text:
Why does silicon not react with dioxygen?

Why does silicon not react with dioxygen?

I'm doing an organic lab experiment. I'm heating up silicon, carbon, and sulfur individually then placing them in pure $\ce{O2}$. I don't really know about carbon chemistry but I thought the reaction might be related to the electronegativity of silicon being lower than that of sulfur/carbon causing the bond to be more polar or something?

Elemental sulfur commonly occurs in form of $\ce{S8}$, with relatively low melting point and, more importantly, relatively low bond energies. So, it easily breaks.
Carbon commonly occurs as graphite or structures composed of graphite-like fragments, but usually something we call carbon contains up to 10% of other elements, meaning it can produce some combustible gases on heating.
However, more importantly, both elements mentioned above produce gaseous oxides on oxidation, that leave the surface of the reacting compound, exposing it to further reaction.
Silicon, on the other hand, produces solid oxide on heating, that prevents further reaction. It is still possible to make a very fine dust of silicon to react with oxygen, but it is not an easy feat. For reaction to occur to bulk silicon, the products must be removed from the surface. Given that, bulk silicon will not give silicon oxide, but it can react with chlorine in reasonable conditions, giving volatile silicon tetrachloride.
Why silicon forms solid oxide with high melting point is another matter, with answer considerably more obscure. The simplest approach to rationalize it would be to consider atomic radius of carbon, sulfur and silicon and consider bonding in alternative structures for this compounds.
Carbon dioxide is an oxide with small central atom, that can form $\pi$-bonds effectively with oxide, so it does not form polymeric structure.
Sulfur dioxide is an oxide of larger atom, but it has a lone pair on central atom, so polymeric form would be considerably hindered. (on contrary, sulfur trioxide forms polymeric forms, but still has gaseous form  which is only marginally less stable).
Silicon dioxide in monomeric form has two $\pi$-bonds, and this is uncharacteristic for third row elements, and silicon is significantly larger than carbon, having enough space for four oxygens around it. So, it easily and effectively forms 3-d network of $\ce{SiO4}$ units with shared oxygens, that is quite hard to melt and vaporize, and, once formed on the surface of silicon, it prevents further reaction.
Still, formation of silicon dioxide films on surface of bulk monocrystalline silicon is essential for microelectronics.

In order for a reaction to occur, bonds in the reactants have to break so that they can form new bonds that comprise the products.
Did you use amorphous carbon? Was it monocrystalline silicon?
The article on amorphous carbon says that it is stabilized by dangling pi bonds with hydrogen, suggesting a bond that is easier to break and start an exothermic reaction that will provide further energy to break other, stronger bonds.  The silicon lacks this.
Did you try graphite?  My guess is that it would not react either, nor would diamond.
The sulfur probably reacted due to the weaker $\ce{S-S}$ bonds.
BTW, electronegativity isn't a very good explanation for much other than bond polarity.

At Chernobyl the graphite moderator did light on fire, didn't it?

How hot was that compared to OP's lab experiment?

He didn't say.  But the OP had pure oxygen on hand, that helps combustion, and once started, the heat generated from combustion helps sustain the reaction.

Both graphite and diamond will burn in oxygen if you heat them enough (and you don't need extreme temperatures).


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Beer freezing after being opened
I am watching the Stoke vs Swansea game, & to be honest, it's pretty boring. So much so, that I've started to wonder why it is that my beer has turned into a slushy after being taken out of the freezer.
I bought it this morning, it wasn't very cold, so just before the game I put it in the freezer to achieve optimal freshness. Started off pretty well, but rather than getting warmer as I drink, it's actually continued to freeze.
I am in France.
The windows are open.
It's 22 degrees Celsius.
I have attached a pic of the beer in question.

Have I discovered the holy grail of beer chilling?

Was it liquid when you took it out of the freezer?

Yes. More liquid than it is now.

Like this? https://www.youtube.com/watch?v=Fot3m7kyLn4

When you opened the beer, carbon dioxide was lost to the atmosphere, increasing the mole fraction of water in the beer and raising its freezing point to a point above the temperature of the beer.  Not all of the water froze, leaving it a slush.
Edit:  There's also the possibility that the loss of CO2 from the solution is not the issue.  It could just be that when you opened the beer you disturbed a supercooled solution enough to start it freezing.
2nd edit:  Instead of continuing the comments I'll add here.  The freezing on the outside of the can shows that the temperature of the beer is below the freezing point of water.  Is the temperature of the beer lower than the freezing point of the beer?  If it were, the beer should freeze instantly upon being disturbed.  If the beer is above its freezing point, but the loss of CO2 raises the freezing point it should freeze gradually as it sits out.

Within a minute, ice that wasn't there initially has appeared on the can

3 minutes later, can is COVERED by a layer of ice

If the beer freezes as soon as you open it, then it is most likely a supercooled solution that froze upon being disturbed.  If it takes time to freeze, it is most likely the loss of CO2 from the solution.

Is the freezing in the beer, or is it on the outside of the can?

Its definitely still liquid, but it's starting to freeze slightly

The freezing happened instantly on the can, but the beer takes a few minutes

@brinnb You can take beer out of the freezer and shake it and it will remain liquid. The moment you open it, it will start to freeze up. The release of pressure changes the temperature when water freezes. See the phase diagram for water. The release of $\ce {CO2}$ would increase the freezing point, not lower it.

@LDC3 The effect of pressure on the freezing point is really subtle (13.35 MPa gives a melting point of -1 °C http://www1.lsbu.ac.uk/water/density_anomalies.html#Pmelt). The curve is very steep http://physics.stackexchange.com/questions/60170/freezing-point-of-water-with-respect-to-pressure  The release of CO2 changes the freezing point somewhat, but more important is the heat release, not the change of the freezing point.

As you can see in the phase diagram for water, higher pressures decrease the freezing point.
From Chemwiki:

one is able to melt ice simply by applying pressure and not by adding heat.

When you open the beer, you release the pressure in the bottle, raising the freezing point of the beer, so it starts to freeze. Since the bottle is colder than the freezing point of water, any water condensing on the outside will freeze.
BTW, the release of $\ce {CO2}$ from the beer will actually raise the freezing point, not lower it.

The pressure inside the can/bottle is higher than the atmospheric pressure and due to this higher pressure, the freezing point of the beer increases. When you opened the can/bottle, the pressure of the beer equals the atmospheric pressure, decreasing the freezing point of the beer and hence, starting to freeze.
Cheers!

There are multiple physical forces at work here.  Pick'em out and list them.
You got "gas laws" available from elementary chemistry texts.
You got $\ce{CO2}$ dissolved in liquid - escaping (Mole laws, antifreeze effect, or not; suppression of freezing &/or boiling point(s)).  Note: When gas leaves a container (beer can, bug spray, fire extinguisher, etc.), it cools the container and what's in it, even though other forces resist temperature change, i.e., freezing due to the solubility of the gas in the water/alcohol fraction, et al. of the beer.
Heard of "equilibrium", "function" (Fx), "ambient temp."?  The frost on the outside comes from  atmospheric water (water vapor) collecting and freezing on the cold container.  As the system warms up (out of the frig), this ice will melt.
The aforementioned does not claim to be a complete account of what's going on here.

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How to calculate the volume of dry hydrogen in magnesium/ HCl reaction
I am not sure if the equivalents of $\ce{H2}$ in the equation
$$\ce{Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2}$$
is that of dry or wet hydrogen.
If the $\ce{H2}$ refers to wet hydrogen in the equation, how should I find the volume of dry hydrogen given the pressure of wet and dry hydrogen and the volume of wet hydrogen?

The water is a liquid.  It exerts a vapor pressure.  Some of the pressure in the gas is not due to hydrogen, it is due to water vapor so it is called wet hydrogen.

Not a duplicate, plus the linked question does not have an answer.

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Using combined gas law to calculate the volume of dry hydrogen gas

Volume of wet hydrogen collected at room temperature: $\pu{40.48 mL}$, ambient pressure: $\pu{29.7 inHg}$ ($\pu{754 torr}$), ambient temperature: $\pu{295 K}$ ($\pu{22^\circ C}$), vapor pressure of water at $\pu{22^\circ C}$: $\pu{19.83 torr}$. Partial pressure of dry hydrogen: $\pu{9.87 inHg}$.
Find the volume of hydrogen under standard conditions.

The equation of combined gas law is $p_1V_1/T_1 = p_2V_2/T_2$.
Say if the volume of wet hydrogen gas is $\pu{40.48 mL}$, does this equation still work if part of that volume is water vapor? Or, in other words, does the same volume of water vapor and hydrogen gas exert the same pressure?
One of the correct way to do this problem is find the moles of dry hydrogen gas using the room temperature and volume of wet hydrogen gas, gas constant and room temp. Right here, I am kinda confused why the volume of wet hydrogen if we are trying to find the moles of dry hydrogen gas. Then rest of this solution is plugging the number of moles into the equation along with STP to find the volume of dry hydrogen gas.

Look up Dalton's Law of Partial Pressures: it states that the total pressure is equal to the sum of the pressures of the individual gases. So, any amount of moisture in the hydrogen gas will have an influence on the pressure.

That is what I mean, because the water vapor added, the  equation is not proportional

If water vapor is considered an ideal gas, then the vapor pressure of the water is only dependent on temperature, and not the gas that it is dissolved in. So, from your experiment, you should have the atmospheric pressure reading at the time of the experiment, and the temperature of the experiment (usually room temperature). Now, you just need to look up the vapor pressure of water at the temperature recorded and determine the pressure of the hydrogen.

Something I find contradicting is at constant temperature if the dry hydrogen has a lower pressure than than the wet hydrogen, which is equal to the air pressure, the the volume of dry hydrogen has to much bigger than that of wet hydrogen. It is impossible for dry hydrogen to have a larger volume

It doesn't have a larger volume. When you mix gases, the volume of the 1 atm of the mix (which is determine from the amount of each gas) is made up of the volumes of the individual gases. In your statement, you have different amount of hydrogen (wet and dry) in the same volume so it looks like you need a larger volume when it is cause by the amount you have.

But the dry hydrogen is part of the wet hydrogen( dry hydrogen+ water vapor) shouldn't smaller amount = smaller volume

You're still confused. Dry hydrogen (less pressure, some volume < 40.48mL) + water vapor (pressure and some volume) equals 1 atm and 40.48mL. Wet hydrogen (water is included) = 1 atm and 40.48mL. I believe that you don't see that there are different volumes involved for the same amount of hydrogen.

