Is molarity an intensive property? $$\mathrm{molarity} = \frac{\text{amount of solute}}{\text{volume of solution}} $$ and amount of substance is based on quantity (larger mass means larger amount), so how come it is an intensive property. Shouldn't it be an extensive property?

Concentration is an intensive property. The value of the property does not change with scale. Let me give you an example:

  Let us say you had a homogenous mixture (solution) of sodium carbonate in water prepared from 112 g of sodium carbonate dissolved in 1031 g of water.

The concentration (in mass percent, or mass of solute per mass of solution) is:

$$c=\frac{112\text{ g solute}}{(112+1031)\text{g solution}}=0.09799 =9.799\%\text{ sodium carbonate by mass}$$

The concentration is the ratio of sodium carbonate to the total mass of the solution, which does not change if you are dealing with the entire 1143 g of the solution or if you dispense some of that solution into another vessel.

  If you dispense 11.7 g of that solution into a flask for a reaction, what is the concentration of sodium carbonate in that flask?

It is still 9.799% by mass. The ratio of the mass of sodium carbonate present to the total mass present has not changed. The actual mass of sodium carbonate has changed:

$$0.09799\dfrac{\text{g solute}}{\text{g solution}}\times11.7\text{ g solution}=1.15\text{ g solute}$$

The concentration is a property dependent only on the concentration of the solution, not the amount of solution you have. The concentration of a solution with defined composition is independent of the size of the system.

In general, any property that is a ratio of two extensive properties becomes an intensive property, since both extensive properties will scale similarly with increasing or decreasing size of the system.

Some examples include:

Concentration (including molarity) - ratio of amount of solute (mass, volume, or moles) to amount of solution (mass or volume usually)
Density - ratio of mass of a sample to the volume of the sample
Specific heat - ratio of heat transferred to a sample to the amount of the sample (mass or moles usually, but volume also)

Each of these intensive properties is a ratio of an extensive property we care about (amount of solute, mass of sample, heat transferred) divided by the scale of the system (amount of stuff usually). This is like finding the slope of a graph showing the relationship between two extensive properties. The graph is linear and the value of slope does not change based on how much stuff you have - thus the slope (the ratio) is an intensive property.


Why can't a galvanic cell be a single cell? I get that galvanic cells require a salt bridge to maintain neutrality so the cathode doesn't become saturated with electrons, but why is a two-cell setup required? Wouldn't the spontaneous reaction occur and neutral conditions be met if Cu and Zn electrodes were placed in a solution of NaCl?

And vice versa, couldn't an electrolytic cell work just fine with a salt bridge?

This website (and many others) leads me to believe that galvanic cells must have a salt bridge and electrolytic cells must be a single cell. My intuition says that's not right though. Is it because you don't want undesirable compounds reacting at the electrodes? I'd think you'd want the same for both EC cell types if that was the case and electrolytic cells would use salt bridges as well (eg proton exchange membranes).
http://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Electrolytic_Cells

Perhaps this isn't the full answer, but it probably plays a role. It is critical for galvanic cells that there not be any electronic conductivity through the cell between the two electrodes. If there is a path for electrons, they will take it which means the cell will spontaneously discharge and not do any useful work through the external circuit. Pure water is a very poor conductor but becomes conductive once ions are dissolved.

By comparison, the construction of a lithium ion cell is close to what you describe for a single chamber galvanic cell. There is a separator involved but mostly to keep the two electrodes from coming into direct contact. One reason this is possible is that the electrolyte is an organic solvent (usually ethylene carbonate, dimethyl carbonate, etc) and doesn't conduct electrons, even when the appropriate salt (eg. $\ce{LiPF6}$) is dissolved in it. There is no salt bridge in this cell, though it is clearly galvanic.

Regarding the electrolytic cell, I don't see any particular reason that you couldn't build one with a salt bridge, provided the salt-bridge let you balance the charges generated on each side. Maybe they're all presented as one beaker just because it's easier to make and there's not really any advantage to a salt bridge in this case.


How can I make a lot of microscopic rods of metal? I want to make around 40-200 g  of metal cylinders that would ideally have a length of around 1-100 μm in size (smaller preferred). The rods will be made of some ferromagnetic material such as Iron or Nickel.
The reason for this is to make a magnetic field display where the tiny metal rods are submersed in a viscous liquid which is trapped between two clear sheets of material (likely glass or acrylic).
Since these are rods and not balls, this should allow them to stick end to end without bunching up so much and better visualise the magnetic field lines.
Although i'd prefer The rods to be smooth, they don't necessarily have to be a cylinder shape, they just have to have a length to width ratio higher than 2:1 (although i'd prefer a bigger aspect ratio than that).
It is preferred, but not necessary that the rods are all the same length with minimal variation in size.
Last preference is for the metal to look shiny rather than black as this helps a lot with visualising the field lines. I'll take what I can get though.
Haven't found anything online like this but I also don't really know what to search for.
I thought of using thin iron/steel wire and cutting it up but this process would most likely be too difficult and tedious to carry out unless I could construct some machine or jig to do it for me. I also have struggled to find any sort of ferromagnetic wire smaller than 0.1mm which would still be too thick for what I want.
I also thought of long metal shavings but they tend to be twisted and irregular (and possibly too flat) to work well for this.
Perhaps I could use some sort of micro 'cheese grater' to grate pieces off of a metal block?
Could these rods be synthesized via electrolysis somehow?
I'm at a loss so any help or pointers on how to go about this would be greatly appreciated!

Here is a relevant 2013 paper: Synthesis of pure iron magnetic nanoparticles in large quantity, to quote:

Free nanoparticles of iron (Fe) and their colloids with high saturation magnetization are in demand for medical and microfluidic applications. However, the oxide layer that forms during processing has made such synthesis a formidable challenge. Lowering the synthesis temperature decreases rate of oxidation and hence provides a new way of producing pure metallic nanoparticles prone to oxidation in bulk amount (large quantity). In this paper we have proposed a methodology that is designed with the knowledge of thermodynamic imperatives of oxidation to obtain almost oxygen-free iron nanoparticles, with or without any organic capping by controlled milling at low temperatures in a specially designed high-energy ball mill with the possibility of bulk production. The particles can be ultrasonicated to produce colloids and can be bio-capped to produce transparent solution. The magnetic properties of these nanoparticles confirm their superiority for possible biomedical and other applications.

So, to answer your question, it actually may be feasible based on more recent work. Note: the article appears in the applied physics section of Journal of Physics, reflecting the likely significance of this topic in this area.


Why is lithium plating irreversible in Li-ion cells? Several sources (1, 2) suggest that charging Li-ion batteries in low-temperature conditions result in lithium plating. This seems pretty logical since low temperature reduces ion mobility, and provided there exists a high enough charge current, lithium ions don't have time to intercalate into carbon layers of the anode and are deposited on top.
What I fail to understand is why this process is irreversible (Wikipedia is careful to say "may not be removable"). This is counterintuitive to me. Since the plated lithium ions don't have to overcome any van der Waals forces, they should be the first to go when the battery is supplying current. So as long as the metallic lithium does not grow dendrites, thereby damaging the internal structure of the cell, a little plating should be mostly harmless.
Does my reasoning make sense? And if not, what did I get wrong?

It's true that when the battery is re-discharged, it's thermodynamically 'easier' to oxidize lithium than lithiated graphite. However, keep in mind that the lithium will react with electrolyte to form an SEI (solid electrolyte interphase) layer. This process consumes lithium ions, so on recharge you don't get all the lithium ions back. This is why the process is considered "irreversible."


Polar vs. Amphipathic How can I tell if a molecule is purely polar or amphipathic?
For example, fatty acids are amphipathic because they have a polar carboxyl group and a non polar chain of hydrocarbons. However, sometimes there are molecules that have hydrocarbon chains and side polar groups but are categorized as polar.
What is it that I should look for?
Thank you so much!

The answer is that it is a sliding scale. There is no such thing as "purely polar" or "purely nonpolar." Water is very polar and hydrocarbons are very nonpolar, but there is no clear dividing line between the two categories.

Further confusion in biological molecules such as amino acids can occur, because different individuals categorize them in different ways, considering either just the side chain or the whole molecule. For example the amino acids have the formula $\ce{H2N-CHR-COOH}$ and in the simplest case, Glycine, R=$\ce{H}$. This is clearly a very non polar side chain, but the molecule as a whole is mostly polar, as the side chain is such an insignificantly small part of the molecule. Thus authors who categorize amino acids by side chain (mainly chemists) put Glycine into the non polar category, while authors who categorize amino acids by the whole molecule (mainly biochemists and biologists) put it into the polar category. When it comes to Alanine, R=$\ce{CH3}$ all authors agree it has a non polar side chain and most seem to categorize it as a non polar molecule.

A simple search of the internet shows that there is very little agreement about the amino acid Tyrosine, R=$\ce{CH2-(C6H4)-OH}$. It is classified as polar, non polar, having a polar side chain and having a non polar side chain by different authors. Clearly it has some characteristics of both, having both a benzene ring and a hydroxyl group. The correct definition would be "moderately polar."

You might hope that it is possible to give molecules a number which describes their polarity. This can be done to some extent. Googling "solvents by polarity" will give you lists for common solvents. This is, however, complicated if the molecule of interest happens to be acidic or basic. An acidic molecule will be polar in acidic solution but will become charged (and therefore much more polar) in basic solution, due to loss of $\ce{H+}$. Similarly, a basic molecule will be polar in basic solution but will become charged (and therefore much more polar) in acidic solution due to gain of $\ce{H+}$.

An amphipathic or amphiphilic molecule has a part that is highly polar and a part that is highly non polar. These terms also generally mean not just a molecule that is moderately polar, but one in which there is some distance between the polar and non polar parts, so the molecular weight would be at least into the hundreds.

The example you give of a fatty acid is a good one. (The deprotonated, negatively charged fatty acid anion in basic solution would be an even better one.) Another example would be many cell membrane proteins, which have a non polar, hydrophobic part (that stays in the cell membrane) and polar, hydrophilic parts (that extend out into the aqueous media inside and outside the cell.)

In summary, different substances should not be pigeonholed as polar or non polar as it would require that you draw an arbitrary line, but in general the polarity of two different substances can be compared fairly objectively. And pH may make a difference.


How to calculate molecular volume from the wavefunction? Very short question but maybe not a short answer !   As the probability of presence of electrons in a specific spatial position is defined as square of the wave function, I am wondering if there is a technique for molecular volume calculation based on the wavefunction.
Is there such a technique?

Update:  I could run the volume keyword in Gaussian.  This is the output for methane , but I can not find the volume in it !

the volume seems to be already in the output:

$\text{Molar volume}=305.994~\mathrm{bohr^3/mol}\ (27.304~\mathrm{cm^3/mol})$

But of course, this is defined with the volume inside a density contour of $0.001~\mathrm{e/bohr^3}$.  That may be different from what you measure, because volume is not a quantity with well defined meaning, and what you find in an experiment depends on how you measure it.

Mathematically any atom or molecule have a volume of infinity since the electron spreads to the entire space, but this is clearly not the physically meaning of volume as we usually know it.


