Difference between free energy of formation and free energy of combustion I was wondering what is the main difference among those two values.

I think free energy of formation is the change in the Gibbs free energy when a compound is formed from its constituent parts (assuming that the system is in the standard state of 298K, 1atm pressure). What it means is how much work the reaction is capable of doing if it does the maximum possible work it can do (in case it's a negative number) or how much work you need to do on the system to make the formation reaction happen.

I never heard of free energy of combustion yet, but I guess it must be about the change in Gibbs free energy during a combustion reaction. So I think it's the change in Gibbs free energy for a combustion reaction when the reaction goes from the reactants to the products, and should have the same interpretation as the change in free energy for formation in general (probably the delta Gibbs value is going to be negative indicating that the reaction released a lot of energy that can be used for work).

Hope this helps...
(I'm a newb so hope others can verify my answer).


Where can I find an extensive list of the preferred chemical nomenclature? I'm writing text to speech software, and as it stands, I already have a method for reading out mathematical formulas accurately, which I have been using with chemicals as well. Thus, something like:

H2O would be read more or less as "Ayche, sub two, O".

and

CO2 would be read more or less as "Sea, O, sub two"

That is fine here and there, However in a lot of texts this becomes extremely tedious, for example:

  In tissue, cellular respiration produces carbon dioxide as a waste product; as one of the primary roles of the cardiovascular system, most of this CO2 is rapidly removed from the tissues by its hydration to bicarbonate ion. The bicarbonate ion present in the blood plasma is transported to the lungs, where it is dehydrated back into CO2 and released during exhalation. These hydration and dehydration conversions of CO2 and H2CO3

https://en.wikipedia.org/wiki/Bicarbonate_buffer_system

In this paragraph alone, you would hear: "Sea, O, sub two" three times. While I am not a chemist, I would assume that most listeners would prefer to hear would prefer to listen to a nomenclature that actually gave a name to the formula in cases like these, thus instead of hearing "Sea, O, sub two" and  "Ayche, sub two, O". -- you would hear the preferred chemical nomenclature,

Carbon Dioxide

and

Water

or

Oxidane

or

Dihydrogen monoxide

Is there a big table or list or resource available of the preferred nomenclature? And how often does this nomenclature depend on context, in the sense where the nomenclature would change depending on whether this was medicine or astronomy (etc...)?

Here's a pretty extensive list on Wikipedia.

  Dictionary of chemical formulas

I doubt this nomenclature would change much depending on context, given that there is a widely used nomenclature system (IUPAC) (see here).


How is silica gel made? How is silica gel prepared from sand (after extracting it). How many ways can it be used? Is there a way to make it without an oven?

Sand may be dissolved in concentrated and hot solutions of NaOH. The dissolution is slow, and pressure may help this action. At the end of the reaction, a viscous solution of so-called sodium silicate is obtained. The formula of the solute is not well defined, and can be described as $(Na_2SiO_3)_n$. It is made of a long polymeric chain similar to the formula $HO-Si-O-(Si-O)_n-Si-O-Si-OH$, with supplementary negatively charged Oxygen atoms above and under each Silicon atoms. The whole chain is negatively charged. And of course, there are Sodium ions $Na^+$ in the vicinity of each negatively charged Oxygen atoms. To make the description more complete, one should mention that from time to time there are side branches of Si-O-Si-O chains, which get fixed on the Si atoms of the main chain. Furthermore there are no $Si=O$ bonds in this formula.

If you add some $HCl$ solution, you produce the precipitation of an insoluble gel whose formula has the same skeleton as the previous solution $HO-Si-O-(Si-O)_n-Si-O-Si-OH$. But the Oxygen atoms above and under the main chain are now holding a H atom, and the whole chain is neutral.

This gel can be filtrated and washed with large amounts of water. This removes the ions $Na^+$ and $Cl^-$. The result is a gel containing more water than silica. Usually this gel is then heated and the water is evaporated. An amorphous white substance is obtained with a formula between $H_2SiO_3$ and $SiO_2$. This is the common "silicagel" used in the labs for dehydrating purposes, because it has a tendency to adsorb water from the atmosphere.

The fresh gel can also be dried in a vacuum, instead of by heating. In this case, the silica gel is transformed into an extremely light substance, lighter than snow, which has a tendency to float in the air. It is a dangerous substance if you breeze it, because it will stay in the nose and the lung, without being able to get out. It may induce cancers, like asbestos.


Why does covalent bonding not break down if observer effect can be applied to atomic electrons? The observer effect in quantum mechanics states that when unobserved, quantum particles such as electrons can simultaneously occupy two different states. In an atom of any element, where there are electrons around the nucleus, this effect is hinted at in some publications:

Electrons can occupy several energy levels, or orbitals, simultaneously. (Scientific American.(2012, October 9). Bringing Schrödinger's Cat to Life. Retrieved from https://www.scientificamerican.com/article/bringing-schrodingers-quantum-cat-to-life/.)

This idea is so weird. It is equivalent of saying that when unobserved, all the electrons of an atom can both be in their normal configuration and be in a totally different configuration. Only the act of observation forces the electrons into their normal configuration. Take the example of neon atom. According to the observer effect, when unobserved, its electron configuration can both be 1s4 2s3 2p3 (one of many possible alternative configurations) and 1s2 2s2 2p6. Only when observed will the electron configuration collapse into its normal version of 1s2 2s2 2p6.
I don't know if this understanding of observer effect on atomic electrons is valid. If it is, there are at least two serious problems. First, the Pauli exclusion principle will break down, because the alternative electron configuration allows more than 2 electrons in the same orbital (specified by the principal quantum number, the angular momentum quantum number, and the magnetic quantum number). Second, the foundation of covalent bonding is that when two or more atoms pool together electrons in orbitals that overlap, the total energy of the system is reduced to the minimum. If the observer effect on atomic electrons is valid, the energy required to maintain the system may get even lower than that in covalent bonding, and then covalent bonding will break down. All that's needed is for each atom to achieve an alternative electron configuration.

Thanks, everybody. I now understand this question better. Yes, the observer effect does apply to atomic electrons by allowing each of them to occupy several orbitals, subshells, or even shells at the same time. However, if we look at all the electrons and summarize their states into an electron configuration scheme, the scheme will still be the same as that obtained through the aufbau principle.

I think part of your confusion stems from neglecting the fact that the electrons in a single molecule are all entangled with each other, so while an individual electron can occupy many states simultaneously, in each of those states it is entangled with all the other electrons such that the sum of states remains the same.
To start with a simple example, consider an atom of helium in its singlet ground state. Prior to measurement, neither electron has a defined spin. Both are a superposition of $+1/2$ and $-1/2$ states, but that does not mean that the helium atom can have $+1$ or $-1$ spin. The net spin is still 0. If we measure one electron's spin to be $+1/2$, the other will necessarily have $-1/2$, even though neither had defined spin prior to measurement.
We can extend this to multi-atomic molecules. Consider ground state methane. We can calculate delocalized molecular orbitals that collectively describe the charge density at any point in the molecule, with each of those orbitals being composed of a standard hydrogenic orbital of carbon (eg 2s or 2p) and athe 1s orbitals of all four hydrogen atoms. Alternatively, we can describe the same charge density distribution with four localized orbitals, each comprised of a single hydrogen 1s orbital and a hybrid sp3 orbital on carbon. Prior to making an observation, neither of these descriptions is correct, but neither is incorrect. Both correctly describe the charge density of the molecule, but both incorrectly assign electrons to specific states. Instead, the electrons are collectively in both states (and an infinite number of others that describe the same density). Because every contributing state correctly describes the charge density, there is no change in the covalent structure resulting from the superposition.


First or second order reaction I have that $100$ mM of K$_3$Fe(CN)$_6$ is dissolved in equimolar of the organic ion (which I assume is HCN). In the solution Fe(CN)$_6$$^{3-}$ reacts to form Fe(CN)$_6$$^{4-}$.
The formation of Fe(CN)$_6$$^{4-}$ as a function of time is measured in a Na$_2$SO$_4$ (which does not take part in the reaction) solution of different concentrations.
One of the concentrations of Na$_2$SO$_4$ = $0$ and I want to determine the rate constant from this and:
$$\begin{array}{c|c}
t/\pu{min} & \ce{[Fe(CN)_{6}^4-]}/\pu{mM} \\ \hline
10\ & 20.3\\
20 & 33.7\\
30\ & 43.3\\
40\ & 50.4 \\
50\ & 56\\
60\ & 60.4 \\
\hline
\end{array}$$
I assumed that the reaction would be second order since the reactants are equimolar and so I plotted using the equation:
$$ \frac{1}{[A]}= kt + \frac{1}{[A_0]}$$
When I plotted using this I get that the linear equation is: $−5.78×10^{−4}*x + 0.0463$
and R$^2$ = $0.771$.
So I tried using a first order equation and got that the linear equation is: $0.0204*x + 3.01$
and R$^2$ = $0.878$.
Doesn't this mean that my reaction should actually be first order? I don't see how it could be that and would appreciate it if someone could tell me if it is first or second or what I am doing wrong. Thank you!
EDIT
I did plot the curves and now when I look at them I do think that the first order line is a bit curved while the second order is more straight. It is really hard to see and my professor usually looks at the R-value which is why I thought that would help me determine the order here. But does that mean the reaction is second order?
Here are my graphs:

Use the 2nd order scheme $2A \to B$ with initial amount of $A=A_0$ so $B=A_0-A$ and B is the product you observe.  The rate of change is $\displaystyle \frac{1}{2}\frac{dA}{dt}=-k_2A^2$ and I'm not going to use [ ] for concentrations just $A,\,B,\,A_0$. Also the 2 in the equation is incorporated into the rate constant just for convenience.
Solving the equation gives $\displaystyle A=\frac{A_0}{A_0k_2t+1}$ and so the product follows $\displaystyle B=\frac{A_0^2k_2t}{A_0k_2t+1}$. This is shown as the red line in the figure below. I just guessed the rate constant a few time to fit  and as you can see the fit is good. You can also linearise this equation by isolating $A_0k_2t$.
The first order plot is also shown (grey line) and this is poor. $B=A_0(1-e^{-kt}$)


Geometry of Ni(acac)2 The complex $\ce{Ni(acac)2}$ is paramagnetic as expected , i.e it has two unpaired electrons giving rise to a magnetic moment of $ \sqrt{8}$ BM but doesn't have the tetrahedral geometry consistent with the CFT assumptions , why ?

well if the above thing is true then what is the actual structure of the complex ? please help !

source: Inorganic Chemistry: Principles of Structure and Reactivity, 4e

It is a good question, it is a gemstone hiding in the mud.

I have searched the Cambridge database, all the bis-acetylacetonatonickel complexes which have four coordinate nickel centres are square planar. There are three common coordination geometries for nickel(II) which we need to consider.

Tetrahedral, square plannar and octahedral. Here is a diagram (not to scale) of the d oribital energies in the three types of complex.

Nickel(II) has eight d electrons, if we pack them into the octahedral or tetrahedral complex we will end up with two unpaired electrons. But if we stick the electrons into a square planar complex then if the difference in the energy between the two uppermost d orbitals in energy is sufficient then we will get a low spin $\mathrm d^8$ complex. To get spin-pairing, the energy difference between the orbitals must be greater than the energy required to spin pair two electrons.

