4^2- -> 2CO2 + 2e^-}$

While this is easy to understand, I have found that in practice, a majority of mistakes happen here. Be careful when counting charges - remember to multiply by the coefficients, and remember that the charges don't have to add up to zero yet - they just have to have the same sum on both sides.

Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel:

$\ce{10e^- + 16H^+ + 2MnO4^- -> 2Mn^2+ + 8H2O }$

$\ce{5C2O4^2- -> 10CO2 + 10e^-}$

This part is also prone to error - be careful and remember to multiply every species by the coefficent.

Add the equations together

$\ce{16H^+ + 2MnO4^- + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2 }$

If the electrons don't cancel out here, then you made a mistake - go back and check each step.

If the solution is basic, at this point you would add $\ce{OH-}$ to both sides of the equation until the 16 protons were eliminated. We can add anything we want as long as it's to both sides (since this is, after all, just algebra). Each $\ce{OH^-}$ will react with a proton to form one water molecule, and so the waters on the right would cancel out some of the waters on the left. The result would be a similar equation, but with $\ce{OH^-}$ instead of $\ce{H^+}$.

Try out your problem using these steps, and if you get stuck, update your question.


SN1 vs SN2 in low concentration of nucleophile Why is SN1 mechanism favoured by low concentration of nucleophilic reagent? It is a first order reaction so shouldn't the reagent be in excess? Also higher the concentration of reagent faster the reaction right?

Let us assume that the competing reaction is a Sn2 based pathway.

Sn2=k[substrate][nucleophile]

Sn1=k'[substrate]

Relative rate of products formed=(ie.. Sn1 to Sn2 product)=k'/(k[nucleophile]

Sn1 Reaction doesnot depend on concentration of the nucleophile , so more the concentration of nucleophile, the relative rate of Sn1 product diminishes.

Hence ,when the competing reaction is a Sn2 pathway, lesser concentration of nucleophile would favour the Sn1 pathway.

Hope this helps.


Transition state optimisation on the surface of periclase I want to model a reaction catalysed by periclase ($\ce{MgO}$) using DFT.
I have a good guess on the transition state (TS) of the reaction that goes in gas phase/solvent (produced using MOPAC).

The proposal for this reaction goes by the Eley–Rideal mechanism and I have a good guess on how the reactant adsorbs.

Now comes the question: I would like to optimise the TS on the surface using DFT (either ORCA or NWChem), but in tractable time.

I am tempted to use a simple, fixed model for the surface, e.g., a set of fixed atoms in bulk arrangement.
Can this work?
How am I supposed to find a structure with just one imaginary frequency using a partially fixed model (maybe using some clever vibrational projections)?

Even though NEB could be applied, the implementation of it in NWChem does not work with fixed atomic coordinates.
Besides, I am confident on the TS I found without the catalyst, so I would like to use that instead of starting all over again.

Edit: what about using a very small cluster model, say, 8-16 atoms, and let it relax? Do you think it could work (in the sense of giving a reliable TS in terms of geometry and energy)?

To study surface reactions I recommend the growing string method (GSM) for surfaces. Here is a nice quote from the paper which developed it (Ref. 1)

  GSM’s efficacy was confirmed by comparison with CI-NEB on
  an extensive set of reactions characteristic of modern surface chemistry studies. In these cases, GSM reduces the computational cost (in terms of gradient computations) by about 45% on average over CI-NEB. In addition to high efficiency, GSM has the advantage of operating in single-ended way to enable explorative study of chemical reactions.

Key to the success of GSM is the use of the efficient hybrid delocalized coordinate system which greatly improves optimization and interpolation steps in comparison to NEB which uses Cartesian coordinates.

The exact code to reproduce their work is not publicly available, although you may be able to request a copy. However you should be able to reproduce it and more, with this python version of the code that I have developed.

Also, as others have pointed out, you should consider a higher level of theory like periodic wavefunction DFT (e.g. as implemented in VASP) .

References:

Jafari, M, Zimmerman, P. M.. "Reliable and efficient reaction path and transition state finding for surface reactions with the growing string method", J. Comput. Chem. 2017, 38, 645– 658. https://doi.org/10.1002/jcc.24720


What happens chemically to 5'-radiolabeled aqueous thyroxine (thyroid prohormone) when its I-131 undergoes beta+gamma decay? Related to this question about $\ce{^{14}C}$ beta decay in $\ce{CO2}$, what happens when the $\ce{^{131}I}$ in radiolabeled thyroxine (thyroid prohormone, or "T$_4$") undergoes beta-plus-gamma decay to form $\ce{^{131}Xe}$?

In order to narrow the scope of the question to something tractable, assume that despite its low aqueous solubility the thyroxine is in dilute solution in deionized water, and that it bears a single $\ce{^{131}I}$ at the $5'$ position (click image to enlarge):

As well, as noted above, assume that the $\ce{^{131}I}$ undergoes its most common mechanism of tandem electron & gamma decay:

$$
\ce{^{131}_{53}I -> ^{131}_{54}Xe + \beta^-~+ \gamma}
$$

I assume the resulting chemical species remaining will start out as a (very transient) xenon-bearing cation:

What happens next?