LDC3 is all your saying that the volume of water in wet hydrogen gas is negligible?

Because  the volume of wet hydrogen could be seen as the volume of dry hydrogen??

@user1.618 No, I'm not saying that. But the amount of water vapor is small. We know we have (754.38mm Hg - 19.83mm Hg) for the pressure of dry hydrogen and 19.83mm Hg for water vapor. For wet hydrogen we have 754.38mm Hg and 40.48mL. The pressure for water vapor is about 2.6% that of hydrogen, that is not negligible.

But then why do we use the volume of wet hydrogen gas in the equation??

Because we can use $PV=nRT$ to determine the amount of hydrogen. $754.38mm Hg \times 40.48mL = n \times R \times 295K$. You will need to look up R with the correct units. This give the amount of wet hydrogen and since the law of partial pressure ($P_T=P_H+P_W$) infers that amount of the gases is $n_T=n_H+n_W$. When we get the correct amount of hydrogen, then we can determine the volume at STP.

@user1.618: The volume of the wet gas is used because that's the volume all the gas is in.  As LDC3 says, the pressure of hydrogen is ambient pressure - water vapor pressure.  An alternative way of thinking about the problem is: let both water vapor and hydrogen be at ambient pressure.  Then together they must take up the observed volume.

We don't have the moles for water!

@user1.618: If you like you can calculate moles for the amount of water, you know the pressure and the volume of the water vapour.  But you don't have to, this number isn't needed.

@user1.618 We can figure it out both, after all we have Dalton's Law of Partial Pressure. Read the first part here: http://www.chm.davidson.edu/vce/gaslaws/daltonslaw.html

Consider the composition of the gas in your eudiometer.  Most of it is hydrogen, but some of it is water vapour.  In other words, not all the pressure is exerted by hydrogen, part of it is due to water.  You could ask yourself: what would happen to the volume of gas produced, had you passed it through a drying tube.
You should have been given vapour pressure of water in a table.  You should also have taken down value of air pressure and ambient temperature.

The temperature is kept the same, I know the pressure and the vaole of wet H2. But because part of the hydrogen is water vapor, the proportion cannt be established

How would you determine the proportion of water vapour in your hydrogen?  Because that is what you need to know to solve the problem.

WAIT! I got it;
the "volume of the wet hydrogen gas" is ALSO the volume of the DRY hydrogen gas. That's because Dalton's Law of partial pressure assumes/ is for CONSTANT volume. IE * at constant volume *: the total pressure of wet hydrogen at x mL = vapor pressure at x ML + dry hydrogen pressure at x mL. It's x mL for both.
Another way to think about it is IF the volume of the dry gas were less, then that would mean the pressure of the dry gas is more than the one we calculated for using Dalton's law of partial pressure * at constant volume * (Boyle's Law) and we are keeping the pressure that we calculated.


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Doesn't the fact that the lone pair on the nitrogen in amides occupy a p orbital contradict Hund's rules?
Since the $\ce{sp^2}$ hybridized orbitals are lower in energy than the p orbital shouldn't the $\ce{sp^2}$ orbitals fill first? Why is this not the case - two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are singly occupied? If you look at Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these orbitals. However, the higher energy p orbital is empty - as would be predicted is Hund's rules are followed. Is it something to do with the conjugation in amides?

two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are
  singly occupied

Not really, hybridization occurs in response to bonding, so we can only talk about an atom's hybridization when it is in a molecule.  In amides the nitrogen $\ce{sp^2}$ orbitals form molecular orbitals (MO's) with whatever else they are bound to.  In the case of a simple amide with the nitrogen bound to hydrogen, the MO would involve a nitrogen $\ce{sp^2}$ orbital and a hydrogen 1s orbital and 2 electrons would occupy this orbital (one from the nitrogen $\ce{sp^2}$ orbital and one from the hydrogen 1s orbital)  - so what was the $\ce{sp^2}$ orbital will be filled with 2 shared electrons when a bond is formed.  Therefore the 3 $\ce{sp^2}$ orbitals, now MO's, and the p orbital on nitrogen will each be filled with 2 electrons in the molecule.

Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these
  orbitals

Carbenes can exist in either a singlet or triplet state, as shown below.  There are 2 possible triplets depending upon how the carbene is hybridized, a bent $\ce{sp^2}$ hybridized carbene, or a linear $\ce{sp}$ hybridized carbene. In the case of methylene ($\ce{:CH2}$), and actually for most carbenes, the triplet states are lower in energy due to Hund's rule (don't pair electrons unless you must); and the triplet has one electron in the p orbital and one in the $\ce{sp^2}$ orbital.


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How do we experimentally know the strength of bonds?
How do we experimentally measure the strengths of covalent versus ionic bonds? Also, for a given molecule, how can we know if two pairs of atoms is covalently or ionically bound?
Related question: Textbooks say that in solution of water, two neighboring water molecules (call them A and B) will be 'attracted' because the O in molecule A will attract the H of the 'nearby' molecule B. It is said that H will alternate states between A and B. How do we know this is true experimentally and how can it be measured?

You measure the energy of any bond by carrying out reactions and seeing how much energy (usually in the form of heat) has to be invested.  For example, if you react one mole of $\ce{C}$ with one mole of $\ce{O2}$, you get $\ce{CO2}$ and heat.  Measure the energy with a calorimeter and divide by how many atoms and you get the energy you invested to produce one bond.  Not coincidentally, bonds are generally measured in units of energy/mol (e.g. kJ/mol).
Whether a bond is ionic or covalent has to do with whether the bonding electron resides on one atom or both.  A perfectly covalent bond finds the electron half on one atom and half on the other.  A perfectly ionic bond has the electron 100% on one atom and not at all on the other.  Real systems are a mix of both.  When we say a bond is ionic (such as $\ce{NaCl}$), we are really referring to the fact that we have measured a circumstance where the electron truly favors one atom.  There are various ways to measure this, one of the most common is to melt the solid and see if it conducts electricity.  In the case of $\ce{NaCl}$, the bond is so ionic that a small voltage will cause the ions to wander from their neighbor.
With a covalent bond such as you find in $\ce{SiO2}$, atoms can't wander without breaking the bond, so a small voltage doesn't cause a current.  A large voltage may break the covalent bonds and cause a current, but it also changes the substance (in this case releasing oxygen gas and producing silicon metal).


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What's the electronic configuration of Nickel and Oxygen in RNiO3 (R=Rare-Earth)?
Nickel's electronic ground state is $\mathrm{3d^8 4s^2}$ (there's a dispute on this, but let's not worry about it for now). Oxygen's is $\mathrm{2s^2 2p^4}$. In a Rare-Earth based Nickelate with chemical formula $\mathrm{RNiO_3}$, if we assume simple ionic bonds, then what should the electronic state be?
Literature indicate that the "classical" case is $\mathrm{Ni^{3+} O^{2-}}$ corresponding to $\mathrm{3d^7, 2p^6}$. But this means that Oxygen has gained 2*3=6 electrons. But how can this be possible if Nickel has donated only 3 electrons?

What's the electronic configuration of Nickel and Oxygen in RNiO3
  (R=Rare-Earth)?

As an example, let's look at $\ce{LaNiO3}$.  Each oxygen has an oxidation state of -2, nickel has an oxidation state of +3 and lanthanum has an oxidation state of +3.  Overall the molecule is neutral.
$$\ce{La(+3)~ plus~ Ni(+3)~plus~ 3~oxygens (-6)~=~0}$$

Why, then, the rare earth is barely mentioned, if at all, when referring to the electronic configuration? Is it simply "too clear" to be mentioned?

@student1 Rare earths usually have oxidation state +3. Exceptions do exist, but they are exactly this: exceptions. The most known is probably Cerium (it has oxidation state +4).

Thanks. In this configuration: $Ni^{2+} O^{-}$, why do we write $3d^8 \underline{L}$? What does it mean?

I'm not familiar with this notation, is there a specific web page that you are referring to? Perhaps it represents nickel being $\ce{3d^8}$ after giving up two electrons and becoming $\ce{Ni^2}$, and the L represents any rare earth lanthanide ion.

Never mind, I found it refers to "a ligand hole on an oxygen orbital".


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How to assign the signals in the H-NMR of Aspirin?
I need to assign the H-NMR of aspirin, below is aspirin with the H's labelled.

The peaks I have are:

2.30ppm   (I this is a singlet and would be F)
7.07ppm   (I think this is E)
7.29ppm   (I think this is D)
7.53ppm   (I think this is C)
8.05ppm   (I think this is B)
11.44ppm (this is a singlet and would be A)

For B,C,D,E I need to say what kind of splitting pattern there would be and how many coupling constants are present and there approximate value. I think I know the assignments of them but I don't know the splitting pattern or coupling constants.
Would E and B be doublet of doublets because they couple with D and C so they would have ortho and meta coupling?
Would D and C be coupling with each other and B and E so would they be doublet of doublets of doublets, with two ortho and one meta coupling?

Here is a link to the actual assigned proton nmr spectrum for acetylsalicylic acid.  Your peak assignments are correct.

Would E and B be doublet of doublets because they couple with D and C
  so they would have ortho and meta coupling?

Yes, that is correct. Here is another link with more detail on the coupling constants (see p. 13 for example)

Would D and C be coupling with each other and B and E so would they be
  doublet of doublets of doublets, with two ortho and one meta coupling?

Conceptually you are correct, however in the case where two coupling constants are very similar, the expected quartet collapses to a triplet. Since $\ce{J_{DC}}$ is very close in magnitude to $\ce{J_{BC}}$ the expected quartet appears as a triplet.  Therefore, proton C appears as a triplet from these two protons and then each of these lines is further split into a tight doublet by the smaller coupling to proton E.  The overall pattern appears as a triplet of doublets.  The same reasoning applies to proton D, and again a triplet of doublets results.