There is a whole continuum of complexes between perfect octahedral and perfect square planar. If the distance between the metal and the two ligands on one axis is increased (tetragonal distortion of an octahedral complex) then the complex is part way between the two extremes. There is a good undergraduate lab experiment based on the early work of David Goodgame which is all about the nickel(II) complexes of N,N-diethylethylenediamine. I did this experiment as an undergraduate, I supervised students doing it as a PhD student and as an academic I have used it in a lab class. Trust me, it is good!

What the person asking the original question should have been thinking about is the question of square planar vs octahedral (or tetragonally distorted octahedral).

If I was to take an acac-type ligand with bulky groups in place of the methyls such as tert-butyls then I am sure that I would have an isolated nickel. This nickel could well be tetrahedral. But in fact it is square planar (F.A.Cotton, J.J.Wise, Inorganic Chemistry, 1966, 5, 1200, CCDC code DPIMNI)

If we look at a nickel(II) complex of an acac where the methyl groups have been replaced with phenyl groups, the closest oxygens to the nickel other than the four we expect are $\pu{4.4 Å}$ from the nickel. They are a long way in terms of angle away from the $z$ axis of the nickel. I think it is safe to say that they are not binding to the nickel. Here is a picture of the nickel complex. (D.V.Soldatov, A.T.Henegouwen, G.D.Enright, C.I.Ratcliffe, J.A.Ripmeester, Inorganic Chemistry, 2001, 40, 1626.)

Just in case anyone is not sure it is very much a square planar complex.

But if you look at the crystallographic literature you will see G.J.Bullen, R.Mason, P.Pauling, Inorganic Chemistry, 1965, 4, 456. Here is the answer: what happens is that the nickel acac complex forms a trimer and the bridging oxygens make all three nickels have an octahedral environment. The CCDC code is ACACNI01 for the best determination which was M.B.Hursthouse, M.A.Laffey, P.T.Moore, D.B.New, P.R.Raithby, P.Thornton, Dalton, 1982, 307.

Also look out for hydrated complexes such as AQACNI (R.L.Harlow, C.E.Pfluger, J. Coord. Chem., 1973, 2, 289).

By the way one of the lectures I had as an undergrad mentioned this problem of waters adding to the coordination sphere or oligomers forming, I think it was in a lecture by David Goodgame.

While I have never seen a 4-coordinate nickel bis-acac complex which is anything but square planar, if you look at the coordination chemistry of a related phosphorus(V) ligand (dithioimidodiphosphinate) you can see both tetrahedral and square planar nickels.

If we look at YUSFIB which is for the tetraphenyl ligand (P.Bhattacharyya, J.Novosad, J.Phillips, A.M.Z.Slawin, D.J.Williams, J.D.Woollins, Dalton, 1995, 1607) we can see that the nickel is tetrahedral. Also in the tetraisopropyl ligands complex (D.Cupertino, R.Keyte, A.M.Z.Slawin, D.J.Williams, J.D.Woollins, Inorganic Chemistry, 1996, 35, 2695) it is tetrahedral as can be seen in the following two pictures.

But the THF solvate of the dppaS2 complex of nickel is square plannar. (E.Simon-Manso, M.Valderrama, D.Boys, Inorganic Chemistry, 2001, 40, 3647.)


Does the initial rate in the reaction of magnesium powder with nitric acid depend on the volume or the concentration of the acid? For the question regarding the initial reaction rate, I don't quite understand why the answer is D.

If I calculated the number of moles from the volume and conc, I would get A=0.4 moles; B=0.2 moles; C=0.2 moles; D=0.1 moles

As original equation is 0.2 moles of NH3, at 25 degree Celsius, I chose C. But why is the answer D, given that it is 0.1 moles not 0.2 moles.

Thanks in advance and sorry for any wrong tags.

The magnesium is present in form of a solid powder. Therefore, the reaction of magnesium with acid can only occur at the solid–liquid interface between magnesium and the aqueous nitric acid solution. Accordingly, the initial reaction rate depends on the surface area of the solid–liquid interface. For the experiments described in the question, the same mass of magnesium powder is used; therefore, the available surface area of the magnesium powder is approximately the same, too. Hence, as long as the added volume $V$ of the nitric acid solution is sufficient to cover the entire available surface of the magnesium powder, the initial reaction rate does not depend on the added volume of the solution. (Assuming a reasonable geometry of the reaction vessel, a volume of $V=100\ \mathrm{ml}$ should be more than enough to cover a mass of $m=5.0\ \mathrm g$ of magnesium powder.)

Furthermore, the reaction rate depends on the concentration $c$ of the acid at the solid–liquid interface. Therefore, the initial reaction rate depends on the concentration of the added nitric acid solution.

And finally, the reaction rate also depends on temperature $T$. Therefore, the initial reaction rate depends on the temperature of the reactants, i.e. the magnesium powder and the added nitric acid solution.

Hence, in order to obtain the same initial reaction rate, the concentration of the nitric acid solution $(c=1.0\ \mathrm{mol\ l^{-1}})$ and the temperature $(T=25\ \mathrm{^\circ C})$ should be the same during both experiments.


Ethylene hybridization In the carbon atom, there are electrons in 2s, 2Px, 2Py, and 2Pz. Out of these orbitals, only 2Px, 2Py, and 2s are hybridized. The 2Pz orbital is left. When making a bond there is a pi bond between two carbon atoms which is formed by the 2 Pz orbitals of the each carbon atom.  In $\ce{CH4}$, 2Pz orbital also hybridized.

The orbital 2Px and 2s always make sigma bonds. Each carbon atom connects with two hydrogen atoms in $\ce{C2H4}$. Therefore I like to imagine that 2py and 2pz orbitals can make sigma bonds with 2hydrogen atoms. Then only left orbitals are 2px and 2s. They can make double bonds between two carbon atoms. 2px and 2s always form as sigma bonds and then the double bond between carbon atoms, one is sigma bond and another also a sigma bond.

I know that according to the VSEPR theory there should be one sigma bond between two atoms. But atoms are hybridized for their comfortability. But according to my above imagination making 4 sigma bonds is more comfortable than making 3 sigma bonds and one pi bond for the $\ce{C2H4}$ molecule. Therefore, why is it not trying to make such a bond?

Suppose the carbon was $sp^3$ hybridised. The carbon would have 4 hybridized $sp^3$ orbitals which would point towards the corners of a tetrahedron; they are $109.5^\circ$ degrees apart from each other.

You have to form two bonds between the same two carbon atoms. Suppose you use one $sp^3$ hybridised orbital to form one $\sigma$-bond. How are you going to form another bond? The other bonds of the carbon are $109.5^\circ$ apart. This is not feasible.

The other option for the carbon is to assume $sp^2$ hybridization and have three hybridized orbitals stay $120^\circ$ apart. The one remaining $p$-orbital can be in any direction ($x$-axis or $y$-axis) except along the bonding axis ($z$-axis). In this way, you can form a double bond using one $sp^2$ hybridized orbital and one $p$-orbital.


Finding pKa from equivalence point on titration curve Is there a mathematical proof/explanation of why $\mathrm{p}K_\mathrm{a}$  corresponds to the $\mathrm{pH}$ at $\text{Volume of titrant}/2$  at the equivalence point?

A concise proof will suffice.

You've got a weak acid, since you're contemplating a positive pKa, which means when you're halfway to the end point you're in the buffer region and you can use the Henderson-Hasselbalch equation:

pH = pKa + log [A-]/[HA]

You've titrated half your initial HA, so half of it is still around and half got turned into A-, which means [A-] = [HA].  For example, if you started with 0.1 mol/L of HA, you now have 0.05 mol/L HA and 0.05 mol/L of A-.

That means [A-]/[HA] = 1, and the log of 1 is zero.  The HH equation then tells you pH = pKa.


Separating Enantiomers I understand that you can react enantiomers with a chiral molecule to resolve the enantiomers of interest into diastereomers.

However I do not understand how exactly the added chiral molecule can make the enantiomers of interest into diastereomers.

Can someone please explain the process/mechanism ?

For example: here is a picture from my textbook.

How are those two salts Diastereomers, even though they are mirror images of each other.

Enantiomers are mirror reflections, they have same physical and chemical properties. This like left and right hand.

Diastereomers have different physical and chemical properties. You can separate them on chromatography column or by crystallization.

In your example upper and bottom products are not mirror reflections.
Mirror relection of upper product looks like:

Just play with the 3D models in any drawing software to get the point.


13C NMR of bromobenzene ipso carbon shielding I've got a $\ce{^{13}C}$ NMR spectrum for bromobenzene. The ipso carbon is significantly more shielded (appearing upfield) than the two ortho carbons. Why is this? Wouldn't the electron-withdrawing effects of the halogen be enough to deshield the ipso carbon?

Does it have to do with the fact the $\ce{Br}$ is a deactivating group? I see that on an NMR database site, the ipso carbon for the $\ce{^{13}C}$ spectrum of fluorobenzene is deshielded. Does it have to do with the size difference between fluorine and bromine?

The principle influence on chemical shift for 13C is the paramagnetic term, which essentially describes the electron density in the p orbitals, and this largely mirrors the influence of electronegativity that is seen in 1H chemical shifts.

However, there is something called the 'heavy atom effect' where the effect of substitution runs counter to electronegativity. This is the case for Bromine, Iodine, Tellurium etc. This comes from an increased level of diamagnetic shielding introduced by the electrons from the halogen atom.

To understand why this is the case, we can consider that the overall shielding on the ipso carbon relative to a methyl susbituent. Even though the bromine is more electronegative than the methyl group, it is so large that it still contributes significant electron density around the ipso carbon nucleus, and hence provides greater overall shielding than a methyl group.

While trends in electronegativity are useful tools to justify observations on chemical shift, it is always good to relate the effects of electronegativity, and other properties, on what is happening to the electron density around the nucleus on question. Ultimately, chemical shift is a measurement of electron shielding around the nucleus.


How to explain Gibbs free energy is a pressure-dependent state function? I'm aware that through deriving Gibbs free energy to infinitesimal changes, we could get the formula: $\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$, giving that Gibbs free energy is pressure-dependent.

However, while dealing with the definition of Gibbs free energy: $ΔG = ΔH - TΔS$ (and $ΔH$ must be held at constant temperature, am I wrong?) most textbooks stated that it is held under constant pressure.

I still couldn't understand the relationship between Gibbs free energy and pressure and why: $\mathrm dG = V\,\mathrm dp$.

$G$ is defined as $H - TS$, and $H$ is $U + PV$.

Since in closed system you have: $\mathrm{d}U = T \mathrm{d}S - P \mathrm{d}V$, it follows that:

$\mathrm{d}H = T \mathrm{d}S + V \mathrm{d}P$
$\mathrm{d}G = -S \mathrm{d}T + V \mathrm{d}P$


Why won't increasing the temperature increase the solubility of a gas in a liquid solvent? If I raise the temperature of a solution of a gas in a liquid solvent, why won't the solubility increase?

Increasing the pressure will increase the solubility, but increasing the pressure doesn't mean increasing the temperature?  What is wrong with this argument?