Is there an experimental way of measuring the critical concentration of pigment in paint? With an assigned, generally paint-based, experimental project I've been looking up CPVC or critical pigment volume concentration in binders. CPVC defines the point of addition of pigments after which follows a severe drop in blistering and increase in rusting on the applied surface due to the pigment not while binding with the binder having voids in it. But going past the critical point can also make the paint fragment and crumble or maybe even lose just some of its pigment, in conclusion lose efficiency.
Paint manufacturers use this to make paints a certain type by knowing the CPVC of the mixture with which they can adjust to make the fitting paint for the job but the way they know is by applying the general formula for determining the point which is

$$CPVC = \frac{1}{ 1 + OA * ρ / K }$$

With the 'ρ' standing for the specific gravity of the pigment, 'K' specific gravity of the binder and 'OA' which is the oil absorbent number of the pigment depended of the exact pigment and the binder. But excluding it is there a way that a clear mixture of paint without any additions like extenders for example and just acrylic polymer binder and an inorganic pigment type can be used to calculate the near CPVC of it. Can it maybe be done by centrifuging a dry paint sample in a centrifuge tube and later sampling the super natant that will contain an ammount of pigment and binder that would dissolve in it  and fall apart from the rest of the paint and quantitatively analysing it in a high performance liquid chromatograher? By increasing temperature or the pH it could also affect the stability (the CPVC) so can that method be used effectevly? Is there a better one and not by just assuming the CPVC by observing the paint on a wall?
Any help could be really helpful sense im young and do not know the process well, only by researching the concept itself, and it's hard to find details and think of a valid project idea.

An image of the CPVC effect is shown in the diagram below. The red color is the binder, the black is the pigment and the white is air (because there is not enough binder to coat all the pigment). The lines depict the change in properties you would expect, and you already have the formula to correlate the paint properties with the raw material properties.

BTW, economics generally suggests maximizing the inexpensive pigment/filler concentration, but adding too much increases permeability to water, which is bad in the case of metals which might corrode. On the other hand, blistering can be caused by corrosion of the substrate which then disadheres some of the film if it is too rich in binder; blistering is minimized by allowing the film to break a little. Of course, then corrosion continues - you just don't see it as much until it is too far gone. The best paint will protect from corrosion with low permeability and without disadhering.
So, how to measure a property related to CPVC if you can't pick one of these? First, you need some of the paint film; I assume you can start with the plain binder and add different amounts of the pigment/filler. Mix well, do not incorporate unnecessary air bubbles when mixing. Cast a film of constant thickness on aluminum foil by using a threaded rod; let the paint dry.

Then do your experiments!

Flexibility. If you bend the painted aluminum foil over a rod of defined radius, the rich binder film will not crack, but as more filler is added more and more cracking tendency will be noticed.

Permeability. If you put a drop of water (or perhaps alcohol) on the dried film, it will wet and spread. You may be able to differentiate between different levels of added pigment. If the aluminum foil substrate does not give good results, you can apply the paint to a decent grade of paper. Then a drop of water or other solvent might wick all the way thru and appear on the other side of the paper.

Glossiness. Ideally, you should have some kind of electronic measurement, but ranking the specimens subjectively (with your own eyes) is better than nothing. In practice, flat paints have a high pigment concentration and are not durable when washed; semi-gloss paints have less pigment, more binder, and can be washed more often. (I always wondered why they didn't go all the way to "full gloss", but I guess the binder was never so good as to give much better performance.)

Washability. Apply a water stream to the painted aluminum foil (like a dripping faucet: constant height, constant volume per minute), watch the paint film disappear.

And try coming up with some more tests!


Does Potassium Superoxide also absorb Carbon Monoxide? Potassium Superoxide ($\ce{KO2}$) is used as an oxygen provider and carbon dioxide scrubber in life support systems
$$\ce{2KO2 + H2O -> 2KOH + O2} \\
\ce{2KOH + CO2 -> K2CO3 + H2O}$$
I encountered a question that asked if Potassium superoxide can also serve as a scrubber for $\ce{CO}$. I thought of a reaction where $\ce{KO2}$ oxidized $\ce{CO}$ to $\ce{CO2}$ and the $\ce{CO2}$ was absorbed as previously mentioned
$$\ce{2KO2 + 2CO -> K2O2 + 2CO2}$$
Another possible reaction would be reduction of $\ce{KO2}$ to $\ce{K2O}$
$$\ce{2KO2 + 3CO -> K2O + 3CO2}$$
I could not find these reactions or similar ones mentioned in any literature online. This article describes the working of $\ce{KO2}$ in detail but I could not find a mention of carbon monoxide in there.
Can $\ce{KO2}$ also be used to scrub $\ce{CO}$? If so, how?

As the other answer and comments imply, potassium superoxide is not a good reagent for absorbing carbon monoxide.  Cuprous chloride solution works better, as in this study[1] where the salt is dissolved in an ammonia-bearing solution.
Reference:

R. V. Gholap and R. V. Chaudhari, "Absorption of carbon monoxide with reversible reaction in cuprous chloride solutions", Ind. Eng. Chem. Res. 27, 11, 2105–2110 (1988).


How are piezoelectric crystals made? I want to know how the powdered formed of piezoelectric material is converted to crystal like used in quartz watches. I know that some more chemicals are mixed with the material and then it is pressed in form of tuning fork (for clocks) at particular temperature and pressure. So what are the chemicals, temperature and pressure which are used. How they are made to vibrate at that particular frequency?

Good quality quartz crystals for use in resonators can be found in nature, such as in Minas Gerais, Brazil. They can also be produced by crystallization in a hydrothermal process, simulating in the autoclave the same process that occurred over geologic epochs.

You can easily grow your own Rochelle salt (potassium sodium tartrate, $\ce{KNaC4H4O6·4H2O}$) piezoelectric crystals. Rochelle salt was commonly used in inexpensive crystal headphones and phonograph cartridges. The raw materials are household chemicals and the directions are straightforward. Again, these crystals are grown, not hot pressed.

Tuning forks watches such as the Bulova Accutron did not use crystals, but rather metal tuning forks, made from NiSpan C, with a low coefficient of thermal expansion, similar to invar. These did not rely on the piezoelectric effect, but on magnetic induction.


Difficulty in cis, trans isomers in Alkene I have a difficulty in appointing cis trans isomer to an Alkene compound .

I have looked it up and drawn out a few -

Which "point" should I look at to understand in order to allocate cis and trans ? I know it's between the first and last line . Thanks!

You may think of the two lines representing the double bond like the bird's eye view on a door of a saloon:  If the two wings, the one on the left of the double bond, and the one of right flap in the same direction, then the double bond is in cis-configuration.  On the contrary, if the two flap in opposing directions, than the double bond is in trans-configuration.

For future reference -- taking into account the third picture shown by you -- cis and trans may be ambiguous; and triple and quadruple substituted double bonds' configuration are described only with E- or Z-configuration (cf. here).  Eventually, as correctly pointed by @Zhe, in your last example, the double bond is neither E-, nor Z-configurated.


Modeling Belousov-Zhabotinsky reaction using ordinary differential equations Is it possible to model the BZ reaction using first order ODEs? If so what would these equations be?

Or, is there something about the reaction that needs higher orders?

Maybe related: I had the chance to perform this very reaction about 3 years ago, for an undergraduate lab course, and there were many things which confounded me back then too. I have a question on the topic here. It contains an experimental procedure as well, in case you are interested.

  Is it possible to model the BZ reaction using first order ODEs? If so what would these equations be?

TL;DR: Yes.

Many variants of the reaction exist. The only key chemical is the bromate oxidiser. The catalyst ion is most often cerium, but it can be also manganese, or complexes of iron, ruthenium, cobalt, copper, chromium, silver, nickel and osmium. Many different reductants can be used.

Similarly, there exist many models to describe this reaction. One that is the simplest, and I particular like is called the "Oregonator" (discussed herein). The precise mechanism/model, as far as I know, is still an open question.

We characterise the reaction using the following equations:

$$  \ce{A + Y ->[k_1]  X } \tag{1}$$
$$  \ce{X + Y ->[k_2]  P} \tag{2}$$
$$  \ce{B + X ->[k_{34}]  2 X + Z} \tag{3} $$
$$  \ce{2 X ->[k_5] Q} \tag{4}$$
$$  \ce{Z ->[k_6] fY} \tag{5}$$

This model is a simplified version of the FKN model, by applying the steady state approximation, and partial equilibrium approximation to the FKN model.

Overall, $$\ce{fA + 2B -> fP+Q }$$ wherein $f$ is a suitable stoichiometry co-efficient. For any, $f$

$$\begin{cases}
     (f-1)A + 2B \rightarrow (f+1)P & f > 1 \\
      2B \rightarrow 2fP+(1-f)Q &f < 1 \\
      B \rightarrow P &f=1
   \end{cases}
$$

Now, one can simply write down the rate equations

$$ \frac{\mathrm{d}X}{\mathrm{d}t} = k_1 AY  - k_2XY + k_{34}BX -2k_5X^2$$
$$\frac{\mathrm{d}Y}{\mathrm{d}t} = -k_1 AY - k_2XY + k_6Z$$
$$ \frac{\mathrm{d}Z}{\mathrm{d}t} = k_{34}BX - k_6Z$$

The system evolves with 2 different time scales, and we can expect it to show relaxation oscillations. We can make this more apparent via some mathematical shenanigans, which you may perform as an exercise. I might revisit this later in the day, and write a bit more about them.


pH of rain with and without thunderstorm I just came to know that pH of rain water will be lesser when it is accompanied by a thunderstorm than when it is not accompanied by thunderstorm.
Why would it be so?
Will the temperature increase due to excessive friction (of air) and lighting and thus more ions of water will dissociate reducing the pH?
Or will it be because of production of acids in air during thunderstorm?
Help.

Though I've never heard of that statement before, it's possible that nitrogen oxides formed by electrostatic discharge (not only lightning, but even glow discharge) dissolve in the rainwater, producing nitrous and nitric acids.

There is a 1997 citation describing comparison of pH in rain accompanied (or not) by lightning.

That said, presence of fly-ash from coal burning, sulfur oxides from volcanic activity and carbon dioxide from natural as well as anthropogenic sources probably have greater influence on pH.


Atropisomerism of naphthyl alcohol In March's Advanced Organic Chemistry (6th ed.), it's stated that the above two are atropisomers. I don't get why is it so. How is the topmost $\ce{C}$ locked in its place? Surely, there's steric hindrance due to presence of two methyls but how does that prevent rotation? It can certainly rotate with the methyls being "rigid" in their positions. Also, are they chiral? I believe they aren't because they are superimposable mirror images.

The hindered rotation is due to the hydrogens at the naphthyl moiety. The following shows the rotation calculated at the DF-B97D3/def2-SVP level of theory. To better visualise this, I have chosen a mode with large atoms (not actually using the van-der-Waals radii, because that looked very strange; click here for a ball and stick version):

We can observe three minima and three transition states corresponding to that rotation, the depiction you have posted is slightly inaccurate as it shows one local minimum and a transition state.
From the three states that should be observable, the global minimum is achiral (1), while the other two are enantiomers (2, 3). The latter are connected via a low lying transition state (TS-2-3); probably because of the hydrogen bonding interaction guiding it. The global minimum on the other hand is separated via a very high energy barrier. These transition states (TS-1-2, TS-3-1) are again enantiomeric to each other.