Your chemical shift analysis is correct. There are two schools of thought concerning how to describe splitting patterns, and your analysis for splitting patterns is perfectly acceptable, and correct.
The first school of thought is to label a splitting pattern based on the appearance of the peaks, which is why Ron describes your aromatic peaks as doublets of triplets, and even quartet. I'm not saying this is wrong; this is just one method used. This method can cause problems for the inexperienced spectroscopist, as the appearance of a spectrum is very much dependent on the strength of the magnetic field, choice of solvent, temperature, sample preparation, quality of shimming etc. This is why experimental conditions should always be reported when publishing NMR data - and something students should be entitled to ask of their teachers when given a spectrum to decipher. What one person reports as a triplet on a 60MHz spectrometer in water might be undeniably a doublet of doublets on a 800MHz in acetone. Second order splittings in aromatic systems are a great example of this. Although common on low field systems, second order splittings are far less common on high field instruments, and many AB quartets collapse to AX spin systems at high field and couplings can be analysed as first order systems. Of course, this method makes very good sense for a number of systems. For instance, labile amine or alcohols are almost always reported as broad singlets, even though they are likely to be possess 3J couplings that are not observed through exchange decoupling.
The second school of thought (which your original descriptions of ddd adhere to) is to look at the molecule and predict the splitting that would be expected (two ortho couplings and a meta couplings for example) and describe that. You can then rationalise quite easily why a ddd looks like a triplet (here, the two ortho couplings are almost identical and fall within the natural linewidth, and so cannot be resolved and appear as a triplet). So your original suggestions are perfectly acceptable. Without seeing the spectrum you have been given, it is hard to say, but a well shimmed sample of aspirin in methanol even at medium fields clearly shows the two ortho couplings aren't quite identical, and the peak for (what you label) spin E does not look like a true triplet, with the centre line not much taller than the outside lines.
My recommendations for reporting splitting patterns and, especially, if you are reporting coupling constants, is to always report descriptions of splittings in accordance with the reported couplings. A doublet of triplets arises from an AMX2 system, and cannot have three coupling constants for spin A; only JAM and JAX. A doublet of doublets of doublets arises from an AHPX system, and will have three distinct couplings for spin A; JAH, JAP, JAX. It is perfectly acceptable for any of these to be equal, such as JAP=JAX. So, go with what you have originally suggested, and the only matter for you to consider is what typical values are for ortho (~7-8Hz) and meta (2-3Hz) couplings (and possibly para couplings - which you don't observe in aspirin, but can be ~1Hz)

This is a very fine answer, I'd loved to have had that around when I was studying NMR.


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How do I find mass equivalence of a photon?

The problem has a flame emitting a wavelength of $589\ \mathrm{nm}$, and asks for the mass of one photon of that wavelength. It tells me that $1\ \mathrm J = 1\ \frac{\mathrm{kg\cdot m^2}}{\mathrm s^2}$.

I converted to $\mathrm J$ using $E=\frac{h\cdot c}{\lambda}$. I'm not quite sure what to do past that. I have $3.36\cdot 10^{-19}\mathrm J$. I know the answer is supposed to be $3.75\cdot 10^{-34}$.

Then the answer should be 3.75x10^-36 not to the -34. If that is what the book answer was then I don't know how they got that.

A  photon's rest mass is always 0.

$$E = \frac{h\cdot c}{\lambda}$$
$$1\ \mathrm J = 6.24150934 \times 10^{18}\ \mathrm{eV}$$
then you have
$$E = mc^2 = \frac{h\cdot c}{\lambda}$$

A photon doesnt have rest mass.

However, what is asked here is the total mass including that imparted by the motion of the photon. Actually a proton at rest makes no sense anyway. We always observe them moving at the speed of light.

What do you mean by " the total mass including that imparted by the motion of the photon"?

@JunSeo-He I think he's referring to "dynamic/relativistic mass", but you probably already know that. What exactly that means is another story, and somebody more learned than I should handle it.

Simply calculate the energy: $E=h\nu$
Then solve for momentum: $p=E/c$
Then divide by the speed of light: $p=mv$ or $m=p/v$
With the speed of light for $v$ so:  $m=p/c$

A photon doesnt have rest mass.

However, what is asked here is the total mass including that imparted by the motion of the photon. Actually a proton at rest makes no sense anyway. We always observe them moving at the speed of light.

What do you mean by " the total mass including that imparted by the motion of the photon"?


text:
What is the purpose of ammonium chloride in anti hydrogenation of alkynes?
In my organic chemistry textbook, lithium is used to create a free radical of the alkynes, to then allow the anti addition of hydrogen to get an alkene.
"The Dissolving Metal Reduction of Alkynes" is a reaction that then uses $\ce{NH4Cl}$. Why put $\ce{NH4Cl}$?

I know that in the Birch reduction there is an alcohol (usually t-butanol) added to protonate the radical anion and allow the reaction to proceed, the ammonia solvent isn't sufficiently acidic to do that.  I don't think you need to do that for alkynes.  As far as I know the ammonium chloride is to quench excess sodium at the end of the reaction.

@AbelFriedman maybe jerepierre's answer is plausible?

@yolo123: Completely agree.

What about quenching sodium?

Hmm, do you think that NH4Cl could be added for a reason relating to the reactivity of the double bond formed? (I know double bonds can be quite reactive sometimes, that's why I'm asking.)

I think the ammonium chloride is used to protonated the amide ion. Protonation generates ammonia, which can easily be evaporated away.

Products from reaction step (1) are the trans-alkene and 2 equivalents of the amide $\ce{LiNHEt}$, a strong base. Step (2) is the workup of the reaction mixture with aqueous $\ce{NH4Cl}$ solution, which serves several purposes. First, it quenches the lithium amide and unreacted lithium metal.
$$\ce{LiNHEt + NH4Cl ->~ EtNH2 + NH3 + LiCl}$$
$$\ce{Li + H2O ->~ LiOH (aq) + \frac{1}{2} H2\uparrow}$$
Secondly, the trans-alkene is usually less water-soluble than ethyl amine and the lithium and ammonium salts, and can be separated from the workup mixture by precipitation or extraction with an organic solvent.


text:
Interpreting a Formula Tattoo (Bipolar?)

Found This Online. Something to do with bipolar disorder?

Welcome to Chemistry.SE! If you could add the details, where you found the formula we can understand the context and maybe give a better answer to your question.

I found it on a website showing bipolar tattoos/symbols etc. I thought maybe it had to do with Serotonin? Have zero knowledge of chemistry, so i thought i would ask some people who do!

This is lamotrigine, a sodium channel blocker prescribed against epilepsy.  Apparently it's also prescribed for bipolar disorder.


text:
Relationship between auto-ignition temperature with pressure and fuel
Experimentally it appears that increase in pressure, fuel concentration, and chain length causes AIT to decrease until a minimum where even further increase in these parameters increases the AIT. Additionally, it seems that AIT is lowered by increasing the vessel volume.
From the last page of this paper:

Semenoff's equation seems to indicate that it is possible to lower the AIT to arbitrarily low values by increasing the pressure. But the higher the total pressure of the fuel/air mixture, the higher is also the partial pressure of the fuel required to reach the most critical fuel concentration. At some temperature, the required partial pressure will become equal to the fuel vapor pressure. At even higher pressures and lower temperatures, it is impossible to reach the critical fuel concentration, and the depression of the AIT is expected to stop. A similar mechanism has been employed to explain the reincrease of the standard AIT with chain length for very long chain molecules /5/.

Edit: Article title is Autoignition Temperature for Mixtures of Flammable Liquids at Elevated Pressures by Brandes, Hirsh, and Stolz
1) From this, it seems that pressure and fuel concentration effects are related. Is the critical fuel concentration related to the equivalence ratio and the ability to reach this equivalence ratio based on the saturation pressure of the fuel?
2) I do not get the relation between the above explanation and the last sentence regarding re-increase of AIT for longer chain molecules?
3) I'm not sure why larger volumes leads to lower AIT as stated in ASTM E659 Appendix X2.

1) One could argue that yes. All other things being equal, larger molecules will clearly have lower vapor pressures at the same temperature so it is reasonable that there is a limit at which the fuel-air mixture is too lean to support autoignition.
2) After reading the paper I can only conclude that they are referring to the above idea that autoignition temperatures increase for larger compounds due to the need to increase the vapor pressure to reach an appropriate fuel concentration that supports autoignition.
3) Larger volumes have lower surface-area-to-volume ratios. The lower surface area means a the walls of a vessel play less of a role in the combustion chemistry leading to a lower AIT. These surface effects are detailed by Babrauskas with references to those who have observed this.

Babrauskas, V.
 of Fire Protection Engineers, S. (Ed.)
 Ignition Handbook
Fire Science Publishers,
2003

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What is the entropy of mixing of two ideal gases starting out with different pressures?
The entropy of mixing of ideal gas is given by this equation:
$\Delta S_{mix} =  -nR(x_1\ln x_1 + x_2\ln x_2)$

Does this equation works only when the initial conditions of both compartments are at the same pressure?
If ideal gas 1 at initial conditions $n_1,P_1,V_1,T$ and ideal gas 2 at $n_2,P_2,V_2,T$ are mixed, which equation should I use to obtain $\Delta S_{mix}$ given that $P_1 ≠P_2$? Assume $V_1 + V_2 = V_{tot}$, and $T_1 = T_2 = T$.

To 1.: No you can't use this equation. Pressure has an effect on the entropy. To 2.: Have you tried anything to solve this yourself? Do you have an idea how to go about it?

@Philipp I tried dS = -RdP/P and integrate from P1 to Pavg and P2 to Pavg and summing the entropy  change.