A gas has bigger entropy when it's free, than in a solution. By increasing the temperature the molecules have bigger energy and they tend to be in a state with more freedom. Therefore, the solubility would decrease.


Reaction of coumarin with phenylmagnesium bromide I came across this problem..

I thought that the Grignard reagent, PhMgBr, would cleave the lactone and then subsequently add to the ketone formed, giving an allylic tertiary alcohol after aqueous workup. Treatment with acid would then lead to intramolecular etherification, giving 2,2-diphenyl-2H-1-benzopyran:

But then our chemistry professor gave out a different mechanism for this reaction as follows.

What is the correct product?

In terms of regioselectivity, at least for the first addition of the Grignard reagent, I suggest to have a look on the cumarin, that may be seen as an α,β-unsaturated carbonyl compound. Hard nucleophiles (in terms of HSAB principle) like $\ce{MeLi}$ would almost exclusively react with the carbonyl carbon, as you drew in your reaction equation.  Soft nucleophiles, such as cuprates, would prefer the 1,4-addition.

Going from $\ce{MeMgBr}$, to $\ce{EtMgBr}$, $\ce{i-PrMgBr}$, and $\ce{t-BuMgBr}$ gradually lowers the polarity of the bond between the metal (magnesium) and the directly attached carbon atom.  Hence, along with this line (in other words, as a trend), the preference to react in 1,4-position  (more like Michael reaction) increases on expense of the reaction of this nucleophile on the carbonyl carbon.  I speculate the bond in the Grignard reagent, between $\ce{Mg}$ and the phenyl group in $\ce{PhMgBr}$, is even less polarised, than in the instance of $\ce{t-BuMgBr}$, and hence conclude at least a significant portion of the product of the first addition formed will be the one of the 1,4-addition (second drawing).

As @Zhe, I however still have a difficulty to understand the subsequently suggested second addition of $\ce{PhMgBr}$ in the lower drawing, because the activation of the double bond once in conjugation with the carbonyl group is gone.


What is the SNi mechanism of the reaction of moist silver oxide with a halide? I was told by my teacher that the following reaction proceeds through SNi:

Moist $\ce{Ag2O}$ is basically $\ce{AgOH}$, what role does it play in this reaction's mechanism? I think this is what happens, but I am not sure:

$\ce{AgOH}$ attacks $\ce{Cl-}$ and during this process the $\ce{Ag-O}$ bond breaks and an intimate ion pair is formed between the carbocation and $\ce{Cl-}$.
Since a partial bond still exists between $\ce{Ag}$ and $\ce{O}$, $\ce{O}$ is forced to attack from the same side from where $\ce{Cl-}$ left and the reaction proceeds to completion without Walden inversion.

Is my reasoning correct? Or is there a different mechanism?

If you suspect that there is something different going on here, and I believe you are right.
My take, in the presence of lab light, $\ce{Ag2O}$ and $\ce{H2O}$ actually results, in the formation of the hydroxyl radical ($\ce{.OH}$) and the hydrogen atom radical ($\ce{.H}$) via the creation of a powerful photocatalyst couple $\ce{Ag2O-Ag}$ as outlined in this article: Ag2O as a New Visible‐Light Photocatalyst: Self‐Stability and High Photocatalytic Activity.
The photoelectric introduced electron hole $\ce{h+}$ and electron $\ce{e-}$ can act as follows in the present of water:
$\ce{H2O <=> H+ + OH-}$
$\ce{H+ + e- -> .H}$
$\ce{OH- + h+ -> .OH}$
The hydrogen atom radical is likely responsible for the extraction of the chlorine atom from the organic forming a product of $\ce{HCl}$.


To model early earth chemical reactions which types of reactions should be included? I'm currently working on building a simulator for the early earth to examine the spontaneous occurrence of food generated autocatalytic sets. No one in my group is a chemist, so the bottle neck for the project is implementing some sort of realistic chemistry in our model.

We are considering using the software outlined in this paper:
http://arxiv.org/abs/1304.1356

This will allow us to create a reaction network from an initial set, based on a set of general rules. The paper demonstrates well how the formose reactions are executed, which we intend to include. What other reaction groups would it be important to include? Our initial set of molecules would be comprised of the food molecules for each set of reaction rules included, we don't intend to start from H,O,N,C for example, so fundamental chemical rules should not be too crucial, as they should be generalised into the major reaction groups we will use.

For the reactions look up Stanley Miller's experiment. The experiment used water (H2O), methane (CH4), ammonia (NH3), and hydrogen (H2). I think you should start with that.


If metals want to "lose" electrons, then why will copper ions take electrons from a zinc electrode? According to my book, metals want to "lose" electrons to achieve noble gas configuration. But at the same time, copper ions in a galvanic element are going to gain electrons from a zinc electrode. Why is it so? Is it because one of them HAS to gain/lose electrons, and zinc wins out in terms of achieving noble gas configuration?

Typically, when one uses terms such as "likes to lose" or "likes to gain", we should think of that phrase as "likes to lose relative to...".  In your case, when we think about electron configurations, it might be more helpful to think "metals like to lose rather than gain electrons to achieve a noble-gas-like configuration".  When comparing two metals, both of which tend to lose electrons, we need to look at different concepts to answer the question.

We can explain the transfer of electrons by looking at the the metal activity series such as the one shown here.  The way one reads this table is that a Galvanic cell created with two metals will have the metal higher on the list as its anode (where oxidation occurs) and the other metal will be the cathode.  Thus, we would expect Zn to be oxidized to Zn(II) and Cu(II) to be reduced to Cu.  These tables are based on standard reduction potentials:

$$\ce{Cu^{2+} + 2 e^- -> Cu(s)}\ \tag{E = 0.34 V}$$
$$\ce{Zn^{2+} + 2 e^- -> Zn(s)}\ \tag{E = -0.76  V}$$

By flipping the 2nd reaction (thus changing the sign of E) one obtains a cell potential of about 1.1 V; a positive number meaning that this reaction will happen spontaneously:

$$\ce{Cu^{2+} + Zn(s) -> Cu(s) + Zn^{2+}}\tag{E = 1.1V}$$

The standard reduction potentials tell us that the electrons would rather be on copper than on zinc.


Which resonance structure to use for writing curved arrow mechanisms? If a structure with delocalized electrons participates in a reaction, which resonating structure do you use to write down the mechanism? Is using either equivalent? Is it preferred to use the resonance hybrid? If so, what are the rules of using curved arrows for partial bonds?

Any resonance structure can be used. Why? Resonance structures represent multiple views of the same compound, so it must be possible to start with any valid resonance structure. Because resonance structures consider different positions of electrons, the sets of arrows needed to describe the mechanisms of different resonance structures will be different.

While it's possible to use any resonance structure in a mechanism, selecting one resonance structure can be used to improve communication or emphasize particular property/reactivity of a compound.

As an example, below is a figure describing the formation of an enolate and its attack on an aldehyde (aldol reaction) using different resonance structures. For clarity, all of the curved arrows are describing reactions, not the relationships between resonance structures.

We know that carbonyls such as 1 have an important resonance structure which places a negative charge on oxygen and leaves a positive charge on carbon (2). While 1 is the most important contributor, 2 emphasizes the carbocationic nature of the carbonyl carbon. I like this structure, because I know some reactions of carbocations: SN1 and E1. This helps explain to me why the alpha-position is acidic. Deprotonation of 2 next to the carbocation, just as in E1, gives enolate 3. Thinking of the carbon-oxygen pi-bond as a leaving group, deprotonating 1 at the alpha- position gives the same enolate 3 but by a different set of arrows compared to 2 to 3.

Note that there is a third way (not shown) to describe the same deprotonation where a base deprotonates the alpha-hydrogen and the electrons are simply localized on the alpha-carbon to give structure 4.

Structures 3 and 4 are resonance structures, and again these can both be used to describe further chemistry. The blue set of curved arrows from 3 to 5 contains one additional electron pair movement to the red set from 4 to 5. Both are valid descriptions of the attack on an aldehyde, but emphasize different aspects of enolate reactivity.

Keep in mind that bond-line drawings, hybrid drawings, and curved arrows are just simple representations of molecules. Thankfully they work very well and allow us to reduce the complexity of electronic structure to a manageable level.

Finally, as @orthocresol suggests in the comments, using hybrid drawings is difficult. This is because bond-line drawings represent electron pairs as localized and curved arrows work best when showing the action of electron pairs.


Why is Sulfur trioxide a neutral electrophile? My chemistry book says, an electrophile is a reagent which is atrracted to electrons, therefore positive ions and reagents with incomplete octets act as electrophiles, Sulfur trioxide belongs to neither category, so how does it act as an electrophile?

Sulfur is bonded to three oxygens, twice each. The sulfur in the middle technically has a formal charge of zero, but the oxygens it is bonded to are extremely electronegative (they're electron hogs), and so the sulfur atom in the middle has a partial plus charge on it. Ergo, it will be willing to accept electrons to try to compensate.


Reaction between a carbene and n-pentane Does the following reaction occur when n-pentane reacts with methylene or does it form some other simpler products due to the reactivity of the carbene?

$\ce{CH3-(CH2)3-CH3 + :CH2-> CH3-(CH2)4-CH3}$

EDIT:

Do higher alkanes (pentane, hexane, octane....) show a similar reactivity with methylene or do the products depend upon the alkane?

n-Hexane is one of the 3 products formed. Carbenes are high-energy, reactive intermediates.  Due to their extreme reactivity they usually react indiscriminately. In the case of methylene (the carbene in your diagram), it will insert into each of the different $\ce{C-H}$ bonds that are present in n-pentane.  Because of its reactivity, the 3 products will be formed in a statistical ratio that is roughly equivalent to the number of different hydrogens present in the starting n-pentane, 6:4:2.

If you'd like to learn more about carbenes, for example

singlet and triplet carbenes
their addition to carbon-carbon double bonds

then take a look at these earlier answers - ref_1, ref_2


Does Gallium (liquid or solid) corrode all forms of brass? I know that it corrodes copper, but does it do the same to all brass or bronze compositions?

From here

  Gallium is corrosive to all metals except tungsten and tantalum, which have a high resistance to corrosion.

Beta-Brass is embrittled by Gallium

Due to its highly reducing character it is highly corrosive material.


Lattice enthalpy and ionic charater As we move down in sulphates of group 2 their solubility decreases because there is a drastic decrease in hydration enthalpy as compared to decrease in lattice enthalpy.

Does the decrease in lattice enthalpy means decrease in ionic character?
But according to Fajans' rules the ionic character should increase or covalent character must decrease  down the group.

Kindly explain this.

Lattice enthalpy is given by the Born-Lande equation:
$${\displaystyle E=-{\frac {N_{A}Mz^{+}z^{-}e^{2}}{4\pi \epsilon _{0}r_{0}}}\left(1-{\frac {1}{n}}\right)}$$
This seems very complicated, but removing some constants, we can reduce it to say:
$$E \propto \frac {z^+z^-}{r_0}$$
here, $z^+$ and $z^-$ are charge on cation and anion and $r_0$ is distance to closest ion.