When the beta decay occurs it is likely that the C-I (or now C-Xe) bond will break. . The electron emitted in the beta decay will fly away from the molecule, the positively charged xenon atom will now pick up an electron from the aqeuous solution.

The Szilárd-Chalmers effect needs to be considered, it is typcial that when a radioactive decay occurs that the bonds to the atom at which they decay is will be broken. The recoil energy from the beta emission should be able to break the C-I bond.

I suspect that the bond will break to form a carbon centred aryl radical. This is a very high energy organic radical which will be likely to react with dissolved oxygen to form a Ar-O-O. radical. This in turn will abstract a hydrogen from another organic molecule to form Ar-O-O-H.

When this decomposes the O-O bond will break thus forming ArO. and HO. The ArO. will abstract a hydrogen from another organic solute to form a phenol. I know that 8-hydroxy-20-deoxyguanosine (8-OHdG) in urine is associated with exposure to radiation in humans. See S. Kato et. al. JOURNAL OF PEDIATRICS, 2015, volume 167, pages 1369 onwards. Also see the work of Martin Tondel et. al., ARCHIVES OF ENVIRONMENTAL CONTAMINATION AND TOXICOLOGY, 2005, volume 48, pages 515-519.

This work on this DNA marker for radiation damage does to my mind support the hypothesis that the autoradiolysis of the thyroxine will create hydroxy compounds. When S. Cefova et. al. irradated emulsions of nitrobenzene and water they did observe the formation of nitrophenols which lends further weight to the idea that the product will be a phenolic compound.


Colour of the transition metal complexes I read in my Chemistry AP class that the colour of the transition metal complexes is due to the wavelength of the visible light that is left unabsorbed, when the d subshell is split due to the repulsion between the ligand electrons and d subshell electron in the transition metal ion. When white light is passed through the complexes, the wavelength that corresponds to the energy ($\Delta E$) required to promote electrons in the split d subshell is absorbed. The part of the visible light wavelength that is left unabsorbed becomes the colour of the transition metal complex.

I was intrigued by a concept of absorption of light, relating it to absorption and emittion spectra. My question is: If we keep that complex isolated in the dark, where no light could reach it would the colour of the complex still be blue? I understand if that is done it wouldn't be visible to the naked human eye. I wonder if there is a way to find that out. Secondly, I was wondering if the colour elements obtain their colour in a similar way?

Most colour in chemistry comes from selective absorption of light by electronic transitions of the appropriate energy

Colour is not limited to transition metal complexes. They have a particularly rich range of colours as the electronic transitions often match the energy range of visible light but also because they are easily altered by the nature of the ligands surrounding the metal (for example, Cobalt complexes can be almost any colour of the spectrum depending on the ligand).

But plenty of other compounds are coloured. Most dyes are purely organic and the colour comes from the absorption of light in transitions between molecular orbitals made from unsaturated chains in the molecules. Many elements are coloured because they, too, have some electronic transmission that matches the energy in some part of visible light. The mechanism for colour is the same in all these cases.

There are other ways to generate colour but they are less common. Some colours come from interference not selective absorption of light: some physical structures that contain features close to the wavelength of light  can reflect light but in a way that causes selective interference that cuts out some colours. This is why oil films on water produce rainbow colours but is also the source of unusual colours in some fruits, butterflies and birds.

Are things coloured if the light isn't present? That is a philosophical question like "does a tree falling the the forest make a sound if nobody hears it". From a chemical point of view the feature that creates colour is still present whether or not there is light. But in the absence of light nobody is going to notice it.


What word, or phrase, do chemists use to describe the color saturation of a solution? Outside of chemistry, I've used the word saturation to talk about how blue (or how green, red, etc.) something is. In the lab that I'm writing, I have three undesirable options:

use saturation and risk ambiguity with the other sense of the word (I could include a footnote that specifies the intended sense of the word; however, my experience has taught me that TA's rarely read footnotes.),
use an awkward construction (e.g. the degree/amount of blue exhibited by the solution), or
use an unusual term or a neologism (i.e. blueness).

None of those options are especially appealing.

Is there a word that chemists use to talk about how blue (or how green, etc.) a solution is?

Using a spectrometer, you could accurately measure the degree to which light is absorbed by a solution ("absorbance" or "optical density") at a certain wavelength, and in a cell of certain dimensions. So, if you had a blue dye dissolved in water (which might appear blue because it primarily absorbs orange light), the absorbance (measured using orange wavelengths of light) would be proportional to the concentration of dye in water.