A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}} S = - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \big|_{p, \{ n_{i} \}}$ later): for isothermal and isobaric conditions it is
\begin{equation}
  G = \sum_{i} n_{i} \mu_{i} + \mathrm{constant} = n_{\mathrm{tot}} \sum_{i} \underbrace{x_{i}}_{= \, \frac{n_{i}}{n_{\mathrm{tot}}}} \mu_{i} +  \mathrm{constant} \hspace{4em} (1)
\end{equation}
with $n_{i}$, $\mu_{i}$, and $x_{i} = \frac{n_{i}}{n_{\mathrm{tot}}}$ being the amount, the chemical potential, and the mole fraction of the $i$th component in the system, respectively, and $n_{\mathrm{tot}} = \sum_{i} n_{i}$ being the total amount of gas in the system.
So, now we need an expression for the chemical potential of an ideal gas.
It can be shown that this is given by
\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\biggl(\frac{p_{i}}{P^{0}}\biggr) \hspace{18em} (2)
\end{equation}
where $\mu_{i}^{0}$ is the chemical potential of the $i$th component in a given standard state, $R$ is the universal gas constant, $T$ is the temperatur, $p_{i}$ is the partial pressure of the $i$th component, and $P^{0}$ is the standard pressure (i.e. the pressure connected to the standard state).
For ideal gases we can apply Raoult's law stating that
\begin{equation}
  p_{i} = x_{i} P_{\mathrm{tot}}
\end{equation}
where $P_{\mathrm{tot}}$ is the total pressure of the system.
Using this with equation (2) we get
\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\biggl(\frac{x_{i} P_{\mathrm{tot}}}{P^{0}}\biggr) = \mu_{i}^{0} + R T \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{tot}}}{P^{0}}\biggr) \hspace{6em} (3)
\end{equation}
Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.
\begin{equation}
  P_{\mathrm{tot}} = \sum_{i} p_{i}
\end{equation}
Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process.
At the initial state the chambers are seperated (whereby chamber $i$ contains only one sort of gas labeled by $i$) and so these subsystems are completely independent from each other.
Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the pressure $P_{\mathrm{tot}} = P_{i}$ in the system (whereby $P_{i}$ is the ambient pressure in chamber $i$), since $x_{i} = 1$ (each chamber/independent subsystem holds just one kind of gas).
According to equation (1) the Gibbs free energy of the $i$th chamber is then given by
\begin{align}
  G_{i} &= n_{i} \mu_{i} = n_{i} \biggl( \mu_{i}^{0} + R T \underbrace{\ln 1}_{= \, 0} + R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \\
        &= n_{i} \biggl( \mu_{i}^{0} + R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr)
\end{align}
and the Gibbs free energy of the total system at the initial state is
\begin{align}
  G_{\mathrm{ini}} &= \sum_{i} G_{i} \\
                 &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln\biggl(\frac{P_{\mathrm{tot}}}{P^{0}}\biggr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + \sum_{i} x_{i} R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + \sum_{i} x_{i} R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \ .  \hspace{7em} (4)
\end{align}
For a multicomponent system with more than two components I would simply use this equation. But for a system with exactly two components we can introduce some substitutions that will simplify the final equations a little. For that purpose we want to establish a relationship between the initial pressures $P_{1}$ and $P_{2}$ and the final pressure $P_{\mathrm{fin}}$. In order to get there we use the ideal gas law and the additivity relations for the volumes and amounts: From
\begin{equation}
  n_{\mathrm{tot}} = n_{1} + n_{2} \ , \qquad V_{\mathrm{tot}} = V_{1} + V_{2} \\
  P_{1} V_{1} = n_{1} R T \ , \qquad P_{2} V_{2} = n_{2} R T \ , \qquad P_{\mathrm{fin}} V_{\mathrm{tot}} = n_{\mathrm{tot}} R T
\end{equation}
it follows that
\begin{align}
  P_{1} V_{1} + P_{2} V_{2} = P_{\mathrm{fin}} V_{\mathrm{tot}} \ .
\end{align}
Now, we can express $P_{2}$ in terms of $P_{\mathrm{fin}}$ by introducing the deviation factor $\alpha$ which is defined by $P_{2} = (1 - \alpha) P_{\mathrm{fin}}$ and use this in the previous equation
\begin{align}
  P_{1} V_{1} + P_{2} V_{2} &= P_{\mathrm{fin}} V_{\mathrm{tot}} \\
  P_{1} V_{1} + (1 - \alpha) P_{\mathrm{fin}} V_{2} &= P_{\mathrm{fin}} V_{\mathrm{tot}} \\
  P_{1} V_{1} &= P_{\mathrm{fin}} \bigl( \underbrace{V_{\mathrm{tot}} - V_{2}}_{= \, V_{1}} + \alpha V_{2} \bigr) \\
  P_{1} &= \biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) P_{\mathrm{fin}}
\end{align}
which gives us the relation between $P_{1}$ and $P_{\mathrm{fin}}$. Introducing these equations into equation (4) yields
\begin{align}
  G_{\mathrm{ini}} &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl(\frac{P_{1}}{P^{0}}\biggr) + x_{2} R T \ln\biggl(\frac{P_{2}}{P^{0}}\biggr) \biggr) \\
  &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\frac{\bigl( 1 + \alpha \frac{V_{2}}{V_{1}} \bigr) P_{\mathrm{fin}}}{P^{0}} + x_{2} R T \ln\frac{(1 - \alpha) P_{\mathrm{fin}}}{P^{0}} \biggr) \\
  &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + \underbrace{(x_{1} + x_{2})}_{= \, 1} R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \\
 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \ .
\end{align}
At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before.
Also, the ambient pressure in the system changes to $P_{\mathrm{tot}} = P_{\mathrm{fin}}$.
Now, the complete system of connected chambers must be treated via equation (1)
\begin{align}
  G_{\mathrm{final}} &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} R T x_{i} \ln x_{i} + \sum_{i} x_{i} \underbrace{R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr)}_{= \, \mathrm{constant}} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \ .
\end{align}
Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:
\begin{align}
  \Delta_{\mathrm{mix}}G &= G_{\mathrm{final}} - G_{\mathrm{ini}} \\
                       &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                       &\quad - n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \\
                       &= n_{\mathrm{tot}} R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) - n_{\mathrm{tot}}  R T \biggl( x_{1} \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} \ln(1 - \alpha) \biggr) \qquad (5)
\end{align}
Now you see why we went through those substitutions for $P_{1}$ and $P_{2}$. This way the $n_{\mathrm{tot}} R T \ln\Bigl(\frac{P_{\mathrm{fin}}}{P^{0}}\Bigr)$ terms cancelled in equation (5).
Differentiating equation (5) with respect to the temperature we get the entropy of mixing since $\frac{\partial G}{\partial T} \big|_{p, \{ n_{i} \}} = - S$
\begin{align}
  \Delta_{\mathrm{mix}} S &= - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \bigg|_{p, \{ n_{i} \}} \\
   &= - n_{\mathrm{tot}} R \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + n_{\mathrm{tot}}  R \biggl( x_{1} \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} \ln(1 - \alpha) \biggr)
\end{align}
The first term $- n_{\mathrm{tot}} R \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right)$ is just the usual entropy of mixing. The second term $n_{\mathrm{tot}}  R \Bigl( x_{1} \ln\Bigl( 1 + \alpha \frac{V_{2}}{V_{1}} \Bigr) + x_{2} \ln(1 - \alpha) \Bigr)$ describes the contribution of the pressure equilibration to the entropy.

Thank you for your answer. Is it possible to use $\Delta{S} = -n_1R\ln\frac{P_{fin}}{P_1} -n_2R\ln\frac{P_{fin}}{P_2}$ and summing this entropy change and that of the one of mixing?

@t.c Where does this equation for $\Delta S$ come from.

I used the ideal gas entropy change equation $\Delta{S} = C_p\ln\frac{T_2}{T_1} - R\ln\frac{P_2}{P_1} = - R\ln\frac{P_2}{P_1}$(at const T). Then I substitute in the initial pressures and the final pressure $P_{fin}$

@t.c I don't think you can use this formula for a mixing process. It doesn't account for the mixing contribution.

@Philipp You can use the formula the OP provided. It can be found in most pchem textbooks. There's no need to account for the mixing per se. That's because, at any given T, the chemical potential of an ideal gas is determined only by its own partial pressure, which is independent of the presence of any other gas(es).  For example, if you take two ideal gases, each in separate 2 L containers, and mix them together (at constant T) into a single 2 L container, their partial pressures are unchanged, and thus $\Delta S  = 0$ for this process.

@Philipp More importantly, did you do any error-checking of your value for $\Delta S_{mix}$ (e.g., checking the value it gives for simple systems with known solutions, e.g., where n1 = n2 & V1 = V2)? That's essential w/ complicated expressions. I ask b/c I think you have an error. If I start with $p_2=(1-\alpha) p_{final}=> \frac{n_2 R T} {V_2}=(1-\alpha)  \frac{n_2 R T}{(V_1+V_2)}$, and solve for $\alpha$, I get $\alpha=-\frac{V_1}{V_2}$. When I subsititue that back into your expression for $\Delta S_{mix}$, specifically into the term $\ln(1+\alpha\frac{V_2}{V_2})$, I get $\ln(0)=-\infty$.

Introducing $P_{tot}$ as the initial total pressure makes absolutely no sense. You cannot calculate a total pressure, since you have here two isolated compartments. Instead:
$$G_{ini}=n_1\big(\mu^\circ_1+RT\ln(p_1/p^\circ)\big)+n_2\big(\mu^\circ_2+RT\ln(p_2/p^\circ)\big)$$
$$G_{fin}=n_1\big(\mu^\circ_1+RT\ln(x_1\cdot p_{fin}/p^\circ)\big)+n_2\big(\mu^\circ_2+RT\ln(x_2\cdot p_{fin}/p^\circ)\big)$$
$$\implies G_{fin}-G_{ini}=n_1RT\ln(1/x_1\cdot p_1/p_{fin})+n_2RT\ln(1/x_2\cdot p_2/p_{fin})$$
The derivative of this with respect to $T$ can be reformed to theorist's solution.

Nvm, I got tripped by the $P_{tot}$ part. It's fine. This solution is equivalent to theorist's solution. If we switch back from $\alpha$ to $p_1$, $p_2$ and $p_f$ with
$$1+\alpha \frac{V_{2}}{V_{1}} = p_1/p_{fin}$$
$$1-\alpha = p_2/p_{fin}$$
we get the solution I commented before. Still don't see the point in $\alpha$ though. It doesn't simplify the final equation at all.