Coming to your question, In the sulphate series of Alkaline earth metals, the charge remains the same for all members: what changes is the distance to closest ion, which is smaller for small ions such as $\ce{Be^2+}$, thereby causing their lattice energies to be high.

The decrease in lattice enthalpy may not neccessarily mean a decrease in ionic character. If two lattices have similar values of $r_0$, then a higher lattice enthalpy for one of them may imply larger ionic character due to large charges on either of the atoms.


What happens when an ionic compound is boiled? Do ionic compounds such as NaCl become individual ions such as one Na ion and one Cl ion or are they hold together by ionic bonds even when they are boiled. Thanks in advance.

Interesting question with no generic answer. Boiling the salts, or ionic compounds, makes them a plasma or freely moving ions is nothing but a misconception. The temperature is too low to convert a substance into a plasma.

Let us start with ionic liquids which are room temperature cations and anions but made up of organic molecules. You heat them, to the state of boiling in the absence of oxygen. They decompose into various products. So nothing special, this is as expected for organic compounds.
Heating of non-decomposable salts at high temperatures, for example, NaCl, KCl: What will happen when we heat them to the point of boiling? Answer is nothing. They remain in vapor phase as NaCl "molecules". The vapors behave as gases. People have developed equations of state. Interestingly enough $\ce{Na2Cl2}$ (dimers) are found in gas phase.

See: Critical point and vapor pressure of ionic fluids including NaCl and KCl, Chemical Physics Letters, 105, 1984, 484-489

Ionic materials, KCl or NaCl, when introduced into flames or plasma convert into gas phase "atoms". For example $\ce{Na+}$ would be converted into a neutral Na atom. This is utilized in a completely different field in chemistry, called flame atomic absorption or flame atomic absorption spectroscopy.

In short, these complexities make science very interesting.


Why can we not compare the coupling constants to chemically equivalent (but not identical) protons to determine magnetic equivalence? I have been learning about magnetic inequivalence today, and I find it quite hard to undersand. See, for example:

In this case, according to the book I am using, $\ce{H_\mathrm{a}}$ and $\ce{H_\mathrm{b}}$ may be chemical equivalents, but they are not magnetic equivalents, due to the fact that $J_\mathrm{{\ce{a-c}}}$ and $J_\mathrm{{\ce{b-c}}}$ are different. Which I understand, but should we not consider that there is a $J_\mathrm{{\ce{b-d}}}$ which is equal to $J_\mathrm{{\ce{a-c}}}$?
I do not understand why they are magnetically different, but both have to chemical equivalent protons, one in ortho and one in para position.

The problem is that from the perspective of $\ce{H_{d}}$ and $\ce{H_{c}}$, $\ce{H_{a}}$ and $\ce{H_{b}}$ are different (you can tell which is which by their coupling constants). This difference means that there is a magnetic inequivalence between the two hydrogens $\ce{H_{a}}$ and $\ce{H_{b}}$, so there will be coupling between them.


Molecular Orbital Mixing to get the Molecular Orbitals of a large molecule I am trying to get the molecular orbital for arbitrarily placed nuclear centres (identical). In other words , I am trying to solve the following one electron hamiltonian.
$$h=-\dfrac{1}{2}\nabla^{2}-\sum_{i=1}^{N}\dfrac{1}{|\vec{r}-\vec{R_{i}}|}$$

I assumed the charge on each nuclear centre is one and to solve the problem I took the following linear combination as the trial function.

$$\psi(\vec{r})=\sum_{i=1}^{N}\alpha_{i}1s(\vec{r}-\vec{R_{i}})+\sum_{j=1}^{N}\beta_{j}2p_{x}(\vec{r}-\vec{R_{j}})+\sum_{k=1}^{N}\gamma_{k}2p_{y}(\vec{r}-\vec{R_{k}})+\sum_{m=1}^{N}\delta_{m}2p_{z}(\vec{r}-\vec{R_{m}})$$

of course, I start with s orbitals and then I proceeded with the above linear combination. My question is that if there are some symmetries in the structure. Does variational principle take care of that and gives the right $$ \alpha, \beta,\gamma,\delta$$, the coefficients of the trial wave functions? Do I need to impose the symmetry of the molecule (if there is any) on the trial wavefunctions? Does it give totally wrong answer if I take the above trial wave functions without worrying about the symmetry of the molecule?Can someone throw some light on it? Also, I am curious to know how do I apply MO theory for polymer like chain molecules? Please suggest some good reference to understand that

Running the calculation without imposing symmetry constraints should give the same ground energy as one with them because the true electronic ground state of the system (within the Born-Oppenheimer approximation) always has the same symmetry as the nuclei. The variational principle always puts an upper bound on your ground state energy, so in principle it will work just as well. Using symmetry, however, could speed up your calculation significantly because the symmetry operators commute with you Hamiltonian. This gives you an additional set of constraints on the coefficients. It is worth mentioning that single reference methods like Hartree-Fock and DFT will sometimes give lower ground state energies without symmetry constraints because the true ground state is a symmetric superposition of several configurations, but the single reference methods can only examine a single one.

For polymer molecules and other 1D infinite systems you would use periodic boundary conditions. Just make the box really large in the two dimensions you don't want to be periodic (so the system doesn't 'see' itself) and only evaluate non-gamma k-points along the periodic vector.


Is there an analytical chemistry alternative to the Scoville scale for the hotness of chilli? The traditional way of measuring the perceived heat of a chilli is the Scoville scale which involves panel testing of diluted extracts taken from chilli. But panels are very variable and the results appear to be a little unreliable.

But we know the class of chemicals that creates the sensation of heat: capsaicinoids. Can't we just measure those directly?

Is there a better analytical method for measuring the heat of a chilli?

Matt did find a method that uses HPLC on the peppers alone, not in a dish.

So given that the capsaicinoids is a class of compounds, each different capsaicinoid would have a different perceived hotness. So a chemical test would measure the concentration of each type capsaicinoid and then use a weighted average over the human perceived hotness for each member of the capsaicinoid class.


ATP chemical structure: phosphate groups with Oxygen anions or with OH groups? I've googled repeatedly (with multiple search queries) to try and get an answer to this question but
nothing comes up.
Embarrassingly, I seem to be the only person who is confused by:

ATP with oxygen anions.

ATP with $\ce{OH}$ groups.

I looked at the $3$D view on PubChem  - when I click 'show oxygens' they are indeed  present as per image 2.

The chemical formula for ATP is $\ce{C10\underline{\mathrm H_{16}}N5O13P3}$: $16$ hydrogens, which means image $2$ fits the bill better than image $1$.
However, image 1 is more commonly found in tutorials about ATP.
Which representation is it best to go with?

in vivo : ATP with oxygen anions.
in vitro : Depends on pH of solution
IN VIVO :
DNA = Deoxy Ribonucleic Acid.
DNA is acidic , because its monomers contain three phosphates bound together and this so-called triphosphoric acid is extremely acidic.
In living organisms, DNA is always accompanied by histones(positively charged proteins) or sometimes by Mg+2 complexes. Therefore the dNMPs ( deoxyribo Nucleoside Mono Phosphates) that constitute the genetic material , are negatively charged .
For rNMPs (ribo Nucleoside Mono Phosphates) the negative charged form is also more preferable.
Generally speaking in living creatures , ATP = -4 , ADP = -3 , AMP = -2 charged.
(It is worth mentioning that Pi which is a by-product in cellular energetics is HPO4(-2). )
You can see in Molecular Cell Biology Lodish et al 8th ed section 12.1 that the overall reaction of cellular respiration is:

Notice the charges.
IMPORTANT : Sometimes during the metabolism of these compounds Proton(H+) may be added to oxygens of phosphate groups only because of the mechanism of the reaction which is temporary. But usually the charges are as I mentioned before.
IN VITRO :
If you purify ATP in a solution isolated from the cell, the number of acidic hydrogens attached to the molecule is related to the pH of solution. you can predict the molar concentration of each chemical species the same way you predict the [H3PO4] , [H2PO4 -] , [HPO4 2-] , [PO4 3-] , [H+] in a solution of H3PO4 and H2O (a typical analytical chemistry question) Either by a complex utilization of Henderson–Hasselbalch equation or by Charge and Mass equations.
Also Look at this awesome picture wiki


Does SCF energy mean same as HF energy? I use Gaussian 09 and Turbomole to do a same calculation (B3-LYP/6-311G) for a same molecule. But the results are not same, Turbomole shows SCF total energy = -772.16125945927, while Gaussian shows SCF Done E(UB3LYP) =  -772.402241773, no UHF energy shows. I use UB3-LYP in Gaussian, while in Turbomole cannot specify that.
I use TmoleX to create the input file for Turbomole. It does not have UB3LYP or RB3LYP in the method part, I can only select DFT > B3LYP as my method. In previous part for molecular attributes, I can select multiplicity: UHF, then generate MOs.
The input file of Gaussian is very simple: UB3LYP/6-311g followed by cartesian coordinates of the molecule.

The energy is not the same in both of these cases. What causes the different results?

Answering your question from the title: no, SCF does not necessarily mean HF. Essentially the same self-consistent field (SCF) procedure is used to solve both the Hartree-Fock (HF) equations and the Kohn-Sham (KS) one, so for the case of DFT, SCF energy means not HF energy, but rather KS energy.

Regarding discrepancies in results obtained with different programs, there might be few reasons out there: different definitions of the B3LYP functional, different harmonics used for 6-311G basis sets (spherical vs. cartesian), different integration grids, different SCF convergence criteria, etc.

I checked the TURBOMOLE documentation and indeed found that to use the very same B3LYP functional that is used in Gaussian (with the VWN3 correlation) you need to use b3-lyp_Gaussian in TURBOMOLE, not b3-lyp. So, first of all, I suggest you try b3-lyp_Gaussian in TURBOMOLE instead of b3-lyp.


Dehydration of carboxylic acids with phosphoric acid To find Z one must find X. Finding Y is of no use to find Z.

I get X as:

After addition of $\ce{H2/Pd-C}$, the unsaturated portion becomes saturated and the carboxylic acid remains unchanged.

However I am unsure what does $\ce{H3PO4}$ do to this (new) carboxylic acid? I was unable to find any information. I did find information regarding dehydration with $\ce{P2O5}$, but unsure what to make use out of it.

After looking at (unofficial/3rd party) solutions, I saw that $\ce{H3PO4}$ protonates the $\ce{-OH}$ part of the carboxylic acid (since it is a much stronger acid than the organic acid?) then $\ce{H2O}$ leaves, rendering a carbo-cation at the carbonyl carbon, followed by EAS at the ring.

However, I do not see how $\ce{H3PO4}$ exactly works with the carboxylic acid. I have merely guessed that the $\ce{-OH}$ portion gets protonated (since, this was the only thing I predicted would happen). I really do not know why this happens and how it happens.

Hints and reference material or examples would be appreciated.

The original question is taken from the JEE (Advanced) [2018] Paper-1.

The $\ce{-OH}$ part will not get protonated, since its lone pair is in resonance with the carbonyl part of the $\ce{-COOH}$ group. Instead, the lone pair of the carbonyl part will give its lone pair to $\ce{H+}$.

Something like this :-

It may look insignificant because the product is same, but this is the correct mechanism.