I have calculated the associated energy barriers in $\pu{kJ/mol}$ at the DF-B97D3/def2-TZVPP level of theory.
\begin{array}{c*{6}{r}}\hline
\text{State} &
\Delta E_\mathrm{o} &
\Delta G^{\pu{50 K}}&
\Delta G^{\pu{75 K}}&
\Delta G^{\pu{100 K}}&
\Delta G^{\pu{298.15 K}}&
\Delta G^{\pu{500 K}}\\\hline
%
\mathbf{(1)}     &
    0.0 &  0.0 &  0.0 &  0.0   &  0.0  &  0.0\\
\mathbf{\text{(TS-1-2)}}     &
   32.9 & 32.9 & 32.9 & 33.1   & 35.7  & 39.7\\
\mathbf{(2)}     &
    0.7 &  0.7 &  0.7 &  0.8   &  1.4  &  2.3\\
\mathbf{\text{(TS-2-3)}}     &
   12.0 & 12.1 & 12.1 & 12.3   & 14.9  & 18.9\\
\mathbf{(3)}     &
    0.8 &  0.8 &  0.8 &  0.8   &  1.5  &  2.3\\
\mathbf{\text{(TS-3-1)}}     &
   32.9 & 32.9 & 33.0 & 33.1   & 35.7  & 39.8\\\hline
\end{array}

From these values we can estimate rate constants for the energy barrier with the Eyring equation, which employs the transition state theory:
$$
k = \kappa \frac{k_\mathrm{B}T}{h}\exp\left\{\frac{-\Delta G}{RT}\right\}$$

I use $\kappa = 1$, i.e. I am ignoring it, to get a guesstimate at which temperature we will be able to distinguish between all, or some of the states.

\begin{array}{c*{5}{l}}\hline
k/\pu{s-1};\ T= &
\pu{50 K}&
\pu{75 K}&
\pu{100 K}&
\pu{298.15 K}&
\pu{500 K}\\\hline
%
\mathbf{(1) \to (2)}     &
   \pu{4.9E-23} & \pu{1.8E-11} & \pu{1.1E-5} & \pu{3.5E6}  & \pu{7.5E8}\\
\mathbf{(2) \to (3)}     &
   \pu{2.6E-1}  & \pu{5.4E3}   & \pu{7.8E5}  & \pu{1.5E10} & \pu{1.1E11}\\
\mathbf{(3) \to (1)}     &
   \pu{2.8E-22} & \pu{5.9E-11} & \pu{2.8E-5} & \pu{6.1E6}  & \pu{1.3E9}\\
\hline
\end{array}

From this we can see, that at room temperature the rotation is pretty much free with a couple million transitions per second. At around $\pu{100 K}$ the is hindered; from the global minimum (1) to the adjacent minima the methyl groups simply cannot pass. Without haven calculated the values, I'd guess that this rotation starts to become hindered at around $\pu{150 K}$.
The other barrier still is easily manageable. That essentially means, that the enantiomers still readily interconvert. You will spectroscopically observe a superposition, which gives rise to the schematic structures that were used in the book. Even at $\pu{75 K}$ this barrier is taken a couple of thousand times per second
Only around $\pu{50 K}$, the interconversion between the enantiomers will also be hindered, and you should be able to distinguish all three states.

TL;DR:
You will start seeing atropisomerism at about $\pu{150 K}$, but at that temperature you'll only see the superposition of 2 and 3 (probably what is depicted in the book, i.e. left in your question) and the achiral, global minimum (on the right).

Side note:
It should be fairly obvious, that non covalent interactions also play an important role in this molecule. The barrier between the enantiomers 2 and 3 is probably significantly lower because of the hydrogen bond. However, one shall not forget, that this interaction is also present in the local minima to a not insignificant amount. The following pictures show the non covalent interactions analysed with NCIPLOT-3.0 and plotted with VMD; green areas are weakly attractive, blue areas are stronly attractive (hydrogen bonds), red areas are repulsive.


How to change the direction of evaporation and thus location of condensation? For a specific region of air, enclosed by a metal box in which a constituent is obviously gaseous $\ce{H2O}$, is there any way in which we can control the location of condensation?

Assume the box is heated, typically we will have condensed $\ce{H2O}$ formed at the ceiling, does anyone have any strategies of condensing water on the inner sides of the box, or maybe at the base?

I feel like I have a limited scope, with the outer-most idea being to stick anhydrous calcium chloride (dessicant) on the inner sides and have a ledge for the water to sit on.

The simple answer is that water evaporate where the concentration in the air is lower than the vapor pressure some liquid water at a given temperature. Then it will condense at a location where the temperature is lower that the saturation temperature of water at that partial pressure.

Industrially, though, one widely used technique to avoid condensation when the surface temperature is low enough for it to happen is to use hydrophobic coatings to prevent the water vapor from nucleating on certain surfaces. A quick google search will reveal lots of these for different applications.


Is it "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function"? If so, why? The recent paper in Nature Independent confirmation of a methane spike on Mars and a source region east of Gale Crater is interesting and the basis is a reanalysis of 2013 data taken by a X spectrometer about the Mars Express spacecraft in orbit around Mars looking down at the Martian atmosphere.
The paper itself is paywalled but the supplemental data describes the reanalysis in great detail.

The Planetary Fourier Spectrometer [38] (PFS) on Mars Express 37 is a double-pendulum infrared Fourier spectrometer optimized for atmospheric studies. It has two distinct spectral channels operating simultaneously and covering the wavenumber range between 200–2000 cm−1 (Long Wavelength Channel, hereafter LWC) and 2000–8300 cm−1 (Short Wavelength Channel, hereafter SWC). Both channels have a sampling step of 1 cm−1 and a spectral resolution of ~1.3 cm−1, when no apodization function is applied, and ~1.8 cm−1 when a Hamming function is applied to the measured interferograms (as in the case of the present work).

Key to the process is an understanding of how the raw data is preprocessed using a Hamming apodization function before spectral fitting.

Step 1: Hamming apodization function is applied to PFS interferograms. It is common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function in order to reduce the amount of ringing present in the resulting instrumental line shape [77]. This reduces the spectral resolution a bit (from 1.3 cm-1 to 1.8 cm-1 in the case of PFS [78]), but also reduces the instrumental noise and the magnitude of the side-lobes in the instrumental line shape, which are a direct result of the finite maximum optical difference in the measured interferograms [79]. The Hamming function H is defined as H(δ/L) = 0.54+0.46cos(πδ/L), where δ is the optical path difference out to a maximum value of L.

Davis, S. P., Abrams, M. C. & Brault, J. W., Fourier Transform Spectrometry, Academic Press (2001).
Giuranna, M., et al., Calibration of the Planetary Fourier Spectrometer short wavelength channel. Planet. Space Sci. 53 (10), 975–991 (2005).
Naylor, D.A. & Tahic, M. K., Apodizing functions for Fourier transform spectroscopy. J. Opt. Soc. Am. A 24, 3644–3648 (2007).

I was surprised to read that raw data is filtered before fitting to spectroscopic models in order to extract concentrations. I'm not an FTIRer, but I would have instead expected that all instrumental errors would be included in the fitted theoretical spectra generation and that the raw data would be fit in its pristine, unaltered form. After all, the only thing you really know for sure when fitting is that the data is the data, it's what you actually measured. Everything else is speculation.
QUESTION: Assuming that the block quote is right and it is indeed "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function in order to reduce the amount of ringing present in the resulting instrumental line shape" why is this considered "safe" to do? Why aren't all instrumental effects incorporated into the fitting function allowing for the raw data to be fit directly instead?

The spectra in question, from Independent confirmation of a methane spike on Mars and a source region east of Gale Crater:

You usually post interesting questions, which appear to be deceptively simple yet they are very challenging. Modern day data acquisition and signal processing is so complicated that it is almost like a black-box. It is amusing when I ask electrical engineers some signal processing questions, they don't know the answers and when I ask mathematicians it is too much applied mathematics for them. I am not an FTIR expert either but as an analytical chemist, I can add some remarks. I think apodization is routinely done in order to decrease noise and oscillations when you do an inverse transform. If you do some mathematical operations in the frequency domain, when you do the inverse transform, the level of noise is crazily high. Of course, when you are looking at very small signals do not wish to have noise.

In general, sometimes I need to fit functions to the data to peak shaped functions. I always get a better fit once I digitally filter the data and fit later (moving average, Savitsky Golay, Hamming in the time domain and so on.) Noise is the biggest enemy of any analytical chemist or a spectroscopist.

With any smoothing process, either in the frequency domain or in the time domain, you tend to loose resolution. There is always a sweet spot for digital filtering or if you have heard the story of Goldilocks who entered the house of three bears..."When she had finished the porridge Goldilocks felt tired and went to find somewhere to sit down. The first chair she found was too big, the second chair was still too big, but the third chair felt just right." The same rule of thumb goes for digital filtering. Too little is useless, too much you loose all resolution and just the right filtering gives you the best signal to noise ratio.


Temperature effects on pKa and pKw I am an industrial electrician apprentice who needs to learn more about pH-measurements and how the temperature effects the measurement itself (I am in charge of all the pH measurements on the powerplant I work at and I have no one to ask about this stuff as no one knows a thing about pH measurements).
So far I have learned how the temperature affects the electrochemical and physical properties of the combination electrode and how this can easily by adjusted for by incorporating an automatic temperature compensation that keeps the NERNST equation updated in real time.
So far so good.
I have, however, also read about the auto-ionisation of water and the constant pKw and how a change in temperature affects the span of the pH scale: With a change in temperature comes a change in hydronium content due to a shift in equilibrium as predicted by Le Chatelier's Principle. However, with a change in hydronium content comes an equal change in hydroxide content and so a change in this constant DOES NOT change the acidity of a solution. In other words, this particular constant decides the SPAN of the pH scale and because of this also the point of neutrality (7 at 25 degree celcius in pure water and less at higher temperatures).
And then there is the so called pKa that says something about the strength of an acid via the acids point of equilibrium in water. This dissociation constant is also temperature dependent although it is my understanding that this constant have nothing to do with the location of the neutrality point. It's simply a constant that changes with temperature and so the solution (unlike with pKw) DOES become more/less acidic as the temperature changes.
Is this correct?
If so, could you please give me an answer to the following questions I have?

Why does a solution become less temperature dependent the more acidic it is? Basic solutions are extremely temperature sensitive, neutral solution still significant sensitive, but highly acidic solutions don't seem give a damn about temperature. Why is that?

It is my understanding that pKw is only important in pure water and highly diluted solutions. But what am I to make about this? If I have a tank of an acidic solution (say pH 1) and the temperature is 50 degrees celcius, does this mean that the point of neutrality hasn't shifted as much away from pH 7 as it would had done in  pure water? Or is it because there is simply so much hydronium present that the pKa constant takes up proportionally more importance than pKw and so the autoionisation becomes a non issue to consider in every practical sense?

Thanks in advance for any help you may provide me with

The background of the topic is, that for the function
$$y = \frac {a(T)}{x}$$
$y$ does depend on $T$, but $x$ does not.
$\mathrm{pH}$ of basic solutions is temperature dependent via $\mathrm{p}K_\mathrm{w}$ temperature dependance.
Additionally, $\mathrm{pH}$ of weak acid/base solutions is temperature dependent via $\mathrm{p}K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{b}$ temperature dependance.
There is also additional, difficult to quantify factor of temperature dependency of activity coefficients.