What is hard and soft water and how does the process of dissolution work? This is a very basic question about solubility, especially solution of table salt in water and solution of sugar in water. I have heard about terms like hard water, soft water, filtering of water to remove impurities etc., but I don't understand these well enough. My questions are:

If table salt is dissolved in water, why do we call it a solution? Don't say that a chemical reaction has taken place and some new product say $\ce{NaH2OCl}$ etc. is formed.
By mere filtering using a cloth or a machine, how can we get rid of minute impurities, (which make water hard) which are not even visible to naked eyes?

Hard water and soft water differ only in the presence or absence of certain impurities:

Hard water: Water with impurities in the form of minerals containing calcium and magnesium added in.
Soft water: Water with these minerals removed.

One thing to note is that hard water is not the slightest bit dangerous. Studies have been conducted by the World Health Organization and the National Academy of Sciences, and no evidence was found linking hard water to and adverse health effects - or any effects at all!

  By mere filtering using a cloth or a machine, how can we get rid of minute impurities, (which make water hard) which are not even visible to naked eyes?

This is impossible. Many devices have been proposed, but none have been shown to work. Quoting from this article,

  There are a variety of devices that claim to manage hard water scale using primarily magnetic or electrical technology. Manufacturers generally claim the devices utilize energy to alter the behavior of compounds or elements within the water. They do not claim water chemistry is altered. In fact, the hardness of the water before and after treatment is not changed. This complicates assessment of their performance. While protocol has been established to assess the effectiveness of ion exchange water softeners (NSF/ANSI 44), at this time, no recognized agency in the U.S. has established protocol to assess the effectiveness of physical water treatment devices. Therefore, questions remain as to their effectiveness.

Chemicals are the only way to go, unfortunately.

More information on hard water can be found here and here.

  If table salt is dissolved in water, why do we call it a solution? Don't say that a chemical reaction has taken place and some new product say $\ce{NaH2OCl}$ etc. is formed.

In case you're interested, the reaction is
$$\ce{NaCl + H2O \to Na+ + Cl- +H2O}$$

It fits the very definition of a solution:

  a homogeneous mixture composed of only one phase.

$\ce{NaCl}$ is separated into ions that separate uniformly throughout the water.


Is there a material that reduces copper melting point and completely dissolves? Is there any material that you mix/alloy with copper that

will result in something that behaves similar to plastic (that is at room temperature it is solid, at maybe 500K or similar it can be easily formed/3D printed)
When heated even more (700K?) it will make the the included copper melt, and dissolves itself, leaving only solid copper.

Is this even theoretically possible? The main goal is to significantly reduce the temperatures involved, 700K is a ballpark number here. Also it doesn't need to be very thick if this makes things easier.

The usual metal 3D-printing processes usually involve metal powder and a plastic binder.  After printing, the green part is sintered.  See this.


How does the relative amount of oxidation of different alcohols affect the amount of energy released? I am investigating the enthalpy of combustion of different primary alcohols and a website I found briefly mentioned how the relative amount of oxidation of different alcohols affects the amount of energy released. I was wondering if anyone could possibly elaborate on this.

I also found a trend where carbon chain length 1 --> carbon chain length 2 has a great increase in experimental heat of combustion but then there is a smaller increases between carbon chain length 2 and 3. Would there be an explanation for this (I have a feeling it relates to incomplete combustion)?

Thanks!

It doesn't have to do with incomplete combustion, as the enthalpy values you've seen reported are for the enthalpy of complete combustion.

It has to do with the alcohol already being partially oxidized, so there is less energy available for release upon the oxidative process of combustion.  In other words, if a particular carbon atom is already partially oxidized, like in an alcohol, then you can't get as much energy from combustion as compared to a more reduced carbon, like that in a methyl group for example.

This effect is most pronounced for the low-molecular weight alcohols.  Methanol has 1 oxygen per carbon atom, while ethanol has 0.5 oxygens per carbon atom.  That is a much larger relative difference than there is between a $\ce{C10}$ alcohol and a $\ce{C11}$ alcohol.  In other words, the larger a mono-hydroxy alcohol is, the more it resembles a pure alkane with respect to the enthalpy of combustion.


Synthesis of copper selenide If I react selenium powder with copper chloride powder above the melting point of selenium (221 °C), will I end up with a chunk of copper selenide, with the chlorine given off as gas, or will I lose some or all of the Se as as selenium chloride gas?

There are many ways to synthesize copper selenide. It generally has two forms $\ce{Cu2Se}$ and $\ce{CuSe}$ but actually it is a non-stoichiometric compound. They are grown as nanoparticles for various applications like making semiconductor, optoelectronic components, sensors etc.