@theorist Your example doesn't invalidate the solution. While $$\lim\limits_{n_1\to0}\ln(1+\alpha\frac{V_2}{V_2}) = - \infty$$
you forget about the $x_1$ part. If we combind both:
$$\lim\limits_{n_1\to0}\ln(1+\alpha\frac{V_2}{V_1})-\ln(x_1) = 0$$

I would like to give an alternative derivation for theorist's answer. You can get the same answer with statistical mechanics using a lattice model.
Let us discretize the container into $M_1$ respective $M_2$ cells. The volume of a cell is $V_{cell}$. Each gas particle fits exactly into one cell and can hop from cell to cell in a discrete manner. Since we are dealing with an ideal gas, the particles ignore each other. In other words, all particles can in principle occupy the same cell, i.e. all particles can always choose from all the cells. We therefore get the following number of possible states $\Omega$:
$$\Omega_1 = \frac{(M_1)^{N_1}}{N_1!} \\ \Omega_2 = \frac{(M_2)^{N_2}}{N_2!} \\ \Omega_{mix} = \frac{(M_1+M_2)^{(N_1)}}{N_1!}\cdot\frac{(M_1+M_2)^{(N_2)}}{N_2!}$$
where $N_i$ is the respective number of particles. The division by the factorial is the correction factor for the fact that the particles of the type $i$ are indistinguishable. If the states are degenerated, i.e. have the same energy, we can use Boltzmann's entropy formula:
$$S = k_B\ln(\Omega)$$
$$\implies\Delta S=k_B\ln(\Omega_{mix})-k_B\ln(\Omega_1)-k_B\ln(\Omega_2)$$
When we plug in all $\Omega_i$ and summarize with logarithmic identities, we obtain:
$$\require{cancel}\Delta S=N_1k_B\ln\left( \frac{M_1+M_2}{M_1}\cdot\frac{\cancel{N_1!}}{\cancel{N_1!}}\right)+N_2k_B\ln\left(\frac{M_1+M_2}{M_2}\cdot\frac{\cancel{N_2!}}{\cancel{N_2!}}\right)$$
Finally, when we plug in $M_i = V_i/V_{cell}$ and $N_ik_B=n_iR$, we obtain theorist's solution:
$$\Delta S = n_1R\ln\left( \frac{V_1+V_2}{V_1}\right)+n_2R\ln\left(\frac{V_1+V_2}{V_2}\right)$$
Bonus: If we consider the constant pressure case for an ideal gas by plugging in $V_i=n_i\cdot RT/p$:
$$\Delta S = n_1R\ln\left( \frac{n_1+n_2}{n_1}\right)+n_2R\ln\left(\frac{n_1+n_2}{n_2}\right)$$
This expression can be reformed to OP's formula by applying logarithmic identities and $x_i=n_i/n_{tot}$:
$$\Delta S = -n_{tot}R(x_1\ln x_1 + x_2\ln x_2)$$

I think there's a problem with your derivation.  When I subsititue your expressions for $\Omega_1$, $\Omega_2$, and $\Omega_{mix}$ into your expression for $\Delta S$, I don't get my solution; instead, I get $R \log \left((V_1+V_2)^{n_1+n_2}\right)-R \log \left(V_1^{n_1}\right)-R
\log \left(V_2^{n_2}\right)$.  Also, note that the number of cells in volume 1 (V1) is not the same as V1 (same for V2 and total), and that you should start with $S = k_B \ln(\Omega)$.

@theorist $log(x^y)=y\cdot log(x)$ and $log(x)-log(y)=log(x/y)$

Actually, I was using those identities; the problem is I didn't combine terms correctly.  So yes, I agree, you do get the correct result.  But I'm still not quite comfortable with the derivation.

It's actually the beauty of statistical mechanics that you can rationalize thermodynamic relations with relative simple models and statistics.

My concern isn't the simplicity, it's the conceptual issues with the derivation. For instance, when you write your expressions for omega, there the V's aren't volume, they're the number of cells.  But when you write your final answer, you've switched the meaning of V (without explanation) from no. of cells to volume. And in determining omega, it should be the number of cells raised to the number of particles, not to moles. [continued]

You still get the right answer in spite of the first problem because of cancellations, and in spite of the 2nd problem b/c you're using a non-standard expression for entropy—$R \ln (\Omega)$ instead of $k  \ln (\Omega)$—but this is not a good way to go about things.  Remember that the purpose of these answers is making things understandable for readers, some of whom are new students of the material, so they can see not only the answer, but how you got to the answer.  And my concern is that there is a problem here with how you've gotten to the answer that could create confusion. [continued]

Contrast your derivation with, for instance, that in in Hill's Introduction to Statistical Thermodynamics (see eqn. 4-28 on p. 80). I found Hill's book to be a nice, easily accessible supplementary text when I first took stat mech; and my students found it helpful when I later TA'd it as a grad student.

Let me put your mind at ease by providing a proper derivation ;)

There's a simple way to solve this:  We know that, for an ideal gas at constant T:
$$\Delta S = -nR \ln \frac{p_f}{p_i}= nR \ln \frac{p_i}{p_f}$$
And since ideal gases ignore each other, we can calculate the entropy change for gas 1 and gas 2 separately, and simply sum the two. [At any given $T$, the chemical potential of an ideal gas is determined only by its own partial pressure, which is independent of the presence of any other gas(es).]  Thus:
$$\Delta S = \Delta S_1+ \Delta S_2 =  n_1R  \ln \left( \frac{p_{1, initial}}{p_{1, final}} \right )+n_2R\ln \left( \frac{p_{2, initial}}{p_{2, final}} \right),$$
where $p_{1, final}$ and $p_{2, final}$ are the respective final partial pressures of gases 1 and 2.  If we don't know the pressures, we can substitute the following into the above expression:
$$p_{1, initial} = \frac{n_1 R T}{V_1}$$
$$p_{1, final} = \frac{n_1 R T}{(V_1+V_2)}$$
$$p_{2, initial} = \frac{n_2 R T}{V_2}$$
$$p_{2, final} = \frac{n_2 R T}{(V_1+V_2)}$$
After simplifying, this gives us the following:
$$\Delta S = n_1R \ln  \left( \frac{V_1+V_2}{V_1} \right )+n_2R\ln \left(\frac{V_1+V_2}{V_2} \right)=n_1R \ln  \left( \frac{V_{total}}{V_1} \right )+n_2R\ln \left(\frac{V_{total}}{V_2} \right)$$

One can also formally consider 2 steps like 1/ Each gas reaching independently the final gas mixture pressure by isothermal expansion/compression 2/ Performing isobaric mixing. So entropy change computation would be the sum of the OP mentioned mixing change and isothermal expansion/compresssion changes.

How does it come that the two entropy terms aren't weighted by the respective amounts of substance?

@Kexanone Good catch! I inappropriately combined two terms. I just fixed it.


text:
Convert hexane to 1,6-dibromohexane
Is it possible by any means or by using some reagents to convert
hexane to 1,6-dibromohexane? The yield should also be good.

In case you need 1,6 disubstituted hexane, consider either a dirt cheap adipic acid as a starting material or coupling of two $\ce{C3}$ fragments, porduced from allilic derivatives.

In principle, alkanes may be converted to alkyl bromides by photoinitiated bromination.
However, taking statistics (2 $\ce{CH3}$ vs 4 $\ce{CH2}$ groups) and stabilities of the radicals resulting from H abstraction by a bromine radical from your starting material into account, this approach is pretty worthless and will most likely yield 1,6-dibromohexane as a minor product in a complex mixture.
Consequently, your reasonable options are a) a different starting material (and a different synthesis) or b) simply buy 1,6-dibromohexane (around 20 Euro for 100 g).


text:
Acidic conditions, high heat, and return to carboxylic acids
Apparently carboxylic acid derivatives such as cyanohydrins can be taken back to "daddy" - a carboxylic acid - under the right conditions. This involves acid, water, and "high heat." What, however, is the mechanism of a cyanohydrin going back to daddy?

Cyanohydrins are typically formed from aldehydes and ketones and can be hydrolyzed to alpha-hydroxy carboxylic acids.  Is it the mechanism of formation of the alpha-hydroxy carboxylic acid that you are asking about?

Yes I believe so

In the cyanohydrin reaction aldehydes or ketones can react with HCN to form a cyanohydrin.  Inorganic cynanides and acid can be used in place of HCN, to generate the required cyanide anion.   The reaction proceeds by typical addition of the nucleophilic cyanide anion to the carbonyl carbon to form the tetrahedral intermediate, which can ultimately be protonated to yield the corresponding cyanohydrin.

Once formed, the nitrile group of the cyanohydrin can undergo acidic (or basic) hydrolysis by the same mechanism as any other common nitrile.  The only difference in the case of a cyanohydrin is that the "R" group contains a hydroxyl substituent on its alpha carbon.

A useful variant of cyanohydrin formation is the Strecker amino acid synthesis.  Instead of forming a cyanohydrin, an aminonitrile is formed and the nitrile is hydrolyzed to the corresponding alpha-amino acid.  A detailed mechanism can be found in the above link.


text:
Concept of "non-superimposable mirror image" in chirality
I didn't understand the fact that chiral objects are non-superimposable on their mirror images. I mean, if you put your right hand (apparently an achiral object) in front of a mirror, the mirror image will be like your left hand, but, when you make your hand touch the mirror, isn't it superimposable on its mirror image (and hence achiral)? Doesn't that mean that everything is achiral?

The mirror image of your hand does not superimpose on your hand. If the palm of your hand is facing away from you, the mirror image palm will be facing towards you. If you rotate the mirror image so that the palm faces away from you, it's thumb will be on the opposite side. Thus your hands are chiral.

if you put your right hand (apparently an achiral object)

No, your right and left hand are both chiral objects, as we will see below.

in front of
  a mirror, the mirror image will be like your left hand,

Yes.

but, when you
  make your hand touch the mirror, isn't it superimposable on its mirror
  image

No, that mirror image is not superimposable.  Just because the bottoms of your right hand and left hand can overlay one another does not make them superimposable. If you could take that 3-dimensional  mirror image out of the mirror and place it next to your hands and compare it to your right and left hand, it would look like your left hand.  And, just like your real left hand, you couldn't pick the mirror image up and place it next to your right hand, look at them both and say, "they are the same" because, for one thing, the thumbs are on opposite sides.  Therefore, they are non-superimposable mirror images.

Would our hands have been achiral if they were just plane having outline as our original hands and with all fingers pointing in same direction?

Our hands would be achiral if they were 2-dimensional (planar) and could still be manipulated through 3 dimensions.  If these planar hands existed in a 2-dimensional world, they would still be chiral.

Your hands are chiral, that is why you need two different leather gloves, one that only fits your right hand, and one that only fits your left hand.  If your hands were superimposable, then you would only need one kind of glove and it would fit both hands.

I have to say that when they taught Chiral molecules to me in Chemistry, they always used confusing molecules to demonstrate. The easiest to visualise the mirror symmetry is with spirenes (spiral molecules).
Think of a spiral staircase, it can spiral clockwise, or it can spiral anticlockwise, but if you tried to put one of each in the same space (superimpose them), nobody could climb the stairs !

You're talking about molecules like CHFClBr, right? They use those because they are the simplest molecules that are chiral - not because they are the molecules for which the chirality is most visualizable.