The $\mathrm{pH}$ of strong acid solutions ( like hydrochloric or sulphuric acids ) of molar concentration $c \pu{[ mol/L]}$ is approximately(*)
$$\mathrm{pH}=-\log{c}$$
( For sulphuric acid, it is more complicated due 2 acidic hydrogens, where the second one is not strongly acidic ( in sense it has considerable $\mathrm{p}K_\mathrm{a}$ )).
The $\mathrm{pH}$ of strong base solutions ( like sodium hydroxide) is approximately
$$\mathrm{pH}=\mathrm{p}K_w(T) + \log{c}$$
because concentration of hydroxide ions $\ce{[OH-]}$ is the same as the concentration of e.g. sodium hydroxide. Concentration of hydronium ions $\ce{[H3O+]}$ then follows the above $y = \frac {a(T)}{x}$, i.e. $\ce{[H3O+]} = \frac {K_\mathrm{w}(T)}{\ce{[OH-]}}$
The $\mathrm{pH}$ of weak acid solutions ( like acetic acid ) is approximately
$$\mathrm{pH}=0.5 \cdot (\mathrm{p}K_\mathrm{a}(T) - \log(c))$$
The $\mathrm{pH}$ of weak base solutions ( like ammonia or sodium carbonate ) is approximately
$$\mathrm{pH} = \mathrm{p}K_\mathrm{w}(T) - 0.5 \cdot (\mathrm{p}K_\mathrm{b}(T) - \log {c})$$
The neutral $\mathrm{pH}$ is then
$$\mathrm{pH}_\mathrm{neutral}=\frac {\mathrm{p}K_\mathrm{w}}{2}$$
Note that if there was an OH- electrode, situation would be the opposite. Strongly acidic solutions would have temperature dependent pOH and strongly bases would not.

For the strong acid/base $T$ dependency, see the table:

t[deg C]
pKw
Solution
pH
pOH

25
14
0.1 M HCl
1
13

25
14
0.1 M NaOH
13
1

50
13.27
0.1 M HCl
1
12.27

50
13.27
0.1 M NaOH
12.27
1

(*): The equations are not exact, as there plays the role also the ionic strength, activity coefficients and for weak acid/base cases the simplification of $\ce{pH}$ calculation.


Homemade preservative for natural aloe-vera gel? I suppose this question can be extrapolated to food in general, but it is the case that I would like to know if there are any methods that would help me to keep the aloe gel fresh, i.e. avoiding it to rot.

I mostly squeeze the gel and pour it into a small jar. So far I have tried the following methods, being some of them obvious, but would appreciate corroboration anyway:

Sterilize the container and the tools, e.g. boiling or steaming.
Keep it in the fridge.
Stir with C and E vitamin powder to avoid oxidation.

I know little about chemistry, but have heard about pasteurization and some other methods that I couldn't afford at home so, even if you mention those, I will be mainly interested in homemade methods.

PS: I am forced to choose the tags for the post but I am not sure if they are the ones that best suit the question. Please, let me know if I should change them.

I suffer from chronic back pain and have tried just about everything on the market.Now I grow and harvest my own Aloe Vera gel,
put the gel in a 50ml roll-on bottle 1/2 full with gel, a 1/2 tsp vitamin C powder
and 10 drops vitamin E oil then add 98.9% DMSO by syringe
10ml at a time till the strength is good enough for me. I personally use 25ml
DMSO with the Aloe gel for remarkable pain relief


Freezing point of oxygen-18 water The freezing points of heavy water (3.8 °C) and tritiated water (4.49 °C) both seem to be well-known. I can't find anywhere that gives the freezing points of $\ce{H2^18O}$, $\ce{D2^18O}$, or $\ce{T2^18O}$, though. Are any of these values known? If not, is it at least known whether they'd be higher than with ordinary oxygen?

In this paper [1] from 1963, various properties of heavy-oxygen water are measured including the melting point of both $\ce{H2^{18}O}$ and $\ce{D2^{18}O}$.

The melting temperatures were measured at:
$$\ce{H2^{18}O}:\ T_\mathrm{mp}=\pu{0.28\pm0.02^\circ C}$$
$$\ce{D2^{18}O}:\ T_\mathrm{mp}=\pu{4.02\pm0.02^\circ C}$$

So, as some commenters mentioned, the melting (or freezing) point of the heavy-oxygen water is slightly larger when it is protium on the heavy-oxygen, and quite a bit larger when deuterium is present. This basically means that the liquid has become more structured on average, which is to say that the average lifetime of a hydrogen bond is slightly longer mostly due to an increase in the average strength of a hydrogen bond.

I would be very surprised if there has been a measurement of $\ce{T2^{18}O}$ because tritium is very rare and not very easy to isolate. Additionally, heavy-oxygen is also quite uncommon, so purifying the $\ce{T2^{18}O}$ to make any reliable measurements would likely be very difficult. I would be very willing to guess, however, that the melting temperature is $\ce{T2^{18}O}:\ T_\mathrm{mp}\approx\pu{4.7^\circ C}$ by just noting that the isotopic substitiution effects seem to be roughly additive.

Extra Credit: Trying to Interpret the Data:

As a note which is really not what you're asking but I think is really interesting, is why there is such a sharp increase in freezing temperature for deuterium (and hence why heavy-oxygen has these effects only on a smaller scale). Indeed, it is true that the $\ce{O-D}$ vibrational frequency in $\ce{D2O}$ is quite a bit smaller than that mode in $\ce{H2O}$, but it really isn't obvious that lowering this frequency should make $\ce{D2O}$ freeze at a higher temperature. For instance, all else being equal, a larger frequency in $\ce{H2O}$ would predict a shorter hydrogen-bond distance because the vibrationally averaged bond length will be longer (because the long bond lengths contribute more than the short ones because the potential is Morse-like). But, what I just said about the vibrationally averaged bond length did not allow the $\ce{O-O}$ distance to relax in $\ce{D2O}$, so if the $\ce{O-O}$ distance shrunk in $\ce{D2O}$, then we could say it's just the lower zero-point energy because then the shorter $\ce{O-O}$ distance would explain the stronger hydrogen-bonding we seem to be observing in $\ce{D2O}$. Well, very interestingly, the $\ce{O-O}$ distance in liquid $\ce{D2O}$ has been both theoretically and experimentally observed to increase slightly or stay exactly the same. See ref. [2] and experimental references therein. So both geometric parameters point towards water actually having a higher melting point even though this obviously isn't true.

What's the deal then? Well most of the answer seems to be that ordinarily we think of nuclei as being classical, i.e. the atom can be located precisely, but this type of reasoning is quite bad for hydrogen. The important point then is still that deuterium is heavier than hydrogen, though not because of the explicit contribution which the frequency lowering would have on the geometric parameters if everything else were the same. Rather, a hydrogen atom is much more spread out than a deuterium atom due to the fact that lighter atoms will essentially behave more quantum mechanically. In the sense of being more wavelike.

The fact that this delocalization effect for hydrogen is most important was demonstrated in ref. [3], where the authors find that the hydrogen-bond length cannot be directly related to the vibrational frequency in a one-to-one sort of way. This is due to a large sensitivity to the $\ce{OHO}$ angle. This then explains what we were missing up above because even though the average bond length in $\ce{H2O}$ is larger than the in $\ce{D2O}$, this sensitivity to the hydrogen-bonding angle greatly decreases the hydrogen-bonding strength in liquid $\ce{H2O}$ compared to $\ce{D2O}$ which does not delocalize as much and hence does not suffer from this angle problem.

As a final fun point, if one does a simulation of liquid $\ce{H2O}$ using classical nuclei, and then the same simulation is done using path integral methods to include nuclear quantum effects, the change in freezing point is a decrease of about $\pu{4^\circ C}$, which can only be due to this delocalization effect I describe.

I basically included all this to say that interpreting what this heavy-oxygen change really means can actually be pretty complicated. Whatever it is, there must be some structural change, and one would always expect isotopic substitution for a heavier isotope to make liquids more structured both from the frequency effects and the nuclear quantum delocalization.

Sorry if that was overkill :)

[1] Steckel, F., & Szapiro, S. (1963). Physical properties of heavy oxygen water. Part 1.—Density and thermal expansion. Transactions of the Faraday Society, 59, 331-343.

[2] Chen, B., Ivanov, I., Klein, M. L., & Parrinello, M. (2003). Hydrogen bonding in water. Physical Review Letters, 91(21), 215503.

[3] Rey, R., Møller, K. B., & Hynes, J. T. (2002). Hydrogen bond dynamics in water and ultrafast infrared spectroscopy. The Journal of Physical Chemistry A, 106(50), 11993-11996.


acyclically stereocontrolled reaction: organolithium and 2-chiral compound Aoyagi et al. (J. Am. Chem. Soc., Vol. 115, No. 24, p 11393) reported that the addition of 2-lithio-1,3-dithiane to the alpha-chiral (S)-2-Acetylpiperidine gives the (2 S,3 S)-configured product. It is stated that the diastereoselectivity of this 1,2-addition (dr = 6.2) is predicted by Cram's model involving complexation of the Li-ion by the piperidine-N. I tried to reproduce the diastereoselectivity in the following scheme:

As you can see I come up with the wrong product by applying Cram's rule with a O,N-complexed Li-ion (right side of the scheme). By applying the non-chelate Cram rule (left side; biggest substituent oriented anti to the carbonyl, attack via smallest group) I get the correctly configured (2 S,3 S)-product.

Where is the mistake?

I have just had a look at the paper and actually they reported that their main product was S,R:

"...treated with excess 2-lithio-1,3-dithiane in THF at -50°C; this treatment generated the tertiary alcohols 22a and 22b in 72% combined yield with 6.2:1 diastereoselectivity as predicted by Cram's model based on participation of the piperidine nitrogen atom, giving a rigid cyclic complex with the lithium compound".

Compound 22a: (2S)-2-[(R)-l-(1,3-Dithian-2-yl)-l-hydroxyethyl]piperidine

Compound 22b: (2S)-2-[(S)-1-(1,3-Dithian-2-yl)-l-hydroxyethyl]piperidine

I do not find any mistake.


Hock rearrangement of hydroperoxides Which hydroperoxides undergo a Hock rearrangement?

e.g. Cumene hydroperoxide undergoes a Hock rearrangement to yield acetone and phenol

Would the following hydroperoxide of ethylbenzene also undergo any similar rearrangement? How does one decide if a Hock rearrangement is likely?

The main parts of the mechanism that will affect the rate of the reaction are the following steps (full mechanism on Wikipedia): abstraction of a hydrogen to form a benzylic radical and the rearrangement of the of the protonated peroxide.
These steps will be slower as they involve the formation of the higher energy species when ethylbenzene is used.

Secondary benzylic hydrogen have a higher bond bond dissociation energies ($357.5\: \mathrm{kJ\: mol^{-1}}$) than tertiary benzylic hydrogen ($353.3\: \mathrm{kJ\: mol^{-1}}$)$^{[1]}$. This will diminish the rate at which the initiation step proceeds, though a difference of $4.2\: \mathrm{kJ\: mol^{-1}}$ shouldn't make or break the reaction.

The formation of a primary carbocation would otherwise be highly unfavorable, except for that it is bonded to an oxygen atom. A lone pair on the oxygen can therefore be shared with the carbocation, thereby stabilizing it. This step will have a higher activation energy, but likely all that will be necessary for this reaction to occur is a little more heat and time.

[1] Baciocchi, E.; D'acunzo, F.; Galli, C.; Lanzalunga, O. Tertiary : Secondary : Primary C–H Bond Relative Reactivity in the One-Electron Oxidation of Alkylbenzenes. A Tool to Distinguish Electron Transfer from Hydrogen Atom Transfer Mechanisms. J. Chem. Soc., Perkin Trans. 2. 1996, 133–140.


Will alkoxide ion in protic medium attack on double bond or carbonyl group? I want to know what will happen next here-
I thought that the alkoxide ion would attack the aldehyde to form a secondary alcohol, but apparently, as the two pi-bonds are in conjugation, so, the ion attacks the carbon of the C-C pi-bond. Why is this so?