What element or compound could be used to make the ocean more basic? The ocean is becoming acidic. This is not a natural evolutionary change but happens through human negative effect on the environment. What element or compound could be used to create a more base ocean? How much of it would be needed? How could it be distributed?

The ocean is becoming more acidic due to the dissolution of excess $\ce{CO2}$ from anthropogenic activities to form carbonic acid, $\ce{H2CO3}$, which even though mildly acidic is decreasing the pH of the ocean as you rightly point out.

So if we're trying to reduce the effect of this $\ce{CO2}$ on the ocean, we might consider removing it via some sink. One of these theoretical methods could be to react a base, such as $\ce{CaCO3}$,  with this carbonic acid in a conventional neutralisation. In symbols:

$$\ce{H2CO3 + CaCO3 -> Ca(HCO_3)2}$$

This could work, and we could derive everything we wanted to know about how much base to use from information such as the amount of residual $\ce{CO2}$ left over from emissions after photosynthesis has been taken into account (and other sinks), and then assuming that all other $\ce{CO2}$ will be taken up by the sea. From here we can calculate the amount of $\ce{H2CO3}$ present in seawater from the equilibrium constant between $\ce{CO2}$ and $\ce{H2CO3}$.

But we won't do that!

Notwithstanding these nonsense assumptions which could grossly overpredict/underpredict the amount of base required, neutralisation reactions don't deal with the excess carbonate effectively. What you really want is a permanent sink (a sink so stable its contents are practically constant through time) at the bottom of the sea. See the long-term carbon cycle for more information on this.

So your question becomes: how do we get the $\ce{CO2}$ dissolved in the ocean to the deep-ocean sediments? And how do we do it 'permanently'?

An interesting paper came out in the Journal of Global Biogeochemical Cycles in 1990 trying to address the question about why phytoplankton in the Antarctic don't use the myriad nutrients in the Southern Ocean to photosynthesise. One might expect that it could be temperature or sunlight which limit photosynthesis in the Antarctic, but this actually isn't the case. The limiting factor in the Antarctic ocean is actually a lack of iron.

It has since be proposed to increase the amount of iron in the ocean (simply by adding it ourselves) which would promote the photosynthesis of phytoplankton in such locations. When these organisms die, they sink and take their biomass with them which contributes to the ocean sediments.

Both neutralisation and iron supplementation may work to offset ocean acidification, however we really don't know the full consequences of these bioremediation efforts—there are simply too many variables involved.


Obtaining calcium iodide from calcium and iodine According to:

https://www.webelements.com/calcium/chemistry.html

Calcium reacts with iodine upon heating as follows:

$\text{Ca}_{\,(s)} + \text{I}_{2\,(g)} \to \text{CaI}_{2\,(s)}$

I am effectively attempting this by dropping elemental calcium and iodine into a flask and putting it under a Bunsen burner. I have a tube leading from the flask to a beaker of water to safely condense escaping iodine gas.

However, this reaction is not occurring at any appreciable rate. I am wondering if I should be using powdered calcium instead of these larger pebble-sized chunks or if a Bunsen burner is insufficient to breach the energy barrier. Are these measures or others sufficient to increase the rate of reaction?

Looks like the Bunsen burner is not providing enough energy for the reaction to occur. AFAIK, you need around 200-400°C of heat for the reaction to occur. Yes, you can always powder the calcium metal to increase surface area and see if the reaction is occurring or adopt simple(chemical) methods. You can react any calcium salt with hydroiodic acid to form calium iodide. I am using calcium carbonate as example[1]:

$$\ce{CaCO3 + 2 HI → CaI2 + H2O + CO2}$$

Or alternatively make a iodide of iron and make it react with a calcium salt[2]. There are many more ways to make calcium iodide in [2]. You can look at them but these two methods is by far the most simple methods.

References:

https://en.wikipedia.org/wiki/Calcium_iodide
Caty J. Braford;  H. A. Langenhan; A preliminary report on the composition of syrup of calcium iodide N. F. V. DOI:https://onlinelibrary.wiley.com/doi/pdf/10.1002/jps.3080180218
http://albumen.conservation-us.org/library/monographs/sunbeam/chap10.html


Calculate Kc with dissociation degree If I have the reaction
$$\ce{2HI (g) ⇌ H2(g) + I2(g)}$$
and know that at $T = \pu{448 °C}$ the dissociation degree $α = 0.2198,$ how do I calculate $K_c?$
I thought that it would be something like this:
$$
\begin{array}{cccCc}
\ce{&2HI &<=> &H2(g) &+ &I2(g)} \\
&2n && 0 && 0 \\
&2n(1 - α) && αn && αn \\
\end{array}
$$
Then
$$n_\mathrm{tot} = 2n(1 - α) + 2nα = 2n$$
$$p(\ce{HI}) = \frac{2nα}{2n} = (1 - α)p_\mathrm{tot}$$
$$p\ce{I2} = p(\ce{H2}) = αp_\mathrm{tot}$$
$$K_p = \frac{(αp_\mathrm{tot})^2}{((1 - α)p_\mathrm{tot})^2} = \frac{α^2}{(1-α)^2} = \frac{0.2198^2}{(1 - 0.2198)^2} = 0.0794$$
and then from $K_p$ I could calculate $K_c:$
$$K_c = \frac{0.0794}{(8.314 \times 721)^0}$$
I think the last step is where I'm going wrong since it shouldn't be raised to zero. The correct answer is $K_c = 0.01984.$

You don't need gas laws, temperature or $K_p.$
You overcomplicated the problem: $K_c$ can be found by following its definition with an RICE table:
$$
\begin{array}{lccccc}
&\text{R}\ce{&2 HI(g) &<=> &H2(g) &+ &I2(g)} \\
&\text{I} & 2c_0 && 0 && 0 \\
&\text{C} & -2\alpha c_0 && \alpha c_0 && \alpha c_0 \\
&\text{E} & 2(1 - \alpha)c_0 && \alpha c_0 && \alpha c_0 \\
\end{array}
$$
For the sake of consistency, I denoted the initial concentration with $c_0$ and dissociation degree as $\alpha.$
By definition, the equilibrium constant on a concentration basis $K_c$ can be found as follows:
$$K_c = \frac{[\ce{H2(g)}][\ce{I2(g)}]}{[\ce{HI(g)}]^2} = \frac{(\alpha c_0)^2}{(2(1 - \alpha)c_0)^2} = \frac{\alpha^2}{4(1 - \alpha)^2} = \frac{0.2198^2}{4(1 - 0.2198)^2} \approx 0.0198$$


Structural isomerism in iso-butyl and tertiary butyl alcohol What type of structural isomerism exists between iso butyl alcohol and tertiary butyl alcohol?

I think that there exists only positional isomerism between them but the source from where I am studying says that they are both positional and chain isomers.

I think the best way to describe this isomerism is constitutional isomerism because it takes into consideration of both position of functional group and structure at the same time.

  Constitutional isomers may have different carbon skeletons (as in
  isobutane and butane), different locations of a functional group along
  the chain (as in isobutyl alcohol and tertbutyl alcohol), or different
  functional groups (as in ethanol and dimethyl ether)

Positional isomers have the same carbon skeleton and same functional groups but differ in location of such functional groups (carbon skeleton remains unchanged). On the other hand, structural isomers the atoms are in completely different order , they arise because of the possibility of branching

I think the reason why these isomers are considered both structural and positional is that they both have the longest chain of max 3 carbons (hence same carbon skeleton), at the same time different location of -OH group  (a different chain - structural)

Hope this helps

References

Organic chemistry Mc Murry

Chemguide: Structural isomerism


Low-volatile liquids that 'ball-up' on siliconoxide or perfluoroalkanes This is a very simple and short question. I am looking for liquids that are low/non-volatile at 20$^\circ$C (at least substantially less volatile than water) which have a high contact angle with glass and oxidized silicon (in air) or with a coating of perfluoroalkanes (specifically Perfluorodecyltrichlorosilane).

The requirement on the contact angle is say $\theta>70^\circ$, i.e. the liquid should 'ball-up' a bit. The reason I am asking is that I cannot seem to find any (common) liquids that have both the low-volatility and the high contact angle $\theta$. Does anyone know some examples? Or, if no such thing exists, can someone explain what the fundamental reason is that these liquids don't exist?

Here it comes... mercury.  75-wt% Ga 25 wti%, mp. 15.5 C and

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

in general wet glass.  Maybe ionic liquids,

J. Am. Chem. Soc. 129(13), 3804 (2007), DOI: 10.1021/ja070169d
Aust. J. Chem. 57(2) 113-155 (2004)

http://www.researchgate.net/publication/233906460_Wetting_of_surfaces_with_ionic_liquids
http://dx.doi.org/10.1016/j.jcis.2008.10.019
http://ura.unisa.edu.au/R/?func=dbin-jump-full&object_id=63927

http://www.ppmp.pwr.wroc.pl/sorpa/file/eee7c6b05fb9497a9a59507ca0346479/download
[DMIM][Cl], 1,3-Dimethylimidazolium chloride, is reported 105.9 contact angle on glass.  That might be a typo - it's a huge number for the class of compound.

http://english.cas.cn/ST/BR/br_progress/201106/t20110622_71442.shtml


Perturbation theory in Quantum/Computational Chemistry As we know, the common methods for computation molecular properties in modern Quantum/Computational Chemistry is the Hartree–Fock method and Density Functional Theory.
But from university course of Quantum Mechanic we know also, the perturbation theory method. Why is this method not common in quantum chemistry?

Many people have already said this in the comments, but I will put it into an answer. Also, it would not be too hard to find a bunch of references in favor of how common any particular method is in quantum chemistry, so I won't bother to list a bunch of examples of perturbation theory being used.

That being said, if I had to guess, I would bet that second to DFT, second-order Moller-Plesset perturbation theory (MP2) is the most common method in quantum chemistry which includes electron correlation. Also, there are large communities in computational chemistry which virtually ignore the existence of DFT because they are willing to sacrifice computer-time (and possibly some accuracy) to be more certain that their answer is reliable. So, MP2 is a good example of perturbation theory being using in QM.

Also, you will often see people refer to coupled-cluster with singles and doubles excitations and a perturbative inclusion of the triples, CCSD(T), as the "gold-standard" in quantum chemistry. This is a bit more complicated application of perturbation theory, but I would say that CCSD(T) is used in virtually any problem where one can afford to do the calculation. There might be times where one could feel more confident about using DFT with a particular functional over MP2, but you would be hard-pressed to find a good argument to not do a CCSD(T) calculation if you can afford to do so. This is especially true if the alternative is to only use DFT because many functionals are parameterized against CCSD(T) calculations, so getting similar results from CCSD(T) helps validate the use of that functional for the class of problem.

There are also many other electronic structure methods which use perturbation theory such as CASPT2, one can also include perturbation excitations into truncated configuration interaction (CI) calculations, and probably lots of other but less-commonly-used methods. Also, one approach to including spin-orbit coupling in electronic structure is via perturbation theory.

One of the reasons there is such a vast array of methods which use perturbation theory is that they are diagrammatic theories, which means there is a rather simple way of understanding all of the terms physically and the connection between these terms and an algebraic representation of them is quite straightforward within the framework of second-quantization (basically using special creation and annihilation operators).