By direct combination of copper and selenium vapour / direct displacement of cupric salts with hydrogen selenide.(1)(2)
Using copper(II) chloride dihydrate ($\ce{CuCl2.2H2O}$) and sodium selenite ($\ce{Na2SeO3}$) in ternary oil-in-water (o/w) emulsion system. The reducing agent, sodium borohydride ($\ce{NaBH4}$) used to reduce sodium selenite source
at room temperature(3)
Using thiol ligands(4)
Using copper acetate dihydrate and hydrazine hydrate(5)

As for your reaction, chlorine will not get displaced that easily. It needs a counter ion for chloride to make bond. That's why instead of just selenium, hydrogen selenide is used. It can be better if any displacement is avoided and just direct combination of elemenys is done. See method 1.

References

http://copper.atomistry.com/cuprous_selenide.html
http://copper.atomistry.com/cupric_selenide.html
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.452.6282&rep=rep1&type=pdf
https://pubs.acs.org/doi/abs/10.1021/acsami.5b01085
http://www.asianjournalofchemistry.co.in/User/ViewFreeArticle.aspx?ArticleID=25_10_57


Carburizing Calculation The problem is stated as follows:
The surface of a steel gear made of 1022 steel (0.22 wt% C) is to be gas-carburized at $927$ °C. Calculate the time necessary to increase the carbon content to 0.30 wt% at 0.030 in below the surface of the gear. Assume the carbon content of the surface to be 1.20 wt% and $D_{927^{\circ}C}=1.28 \times 10^{-11}\frac{m^{2}}{s}$

The formula is $\frac{C_{s}-C_{x}}{C_{s}-C_{0}}=ERF(\frac{x}{2\sqrt{Dt}})$

The solution manual is setting $C_{x}=0.30$, but should it not be $C_{x}=0.30+0.22=0.52$, since the initial carbon content in the steel is $C_{0}=0.22$?

No.

In the equation $C_x$ simply means the concentration at depth x. It does not mean the change of concentration. There is no need to add its initial concentration.

Besides, concentration cannot be linearly added. (Does adding a 1M solution to a 2M solution give 3M?)


Which compound can exist as stereoisomers? I'm having trouble in understanding how to identify chemical compounds as stereoisomers.

In the question below, D is the correct answer according to the markscheme. I've drawn out the chemical structure but am unable to see what possible stereoisomers it can form.

I also tried visualising the possible isomers on Wolfram Alpha:

http://www.wolframalpha.com/input/?i=CH3CH2CHOHCH3

There are two isomers which stand out to me, which I guess are the stereoisomers?

(S)-(+)-2-butanol
(R)-2-butanol

However, to me it looks like they are just rotated versions of each other.

"D" is the correct answer. 2-Butanol exists as two stereoisomers, R- and S-2-butanol, just as you described.  The "2" carbon in 2-butanol is a chiral carbon atom, it has 4 different groups attached to it (ethyl, methyl, hydroxyl and a hydrogen atom).  The R and S forms of 2-butanol cannot be rotated one into the other, they can only be interconverted by reflecting one in a mirror.  They are non-superimposable mirror images like your right and left hand. Build a model of 2-butanol and then build a second model that is the mirror image of the first - you'll find that you cannot intercovert them by rotation and one is not superimposable upon the other.


“T” in subscript in formulas Does a “T” in a formula subscript indicate all possible numbers of atoms? $\ce{FeO_T}$ is the specific one I came across (in the "bulk chemistry" table here: https://sciences.ucf.edu/class/simulant_lunarmare), it seems to make sense in context but I've not done chemistry since school (tried googling already) so thought I'd better check.

Interesting question.
It looks like this is a rather common notation in geology and refers to the total iron(II) and iron(III) oxide (e.g. ferrous and ferric) content.
For a chemist this is probably somewhat confusing as “total” is normally denoted via subscripted abbreviation “tot”.
Flipping through numerous sources, it appears there is no standardized notation and alongside with $\ce{FeO_T},$ $\ce{FeO_t},$ $\ce{FeOt}$ and $\ce{FeO^\mathrm{t}}$ are being used.

An introductory textbook by Shikazono [1, p. 21] explicitly defines $\ce{FeO^\mathrm{t}}$:

  $\ce{FeO^\mathrm{t}}$: Total iron $(\ce{FeO} + \ce{Fe2O3})$

Reference

Shikazono, N. Introduction to Earth and Planetary System Science: New View of Earth, Planets and Humans; Springer: Tokyo; New York, 2012. ISBN 978-4-431-54058-8.


why distillation of mercury is done at low pressure? While reading about extraction of mercury from it's cinnabar ore (HgS), it's purification required distillation due remove impurities like Au, Ag, etc. This distillation was done at low pressure. Why is it done at low pressure?

There is one obvious reason and one possible confusing factor.

The production of mercury from cinnabar involves heating cinnabar in a stream of air which causes oxidation of the sulphur and produces a stream of mercury vapour which can be condensed to give the metal. This isn't distillation even though it involves condensation of mercury vapour, but a chemical reaction.

But mercury can be purified by distillation because it is very volatile for a metal. But the boiling point at normal pressure is ~360 °C which is inconveniently hot. Using reduced pressure (say around 1/1000th of an atmosphere) can make this practical at closer to 100 °C which is far more convenient.