Your title should read "protic medium" as opposed to "acidic medium". Acidic medium would be EtOH/H+. Your reaction conditions are basic and protic. Nothing much will happen to the unsaturated aldehyde 1. 1.2-Addition leads to hemiacetal 2. Hemiacetals are unstable to base and reverse to the carbonyl compound. Adduct 3 would eliminate to form 1 in the presence of base. If you employed EtOH-d1 as the solvent, you would incorporate deuterium at the γ-position of 1-d2 via enolate 4. If heptanedial 5 were exposed to EtOK/EtOH, an aldol condensation would ensue to form unsaturated aldehyde 1. Under truly acidic conditions, the diethyl acetal (in the red box) of aldehyde 1 would form. Formation of the β, γ-isomer of 1 is unlikely because the α, β-form 1 is much more stable.


Can heat capacity at constant volume for a gas decrease as temperature increases? I found a function for the $C_V$ of a gas which is being heated under constant volume,but the function shows that the $C_V$ is decreasing. I was wondering if this is possible for any gas since I know the $C_V$ is supposed to increase with temperature. Is there any reason why this would happen?

As temperature increases, it activates the rotational and vibrational degrees of freedom, so $\ce{Cv}$ increases with increase in temperature. The graph $\ce{CO2}$ would be:


Extraction of calcium carbonate from liquid I'm using calcium carbonate ($\ce{CaCO3}$) in apple juice to reduce the acidity of the juice.  Once the $\ce{CaCO3}$ reacts with the apple juice and raises the pH, the juice is left cloudy in color BUT has the desired pH of 3.4.  I need a way to extract (separate) the $\ce{CaCO3}$ precipitate from the apple juice which leaves the resulting liquid (apple juice) drinkable still...

The pure juice has an initial pH of 2.9. I want to reduce the acidity and be left with pH of say 3.4. Perhaps by blending with another drink?  I did a test using the pH meter, I started with water (pH 6.0) and slowly added apple juice, and the pH drops to 2.9 adding 1 part juice to say 10 parts of water.  So blending another higher pH drink (tea/water/etc) would totally dilute the apple juice before the pH would reach 3.4. How can I raise the pH and still keep the apple juice pure or close to pure?

Apple juice out of the carton tests at 2.9 pH with digital pH meter.
My tap water tests at 6.0 pH.
Calcium carbonate - pill form from health food store (oyster shell derivative)

The stuff you're seeing is solid calcium carbonate that most likely did not dissolve, or some other insoluble calcium compound that formed by reaction of calcium with whatever is in the apple juice.

You might want to try using sodium carbonate instead of calcium carbonate. Sodium compounds are usually more soluble than calcium compounds, and it's a more common food additive than calcium carbonate.

That said, I wouldn't recommend adding chemicals to your food without knowing exactly what you're doing. Some seemingly safe chemicals may have hazardous impurities and do not qualify as food-grade. If you're doing this because of health problems then you should consult with a doctor on better ways of action than pouring carbonates in your drinks.


Why is a racemic mixture formed in the Diels-Alder cycloaddition? Why is a racemic mixture formed in the Diels-Alder cycloaddition? I note that for both endo and exo products, either enantiomer can be formed. Why is this so?

The concerted Diels-Alder reaction has in principle eight different possible transition states. Depending on the substrates some of them will lead to the same products.

Below you can see an overview of the (non-optimised) transition state arrangements for the reaction of isoprene and propene.

From these images you should be able to determine that the following behave like enantiomers:

(a) and (g)
(b) and (h)
(c) and (e)
(d) and (f)

When you think a little bit longer about it, you will also see that (a) and (h) will form the same product. This is also true for (b) and (g), (c) and (f), as well as (d) and (e).
In total, you will therefore obtain four products two different regio-isomers and their corresponding enantiomer each.


How to efficiently reduce of ilmenite to pure titanium? I am looking for a process in which reduces ilmenite to pure titanium with the highest yield possible. We are to use a process of our choice or one that we have created in conjunction with FactSage to calculate the yield of titanium.

the materials we have available are:

Pure Metals

$\ce{Al, Ca, Mg, Na, Si}$

Pure Non-metals

$\ce{Ar, C, Cl2, H2, N2, O2, S }$

Molecules

$\ce{CH4, CO, CO2, HCl, H2O, H2S, NH4}$

There are a few process that come to mind such as reduction using Magnesium, $\ce{H2SO4}$, high temperature $\ce{CO2}$, but any help on which would be the most efficient combined with the highest yield would be great.

Ilmenite mainly consists of $\ce{TiO2}$ and iron oxides. The idealized formula of ilmenite is approximately $\ce{FeTiO3}$.

Note that the aluminothermic process would yield ferrotitanium alloys, and the reduction with carbon would yield titanium carbide.

The titanium ore can be converted to titanium tetrachloride according to
$$\ce{TiO2 + 2Cl2 + 2C -> TiCl4 + 2CO}$$
or
$$\ce{2FeTiO3 + 7Cl2 + 6C -> 2TiCl4 + 2FeCl3 + 6CO}$$
(If a large content of iron oxides causes problem, it can be partially removed first by partial reduction with carbon.)

Titanium tetrachloride (boiling point: 136.5 °C) can be purified by fractional distillation.

Finally, titanium tetrachloride can be reduced to titanium sponge with magnesium (Kroll process):
$$\ce{TiCl4 + 2Mg -> Ti + 2MgCl2}$$

The remaining $\ce{Mg}$ and $\ce{MgCl2}$ can be removed with hydrochloric acid.


Reaction between bromobenzene and methyllithium to yeld toluene: possible? I was solving some reactivity problems, and I was asked to prepare toluene from bromobenzene.

While the suggested solution of the problem involves the preparation of a Grignard reagent, followed by a reaction with methyl bromide, I was wondering:

Could the same compound be prepared with the straightforward reaction between bromobenzene and methyl lithium?

I tried to search some similar reactions, but I found nothing for both the possible ways.

MeLi certainly does lithium-halogen exchange and in the presence of MeI takes bromobenzenes to toluenes   Synth. Comm. 2001 2323

If you're asking whether the MeBr produced by the exchange can act as the methylating agent? I have never seen this reported but maybe someone else can pull up a reference.


Stereoelectronics of the Houk model In Clayden's Organic Chemistry it gives the Houk model as a way for predicting selectivity for alkene reactions where an existing stereocentre is present (diastereoselectivity).

Unlike the description of the Felkin-Ahn model, there is no real description of how the Houk model works, and I can't find this information in March Advanced organic chemistry either.

My feeling is that there must be some stabilising interaction between the pi system of alkene with a sigma orbital on the adjacent carbon, however this doesn't explain why one would be favoured over another.

1. Origins of the Houk model
The initial idea behind the Houk model was infact provided by Kishi, who proposed that alkenes with a chiral centre on the adjacent carbon adopted a 'reactive conformation' in which the small group eclipsed the alkene in order to minimise unfavourable steric interactions.[1]

Fig1: Kishi's "reactant conformer" model, taken from ref [1]
Houk took the 'reactive conformation' from the Kishi studies and modelled it computationally, showing that the lowest energy ground state conformation for alkenes with substituents at the allylic position was indeed the one in which the small group eclipsed the alkene, as proposed.[*]
2. Allylic strain
The Houk model works largely on steric grounds, balancing A1,2  strain and A1,3 strain, as can be seen from the figure below:

Fig2: The reactive conformation, taken from Organic Synthesis- Strategy & Control, Wyatt and Warren (Wiley)
The 'reactive conformation' is the middle structure (now days often known as the Houk conformation), in which A1,2  strain and A1,3 strain is minimal. In the left hand structure A1,2  strain is present, in the form of a H-H clash (doesn't look too significant, but raises the energy of the conformation significantly enough, similar to in the conformations of ethane). In the right hand structure A1,3  strain is present, this A1,3 strain is far more significant than A1,2, making this conformation sufficiently high in energy as to be barely populated at room temperature.
Based on the arguments above, it's worth pointing out that the Houk model works best for cis alkenes, with trans alkenes often giving poorer selectivity:[2]

Fig3: cis and trans alkenes reacting via the Houk model, taken from Organic Synthesis- Strategy & Control, Wyatt and Warren (Wiley)
This result is consistent with the fact that in a trans alkene, A1,3 strain is already minimised by virtue of the olefin geometry, meaning that two possible 'low energy' conformers exist which have neither A1,3 strain nor significant A1,2 strain.
3. Reactions not following the Houk model
Not all reactions for which the Houk model could be used give the expected 'Houk product'. The most common of these (and indeed the one in Clayden that you mention in the question as not following the Houk model) is when the starting material is an allylic alcohol.

Fig4: Reactions not following the Houk model due to directing effects, taken from Organic Chemistry, Clayden and Warren (Oxford University Press)
In the example above, the hydroxyl group is acting as a directing group, guiding epoxidation with the mCPBA to the same face (the steric argument, which would always predict epoxidation away from a bulky group, does not apply here).

Notes and references
[1]: Houk, K. N.; Rondan, N. G.; Wu, Y.-D.; Metz, J. T.; Paddon-Row, M. N. Theoretical studies of diasteroselective hydroborations. Tetrahedron 1984, 40 (12), 2257–2274. DOI: 10.1016/0040-4020(84)80009-5.
[2]: Vedejs, E.; McClure, C. K. Hyperconjugative effects of allylic substituents are not important in osmylations. J. Am. Chem. Soc. 1986, 108 (5), 1094–1096. DOI: 10.1021/ja00265a048.
[*]: The Houk group has been instrumental in providing a bridge between theoretical predictions and experimental results, providing weight to proposed models, and giving confidence in being able to look at a reaction and predict what product it may be.


Acid - Base Equilibrium Suppose we had a closed system proceeding according to,

$$\ce A +2\ce B \rightleftharpoons 2 \ce C$$

And that three experiments produced the following concentration data.

Experiment 1:

\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 0.500\rm~M \\
\ce B & 0.700\rm~M \\
\ce C & 0.900\rm~M
\\\hline\end{array}

Experiment 2:

\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 0.300\rm~M \\
\ce B & 0.420\rm~M \\
\ce C & 0.424\rm~M \\\hline
\end{array}

Experiment 3:

\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 1.40\rm~M \\
\ce B & {x}\rm~M  \\
\ce C & 1.49\rm~M \\\hline
\end{array}

How would you find the concentration of $\ce B$ from experiment 3 at equilibrium?

I assumed that each experiment was in equilibrium so that I could do the following,

$$K_c = \frac{\ce{[C]}^2}{[\ce{A}][\ce{B}]^2} = \frac{[0.900]^2}{[0.500][0.700]^2}$$

$$\therefore \frac{\ce{[C]}^2}{\ce{[A][B]}^2} = \frac{[0.900]^2}{[0.500][0.700]^2}$$

$$\therefore \ce{[B]} = \sqrt\frac{[1.490]^2[0.500][0.700]^2}{[0.900]^2[0.140]}$$

$$\therefore \ce{[B]} = 2.19\rm~M$$

However, $[\ce{B}] = 2.19\rm~M$ is not the answer. Could someone please provide some insight and direction into this question?