Also, I know you're mostly asking about electronic structure, but it should be noted that perturbation theory is often used to add anharmonicity onto a harmonic reference calculation of vibrational frequencies. Probably the most common method of doing this is via VPT2 which is basically the vibrational analogue of MP2.

In summary, perturbation theory is an extremely powerful tool which will almost certainly be applied to any problem where a simpler reference answer can be found. On this note, these days I don't think you could get away with doing only a Hartree-Fock calculation and expect people to believe your conclusions about the properties of the physical system. Also, there are such good, parallel codes for MP2 that if you can do a HF calculation, you can probably afford an MP2 calculation.

As a response to one of the comments, I would argue that CCSD by itself is not much more accurate than MP2 for the increased cost. Both methods can be understood as including double-excitations out of a reference determinant, and my experience is just doing CCSD does not improve much over MP2. Although MP2 and CCSD sometimes seem to be wrong in different ways, so they aren't strictly comparable. This is why people insist on CCSD(T) being the gold-standard and not just CCSD.


What sort of molecules or polymers are used in OLED displays? OLED displays have been emerging as competitors for LCD displays for some time and are now common in mobile phones (see comparison of a Google Nexus 1 and an iPhone here)

The displays use small molecules or polymers that emit light when a current is passed through them (like a semiconductor LED). But what is the structure of the emitters?

A typical organic light emitting diode (OLED) structure consisting of a glass substrate, a indium titanium oxide (ITO) anode, and an organic bilayer diffused into a metallic cathode, is given below:

The organic bilayer consists of:

a hole transport layer (HTL), typically of N, N'-diphenyl-N, N'-bis(3-methylphenyl) 1, 1'-biphenyl-4, 4' diamine (TPD) or N,N′-di(naphthalene-1-yl)-N,N′- diphenylbenzidine (NPB)
an emitting/electron transport layer (EML/ETL), typically of tris(8- hydroxyquinoline) aluminum (Alq3)

Source: http://espace.library.uq.edu.au/eserv.php?pid=UQ:8598&dsID=SMS.pdf


Calculating the density of a magnesium sulphate solution I am trying to calculate the density of a magnesium sulphate solution and came across this paper which determines an equation from experiment to do that. However, I am having trouble understanding the paper, as parameters A, B, C and D are mentioned but I don't understand how to calculate these from Table 2 in the paper. Can someone please explain how this works?
EDIT:
I am adding the data from the above paper:

The paper mentions the equation
to calculate the relative density where the parameters A, B, C and D are said to come from the table. m stands for the molality (I assume mol/kg) t is temperature (°C) and d is in g/cm^3. As you can see there is no mention of A, B, C or D in the table, leading to the confusion.

Each of parameters A-D is a polynomial function of t:
$$A = \text{"m"} + \text{"mt"}.t + \text{"mt2"}.t^2 + \text{"mt3"}.t^3 + \text{" mt4"}.t^4$$
$$B = \text{"m3/2“} + \text{"m3/2t"}.t + \text{"m3/2t2"}.t^2$$
$$C = \text{"m2“} + \text{"m2t"}.t$$
$$D= \text{"m5/2"}$$
The terms in quote means the equally denoted numerical values from the provided table.
The complete final function $1000 \cdot \Delta \rho=f(m,t)$ has for each term $a_{b,c} \cdot m^a \cdot t^b$ the value for $a_{b,c}$ equal to the particular table value named "$m^bt^c$".
So one can the complicated function use in 2 ways:

The full blown function $1000 \cdot \Delta \rho=f(m,t) = \sum {\left( a_{b,c} \cdot m^a \cdot t^b \right)}$ without the need of coefficients A-D.
Temperature specific $1000 \cdot \Delta \rho=A \cdot m + B \cdot m^{3/2} + C \cdot m^2 + D \cdot m^{5/2}$ with precalculated A-D values for given temperature.

BTW, it looks like a very interesting numerical approximation. Maybe even better could be rational functions like Padé approximant, that have smaller deviations and better  extrapolation behaviour, compared to polynomials.

Feedback for the Padé approximation:
For approximating of analytical functions, see e.g Math SE site - how-to-compute-the-pade-approximation. For approximation of experimental discrete data, the best approach is probably numerical mathematics, the least square method combined with function minimum search algorithm. They probably used the same technique for their numerical model with half-integer powers.
Using half-integer power model is unusual. It may work better than polynomials in many cases, but just for positive numbers. Fortunately, they do not calculate with negative Celsius temperature :-) ) Note that Math SE site has even tag pade-approximation . Also, it is included in MATLAB libraries, if applies. See Octave as both online/offline free functional clone of MATLAB.
See also my answer in this Q: Is there a quick way to approximate a logarithm in Nernst equation?.


Can ethanol be oxidized by hydrogen peroxide? Is it possible to oxidize ethanol to acetic acid with hydrogen peroxide and if yes then
under what circumstances? I tried it in room temperature but either concentration was too small (of hydrogen peroxide(3%)) or I couldn't quite precisely read the the results of the universal indicator. I also tried heating it up, but it didn't change anything. The color stayed the same (of indicator). Can someone please explain me if the mistake was in my experiment (if the reaction can happen in room temperature) or the reaction needs some specific catalyst or other conditions.
The reaction would be:

\begin{align}
\ce{\underset{(ethanol)}{C2H6O} + H2O2 &->
  \underset{(aldehyde)}{C2H4O} + 2H2O}\\
\ce{\underset{(aldehyde)}{C2H4O} + H2O2 &->
  \underset{(acetic acid)}{C2H4O2} + H2O}
\end{align}

Or the reaction without the middle part (since aldehyde will try to oxidize faster then ethanol) would be:

$$\ce{C2H6O + 2H2O2 -> C2H4O2 + 3H2O}$$

First things first: Don't mess with higher concentrations of hydrogen peroxide unless you are a trained and well-equipped chemist. We're talking "steel-reinforced gauntlets" here. This chemical is as volatile as nitroglycerine.

At 3% you have mostly de-ionized water and for good reason. Hydrogen peroxide is extremely, violently unstable, and is just as likely to oxidize anything else than what you actually intend to react it with.

Anything above a concentration of 30% is very dangerous. Above 60% is suicidal in ill-equipped scenarios. 100% is a hypothetical, and is in essence a potential component of rocket fuel.

So, yes it will work. You may lose a hand, but it will work.


Why is there no such substance as p-phthalic anhydride Its formula is C6H4(CO)2O.  It is an organic molecule that doesn’t have any branching or functional groups on carbon 4 or 1.

First lets take a look at o-pthalic acid and it's anhydride.

Now as you can see the ring formed in case of o-pthalic anhydride is sterically feasible. There is not very high angle strain and the aromaticity is still intact. Now, take a look at p-pthalic acid.

Now, if you try to dehydrate this molecule as you see there is a very big steric problem. To accomplish this dehydration, the molecule must take a Boat form, which looks something like this, (I am using image of cyclo-hexane just to show you the structure, benzene ring will never do this)

But to accomplish this conformation the benzene ring has to loose it's aromaticity and also create a very unstable system with the pi bond densities repelling each other. So, as this conformation is not Thermodynamically feasible, anhydride formation doesn't take place.


Problems with creating sodium hydroxide from sodium (hydrogen) carbonate If sodium hydrogen carbonate (baking soda, $\ce{NaHCO3}$) is heated to greater than $50~\mathrm{^\circ C}$, it will release carbon dioxide and water to form sodium carbonate:

$$\ce{2NaHCO3 -> Na2CO3 + H2O + CO2}$$

And when the sodium carbonate is heated further to about $850\ \mathrm{^\circ C}$, it will release more carbon dioxide and form sodium oxide:

$$\ce{Na2CO3 -> Na2O + CO2}$$

And according to Wikipedia:

  Sodium oxide (SOX) is a chemical compound with the formula $\ce{Na2O}$. It is used in ceramics and glasses, though not in a raw form. It is the base anhydride of sodium hydroxide, so when water is added to sodium oxide $\ce{NaOH}$ is produced.

So it should be possible to produce sodium hydroxide from baking soda and water, with some heat. I heated some baking soda in a gas flame. I am not sure exactly what the temperature of the flame was, but it was enough for the stainless steel spoon that I used to glow red. (If it is useful, the gas where I live is approximately $50~\%$ hydrogen, $25~\%$ methane and the rest $\ce{CO2}$ and $\ce{CO}$, and it burns completely with a blue flame. I suspect this flame to be at least $1000\ \mathrm{^\circ C}$, based on available information about flame temperatures of different fuels.)

However, $\ce{NaOH}$ is produced industrially via the chloralkali process, so there must be problems with using this method. What are they, and are they significant if trying to produce a small amount of sodium hydroxide for a minor experiment?

You are correct: Taking baking soda and heating it to $850~\mathrm{^\circ C}$ will generate sodium oxide and liberate carbon dioxide and water. To generate sodium hydroxide, all you would need to do is wait for the heated soda to cool down and then add water. Beware! The reaction of sodium oxide and water is strongly exothermic. React the two carefully!

A red-glowing spoon is a certain indicator that the temperature of the flame was larger than $500~\mathrm{^\circ C}$. A blue flame is usually an indicator for complete combustion. In the context of bunsen burners, that is typically larger than $1000~\mathrm{^\circ C}$. I cannot say whether your gas flame reached that temperature or not.

Note that once you have created sodium hydroxide you should handle it carefully. Especially do not attempt drinking it and keep it off your hands. Proper safety equipment (goggles, gloves) would be preferable.

Concerning the industrial process, the question is usually not so much what the problems with a certain method are, but rather what the benefit of a certain method is — especially if that benefit relates to cash. Electrolysis of a sodium chloride solution is a very cheap process with sodium chloride and water both available at almost no cost and electricity also typically cheap. Acquiring sodium carbonate is probably already more expensive than the required sodium chloride and heating to $850~\mathrm{^\circ C}$ is certainly much more expensive than letting a current flow through the solution. Hence why the chloralkali process is preferred in industry.


Do formic acid and phosphorus pentoxide react at room temperature? If I add phosphorus pentoxide to formic acid at atmospheric pressure and 20°C will they react? At what rate will CO be produced?
I have not been able to find a reference discussing this.

As noted, the dehydrating action of H2SO4 on HCOOH liberates CO gas. To quote a source:

The sulfuric acid removes the elements of water (i.e., H2O) from the formic or oxalic acid and absorbs the water produced.

As P2O5 is at least as dehydrating as Sulfuric acid, expect the same per the reaction:
$$\ce{P2O5 + 3 HCOOH -> 3 CO (g) + 2 H3PO4}$$


Heating of Enamel coated cooking utensils and health hazards I've tried to research about the issue of health concerns about enamel coating of cooking utensils (i.e. dutch ovens) and failed. I should prepare a small report about the health concerns if they exist.

Could you please help to answer that question? Any reliable sources would be greatly appreciated, especially scientific articles.

Thanks.

I unfortunately don't have time to dig through these articles in great detail, but they should help get you started. Feel free to comment with more questions.