Why is density fitting not used to reduce four-center integrals further down to two-center integrals? In density fitting, why don't researchers turn all the 4-center integrals into 2-center ones instead of stopping at composing them from 3-center integrals?

I try to answer your question, although I'm not completely sure if I understand what you are asking for.

You want a decomposition like for example:
$$
 (pq|rs) = \sum_P A^p_P A^q_P A^r_P A^s_P
$$
am I correct?

In the framework of density fitting this type of decomposition is not possible without further steps/tricks.
If we look at the two electron integrals
\begin{equation}
(pq|rs)=\int d\pmb{r}_1 \int d\pmb{r}_2 \phi_p(\pmb{r}_1) \phi_q(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \phi_r(\pmb{r}_2) \phi_s(\pmb{r}_2) \quad,
\end{equation}
we can consider this electron repulsion integral as repulsion between two generalized electron densities
\begin{equation}
(pq|rs)=\int d\pmb{r}_1 \int d\pmb{r}_2 \rho_{pq}(\pmb{r}_1)  \frac{1}{\pmb{r}_{12}} \rho_{rs}(\pmb{r}_2) \quad,
\end{equation}
with $\rho_{pq}(\pmb{r}) = \phi_p(\pmb{r}) \phi_q(\pmb{r})$ and $\rho_{rs}(\pmb{r}) = \phi_r(\pmb{r}) \phi_s(\pmb{r})$.
This density maybe approximated with the aid of an auxiliary basis set:
\begin{equation}
 \tilde{\rho}_{pq}(\pmb{r}) = \sum^{N_{aux}}_P d_P^{pq} \chi_P(\pmb{r})
\end{equation}
Hence the name "density fitting". The coefficients $d_P^{pq}$ are determined by minimizing the functional
\begin{equation}
  \Delta_{pq} = \int d\pmb{r}_1 \int d\pmb{r}_2 \frac{\left[ {\rho}_{pq}(\pmb{r_1}) - \tilde{\rho}_{pq}(\pmb{r_1}) \right] \left[ {\rho}_{rs}(\pmb{r_2}) - \tilde{\rho}_{rs}(\pmb{r_2}) \right]}{r_{12}}
\end{equation}
and leads according to Dunlap [1] to:
\begin{equation}
 d_Q^{pq} = \sum_{P} \left(pq|P\right) \left[\pmb{J}^{-1}\right]_{PQ} \quad,
\end{equation}
with
\begin{equation}
 \left(pq|P\right) = \int d\pmb{r}_1 \int d\pmb{r}_2 \phi_p(\pmb{r}_1) \phi_q(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \chi_P(\pmb{r}_2)
\end{equation}
\begin{equation}
 J_{PQ} = \int d\pmb{r}_1 \int d\pmb{r}_2 \chi_P(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \chi_Q(\pmb{r}_2)
\end{equation}
Then we can re-write the electron repulsion integral as
\begin{equation}
 (pq|rs)=\sum_{PQ} \left(pq|P\right) \left[\pmb{J}^{-1}\right]_{PQ} \left(Q|rs\right)
\end{equation}
The framework of density fitting does not give us the possibility to decompose $\left(pq|P\right)$ further,
but it is possible in combination with more general tensor decomposition techniques like the canonical decomposition (CP)[2],
in which we approximate a tensor in the format you were fishing for (as long as I understood you correctly)
\begin{equation}
 X_{i,j,...d} = \sum_{r}^{R} \omega_r  a_{ir}^{(1)} a_{jr}^{(2)} ... a_{dr}^{(d)} \quad,
\end{equation}
where $R$ is the rank of the tensor. Without losing generality, we will in the following absorb the weighting factor $\omega_r$
in one of the factors.