You have an error somewhere in your math.
Assuming you were supposed to interpolate $K_C$ from both experiment 1 and 2, you would do the following:
Experiment1 yields $K_C=3.31$. Experiement two yields $K_C=3.40$. Interpolation yields $K_C=3.35$.
This would give you $$B=\sqrt{\frac{C^2}{A\cdot K_C}}=\sqrt{\frac{1.49^2}{1.4\cdot 3.35}}=0.69$$


Help on alternative terpene names and naming schemes I am a software developer and currently parsing data involving terpene profiles of plant samples. The data in question uses different ways of naming the same terpenes in-between samples and since this is not my field I would like to ask:

if there is an online index showing all possible names for a specific terpene or
if there is a software library for normalizing those names or
if somebody could simply tell me which ones of the attached list are the same terpene and what the recommended naming scheme would be

The list of terpenes I am concerned with is as follows:

3-Carene
α-Bisabolol
α-Humulene
α-Pinene
α-Terpinene
β-Caryophyllene
β-Myrcene
β-Ocimene
β-Pinene
Camphene
Caryophyllene
Caryophyllene Oxide
CaryophylleneOxide
cis-Nerolidol
δ-Limonene
DLimonene
Eucalyptol
Geraniol
Guaiol
Humulene
Isopulegol
Linalool
Ocimene
PCymene
p-Cymene
Terpinolene
trans-Nerolidol
TransNerolidol 1
TransNerolidol 2
trans-Ocimene
γ-Terpinene

(Some of those names are obvious duplicates of others, but I just wanted to give you the full, unaltered list here)

If anyone could point me in the right direction I would be very grateful!

Chemical nomenclature is officially defined by IUPAC (International Union of Pure and Applied Chemistry). You can find most IUPAC names for these terpenic compounds by searching their popular names on Wikipedia, they are quite long and confusing terms though, and can only be interpreted if you have a certain knowledge in organic/inorganic chemistry.

What you are probably more interested in are databases with hundreds of thousands of compounds with unique identifying numbers that are unequivocally assigned to a specific substance. Two of the largest are CAS Index (https://en.wikipedia.org/wiki/CAS_Registry_Number) and ChemSpider (https://en.wikipedia.org/wiki/ChemSpider). I believe working with numbers will facilitate your work a lot, so I would suggest that you look for one of these databases (but be careful, each database uses a different numbering system so you should pick one and stick to it).


Why is the electron-nucleus attraction modelled with only electrostatic interactions? While reading about the structure of an atom, I've encountered (even in some renowned books) the statement that electrons and nuclei are attracted due to electrostatic, or Coulombic, attractions.

However, since the charged particles (electrons and nuclei) are moving, how can the Coulombic force be sufficient to describe their interactions? Surely electrostatics cannot be used in this context. Is it merely an approximation?

If I understand the question correctly, OP is somewhat surprised that Coulomb's law is used to describe the interaction between an electron and a nucleus, although it is usually pictured that electrons are moving and Coulomb's law describes interaction between static particles. Should not then the Lorentz law be used instead Coulomb's one?

First note, that electrons do not move around a nucleus in an atom. At least, they do not do so in the classical sense. Yet, a non-zero orbital angular momentum $\vec{L}$ of an electron in an atom gives rise to the orbital magnetic dipole moment $\vec{\mu}_{L}$ and so the magnetic field is indeed generated by the "moving" electron. Plus, an electron has an intrinsic magnetic moment (spin) which also contributes to this magnetic field. The nucleus possibly possess orbital magnetic momentum as well (if has non-zero angular momentum) that couples with its spin which interacts with that of electron. These interactions splits the energy levels of the atom and the resulting hyperfine structure can be measured using radio-frequency spectroscopy.

And in principle, for a complete description of an atom we could use a model in which we treat the interaction between electron and nucleus taking magnetic interactions into account, i.e. similarly to how it is done in classical electrodynamics with the Lorentz law. There is a problem with the Lorentz law though - it is formulated in terms of forces and velocities and this terms are not even present in the vocabulary of quantum mechanics. So, the magnetic field must appear somehow differently in the Hamiltonian, and to make a long story short the full (non-relativistic) Hamiltonian for one electron in an external electromagnetic field is:
$$
\hat{H} = \frac{1}{2m}(\boldsymbol{p} - e\boldsymbol{A})^2 + e\phi \, ,
$$
where $\phi$ is the scalar potential and $\boldsymbol{A}$ is the vector potential. And even in the absence of an external to the whole atom magnetic field, the nucleus itself gives rise to the vector potential $\boldsymbol{A}$.

So, this is certainly true that electrostatic interaction between electron and nucleus is not the whole story. However, electrostatic interaction dominates magnetic ones by far, so that usually the later are just ignored, unless a very high precision is required.


Esterfication of 2,2,2-triphenylethanoic acid Explain the formation of (1,1,1-triphenylmethyl)methyl ether when a solution of 2,2,2-triphenylethanoic acid in concentrated sulfuric acid is poured in methanol.

I cannot figure out where is one carbon atom lost. Has it escaped as $\ce{CO2}$? Or is there something more strange happening.

I wouldn't call this reaction an esterification. There is no ester functional group in the final compound.

I guess there's some special role of the three phenyl groups but I'm unable to come up with any mechanism. I hope dehydration is involved as one of the reaction steps.

Disclaimer: The question has pretty much been answered in the comments. I'm just making it substantial.

The triphenylacetic acid is dehydrated in the presence of concentrated sulfuric acid to give an acylium ion, namely triphenylacetylium cation, which is stabilized by resonance.

Then, the triphenylacetylium cation undergoes decarbonylation (removal of carbon monoxide) to give triphenylmethyl cation, which is stabilized by the three phenyl groups.

The oxygen in methanol acts as a nucleophile to attack the carbocation, and is deprotonated to give our desired product triphenylmethyl methyl ether.

It is noted that this reaction should be done in a well-ventilated area because of the evolution of toxic carbon monoxide gas.


What is the correct definition of the Gibbs free energy of activation? Is the following correct?

  Gibbs free energy of activation is used in energy profiles where the stabilities of the species are expressed as changes in Gibbs energy, while the activation energy Ea is used in energy profiles where the stabilities of the species are expressed as changes in enthalpies.

There are a lot of possible definitions and it possibly depends very much on the case and the context you are looking at. However, there are the definitions provided by the IUPAC goldbook, which can be considered the official ones.

The first definition you suggested is correct so far and known as (standard) Gibbs energy of activation (standard free energy of activation) $\Delta^\ddagger{}G^\circ{}$:

  The standard Gibbs energy difference between the transition state of a reaction (either an elementary reaction or a stepwise reaction) and the ground state of the reactants. It is calculated from the experimental rate constant  $k$ via the conventional form of the absolute rate equation:
  $$\Delta^\ddagger{}G^\circ{} = \mathcal{R} T \left[\ln\left(\frac{\mathcal{k}_\mathrm{B}}{h}\right) − \ln \left(\frac{k}{T} \right)\right]$$
  where  $\mathcal{k}_\mathrm{B}$ is the Boltzmann constant and  $h$ the Planck constant ($\frac{\mathcal{k}_\mathrm{B}}{h} = 2.08358\cdot 10^{10}~\mathrm{K^{−1}s^{−1}}$). The values of the rate constants, and hence Gibbs energies of activation, depend upon the choice of concentration units (or of the thermodynamic standard state).

Your second definition refers to the definition of enthalpy of activation, $\Delta^\ddagger{}H^\circ{}$:

  The standard enthalpy of activation $\Delta^\ddagger{}H^\circ{}$ is the enthalpy change that appears in the thermodynamic form of the rate equation obtained from conventional transition state theory. This equation is only correct for a first order reaction, for which the rate constant has the dimension reciprocal time. For a second order reaction, for which the rate constant has the dimension (reciprocal time) × (reciprocal concentration), the left hand side should be read as  $k c^\circ$, where $c^\circ$ denotes the standard concentration (usually  $\mathrm{1 mol\cdot dm^{−3}}$).
  $$k = \frac{\mathcal{k}_\mathrm{B} T}{h} \exp\left\{\frac{\Delta^\ddagger{}S^\circ}{\mathcal{R}}\right\} \cdot\exp\left\{\frac{-\Delta^\ddagger{}H^\circ}{\mathcal{R}T}\right\}$$
  The quantity  $\Delta^\ddagger{}S^\circ$ is the standard entropy of activation, and care must be taken with standard states. In this equation  $\mathcal{k}_\mathrm{B}$ is the Boltzmann constant,  $T$ the absolute temperature,  $h$ the Planck constant, and $\mathcal{R}$ the gas constant. The enthalpy of activation is approximately equal to the activation energy; the conversion of one into the other depends on the molecularity. The enthalpy of activation is always the standard quantity, although the word standard and the superscript $^\circ$ on the symbol are often omitted. The symbol is frequently (but incorrectly) written  $\Delta{}H^\ddagger{}$, where the standard symbol is omitted and the  $\ddagger$ is placed after the  $H$.

The activation energy (Arrhenius activation energy) $E_a$ itself is defined as:

  An empirical parameter characterizing the exponential temperature dependence of the rate coefficient, $k$,  $$E_a = \mathcal{R} T^2 \frac{\mathrm{d}(\ln k)}{\mathrm{d} T}$$, where $\mathcal{R}$ is the gas constant and $T$ the thermodynamic temperature. The term is also used for threshold energies in electronic potential surfaces, in which case the term requires careful definition.

And while we are at it, entropy of activation $\Delta^\ddagger{}S^\circ$:

  The entropy change that appears in the thermodynamic form of the rate equation obtained from conventional transition-state theory.

Additional note and general warning
Although there are the official definitions, the term activation energy is often used as a synonym for many kinds of different barriers. One has to be very careful when reading articles, which definition is used.
In computational chemistry articles you may also find the terms electronic energy of activation, zero-point corrected energy of activation, solvent corrected energy of activation, ..., etc., pp., and many more.


Is it correct to say that SOME iodine undergoes sublimation at 1 ATM The triple point of Iodine occurs at 0.12 atm and at standard temperature it is a solid on the phase diagram.

I understand that therefore, when heating iodine at 1 atm, it (at least some of it) will pass through the liquid phase, and it is also obvious from observing heated Iodine that it yields liquid when heated gradually. Also, just as liquids can evaporate below the boiling point, I understand that some Iodine can escape from a sample in the vapor form when the vapor pressure is lower than the atmospheric pressure.

My question is 2 part:

1) Is it correct to call the change form solid to gas below equilibrium vapor pressure "sublimation".  In other words, is sublimation equivalent to the term "boiling" or is it equivalent to the term evaporation in the change of phase from liquid to vapor, or both?  (Is there a separate term, even if out of use,  for the solid to vapor transition below equilibrium pressure?)

2) At 1 atm, when Iodine molecules in a solid sample escape into the vapor form, either at standard temperature, or as the sample is heated, do they all briefly pass through the liquid phase first, either at a vapor pressure below 1 atm or as vapor pressure is raised above 1 atm.  .

Iodine's crystal vapor pressure will of course be <0.12 atm below the triple point pressure as can be seen on both graphs.  A converse but equivalent question would be whether vaporous Iodine at 1 atm can crystalize and deposit without liquifying.

To answer part 1 of your question:

It is correct to call the change form solid to gas below the equilibrium vapor pressure sublimation. And there is no separate term for vaporization of a solid above the equilibrium vapor pressure; it is still sublimation.  Similarly, a boiling liquid is still perfectly correctly said to be evaporating.

To answer part 2 of your question:

At temperatures below the melting point of iodine, it will sublimate until the equilibrium vapor pressure is reached without going through a liquid phase in that process.

Once the iodine is heated to it's melting point, it will continue the same process of vaporizing to it's equilibrium vapor pressure. Because it is now a liquid, we just call that evaporation rather than sublimation.