I think @RogUpton's point is that it's very helpful to also have a visualizable example, not just a simple one. Some students learn better by visualization and it's important to provide multiple examples.


text:
Ultraviolet acid base indicator
Is there a chemical(s) that provides good discrimination between absorbance curves for different pHs in the UV light spectrum (100-400nm?)

Why this unusual wavelength range?  Oxygen absorption cuts in at ~ 200 nm, water stops being UV-transparent at ~ 175 nm.

100 nm is tough! In terms of emission wavelength, the closest light source that comes to my mind is an argon excimer lamp (or laser) with $\lambda_{em}$ = 126 nm.

Let's ask the other way around: Which pH range do you have to cover, which other compounds (with relevant absorption in the UV/VIS) are present in your solutions? You might want to chose your indicator accordingly and perform UV absorption measurements at two different wavelength, i.e. at the absorption maxima of the protonated and unprotonated form of your indicator.

My motivation is curiosity. Universal indicator is a good broad range indicator in visible light but the discrimination between colours at pH 7 is quite poor. The light range doesn't have to be 100-400nm, it could be 200-400 or 700nm+ (infrared) it just needs to be good at highlighting differences in pH. Typically we use visible light indicators because we can see them but if you have a spectrophotometer that does UV, why not use it?

There are some good examples in Multiwavelength Spectrophotometric Determination of Acid Dissociation Constants Pharaceutical Research Volume 16, Issue 3, pp 374-381.
Benzonic acid (pka = 3.98), phenol (pka 9.73), phthalic acid (pkas 2.70 and 4.86), nicotinic acid (pKas 2.1 and 4.63) and p-aminosalicylic acid (pKas 1.79 and 3.58) would be good in the ~220nm-300nm range.

Red cabbage indicator absorbs the uv and visible radiations from520nm to600 nm range.For every pH the colours as well as the absorptions are also different. It shows deep red in pH 1,light red in 2,pink in 3,light pink in 4,5,6,purple in7,light purple in 8,light violet in 9,violet in 10,blue in 11,green in 12,yellow in 13 and  deep yellow in 14.

A bunch of links and references would be nice. Unreferenced material may be challenged and removed.

The main problem with this answer is "uv...radiations from 520nm to 600 nm"


text:
Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that?
I know the history of the discovery of the rare earth elements, that they were discovered near the village Ytterby in Sweden in the 19th century. Four elements are directly named after that village: erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb).
Why are they named like this? It makes sense to name one element after the locality, but why name the rest as misspellings of that village? It's not like we have ermanium, germium and ermium in addition to germanium (although we do have holmium).
This site mentions this:

In 1842, Swedish chemist Carl Gustaf Mosander separated the "yttria" component found in the mineral gadolinite into three segments that he called yttria, erbia and terbia.

Is the any record as to why Mosander chose these names? Because...

As might be expected considering the similarities between their names and properties, scientists soon confused erbia and terbia.

Asimov's guide to science has a nice section on this [a book with several names, see http://en.wikipedia.org/wiki/The_Intelligent_Man%27s_Guide_to_Science ]

The links provided below are to the website "Elementymology & Elements
  Multidict".  It is a fascinating site with information on the
  discovery, naming and various other facts for each element. As the author (Peter van der Krogt) says, "I am not a chemist, but a (map) historian much interested in the origin of names". Enjoy!

A nice piece on the complex etymology for Erbium, Terbium and Yttrium can be found here.  In brief the article says, "All three names were derived from the Ytterby quarry where the gadolinite was originally found in 1787. It is also said that since the original earth was divided into three new earths, Mosander split the name of Ytterby in three parts: ytt, terb, and erb."
As to ytterbium's etymology see here.   Again the history is complex with the name changing as recently as 1925. Another chemist isolated this element and again, wanted to pay homage to the geographic point of discovery, the Ytterby quarry.

Found a source here with some of the history and explanation,

Named after Ytterby, a village in Sweden near Vauxholm. Yttria – earth containing yttrium – was discovered by Gadolin in 1794. Ytterby is the site of a quarry which yielded many unusual minerals containing rare earths and other elements. This small town, near Stockholm, bears the honor of giving names to erbium, terbium, and ytterbium as well as yttrium.
In 1843 Mosander showed that yttria could be resolved into the oxides (or earths) of three elements. The name yttria was reserved for the most basic one; the others were named erbia and terbia.

Another source describes that the two names were derived from the village name, but i don't quite see the relation.

Given that he separated the two new earths from yttria, Mosander derived the names of terbia and erbia from the name of the village Ytterby


text:
How to draw glucose using a Fischer projection?
I asked how to draw glucose as a Fischer projection from its given chair conformation but all I got was an unsatisfactory ad hoc explanation involving rotation of my neck muscles to somehow see where the hydroxyl groups were. How does one actually draw a Fischer projection of a sugar molecule from its chair conformation?

A Fischer projection of glucose would return about a gazillion hits on google. What exactly are you unsure about, if not found in these two simple references 1 and 2?

@long Reference 2 is an excellent description.

I find it a lot easier if you draw the ring as a Haworth projection. I'll post an answer later if no one else has.

Here is a web page that shows how to interpret the position from the structure.
The UP groups go to the left side of the backbone and the DOWN groups go to the right side.


text:
Caclulating the Solubility of Fluorite in River Water
Source: Chapter 4 of Brownlow's 1979 Geochemistry text.

Calculate the solubility of fluorite in river water with the average composition given in table 4-1.  Assume that no fluoride complexes form.

Table 4-1 lists ppm, mEq/L and moles/L values for constituents, anions and cations found in the water sample.  There is no value for $\ce{F}^-$ but there is for $\ce{Ca}^{2+}$. The molarity of $\ce{Ca^{2+}}$ is given to be $\pu{0.375E-3M}$
My attempt::
$$\ce{CaF2 -> Ca^{2+} + 2Fl^{-}}$$
$K_\mathrm {sp}$ of $\ce{CaF2}$= $10^{-10.5}$ (from a table in the book)
$K_\mathrm{sp} = [\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][\ce{F_{\text{added}}}]^2$
$[\ce{F_{\text{added}}}] = [2\ce{Ca_{\text{added}}}]$
$K_\mathrm{sp} = [\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][2\ce{Ca_{\text{added}}}]^2 = 4[\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][\ce{Ca_{\text{added}}}]^2$
Am I on the right track?  I end up with a polynomial to solve for.  The answer in the back of the book is $10^{-4.64}$ which I can't arrive at.
The other method I've tried is to calculate the ppm of fluorite dissolved in pure water ($\pu{15.6ppm}$) and then subtract the ppm value for the river water $\ce{Ca^{2+}}$ given in the table ($\pu{15.0ppm}$) and then convert the result ($\pu{0.6ppm}$) to $\pu{7.69e-6 moles/L}$ which doesn't equal the answer in the back of the book ($10^{-4.64}$), so I believe that I'm missing something conceptually.

The author is Arthur H. Brownlow, not Brunlow, and it could be informative to copy in the page with all the data, there is maybe something missing in the equations.

the ppm of fluorite in pure water (15.6ppm)

the ppm of fluorite in pure water is zero!

$K_\mathrm{sp}=[\ce{Ca_{original}} + \ce{Ca_{added}}][2\ce{Ca_{added}}]^2 = 4[\ce{Ca_{original}} + \ce{Ca_{added}}][\ce{Ca_{added}}]^2$

Yes, this is correct and can be solved, you say you know $[\ce{Ca_{original}}]$ is $\pu{0.375E-3M}$ and  $K_\mathrm{sp}$ is $10^{-10.5}$ so there is one equation and one unknown, just a matter of solving a cubic equation, which can be solved exactly or by approximate methods.

The third-order polynomial has one real root, which has the correct order of magnitude to be a credible solution, and the OP's approach is correct.  But just like him, I find it impossible to arrive at the same number as his book.

@AbelFriedman I agree, either the book has the wrong answer or OP is misconveying the data in table 4-1

@DavePhD - Thank you for your answer.  Regarding the first quote, what I was trying to indicate was the ppm of fluorite dissolved in pure water (with no common ions).  Would it be valid to use the approach I suggest where I subtract ppm values?

@equant oh, I understand what you mean now, but no that would not be valid.  If you take your proposed answer from that method, [Ca2+] and [F-] concentrations, they will not yield the solubility product.  Your first method which gives the cubic equation is correct.

I would start from DavePhD's equation :
$\ce{_{sp} = [Ca_{original} + Ca_{added}][2Ca_{added}]^2 = 4[Ca_{original} + Ca_{added}][Ca_{added}]^2}$ = $\pu{10^{-10.5}}$
But at a difference from him, I would admit that :  $\ce{[Ca_{original} + Ca_{added}] = 0.375×10^{−3}}$ M.
As a consequence, [$\ce{Ca_{added}}$] = $\frac{10^{-10.5}}{4~ · ~0.375~·~10^{-3}}$= $\pu{2.43 10^{-8}}$M
This is the solubility of $\ce{CaF2}$ in the river water containing some calcium ions. It is much smaller than in pure water.


text:
pH Indicators for Hoffman apparatus
I've read that bromothymol blue can be mixed in with the electrolyte in the demonstration of electrolysis using a Hoffman apparatus. The indicator clearly shows the local differences in pH for the anode and cathode. And the blue color provides good visual contrast between the fluid and the gases that evolve. But I've found this indicator tends to break down quickly into a colorless solution while operating the apparatus, and after awhile you lose the indicating effect as well. Does anyone know of a better choice of indicator to use that will last longer?