The overall impression I got from my brief search is that there is definitely a potential health hazard depending on how the coloring was made, but in most modern countries, there are  regulations in place to prevent those kinds of problems. Enough about what I think though, let's get into the references.

According to Wikipedia, the coating on most cookware is vitreous enamel, which is made by heating a mixture of glass powder and pigment to a very high temperature and letting it melt onto a surface (much like glazing in pottery).

The glass itself is pretty much harmless, so the main danger comes from the added pigments. Here, we're stuck, because it really does depend on what was used to make the pigment. If it's a zinc pigment, you're probably fine. If it was cadmium based, you might be kinda screwed.

I've found references to several articles which describe experiments trying to find out what the leach rate of metals from the glass into a solution is (usually by boiling in vinegar), but I can't hunt down the fulltexts right now, aside from the one open-access article from the NIH. Here's what I have:

Full text:Lead, Cadmium and Cobalt (Pb, Cd, and Co) Leaching of Glass-Clay Containers by pH Effect of Food
While this study was on clay containers, it should theoretically apply to glass-on-metal as well. Keep in mind that the containers seem to have been acquired from local craftsmen in Mexico, so the results won't necessarily apply to stuff you find on the supermarket shelves.

Abstract: Hot leaching of ceramic and enameled cookware: collaborative study.
This is what you're looking for, but I can't find the full article, and the authors seem to have been more interested in establishing a method than finding the health effects. If you can find the full paper and cross-reference their results with acceptable limits of exposure (maybe from the EPA), that'd probably be your best bet.

The second link also has other articles on similar problems. If you can't find your information from these two articles, maybe you can find it by following the links on PubMed. Good luck!


Who coined the term "gas evolution reaction"? I found a paper from 1812 which specifically mentions the evolution of a gas [1].
There is a similar reference from 1808, except it uses the term evolution to speak of green light emitted when a material touches a green wire [2].
Is there anyone who knows for sure when the first time “gas evolution reaction” and the “gas evolved” and other such uses for the term “evolved” appeared in chemistry?
References

Henry, W. LX. Additional Experiments on the Muriatic and Oxymuriatic Acids. The Philosophical Magazine 1812, 40 (175), 337–343. DOI: 10.1080/14786441208638245.
Scholes, J. XI. On the Union of Gases. The Philosophical Magazine 1808, 31 (121), 68–70. DOI: 10.1080/14786440808563850.

There is no specific term as "gas evolution reaction" in modern chemistry but the word evolve itself conveys this meaning of emission. I am curious as to why you are interested in this terminology? Just for the sake of historical interest, the unabridged Oxford Dictionary (subscription required) shows that the term dates back to 1770s in relation to gas, heat, light etc. However there is no guarantee that the following references are the earliest. If you have more energy or you are a serious scholar of history, you might need to see German and French literature before the 1770s and what words Lavoisier used when he heated the oxide of mercury and discovered oxygen.

IV. The process of emission.

Chiefly Chemistry. The emission or release of gas, heat, light, etc.; the disengagement of a substance from combination; an instance
of such a process. Cf. evolve v. 5.

1783   Philos. Trans. (Royal Soc.) 73 350   The method put in practice
by Mr. Hutchins to settle the freezing point of quicksilver depends
entirely upon this generation, re-appearance or evolution of heat.
1800   H. Davy Res. Nitrous Oxide 425   This is on account of the
increased evolution of nitrogene from the blood.

Similarly, evolve is even older

1772   [implied in: J. MacLurg Exper. Human Bile 38   The colouring
matter of the Bile discovers also a much stronger attraction to the
mineral acids, and gives indubitable marks of a more evolved
phlogiston. (at evolved adj. 2)].
1786   Edinb. New Dispensatory i. 7   In the incipient state a
quantity of gas is still evolved, and along with it a quantity of
alkaline air.


Gaussian Calculations - What is the Basis? What am I actually changing when I click 631G(d) or 631G(d,p), etc. when I choose a "basis" in restricted Hartree Fock calculations? What is the difference between using an alternative basis?  Many thanks.

The basis set is used to describe the molecular orbital wave functions.

I suggest reading through a number of links:

Wikipedia
The Gaussian manual - Basis Sets - a nice description of the different basis sets, if a bit technical at times.
Naming of quantum chemistry basis sets
GAMESS Crash Course - applies to Gaussian or other packages


Factors affecting mass of a gas-evolving reaction mixture over time Excess magnesium was added to a beaker of aqueous hydrochloric acid on a balance. A graph of the mass of the beaker and contents was plotted against time (line 1).
What change in the experiment could give line 2?

The same mass of magnesium but in smaller pieces
The same volume of a more concentrated solution of hydrochloric acid
A lower temperature

Original question:

The answer is just II. However, I thought that both I and II would answer the question. Why is it just II?

I. False – because the magnesium has a higher surface area in this reaction, the rate will proceed faster, producing a steeper slope in the graph; however, the final mass of the solution will be the same in both experiments. We would note that the lines converge when each system reaches equilibrium.

II. True – the higher concentration of $\ce{HCl}$ means more $\ce{H2}$ gas will be lost from the system at equilibrium. Thus, the line would flatten beneath line 1.

III. False – a lower temperature would mean the average kinetic energy of molecules in the system would be lower; therefore, a smaller fraction of the molecules will have sufficient energy to react with others. In this case, we would note an elongated curve that converges with the first.


Can an EWG behave as an EDG and vice versa? Can an Electron Withdrawing Group(EWG)(like NO2, F, Cl, OH, etc.) behave as a Electron Donating group(EDG) in a compound where a carbocation is present in order to stabilise it?

Does the position where it is present or distance from the carbocation affect it's behaviour as an EWG or EDG?

I feel that it should behave as an EDG as carbocation is electron deficient so it has a greater tendency to accept an electron.

Can we say a similar thing that an EDG behaves as an EWG in presence of carbanion as it has excess electrons?

Of course not. That's why it is called EDG or EWG. It tends to destabilize it more in the absence of other effects like resonance and hyperconjugation.


What makes R-MgX a more reactive Grignard reagent than R-Cu? What makes $\ce{R-MgX}$ a more reactive Grignard reagent than $\ce{R-Cu}$? Has it to do with solvation, bond enthalpy, electronegativity?

The difference in electronegativity between copper (1.9) and magnesium (1.3) is the key difference.

Since copper is more electronegative than magnesium its electronegativity is much closer to that of carbon (2.55).  This results in carbon-copper bonds being less polarized and more covalent than carbon-magnesium bonds.  The electrons in a carbon-copper bond are better shared between the two atoms, whereas the electron distribution in a carbon-magnesium bond places more electron density on carbon.

As a result of this difference in electron density on the carbon atoms the carbon portion of a cuprate is a softer nucleophile, while the carbon portion of a Grignard is a harder nucleophile.  These difference in nucleophilicity produce different reactivity patterns for the two reagents.  For example, the softer cuprate will undergo 1,4 addition to an $\alpha,\beta$-unsaturated carbonyl compound (a softer electrophile), while the Grignard adds directly to the carbonyl (which is a harder electrophile).


Removing prenylated phenolics from clothing, shoes, and upholstery I live in Southern California and sometimes hike in areas that can be overgrown with a plant called purple poodle-dog bush. As I found out the first time I went crashing through a stand of the stuff, it's a horrific skin irritant, much worse than poison oak. It has microscopic hairs that get stuck in your skin and act like syringes. The hairs have chemicals on them that are skin irritants. Wikipedia says:

  This contact dermatitis is due to prenylated phenolics exuded by hairs (glandular trichomes) of the plant. The principal irritants are derivatives of farnesyl hydroquinone and 3-farnesyl-P-hydroxybenzoic acid.

I'm not a chemist, and all the nomenclature is Greek to me. My question is this. When I hike in an area where it will be hard to avoid the bush, I try to wear long sleeves and change my clothes immediately afterwards. Will throwing my clothes in the washing machine remove these chemicals and/or break them down into harmless ones? What if my shoes aren't machine washable? Will the chemicals break down naturally within a certain period of time? Would other methods such as exposure to sunlight help? Are these chemicals soluble in water, so that simple rinsing would help? How about making sure they aren't lurking in the upholstery of my car?

As a starting point, I shall assume that the irritant compounds are freely exposed to the environment (which is not necessarily the case, as I mention further on). As such, one of the conceivably simplest ways to remove them is to wash them off, i.e. disperse them in a solvent and discard the liquid mixture. For several reasons (abundance, extensive solvation ability, non-toxicity, etc.) water is a good choice.

First of all, how well would pure water work at dispersing the prenylated phenolics? Probably not too great. The representative parent compounds of the irritants, hydroquinone and p-hydroxybenzoic acid have modest solubilities in water (59 g and 5 g per 1 L, respectively), and the addition of farnesyl chains in the compounds would quite likely cause a severe reduction in water solubility, unless the farnesyl chains are highly substituted with hydrophilic groups.

Of course, there is no substance that is entirely insoluble in any solvent, though it is very common for certain solutes to require impractical amounts of solvent to disperse completely. It could well be that a washing machine's load of water (of order 100 L) is actually capable of dissolving the farnesylated toxins, especially because natural products are generally present in rather small quantities. However, there are some techniques which can possibly help the process. I list a few in the following:

Use hot water - The majority of substances become more soluble in hot solvents because dissolution very often increases the entropy of a system ($\Delta S>0$), and hence from the Gibbs' free energy relation $\Delta G=\Delta H - T \Delta S$, an increase in temperature leads to a larger decrease in free energy, making the process more favourable. Of course, this only really works well if the water is hot throughout the soaking/washing process; adding hot water and letting it cool off will just cause part of the solute to precipitate out of solution again.
Use a surfactant (detergent) - Since the offending compounds potentially have large hydrophobic chains, they can conceivably removed by the same process we use to wash away oils and fats. An amphipatic molecule contains a hydrophobic part (such as the hydrocarbon chain of sodium dodecylbenzenesulphonate) responsible for interacting with the oils and a hydrophilic part (in SDBS, the sulphonate group), which is well solvated by water. Thus, the surfactant bridges the interaction between oil and water, allowing effective dispersion.
Use a basic solution - In this specific case, the undesirable compounds have some acidic character. In particular, the hydrogen atom in the -COOH (carboxyl) group of the p-hydroxybenzoic acid derivatives are about as acidic as acetic acid. The -OH (hydroxyl) groups in both p-hydroxybenzoic acid and hydroquinone are also slightly acidic, though weaker by some 5-6 orders of magnitude. Thus, mixing in a basic solution may allow an acid-base reaction to take place, forming salts of the organic compounds with significantly increased solubility in water. Many washing powders/liquids are already slightly alkaline, but it is also easy to prepare basic solutions separately. A cheap and very safe base would be sodium carbonate, $\ce{Na2CO3}$, though a few tens of grams may be required to prepare a decent volume of basic medium for soaking. Sodium bicarbonate, $\ce{NaHCO3}$, also works though it is a weaker base and would require a higher amount to reach a given pH value. Caustic soda, $\ce{NaOH}$, is yet another cheap option, but care must be taken to not add too much, as it is a strong base (in contrast to the previous two). I don't know how resistant different fabrics are but, very approximately, any more than a couple of grams of lye per litre of water risks partially hydrolysing the fabric and weakening its structural integrity. Use at your own risk!
Use a bleaching agent (chlorinated or non-chlorinated) - Bleaches work by partially oxidizing organic matter, breaking organic molecules into smaller chunks and making each piece more hydrophilic (due to added oxygenated functionalities). Even if the compounds aren't removed entirely, the reaction products may be much less harmful.
Miscellaneous - You suggest exposure to sunlight. Light is certainly capable of decomposing organic substances, especially UV light, but there's no guarantee that the exposure required to eliminate the irritants won't also break down fabric and bleach dyes. Waiting is also probably not very efficient, as there is no guarantee the irritant compounds are unstable enough to undergo spontaneous decomposition in ambient conditions at a significant rate.