If the iodine is further heated to it's boiling point at 1 atmosphere, then it will continue to rapidly evaporate as it boils away.

Regarding the thermodynamic relationships between these processes as mentioned in one of your comments (which I think you should just edit into the question), what you have said is correct. This is a path-independent thermodynamic process, so with respect to the ΔH of the phase-change processes, it doesn't matter how you get from the solid phase to the vapor phase. ΔH for the direct solid-vapor transition (sublimation) is equal to the ΔH of melting plus the ΔH of evaporation.

I hope I have addressed all your points, please ask for clarification in the comments if I've left anything unclear.


Why is standard reduction potential an intensive property? My book says that

  Since the number of electrons lost must equal the number gained, the half-reactions must be multiplied by integers as necessary to achieve the balanced equation. However, the value of e° is not changed when a half-reaction is multiplied by an integer. Since a standard reduction potential is an intensive property (it does not depend on how many times the reaction occurs), the potential is not multiplied by the integer required to balance the cell reaction.

However, I do not understand why the reduction potential is unaffected by the number of mols. If you have a greater number of mols of the metal, doesn't that allow for a greater potential difference?

Look at the unit volt. 1 volt = 1 joule per coulomb. If the number of moles is doubled, the coulombs are doubled, as is the number of joules. Think of the volt as the driving force behind an individual electron in an electrochemical system.


Can a double bond extend beyond the plane and if so, how would it be drawn? Can double and triple bonds, etc., extend beyond the plane in a Lewis structure, and if so, how are they represented? I.e., how would 1 and 2 in this image be drawn if they were doubles or triples?

Yes, chemical bonds of higher order can be extended from the plane in some rare cases.
There are IUPAC recommendations existing at least for double bonds suggesting to depict them simply as doubled wedge bonds if it's really necessary.
From [1, p. 1930] (emphasis mine):

  ST-1.5 Tetrahedral stereocenters including higher-order bonds

  Tetrahedral configuration requires a tetrahedral arrangement of ligands around a central point. It does not require that all of those ligands be connected to a central atom using single bonds. It is perfectly reasonable to have one or more ligands connected to the central atom using higher-order bonds. This is most commonly seen with chiral sulfoximides, phosphates, and related compounds, each of which is
  commonly drawn with one or more double bonds. Stereocenters of this type should be indicated by the appropriate placement of bold and/or hashed wedged bonds connecting the central atom to the singly bonded ligands. Although wedged and hashed double bonds have appeared in the literature, they are rarely required for chemical clarity, and their use is infrequent enough to be confusing to the casual
  viewer (see ST-2.5).

Further from [1, pp. 1940-1942] (emphasis mine):

  ST-2.5 Trigonal pyramidal Polyhedral symbol: TPY-3 Number of ligands: 3

  […]

  In rare cases, a trigonal pyramidal atom with at least one double bond may be flanked by other stereocenters at the other ends of all of its single bonds. In such cases, one of the plain bonds should be changed to a wedged bond even though it joins two stereocenters. It is also acceptable to use a solid wedged or hashed wedged double bond, but that would not be preferred because wedged double bonds are so rarely seen in the literature.

References

Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. https://doi.org/10.1351/pac200678101897. (Free Access)


The density of an aqueous solution Could the density of an aqueous solution be less than 1g/ml ? For an example, can 0.98g/ml be taken as a density for an aqueous sulphuric acid solution ?

Answer to your question, "could the density of an aqueous solution be less than $\pu{1g/mL}$," is yes. See following examples:

The density of the commercially available 28% ammonia solution ($\ce{NH3}$ 1n $\ce{H2O}$) is listed as $\pu{0.9 g/mL}$ at 25 °C (Sigma-Aldrich Catalog).

MERCK Tabellen für das Labor has listed the density of 35% ammonia solution ($\pu{18 mol/L}$ concentration) as $\pu{0.88 g/mL}$ at 20 °C (Steffen's Chemistry Pages)

As obvious, the densities of various aqueous ethanol solutions are also equal to less than unity since the density of absolute ethanol is $\pu{0.789 g/mL}$ at 20 °C (Wikipedia). By a quick google search, I found the densities of various aqueous ethanol solutions listed in MERCK Tabellen für das Labor  in Steffen's Chemistry Pages, which is pretty interesting.


Why are acid-base reactions so exothermic? If I mix sulfuric acid with sodium hydroxide, the solution really heats up. Could someone explain why acid-base reactions are exothermic from a molecular perspective?

One can think of it as simply because of electrostatic attraction. Basically, an acid produces $\ce{H+}$ ions in water while a base produces $\ce{OH-}$ ions in water. So when you mix an acid and a base (e.g. sulfuric acid and sodium hydroxide) in a beaker, you get a solution containing $\ce{H+}$ and $\ce{OH-}$ ions. So, when these ions are far apart, there exists potential energy between them (because opposite charges attract each and want to come closer together). So when these ions come closer together, the attraction between them becomes strong enough that a bond forms between the ions to form $\ce{H2O}$. During this process, the initial potential energy between the ions decreases and is converted to heat energy. Hence this is why it is exothermic.

So generally speaking, energy is released when a bond is formed, while energy is required (energy is absorbed) to break a bond. So in the case of acid and base reactions, no bond is being broken, bonds are only being formed. So it must be an exothermic reaction then.


Why do DMSO and acetone have such radically different melting temperatures? DMSO and acetone have almost identical structures, except for us replacing the carbonyl carbon with sulfur to obtain DMSO. DMSO's melting point is $19~\mathrm{^\circ C}$, whereas acetone's is $-95\ \mathrm{^\circ C}$.

Is the dominant factor DMSO's larger molecular weight? Probably not. While acetone is $58\ \mathrm{g/mol}$ vs. DMSO's $78\ \mathrm{g/mol}$, 2-pentanone is $86\ \mathrm{g/mol}$, but still has a very low melting temperature of $-77\ \mathrm{^\circ C}$.

The only protons on all of these species are aliphatic, so hydrogen bonding is probably not the dominant contributor to this effect either.

What's going on?

The first thing to remember is that boiling and melting points, while seemingly simple, are incredibly hard to predict. Yes, many teachers and professors give their pupils and students a number of simple prediction rules, but there is a lot of intrinsic balance and experience associated with applying them that they may often seem unintuitive.

Comparing DMSO and acetone, the first striking difference is the replacement of carbon with sulfur. This may seem like a small change but results in radical differences. For example, compare propane with dimethyl sulfide; or the latter with dimethyl ether. $\ce{Me2S}$ is a liquid with a boiling point of around $40~\mathrm{^\circ C}$, the others are gases at room temperature and standard pressure.

There are two factors that explain the difference between $\ce{Me2S}$ and propane/methyl ether — one does better in explaining the difference to propane the other does much better in explaining the difference to dimethyl ether.

Polarity. Sulfur is more electronegative than carbon, thus the $\ce{Me2S}$ molecule is slightly polar while propane is essentially nonpolar. Since all of these molecules exhibit (approximate) $C_\mathrm{2v}$ geometry, the polarity of individual bonds does not cancel out. However, one would expect $\ce{Me2O}$ to be more polar and thus have an even higher boiling/melting point.
London interactions. These are often simplified as ‘rising with molecular weight’, but in fact, the key factor is not molecular weight (which is a nuclear property) but polarisablility of electrons. Sulfur, being larger, having an additional shell and thus being much more diffuse than both carbon and oxygen is much more polarisable. Thus, the significant difference in the boiling and melting points of $\ce{Me2S}$ and $\ce{Me2O}$ can be attributed mainly to the stronger London interactions, made possible by adding eight (core) electrons and thereby moving the valence electrons further away from the core.

Both factors play a role in the difference of melting points of DMSO and acetone, although there is even more. While acetone has a trigonal planar structure featuring a $\ce{C=O}$ double bond, DMSO is trigonal pyramidal with an additional lone pair and formal positive charge on sulfur, to give a $\ce{S^+-O^-}$ single bond. While this feature reduces the London interactions somewhat — sulfur’s higher oxidation state means that the electrons are more stable and the orbitals appear smaller at the same cutoff value — it increases polarity significantly.

Indeed, it is better to compare similar molecules; isopropanol’s structure is much closer to DMSO’s than acetone’s is.


contradiction between Gibbs energy variation and redox potential Let's consider the following two redox reactions;

$\ce{C_2H_4O_2 + 2\;O_2 -> 2\;HCO_3^- + 2\;H^+}$
$\ce{C_2H_4O_2 + 1.6\;NO_3^- -> 2\;HCO_3^- + 0.8\;N_2 + 0.8\;H_2O + 0.4\;H^+}$

Considering that the standard Gibbs energies of formation of the involved chemical species are the following;

$\ce{C_2H_4O_2}$: -393.8 kJ.mol-1
$\ce{HCO_3^-}$: -586.9 kJ.mol-1
$\ce{NO_3^-}$: -111.3 kJ.mol-1
$\ce{H_2O}$: -237.2 kJ.mol-1
$\ce{O_2$, $N_2$, $H^+}$: 0 kJ.mol-1

Then the standard Gibbs energy variation associated to those reactions are respectively -804.4 kJ.mol-1 for the oxidation of acetic acid by oxygen and -816.0 kJ.mol-1 for the oxidation of acetic acid by nitrate.

It appears that the oxidation by nitrate is more exergonic that the oxidation by oxygen in standard conditions, when computed this way.

However, the redox potential of $\ce{0.25\;O_2 + H^+ + e^- -> 0.5\;H_2O}$ is +0.811 V, while the redox potential of $\ce{1/5\;NO_3^- + 6/5\;H^+ + e^- -> 1/10\;N_2 + 3/5\;H_2O}$ is 0.746 V. I haven't found any data for the redox potential of acetic acid, however the redox potential of its electron acceptors indicates that an electron transfer from acetic acid to oxygen should imply a larger Gibbs energy variation than an electron transfer from acetic acid to nitrate (according to the $\Delta G = nFE$ formula), which contradicts what has been previously computed from Gibbs energies of formation.

I am new to electrochemistry, so what did I misunderstand here?

Looking up the redox potentials in tables vs. Hydrogen as standard at zero volts the redox for Oxygen is

$$\ce{O2 +  4H^+ + 4e <=> 2H2O} \qquad \qquad E^\circ = +1.229~\mathrm{V}$$

so the tables you use may be vs. calomel electrode as standard.

The redox potential for the nitrate reaction has to be constructed from others as

$$\begin{align}
\ce{NO3^- + 3H^+ + 4e}\, &\ce{<=> HNO2 + H2O} & E_1^\circ &= +0.934~\mathrm{V} \\
\ce{2HNO2 + 4H^+ + 4e}\, &\ce{<=> N2O + 3H2O} & E_2^\circ &= +1.297~\mathrm{V} \\
\ce{N2O + 2H^+ + 2e}\, &\ce{<=> N2 + H2O} & E_3^\circ &= +1.766~\mathrm{V}
\end{align}$$

The overall half-reaction, after multiplying the nitrate reaction by 2 (to balance) and adding is

$$\ce{2NO_3^- + 12H^+ + 10 e <=> N2 + 6H2O} \tag{1}$$

The redox potential for the reaction $(1)$ is calculated from $n_\mathrm{r}E_\mathrm{r}^\circ = n_1E_1^\circ + n_2E_2^\circ +....$ where the n's are the number of electrons in each half reaction and $n_\mathrm{r}$ for the new reaction.