We can apply this general tensor decomposition to the density fitting representation of the two electron integral.
For the $[J^{-1}]_{PQ}$ matrix we can simply us a SVD
\begin{equation}
 [J^{-1}]_{PQ}  \approx \sum_{r_x}^{R_x} \bar{U}^{P}_{r_x} \bar{V}^{Q}_{r_x}  \quad,
\end{equation}
where the singular values $\sigma^{1/2}$ are absorbed into the vectors
For the 3-index integrals we do a CP decomposition
\begin{equation}
  \left(pq|P\right) \approx \sum_{r}^{R} A_{r}^{p} A_{r}^{q} A_{r}^{P}
\end{equation}
Thus the 4-index integrals can be written as
\begin{equation}
 \left(pq|rs\right)  = \sum_{PQ} \left( \sum_{r_1}^{R_1} A_{r_1}^{p} A_{r_1}^{q} A_{r_1}^{P} \right)
                                 \left( \sum_{r_x}^{R_x} \bar{U}^{P}_{r_x} \bar{V}^{Q}_{r_x} \right)
                                 \left( \sum_{r_2}^{R_2} A_{r_2}^{r} A_{r_2}^{s} A_{r_2}^{Q} \right) \quad.
\end{equation}
In some sense we are done since only two index quantities occur, but we can simplify the expressions a bit.
The sums can be rearranged to
\begin{equation}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2}\sum_{r_x}^{R_x} \left( \sum_{P} A_{r_1}^{P}  \bar{U}_{r_x}^P  \right) \left( \sum_{Q} \bar{V}_{r_x}^Q   A_{r_2}^{Q} \right)
                         A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r} A_{r_2}^{s} \quad,
\end{equation}
and a summation over $P$ and $Q$ yields
\begin{equation}
  \label{eqn:thc1}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2}\sum_{r_x}^{R_x} P_{r_1 r_x} Q_{r_x r_2}
                         A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r}  A_{r_2}^{s} \quad.
\end{equation}
Only $P_{r_1 r_x}$ and $Q_{r_x r_2}$ depend on $r_x$ and via summation we introduce a new coefficient matrix $B_{r_1 r_2}$, which
combines the expansions length $R_1$ and $R_2$.
\begin{equation}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2} B_{r_1 r_2} A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r} A_{r_2}^{s} \quad,
\end{equation}
which is quite similar to the THC format introduced by Martinez and co-workers.[3]
In one additional step we obtain a CP factorization for the full electron repulsion integral
\begin{equation}
  \left(pq|rs\right) = \sum_{P}^{R_1\cdot R_2} \bar{A}_{P}^{p}  \bar{A}_{P}^{q} \bar{A}_{P}^{r} \bar{A}_{P}^{s}
\end{equation}
So finally we arrived at a decomposition in only two index quantities and there are several groups[4,3,5], who looked
at this integral format(s). It is not of widespread use since so far the available algorithms for the required CP decomposition
are quite costly. In combination with density fitting the costs will scale with $\mathcal O(N^4)$, which is not that bad,
but the prefactor is so large that for a long range of molecular sizes a CCSD(T) calculation will finish quicker than the decomposition
of the two electron integral tensor.

The story is a bit different for the intermediate step, which can be identified as the THC format. Martinez and co-workers
are heavily working on using this format and enabled $\mathcal O(N^4)$ scaling implementations for MP2 and CCSD. However,
for MP2 it is to the best of my knowledge not really compatible with DF-MP2 implementations.

So after this long answer the short conclusion: Such integral formats exist in form of the THC format or CP decomposed integrals,
but further work is necessary to increase the efficiency and resolve some other problems.

References:

Dunlap, B. I., Connolly, J. W. D. & Sabin, J. R. On Some Approximations in Applications of X Alpha Theory. The Journal of Chemical Physics 71, 3396–3402 (1979).
Carroll, J. D. & Chang, J.-J.   Analysis of individual differences in multi-dimensional scaling via an n-way generalization of “eckartyoung” decomposition. Psychometrika 35, 283–319 (1970). URL http://dx.doi.org/10.1007/BF02310791
Hohenstein, E. G., Parrish, R. M. & Martínez, T. J. Tensorhypercontraction density fitting. I. quartic scaling second- and third-order Møller-plesset perturbation theory. The Journal of Chemical Physics 137 (2012). URL http://scitation.aip.org/content/aip/journal/jcp/137/4/10.1063/1.4732310.
Benedikt, U., Auer, A. A., Espig, M. & Hackbusch, W. Tensor decomposition in post-Hartree Fock methods. I. Two-electron integrals and MP2. The Journal of Chemical Physics 134 (2011). URL http://scitation.aip.org/content/aip/journal/jcp/134/5/10.1063/1.3514201.
Schmitz, G., Madsen, N. K. & Christiansen, O. Atomic-batched tensor decomposed two-electron repulsion integrals. The Journal of Chemical Physics 146, 134112 (2017). URL https://doi.org/10.1063/1.4979571. https://doi.org/10.1063/1.4979571


Why doesn't silica gel dissolve in ethyl acetate Why doesn't silica gel dissolve in ethyl acetate when something like aspirin does. Both silica gel and ethyl acetate are highly polar. I though like dissolved like.

Because (among other things) silica is not actually polar. Yes, theoretically a molecule of $\ce{SiO2}$ should have some degree of polarity to it. However, silica as a solid does not exist as single molecules.

Silica exists as a network of $\ce{Si}$ cations bonded together by $\ce{O}$ anions. One way of looking at it would be to say that each small bead of silica gel (or any other form of silica, glass or quartz) is one huge single molecule of silica.


What is the empirical formula of the polyethylene? Polyethylene is a polymer of only carbon and hydrogen.  If 2.300 g of the polymer is burned in the oxygen it produces 2.955 g H2O and 7.217 g CO2.  What is the empirical formula of the polyethylene?