I've never used it for that particular experiment, but bromocresol purple has a similar transition range and a very obvious colour change. The trouble with using indicators here is that many are organic compounds and are subject to redox reactions near the potentials at which electrolysis of water is occurring, so they break down or polymerize if they reach the electrodes. Bromothymol blue is known to oxidize at below 1 V and deposit on the electrode. Bromocresol purple tends to polymerize, but it seems to be at a relatively high potential (>1.5 V), so it may fare better.
It may also be possible to limit the diffusion of the indicator to the electrode surfaces. Perhaps using an indicator trapped in polymer beads or a block of agarose gel.


text:
Metal-Ligand-Complex and bond directions: What defines coordinate system of the complex ion?
I have a question concerning metal ligand complexes.
In every chemistry text book I have read which treats metal ligand complexes, following situation is discussed:
When 6 ligands (in octahedral coordination) approaches a metal ion with d-orbitals, 2 of the three orbitals are energetically raised because these orbitals have their lobs in direction of the ligands.
But why is this always true?
Why does always point the lobes of the z² and x²-y² orbital towards the ligands coming from "above" and "below"? Isn't it possible, that the z-axis of orbital system of the metal is a little bit inclined with respect to the octahedral ligand system surrounding the middle metal ion?
The only thing I can think of, is that this is because of energetic reasons. That in total, it's energetically more favourable when 3 of the 5 d-orbitals are arranged, so that electron occupying these orbitals have the lowest possible energies.
But up to now, I haven't found anything discussing this question.
Thanks in advance,
Jürgen

See also: http://chemistry.stackexchange.com/questions/17284/what-are-t2g-and-eg-in-cft

This is not a duplicate question. The other question asks why does the energy of three orbitals ever decrease, because there exists repulsion. This question asks why does ligands always approach z² and x²-y² orbitals; why not z² and xy for example?

The answer lies in symmetry.  When placed in an octahedral field, the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals are directed at the surrounding ligands because they have $e_g$ symmetry.  The remaining three orbitals have $t_{2g}$ symmetry and are directed between the ligands.  This explanation can be found in an undergraduate inorganic chemistry textbook such as that by Miessler, Fischer and Tarr.  (In the 5th edition, that's section 10.2.1 - Crystal Field Theory).


text:
Does it change the compound if the element letters in its molecular formula are switched?
Does it change the compound if the element letters are switched? ex. $\ce{HO}$ or $\ce{OH}$. Are those two different compounds or are they one in the same?

Actually neither HO or OH is a compound.

In some cases it can matter. $\ce{COH}$ usually refers to an alcohol. $\ce{CHO}$ refers to an aldehyde. But since acetic acid is a carboxylic acid, it is written as $\ce{CH3COOH}$

How does the edited version answer the question. It doesn't. He was asking if there was a difference between OH and HO. $\ce{CH3COOH}$ doesn't show any difference in the use of OH and HO. Too bad you have never seen acetic acid written as AcOH.

Apart from affecting organic compounds, it also affects coordination compounds.

$\ce{Hg[Co(SCN)4]}$ is a significantly different compound from $\ce{Co[Hg(SCN)4]}$. The central metal atom and the counter-ions in both cases differ, as does the complex anion.
In fact, in the above compound itself, $\ce{Hg[Co(SCN)4]}$ differs from $\ce{Hg[Co(CNS)4]}$ or $\ce{Hg[Co(NCS)4]}$: this is called linkage isomerism. The order in which ligand atoms are written denotes the atom of attachment to the central metal atom, which happens to differ in this case.
The complex salts $\ce{[Co(NH4)6]^3+[Cr(CN)6]^3-}$ and $\ce{[Co(CN)6]^3+[Cr(NH4)6]^3-}$ - formed by swapping the two radicals - are coordination isomers, and not identical.

Around year 1830, two different compounds (with different chemical and physical properties) having almost the same formulae except from the order of the elements (in the molecule/ion structure) were found. This is when the concept of chemical isomerism has originated.
The compounds are

$\ce{AgCNO}$, silver fulminate (Leibig), and
$\ce{AgOCN}$, silver cyanate (Wöhler).

It may or may not make a difference depending on the intention to explain the structure of the compound.
The trouble is that chemical formulae (like $\ce{CH3COOH}$) are not a very good way to explain the structure of a compound. In some cases all the formula does is to tell us the total composition if the compound in terms of the rations of the elements making it up. But there are plenty of examples where several distinct compounds consist of different arrangements of the same atoms ($\ce{C2H6O}$ for example could be dimethyl ether or ethanol, two very different compounds).
Chemists don't like this ambiguity so often write such formulae with some hints as to the real connectivity of the molecules. In this case the two isomers are usually written as $\ce{CH3OCH3}$ and $\ce{CH3CH2OH}$ which, in this simple case is quite a good hint as to the actual connectivity in the molecules.
The trouble is that, for more complex molecules, writing the structure as a line of text rapidly becomes unwieldy or ambiguous and the structure needs to be explained with a picture that unambiguously shows the connectivity of the atoms in the molecule.
So to know whether switching the position of atoms on a text formula matters, you have to know the intention of the writer. For simple molecules this is usually clear but for complex ones it may rapidly become ambiguous. Chemists have many conventions that are widely understood even when they are not completely unambiguous to a novice. The best approach is to learn the conventions but, when learning, demand clear structural pictures as well. Also, read the formula in context and ask whether it is trying to convey structure as well as composition.


text:
How to calculate the rate of consumption of aluminium with hydrochloric acid?

An experiment is done to determine the rate of the following reaction:
  $$\ce{2 Al (s) + 6 HCl (aq) -> 3 H2 (g) + 2 AlCl3 (aq)}$$
The following data is collected:
  \begin{array}{rr}
\text{time}/\pu{s} & (\text{Mass of beaker + contents})/\pu{g}\\\hline
   0.0 & 270.230\\
  30.0 & 270.200\\
  60.0 & 270.170 \end{array}
Calculate the rate of consumption of $\ce{Al}$ in $\pu{g/min}$.

The answer is $\pu{0.534 g/min}$, but I don't know how to get there.

Hydrogen is generated, while aluminium metal is consumed.  The hydrogen escapes.

Since one of the products is a gas $\ce{H2}$, the decrease in system mass must reflect the escape of hydrogen. Thus, after thirty seconds, the system has lost $\pu{0.030 g}$ of hydrogen, which is $\pu{0.01488 mol}$ of $\ce{H2}$.
Thus, the rate $r$ of hydrogen production is:
$$
r (\ce{H2})
= \frac{\Delta n(\ce{H2})}{\Delta t}
= \frac{\pu{0.01488 mol}}{\pu{30 s}}
= \pu{4.96E-4 mol//s}$$
Now, how can you use this rate and the stoichiometry of the reaction to determine the rate of aluminum consumption?

 $\ce{2 Al (s) + 6 HCl (aq) -> 3 H2 (g) + 2 AlCl3 (aq)}$
 For every three moles of hydrogen produced two moles of aluminium are consumed, therefore $r(\ce{Al}) = -\frac{2}{3}r(\ce{H2}) = \pu{3.31E-4 mol//s}$.


text:
Bond Dissociation Temperatures for Hydrocarbons
I'd like to know the temperatures that various hydrocarbons such as methane, butane, and propane, etc. will dissociate into hydrogen and carbon.  Is there a reference where I can look up these temperatures?
Does pressure or partial vacuum effect these temperatures?

When hydrocarbons undergo thermal decomposition carbon-hydrogen and carbon-carbon bonds are broken.
$$\ce{R-CH3 -> R-CH2. + ~H.}$$
$$\ce{R-R' -> R. + ~R'.}$$
Here is a link to a table that shows the energies required to break these bonds (the bond dissociation energy). These reactions are unimolecular decompositions and are governed by first order kinetics.  The rate of the reaction is given as
$\text{reaction rate} = k[\ce{A}]$
where k is the rate constant and $[\ce{A}]$ represents the concentration of the hydrocarbon.  The rate constant can be described using the Arrhenius equation
$k= A e^{-E_a/(R T)}$
where $A$ is the pre-exponential factor, T the absolute temperature in kelvin, ${E_a}$ is the activation energy, and R is the universal gas constant.  In the thermal unimolecular decomposition of a hydrocarbon, $A$ will have a value around $\pu{10^{12}sec^-1}$ and the bond dissociation energy can be used for the activation energy.
The Arrhenius equation tells us that this decomposition will be occurring at all temperatures above absolute zero.  As we increase or decrease the reaction temperature the rate constant will increase or decrease respectively, but the reaction is always occurring as long as we are above absolute zero.
As to pressure, yes it will affect the reaction rate.  Pressure is directly related to concentration, the $\ce{[A]}$ term in our rate equation.  As we increase or decrease the pressure, we are increasing or decreasing the concentration of $\ce{A}$, and thereby increasing or decreasing the rate of the reaction.


text:
Reduction of acetaldehyde from rectus and sinister faces
I am asked to consider the reduction of acetaldehyde from the re and si faces. I am also asked whether the added hydrogen (labeled H* in my picture) is pro-R or pro-S in both cases. In my attached picture I replaced the added hydrogen (H*) with D to determine the absolute configuration.
If I did this correctly, addition of hydrogen from the si face results in a pro-S product with respect to the added hydrogen (H*). And addition of hydrogen from the re face results in a pro-R product with respect to the added hydrogen. Am I correct?

Re is Rhenium, Si is silicon. I think re and si might be what you mean

Just to keep it slightly more realistic from a chemistry point of view, let's consider the reduction of deuterated acetaldehyde (1-Deuterioethanal):

When we add from the si face, we get an (R) center, and if we add from the re face, it's the other way around (as it should be!) and we have (S) configuration.
Now, what if we reduce "normal" acetaldehyde? We then have to decide which of the proton substituents gets the higher priority when trying to assign pro-R or pro-S. If you decide that the added proton has higher priority (as you did in your question) then you get an equivalent of a pairwise exchange of the substituents (in comparison to the structures that I drew), and the stereocenter is exactly the opposite of what I drew in the figure above.
From a synthetic point of view it is much cheaper to use hydrogen gas for the reduction and have the isotope label already reside on the starting material.

So it appears I did it right?

@Dissenter Yes, it appears so.


text:
Are there some similarities between J-coupling and j,j-coupling?
I'm a bit confused having heard two different terms in different lecture parts:

J-coupling

There is the J-coupling (= indirect dipole dipole coupling = scalar coupling) which arises from a hyperfine interaction between the nuclei and the interaction. The J-coupling is a coupling between the two nuclear spins of two nuclei (e.g. a C-H bond). Because of the Pauli principle the two binding electrons are anti-parallel located so the nuclear spins $I_1$ and $I_2$ also have an anti-parallel nuclear spin.

j,j-coupling

There is as well the jj-coupling (= spin orbit coupling) which describes the coupling of the different $j_i$. This coupling scheme is valid for heavier atoms. (For leight atoms the ls-coupling is valid). In atoms with bigger nuclear charges, spin-orbit interactions are frequently as large as or larger than spin-spin interactions or orbit-orbit interactions.
$\mathbf J = \sum_i \mathbf j_i = \sum_i (\boldsymbol{\ell}_i + \mathbf{s}_i).$ Wikipedia
I don't see any commonalities. Why is the term "J-coupling" similar to the term "j,j-coupling"?