The $\ce{NO_3^- to N2}$ redox potential is therefore $1.246~\mathrm{V}$, calculated as

$$E_\mathrm{r}^\circ = \frac{(4 \times 0.934) + (4 \times 1.297) + (2 \times 1.766)}{10}$$

The two values, $1.229~\mathrm{V}$ and $1.246~\mathrm{V}$, would now seem to agree with your $\Delta G$ values. They may be the same within error as may your $\Delta G$ values.

(Redox Values are from the CRC Handbook of Chemistry and Physics, 75 ed.)


Thermometer fluid is separated, defective? I bought a thermometer from Cole-Parmer having kerosene as the fluid, and the fluid is broken up. The fluid in the bottom part goes up to about 10 °C even though it is 20 °C in the room, then there is a gap followed by fluid between 23 °C and 25 °C, then there is no fluid going all the way to the top where there is another small stretch of fluid. There is a short black rubber sleeve on the top of thermometer.

Does this mean the thermometer is defective or am I using it incorrectly, not “calibrating” it or something?

No, it is not defective, just slightly damaged, possibly from shipping. You should call customer service at Cole-Palmer and see if they will replace (or repair) the thermometer.

If not, there are 2 ways to fix it. I haven't tried either one, so I don't know how well they work.

If there is a bulb at the top and the thermometer is low range (under 150°C), then you would heat the thermometer in an upright position until all the bubbles are pushed up into the bulb.

The other method is to cool the thermometer down until all the liquid is in the bottom bulb. The second method may not work if the liquid freezes before all the liquid has moved into the bulb.


Volume of NaOH needed to precipitate copper(II) but not calcium I have $\ce{NaOH}: C = \pu{0.1M}, V_{\ce{NaOH}}$ that I want to progressively add in a solution (in water) of:
$$\begin{align}
\ce{Cu^{2+}}&: \pu{0.01M}; & K_\mathrm{sp} &= 10^{-18.6} \\
\ce{Ca^{2+}}&: \pu{0.01M}, & K_\mathrm{sp} &= 10^{-5.26}
\end{align}$$
with initial volume: $V_\mathrm{init} = \pu{0.1L}$. I want to add $\ce{NaOH}$ as long as $\ce{Cu(OH)2}$ precipitates, without causing any precipitation of $\ce{Ca(OH)2}$.

Knowing  that the equilibrium will be displaced to the left and I will have precipitation, I have done:
$$
\begin{array}{ccc}
\ce{Cu(OH)2 &->&Cu^2+ &+ &2OH^-}\\
&&0.001&&0.1V_{\ce{NaOH}}\\
&&0.001-x&&0.1V_{\ce{NaOH}} - 2x\\
\end{array}
$$

Also, since I want to achieve $99.999\%$ precipitation of $\ce{Cu(OH2)}$ I have got: $$x = 99.999\cdot10^{-5}$$

Using all that in the expression of $K_\mathrm{sp}$ I get :
$$10^{-18.6} = \left (\frac{0.001 - 99.999 \cdot 10^{-5}}{0.1+V_{\ce{NaOH}}} \right)\cdot\left(\frac{0.1\cdot V_{\ce{NaOH}} - 2\cdot99.999\cdot10^{-5}}{0.1 + V_{\ce{NaOH}}}\right)^2$$

However for this expression I get a value of $V_{\ce{NaOH}}$ that is extremely small. Is there anything wrong with my reasoning and if yes, how can I determine the volume for which i get the precipitation I want?

You need to precipitate $\pu{9.9999\times 10^-4 mol}$ of $\ce{Cu^{2+}}$, hence you need to add $2(\pu{9.9999\times 10^-4 mol}) = \pu{1.99998\times 10^-3 mol}$ of $\ce{NaOH}$, this is:

$\displaystyle{V_\ce{NaOH} = \frac{n_\ce{NaOH}}{c_\ce{NaOH}} = \frac{\pu{1.99998\times 10^-3 mol}}{\pu{0.1 mol L-1}} = \pu{19.9998 mL} \approx \pu{20.0 mL}}$

There are $\pu{1\times 10^-8 mol}$ of $\ce{Cu^{2+}}$ left in solution, i.e. $\ce{[Cu^{2+}]} = \pu{8.33 \times 10^{-8} mol L^-1}$. Then, $\displaystyle{\ce{[OH-]} = \sqrt{\frac
{K_\text{sp}}{\ce{[Cu^{2+}]}}} =  \pu{1.74 \times 10^{-6} mol L-1}}$.

The final concentration of $\ce{Ca^{2+}}$ is $\pu{0.8333 mol L-1}$, so $Q_\text{ps} = \ce{[Ca^{2+}][OH-]^2} = \pu{2.52 \times 10^{-12}} < K_\text{sp}$, i.e. $\ce{Ca(OH)2}$ won't precipitate at this point.


The cutoff keeps defaulting to 0.05 in Jmol I am using Jmol for previewing chemical structures. I need the isoSurface value to be 0.008 for 3d printing purposes.But, the cutoff for isosurface keeps defaulting to 0.05.

How should I change it and can anyone also suggest why this keeps happening?

The syntax is

isoSurface id saSurface radius

So you should use

isoSurface id saSurface 0.008

You can use the optional id for giving it a name.

You may also want to see their reference page

Edit 1: Have you tried using DX files ?
I am not an expert in using DX to render the required results but here is the reference  :

  DX, as Cube, defines a three-dimensional grid of points in space. At
  each point is a number (a "scalar value"). This set of point values is
  then used by Jmol to define an "isosurface" -- the surface separating
  points having values greater than a given cutoff from those that have
  a value less than a certain cutoff. A typical application is molecular
  orbitals. The phase parameter of the isosurface command allows bicolor
  rendering: one color for "points greater than x" and another color for
  "points less than -x". Jmol can read DX files and re-export them to
  much smaller JVXL files.

Quoted from here


How many Au atoms can be extracted from 0.250 L of seawater? Gold is present in seawater to the extent of 0.15 mg/ton. Assume the density of seawater is 1.03 g/ml and determine how many gold atoms could be extracted from 0.250 L of seawater.

($\pu{1 ton}=\pu{2000lbs}, \pu{1kg}=\pu{2.205lbs}, \pu{1 mole}=\pu{6.022E23 atoms}, \text{1 mole of gold}=\pu{197g}$)

I just multiplied everything together using dimensional analysis and got $5.2\times10^{14}$ atoms of Au per liter of seawater, but they want how many atoms of Au in $\pu{0.250L}$ of seawater so I just multiplied $5.2\times10^{14}$ by $0.250$ to get $1.30\times10^{14}$ atoms of Au per $\pu{0.250L}$ of seawater. Can you simply multiply everything together like this? When can you do this and when can't you?

You also need to convert the grams of gold into moles in order to use Avogadro's number.

Atomic weight of gold = $\pu{196.967 g/mol}$
$$
\begin{align}
\frac{\pu{1.5E-04 g/ton}}{\pu{196.967 g/mol}}&= \pu{7.615E-07 mol/ton}\\
\pu{7.615E-07 mol/ton}\times \pu{6.022E+23 atoms/mol} &= \pu{4.586E+17 atoms/ton}\\
\frac{\pu{4.586E{+17} atoms/ton}}{\pu{2000 lb/ton}}&=\pu{2.293E{+14} atoms/lb}\\
\frac{\pu{2.293E{+14} atoms/lb}}{\pu{0.454 kg/lb}}=\pu{5.056E{+14} atoms/kg}&=\pu{5.056E{+11} atoms/g}\\
\pu{5.056E{+11} atoms/g} \times \pu{1.03 g/ml}&=\pu{5.208E{+11} atoms/ml}\\
\pu{5.208E{+11} atoms/ml}\times \pu{250 ml}&= \pu{1.302 E{+14} atoms}/\pu{250 ml}\\
\end{align}
$$
There would be $1.302 \times10^{14}$ atoms of gold in $\pu{0.250 L}$ of seawater.


What is meant by 'stability'? I'm currently learning about substitution and elimination reactions and to do so I need to learn about bases and nucleophiles.

Apparently a good base is an unstable species.

What does stability even mean in chemistry?

I just need to conceptually understand and wrap my head around this concept of a compound or atom or species being 'stable' or 'unstable'. Please don't explain it in complicated terms.

Stability has several meanings. A compound can be stable on its own, under the given conditions such as temperature and pressure. This is the case for a lot of compounds. The other definition of stable could be a compound's inability to react with other molecules (like the inability of carboxylic acids to be oxidized). This is probably what you meant. A good base, like $R-O^-$ is unstable because of the +I effect of the alkyl group, which shifts the electron density towards oxygen, which in turn makes the ion a very good and strong nucleophile. This is the reason why such "good bases" tend to cause eliminations  and substitutions in organic molecules.

P.S.: This is the simplest explanation I can provide. If you have any doubts, ask them in the comments.

P.S.S.: Stability of a compound can also be explained on the basis of what @NotNicolaou said.


Pressure dependency in Haber Bosch ammonia synthesis Haber-Bosch ammonia synthesis reaction:
 $$\ce{3H2 +N2 -> 2NH3}$$

According to the ideal gas law: $pV=nRT$ where constant volume implies $\displaystyle \frac nV=\frac p{RT}$

Then $\displaystyle K_\mathrm c=\frac{[\ce{NH3}]^{2}}{[\ce{H2}]^3\cdot [\ce{N2}]}$

When I substitute $\displaystyle \frac nV=c=\frac {p_{\text{gas}}}{RT}$ I get:

$$ K_\mathrm c=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$

But according to Le Chatelier's principle, increased pressure should result in an increased $K_\mathrm c$, but my $\displaystyle K_\mathrm c\approx \frac 1{p^2}$.

Where is the error? Is is impossible to combine the partial pressures like that because they scale differently with total pressure? Can one derive from the gas law the pressure dependency of this reaction?

Edit after solved:

First mistake was that the formula had $K_c$ (constant) instead of Q,
$$ Q=\frac{\left(\frac{p(NH_3)}{RT}\right)^2}{\left(\frac{p(H_2)}{RT}\right)^3\left(\frac{p(N_2)}{RT}\right)}=\frac{R^2T^2}{p^2}$$
where Q is the current value/ratio of concentrations, so the reaction tries to get Q back to $K_c$ so for greater pressure p, Q is smaller so the reaction goes in the direction of NH3...

According to Le Chatelier's principle, a pressure increase should result in the equilibrium shifting to the right, but it should not - at least to first approximation - affect the equilibrium constant.

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

Using partial pressures, this can be rewritten as

$$K = \left(\frac{RT}{P}\right)^2 \frac{y_{\ce{NH3}}^2}{y_{\ce{H2}}^3 y_{\ce{N2}}}$$

Then for $K$ to be constant, an increase in the denominator, given by the pressure for example, has to be followed by an increase in the numerator, which means the reaction working in the direction of formation of ammonia.


Which substance can be easily adsorbed by charcoal? Which substance can be easily adsorbed by charcoal?

$\ce{CO2}$
$\ce{SO2}$
$\ce{H2}$
$\ce{He}$

I thought the answer was (3) because the intermolecular forces are weak. So charcoal can easily adsorb hydrogen. But answer given is (2) which has strong intermolecular forces. Is my reasoning wrong?

Gas molecules bind to activated carbon as a result of London dispersion forces. Therefore, the larger the molecules, as determined by the size and the number of atoms in it, the better they bind to the carbon.

If you were to rank these in order of highest dispersion forces to lowest, that list would read: $$\ce{SO2>CO2>H2>He}$$