So my method for this problem is
1) find the mass of carbon from CO2
2) find mass of H from H20
3) Find mass of O by subtracting
4) cont. answer with formula from % composition

Therefore:

(7.217 g x 1mole x 1 mole Carbon x 12 grams) /  44g CO2 = 1.96 g
(2.955 g x 1mole x 2 mole Hydrogen x 1g)     /  18g H2O = .328 g

1.96g / 12g = 0.163
.328  / 16g = .0205 (smallest #)

.163 / .0205 = 8 Carbon
.0205 / .0205 = 1 Hydrogen

The answer is CH2 :/, but I am having a hard time finding out where I slipped :(


Which Lewis Structure for BeCl2 is more commonly seen? One is more stable according to its charge. The other one does not satisfy as much as the first does the formal charge rule but has a complete octet.

Beryllium chloride has a significant ionic contribution to its electronic structure. A quick calculation, DF-BP86/def2-SVP, reveals that the charge of beryllium is $\mathbf{q}(\ce{Be})=0.8$ and of chlorine $\mathbf{q}(\ce{Cl})=-0.4$, based on natural bond orbital analysis.

The Lewis structures best describing this are depicted below. The percentage indicates how much Lewis character this model actually describes based on the before mentioned calculation.

Of course this is just an approximation and the truth lies somewhere in between, like Eljee already mentioned. For these bonding situation, resonance is unavoidable. In quantum chemistry these phenomena are usually very well described by valence bond theory.

Another possibility, which is quite complementary, is looking at the structure with molecular orbital theory. Since the molecule has $D_\mathrm{\infty{h}}$ symmetry, you can observe $\pi$ bonds, that are delocalised over the whole molecule. And these are only depicted in the right hand side structure above. Below are the valence orbitals of $\ce{BeCl2}$, DF-BP86/def2-SVP.

So you can see, that one Lewis structure is not enough to actually describe the bonding in that molecule well, but as a first approximation, the left hand structure above should be sufficient.


DOSY-NMR: How to calculate the volume of the molecule from the diffusion coefficient? DOSY-NMR allows for the determination of the diffusion coefficient of the analyte molecule. How do I calculate the volume/radius of the molecule from the diffusion coefficient?

You can try and use the Stokes-Einstein equation for the diffusion coefficient $$D=\frac{k_\mathrm{B}T}{\zeta}$$ where the friction term is $$\zeta = 6\pi\eta r$$ and $k_\mathrm{B}$ is the Boltzmann constant, $\eta$ the solvent viscosity $(1\ \mathrm{cP} = 10^{-3}\ \mathrm{kg~m^{-1}~s^{-1}} = 10^{-3}\ \mathrm{Pa~s})$ and r the molecule's radius . You only have one number in diffusion constant so you can only get radius or volume assuming the molecule is a sphere.

The model assumes that the molecules are larger than the solvent molecules, i.e. solvent is effectively continuous not molecular. There are slightly different expressions for the friction, 4 instead of 6 for example depending on whether 'slip' or 'stick' limits are assumed.


Why is urobilin yellow? The yellow color of urine is due due a compound known as urochrome, or urobilin. Why is this chemical yellow? There isn't any coordination complexes. What else could cause this color?

I have not learned organic chemistry yet so please make it simple.

Many organic molecules, including food dyes, are colored because they contain chromophores. A chromophore is a region in the molecule where the energy difference between two different molecular orbitals falls within the range of the visible spectrum. Light that hits the molecule can thus excite the electrons in the chromophore, resulting in the emittance of a particular wavelength of light.

$\hspace{1cm}$

In urobilin (seen above) specifically, the tetrapyrrole moiety, as a result of its conjugated pi-bond system, is the chromophore that gives this compound its characteristic yellow color.


Why are low spin tetrahedral complexes so rare? As I was going through Concise Inorganic Chemistry by J. D. Lee, I  realised that there are simply no low spin tetrahedral complexes mentioned in the book. Is there any specific condition required for the formation of such a complex?
Usually low-spin complexes are in $\mathrm{dsp^2}$ electronic configuration. But can this kind of orbital form a tetrahedral geometry? Because usually tetrahedron is usually synonymous to $\mathrm{sp^3}$.

Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.

The reason why low-spin $T_\mathrm d$ complexes are rare is because the splitting parameter, $\Delta_t$, is significantly smaller than the corresponding octahedral parameter $\Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)

$$\Delta_t = \frac{4}{9}\Delta_o$$

For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.

Since the splitting $\Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $\mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $\mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.


Chemiluminescence of luminol Recently I have tried to perform an experiment with luminol and hydrogen peroxide (with potassium ferricyanide as the catalyst). Upon googling, I found two ways to conduct this experiment and was wondering which way is more suitable (because I have limited amounts of luminol and do not want to fail the experiment and waste luminol).

1st way: Preparing an alkaline stock solution of luminol and potassium ferricyanide and another stock solution consisting of hydrogen peroxide only.

2nd way: Preparing an alkaline stock solution of luminol  and another stock solution consisting of hydrogen peroxide and potassium ferricyanide.