Unfortunately, all this discussion is of limited use if the initial assumption (freely accessible irritants) is incorrect. It could be that the compounds are stored in cavities in the plant hair's surface or interior, and the hairs themselves are insoluble and impermeable to water, effectively locking the toxins away. In this case, the compounds will likely be removed only slowly, as the hairs are progressively worn out by abrasion and other processes. Thus a physical removal method may be preferable. I have heard that adhesives (sticky tape, waxing strips, etc.) can be successfully used to peel out painful hairs. This is also a useful option for surfaces which cannot be washed easily (such as car upholstery).

In practice, I am not sure how this will all work out. It's been a few days since I wrote my comments, so perhaps you've had time to test something out? If so, I would be curious to hear some feedback!


What is the biggest known difference between rₑ and r₀? What is the biggest known difference between $r_e$ and $r_0$? Where $r_0$ is the average bond length of the lowest vibrational state, and $r_e$ is the location where the potential is lowest.

For a perfect harmonic oscillator, these are the same (notice that the peak probability is exactly at the minimum of the potential):

What is the largest known difference between $r_e$ and $r_0$ in a real molecule?

This is important because the CRC Handbook on Chemistry and Physics almost always gives $r_0$ because that's what can be determined more easily from experiment, but ab initio geometry optimization almost always gives $r_e$, yet many publications do not even say which type of bond length they are reporting.

Let me start by saying that this is a very difficult question to answer because it's not very hard to come up with pathological cases which I will address at the end. Also, as you point out, it is somewhat non-trivial to calculate $r_0$ from theory and also somewhat non-trivial to determine $r_e$ from experiment. That being said, I think I have a good first suggestion that I am guessing will be hard to beat. I will just answer with theory because this data is more readily available.

$\ce{F^-(H2O)}$:

As has been pointed out, this question is very closely related to the anharmonicity of the vibration in question. The $\ce{O-H}$ stretch of $\ce{F^-(H2O)}$ is the most anharmonic known vibration for a water molecule. This has an experimentally observed vibrational frequency shift of about $2000~\mathrm{cm^{-1}}$ which I'm almost sure is the largest known vibrational frequency shift in a molecular complex, where the molecule in question doesn't dissociate.

Now, that being said, this will vary based on the method used, but at the MP2/AVTZ level, $r_e=1.0653$ angstroms while $r_0=1.1248$ angstroms resulting in a difference of $0.0595$ angstroms.

This $r_0$ value is calculated from VPT2, which is probably the easiest way to calculate this quantity, but this can fail for certain pathological cases like those I'll discuss below.

$\ce{H5O2^+Ar}$:

Ok, now I know that 0.06 angstroms doesn't seem like that huge of a difference, but generally both of these numbers will increase together, so it's hard for them to be all that different. Now, another way you could consider finding a very large difference would be to look for a double-well system where the zero-point energy of the vibration in question is larger than the barrier height.

Now, the zundel cation is a proton shared symmetrically between two water molecules. This is a single well system, however, so $r_e$ and $r_0$ will be the same and they have a calculated $\ce{O-H}$ distance of $1.202$ angstroms at the CCSD(T)/aug-cc-pVDZ level of theory.

If, however, we disturb this system ever so slightly, we will break the symmetry and might get the desired behavior. This can be done by tagging the complex with a rare gas atom, which is often done in experiments as it reduces the noise in spectra for reasons I won't get into.

$\ce{H5O2^+Ar}$ has a hydronium like complex with an elongated $\ce{O-H}$ bond length of $1.108$ ansgtroms. It is very likely that the barrier to proton transfer in this case is very small because there is no barrier in the absence of an Argon atom. Therefore, it is likely that the proton will be close to symmetric after vibrational averaging, so if it were perfectly symmetric, we would get an $r_0$ of $1.202$ angstroms again. It is likely less than this, but it is possible this would lead to a difference greater than $0.06$ angstroms which we got from $\ce{F^-(H2O)}$.

I should point out that calculating $r_0$ for this system is pretty difficult because using VPT2 will almost certainly give nonsensical answers because the harmonic reference is a terrible one, so the perturbation theory will break down. The best option is probably to do diffusion monte carlo and extract the average from the wavefunction directly. It's possible this has been done, but a cursory search of the literature didn't turn anything up.

Also, the data for the first part comes from the SI of ref.[1] and the data for the second part comes from ref.[2].

References:

[1]: Boyer, M. A., Marsalek, O., Heindel, J. P., Markland, T. E., McCoy, A. B., & Xantheas, S. S. (2019). Beyond Badger’s Rule: The Origins and Generality of the Structure–Spectra Relationship of Aqueous Hydrogen Bonds. The journal of physical chemistry letters, 10(5), 918-924.

[2]: Park, M., Shin, I., Singh, N. J., & Kim, K. S. (2007). Eigen and Zundel forms of small protonated water clusters: structures and infrared spectra. The Journal of Physical Chemistry A, 111(42), 10692-10702.


Aren't isothiocyanate and thiocyanate ligands monodentate? I will keep it brief; if I generalize what an assignment is telling me, then a metal complex with 2 (iso)thiocyanate ligands would be named "bis(isothiocyanate)...metal(oxidation state) or bis(thiocyanate)...metal(oxidation state). From what I can tell (1) (2), (iso)thiocyanate ligands are exclusively monodentate... so why do they get the bis-tris-tetrakis-etc. prefix? I thought that those were reserved for polydentate ligands (3) or ligands already containing one or more di-tri-tetra-etc. prefixes. Thanks!

The answer to your title question is: Yes, they are monodentate.
But I think you have a misconception about when to use which type of counting: It does not depend on whether the ligand is polydentate or not.
The IUPAC guide states that you use the bis-tris-tetrakis-etc. prefix when you want to avoid confusion:

The  prefixes are ‘di’, ‘tri’, ‘tetra’,etc., for  use  withnames for  simple  entities,  or ‘bis( )’, ‘tris( )’, ‘tetrakis( )’,etc., for names for most entities  which themselves contain multiplicative  prefixes or locants.  Care must  also  be  taken  in  situations when  use  of  a simple multiplicative prefix may be misinterpreted, e.g., tris(iodide) must  be  used  for  $\ce{3I−}$ rather  than  triiodide  (which  is  used  for  $\ce{I3^−}$), and  bis(phosphate)  rather  than  diphosphate  (which  is  used  for $\ce{P2O7^{4−}}$).
Source: https://iupac.org/cms/wp-content/uploads/2018/05/Inorganic-Brief-Guide-V1-3.pdf

So while in the case of thiocyanates and isothiocyanates I don't see that there is any confusion avoided by using bis( ), it is certainly not wrong.


Relation of pressure and temperature for an ideal gas at constant volume I have a tank of ideal gas with volume $3.1\ \mathrm L$; I have $R = 8.314$, and 11 moles. The tank is filled at $23\ \mathrm{^\circ C}$ (not sure why that's relevant) and the tank will rupture at $P = 100\ \mathrm{atm} = 1.013\cdot 10^7\ \mathrm{Pa}$. At what temperature does the tank rupture?

I divided $PV$ by $nR$. However, plugging those numbers in gives

$$\frac{3.1 \times 1.013\times 10^7}{8.314\times 11} = 342374$$
in kelvin. However, my book gives $70.2\ \mathrm{^\circ C}$ as the correct answer. I assume my units are somehow profoundly messed up. Or maybe that mysterious 23 has something to do with it? What am I doing wrong?

When using the value of $R=8.314$, the units are $\mathrm{\frac{J}{K\cdot mol}}$. One joule is equal to:

$$\mathrm{J=\dfrac{kg\cdot m^2}{s^2}=N\cdot m = Pa\cdot m^3 = W\cdot s = C\cdot V}$$

If you want to use volume in liters, you need to use a different value for R.


Distribution of polymer lengths Suppose we have a chemical reaction where a linear chain can form (reversibly) from building blocks (e.g. atoms or molecules):

$$\ce{X_$i$ + X_$k$ <=>[$k_\mathrm{fwd}$][$k_\mathrm{rev}$] X_$i$X_$k$}$$

where $k_\mathrm{fwd}$ and $k_\mathrm{rev}$ are rate constants.

What is the distribution of polymer lengths (at equilibrium)? What are some canonical references that derive these distributions under variations of the above polymerization process (e.g. step versus chain growth)?

Concentrations are a geometric series

First, let's focus on the reactions where one of the reactants is a monomer $\ce{X_1}$:

$$\ce{X_1 + X_n <=> X_{n+1}}$$

and call the equilibrium constant for this reaction $K_{\mathrm{extend}}$.

From this, we can figure out the equilibrium constant for making a polymer from monomers:

$$\ce{n X_1 <=> X_n}$$

$$K_n = (K_{\mathrm{extend}})^{n-1} = \frac{[X_n]}{[X_1]^n}$$

So we can express the concentration of a polymer $X_n$ dependent on the monomer concentration:

$$[X_n] =  (K_{\mathrm{extend}})^{n-1} [X_1]^n = \frac{1}{K_{\mathrm{extend}}}(K_{\mathrm{extend}} \cdot [X_1])^n$$

This is a geometric series (if you take the concentration of polymers as a series, the ratio of neighbors in the series is constant). Because the amount of matter is finite, $K_{\mathrm{extend}}\cdot [X_1] < 1$ for an equilibrium scenario (as opposed to ending up with a single chain containing all the monomers). So the concentration of $[X_1]$ is the highest, and concentrations drop with increase of $n$.

Expression for $[X_1]$

Let $[X]_\mathrm{tot}$ be the total concentration of monomer, no matter whether in a polymer or not.

$$ [X]_\mathrm{tot} = \sum_{j=1}^{\infty} j[X_j]$$  $$ = \frac{1}{K_{\mathrm{extend}}} \sum_{j=1}^{\infty} j (K_{\mathrm{extend}}\cdot [X_1])^j $$

Using $$\sum_{j=1}^{\infty} j \beta^j = \frac{\beta}{(\beta - 1)^2}$$

we get

$$ [X]_\mathrm{tot} = \frac{[X_1]}{(K_{\mathrm{extend}}\cdot [X_1] - 1)^2}$$

which you can solve for the monomer concentration.

Relationship to real polymerization reactions

The polymerization reactions in chemistry and biochemistry that I am aware of are kinetically controlled. There are fibers (such as actin) in biochemistry that are dynamic (i.e. there is polymerization and depolymerization), but the processes are affected by nucleotide binding, complicating matters. I would be interested to hear whether the OP has an example that behaves like this.