Here they replace the ferricyanide for bleach, but the description does say that the luminol solution is stable in base for up to three weeks. So I would follow your 2nd way.


Rock that transfers image This is a question that has been bugging me for over 15 years!
In high school while studying elements our teacher showed us a clear white rock or element.

It was simply flat on both sides but when placed on a sheet of paper would transfer the letters from the page beneath it to the top surface of the rock.

I remember being amazed by this rock, but could not remember what it was.
Now I am working on a creative project and would love to see if I can incorporate the material in my design.

I used F'x answer as a starting point, and I found the mineral. It's called Ulexite.

  The fibers of ulexite act as optical fibers, transmitting light along their lengths by internal reflection. When a piece of ulexite is cut with flat polished faces perpendicular to the orientation of the fibers, a good-quality specimen will display an image of whatever surface is adjacent to its other side (as shown in the photograph).


Rate and order confusion I just learned about rate and order... I just want to confirm if these three ideas are correct:

If the first elementary step is the slow step, will the stoichiometric coefficients of the reactants in the rate determining step (RDS) be EQUAL to the order the concentration of the reactants in the rate equation is raised to?
e.g.
$$\begin{align}
\ce{2A + B &-> C} &(\text{slow})\\
\ce{C + B &-> D} &(\text{fast})
\end{align}$$

Will the order with respect to $[\ce{A}]$ be 2 and $[\ce{B}]$ be 1 if this proposed mechanism is correct?

If the balanced equation is the ONLY elementary step in the reaction mechanism, the stoichiometric coefficients of the reactants will be the same as the order the concentration of the reactants are raised in the rate equation
If the second elementary step is the slow step, will the stoichiometric coefficients of the reactants in the rate determining step (RDS) be different to the order the concentration of the reactants in the rate equation is raised to?

Although Nicolau Saker Neto's answer provides the essential, I wish here to propose two emblematic examples for the situations cited in your question:

First case: two steps process, the first being the slowest.

Reaction between $\ce{H2O2}$ and $\ce{HBr}$:

\begin{align}
\ce{H2O2 + H+ +Br- &-> HOBr + H2O}  && \text{slow}\\
\ce{HOBr + H+ +Br- &-> Br2 + H2O}   && \text{fast}\\
\end{align}

The rate of this reaction, determined by the slow step, equals:

$$
\frac{d[\ce{Br2}]}{dt}=k_{1}\cdot[\ce{H2O2}]\cdot[\ce{H+}]\cdot[\ce{Br-}]
$$

This is therefore a third order reaction.  Stoichiometric coefficients of the slow elementary step yield the specific order in each of the three reacting species.

Second case: two steps process, the first being the fastest.

Reaction between $\ce{NO}$ and $\ce{O2}$:

The initial equilibrium is rapidly established:

\begin{align}
\ce{2NO &<=> N2O2}      && \text{fast}\\
\ce{N2O2 + O2 &-> 2NO2} && \text{slow}\\
\end{align}

In this case, the rate is:

$$
\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot[\ce{N2O2}]\cdot[\ce{O2}]
$$

This is therefore a second order reaction, with specific orders coinciding, as usual, with the stoichiometric coefficients in the slow step. Using the equilibrium constant $K$ of the first step, one can finally express the rate of the global equation as a function of $[\ce{NO}]$ and $[\ce{O2}]$, say:

$$
\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot K \cdot[\ce{NO}]^{2}\cdot[\ce{O2}]
$$

Summarizing, as you correctly say, the stoichiometric coefficient of a given reactant, in the rate determining elementary step ("slow" step), directly yields the order by which the concentration of that reactant is raised in the rate equation.


Why is the boiling point of pentane (36.1 °C) slightly higher than that of diethyl ether (34.6 °C)? Both are chains of the same length, except diethyl ether has an oxygen atom in the middle instead of a carbon. Diethyl ether has a slight dipole from the oxygen atom, so shouldn't the intermolecular forces be stronger than for pentane which give it a higher boiling point?

This is perhaps not a very scientific answer, but consider how branched alkanes have a lower boiling point than straight chain alkanes.  This is due to the inability of the individual molecules to "stack" (if you want to get technical it's to do with surface area).  Now imagine stacking diethyl ether molecules.  The problem arises that two oxygen molecules (2 lone pairs each) on two separate diethyl ether molecules would not want to stack directly on top of eachother as the oxygen atoms have decently high electron densities.  The repulsive force would likely be minimal, as demonstrated by the insignificant difference in boiling points between diethyl ether and n-pentane.  However, this is only an educated guess.

To strengthen my argument, check out Diethyl sulfide, which has a much higher b.p. than both n-pentane and diethyl ether.  This could be due to the larger atomic radius (and therefore larger surface area) of Sulfur in comparison to Oxygen.  The charge has a larger area to be distributed over, resulting in less repulsion between "stacked" sulfur atoms.