I don't believe there are any commonalities, and the terms have been derived independently.
The term J coupling in NMR was first proposed in 1951 by Erwin Hahn in his paper  describing the spin echo experiment, in which two frequencies are observed; one at the chemical shift (which they called delta, as was convention, even back then) as expected, and a second frequency which they labelled as frequency J. There is no explanation for why the choice of J is given. Gutowsky and McCall  who observed the same phenomenon independently did not give it a label, and the term J splitting stuck. J coupling describes the relationship between nuclear spin, the intrinsic angular momentum of atomic nuclei.
The term jj coupling relates orbital angular momentum of electrons. I'm not going to pretend to know anything about the origins of the term, nor anything about its application in physics and astronomy. However, the term J is sometimes used to represent total angular momentum, which is made up of individual components ji. It is the interaction between these individual j terms that give rise to jj coupling.
Other than being types of interactions between entities of angular momentum, the two couplings are quite unrelated, as are the terms used to describe them.


text:
Why would electrode reaction in galvanic cell stop when there's no neutrality in each cell?
I know that the salt bridge produces ions to balance out the positive and negative ions in each solution, but why won't it work without a salt bride? Would it even work at the start since there's few electrons gained and lost and few ions produced?

It would work vanishingly briefly. The separation of charge that occurs when you connect two separated half cells without a salt bridge produces an electrostatic potential difference that opposes the electrochemical potential difference. The cell reaction would proceed until the two voltages balance, and it doesn't take much charge separation to hit the 1–2 volt electrochemical potential of typical cells, so this process is very fast.

Would it be correct to say this is like shorting a battery

@Buraian The effect is actually the complete opposite; more like sawing a battery in half and trying to connect a circuit to the two halves. When you short a battery, both half cells will proceed to react as fast as they can until they are limited by something else, e.g., mass transport.

@MichaelDMDryden It wouldn't occur "vanishingly briefly" -- after all, these things can last a while!

@ManRow If you look at the last paragraph in Theory: "For this protection to work there must be an electron pathway between the anode and the metal to be protected (e.g., a wire or direct contact) and an ion pathway between both the oxidizing agent (e.g., oxygen and water or moist soil) and the anode, and the oxidizing agent and the metal to be protected, thus forming a closed circuit; therefore simply bolting a piece of active metal such as zinc to a less active metal, such as mild steel, in air (a poor ionic conductor) will not furnish any protection. "

@MichaelDMDryden and what does that have to do with my comment?? Did you see me use the word "air" anywhere at all? I was replying to your answer and refuting your claim that a "no-salt-bridge" aka single cell galvanic current would last only "vanishingly briefly". I even gave you an example too -- the sacrificial anodes, like the kinds used on ship hulls, clearly do not function merely only "vanishingly briefly"! (nor do ships often complete their trips "vanishingly briefly" for that matter either)

@MichaelDMDryden you're assuming that "no salt bridge" means two separate / electrically disconnected solutions.. This is often not the case -- "no salt bridge" can also mean a single-cell electrochemical setup with two electrodes simply sharing one and the same solution without any special/extra "membranes" or added "salt-bridges" in between. In fact, this entire industry pretty much relies on that single-cell operation principle as well...

In my answer, I am referring specifically to "separated half cells", as was implied by the mention of a condition of non-neutrality in the title. It is of course possible to create a galvanic cell without a salt bridge or membrane in a single solution (though it generally isn't a good idea if making a battery or performing an analytical experiment).

@MichaelDMDryden Ahh I see -- questions like "What if we remove component X" without also explicitly clarifying "And replace it with X", or Y, or simply "nothing at all", often seem vague to me. It's like in an electronic circuit if someone asks, "What if we remove this resistor?" -- it's not clear if they mean also "disconnect the terminals completely too" or "just short them with a plain wire instead without using any components at all"


text:
MDLCT (Connection Table) Format specification
Is there a specification for MDL Connection Table format somewhere. At some places it is mentioned that the format writes the line length as a character then the actual line characters. If I save a structure using ChemDraw, I see no such thing and the lines are separated using CarriageReturn and LineFeed characters. Can someone please point me to the correct format? Is it that the format is different while SAVING TO FILE and while COPYING TO CLIPBOARD?

Do you mean SDF? MDL uses "Connection Table" or CTfile to represent their general file format, e.g. Accelrys / MDL or Wikipedia

Or do you mean the "Connection Table" format (.ct) output by ChemDraw? This is an old, deprecated ChemDraw format.

Which is the Standard one? Isn't ChemDraw's CT same as MDLCT? Do they have their own CT format?

Here's the MDL spec, I assume it's the original one: https://www.daylight.com/meetings/mug05/Kappler/ctfile.pdf

MDL "CT" is a synonym for part of a MDL Molfile or SD file. ChemDraw has a "Connection Table (.ct)" format, which is different and deprecated.
The specification is online:
http://accelrys.com/products/informatics/cheminformatics/ctfile-formats/no-fee.php
And in any format, something saved to the clipboard should be the same as saved to a file.

I doubt if CT is the same as Molfile. I think CT is part of Molfile and Molfile format contains additional rows of information such as molecule name and description. CT (Connection Table) represents the atom and bond information in a Molfile format.

I still don't know where you're referring to this. Yes, there is a connection table that's part of all MDL formats. It's specified in the "CTfile" specifications

But nothing from MDL ever refers to a "connection table" as a separate format.

+1. Yes CT is part of Molfile format. I was referring to program specific CT formats such as ChemDraw's and then there is one in the Indigo Chemistry Toolkit API that saves CT as (SingleCharForLineLength... then the actual lines follows).

MDL's CT is different from the rest. Certainly ChemDraw's CT is marked as deprecated. Open Babel has an implementation to read it, but I doubt anyone really uses it. I haven't seen the Indigo version to know if it's different.

I know this is an old question but like to provide some detail since this does not seem to be documented officially anywhere.
The MDL connection table format used in MOL files is described in the CT File Formats document referenced in the first answer. This is simple ASCII text with hard linebreaks (LF or CR+LF, depending on platform).
But there is also a special clipboard format MDL_CT created by ISIS/Draw, Accelrys/Draw etc. on Windows. This is not simply a MOL file put as text onto the clipboard. Instead each line of the molfile data is prefixed by a byte holding the number of characters, followed by the ASCII (one byte) characters of the line. The lines are not terminated by line breaks! This is basically the good old TurboPascal string format.
In addition to a single molecule the MDL_CT clipboard format also supports reactions. In that case the data comes from a RXN file, which is converted to the clipboard format the same way a molfile is: prefix each line with a length byte, remove hard linebreaks.

Interesting that the clipboard format is different - didn't realize that, thanks!


text:
What do chemical constituent measurement units 'g%g' and 'µg g⁻¹' mean?
I'm trying to understand the following data being presented in this paper measuring the mineral content of the Cannabis sativa plant (leaves and inflorescences):

I need to interpret these data readings in terms of simple percentages.
What would the first two examples - sodium and potassium - translate to in simple percentage / ppm etc.?

I am not sure if Mathematics would be a better site for this question, (the theme/context is chemistry though the question itself specifically on mathematical units of data measurement), suggestions welcome about that!

I'm not sure I've seen the g%g, but micrograms/gram is interpreted as micrograms of the element per gram of the sample, so in parts per million.  I would lean toward g%g being a percentage, so parts per hundred.

$\pu{g\%g}$ literally means grams per cent grams or grams of mineral per $\pu{100g}$ of sample. So, in population 1 there is $\pu{2.33g}$ of $\ce{K}$ per $\pu{100g}$ of leaves. Not bad when you consider cooked spinach only has about$\pu{0.05\%}$ $\ce{K}$! Eat your greens, kids. About $\pu{2\%}$ $\ce{K}$ in plants is average across a broad spectrum of plants.
$\pu{ug g-1}$ is micrograms of mineral per gram of sample. Population 1 has 23 micrograms of $\ce{Na}$ per gram of leaves.
You should be able to convert these to percents or ppm, as needed, and check in the text as to whether this refers to dry weight of plant or not. I'm guessing with this multivariate array of mineral analyses, the origin of particular cannabis crops could be determined.


text:
Confusion with Law of Equivalence
Law of equivalence states that

One equivalent of an element combines with one equivalent of the other.

Equivalent of $\ce{Na}$ is $22.99$ and that of $\ce{O}$ is $8$. By the law of equivalence, $1$ equivalent of $\ce{Na}$ combines with $1$ equivalent of $\ce{O}$. But in $\ce{Na2O2}$, $$ 23 \times 2 ~\text{parts by weight of sodium combines with } 16 \times 2~\text{parts by weight of oxygen} \\\implies 23~\text{parts by weight of sodium combines with } 16~\text{parts by weight of oxygen}$$ which means one equivalent of sodium combines with two equivalents of oxygen which is contradiction to law of equivalence. So why does this compound not follow the law of equivalence?

The idea of a chemical equivalent relies on a compound's properties with respect to the usual reactions that occur with a few select species that tend to behave in very predictable ways.  In this case, the number of grams of sodium that would react with 8 grams of elemental oxygen, 23g Na, are one equivalent of sodium for oxidation-reduction purposes.
$\ce{Na_2O_2}$ is sodium peroxide.  It contains oxygen (normally a very well behaved element, chemically speaking) behaving in an unusual way, bonding not only with sodium but also with another oxygen atom.  The oxygen can be thought of as part of a peroxide ion, $\ce{O2^2-}$.
If you know about oxidation and reduction, then you'll remember that oxygen usually has an oxidation number of -2, but when it is in a peroxide it has an oxidation number of -1.  Because this particular compound involves the -1 state, it takes twice as much oxygen to react with a given amount of sodium (which in both compounds has an oxidation state of +1.
Finally, and perhaps more important in answering the root source of your trouble here, a scientific law is a thing that describes the behavior of the universe in specific conditions.  Outside of those conditions the law does not apply.  Even inside those conditions laws are often incorrect to some degree (though still useful approximations.)  It's not unexpected that we would find some exceptions to this law.

Sir,what is the equivalent of sodium in sodium peroxide? My book says sodium doesn't exhibit variable equivalent weight so eq.wt of $\ce{Na}$ is $23$ . But to me, it is $11.5$ . Which is right?

@user36790 One equivalent of sodium would still be 23 grams.  In this case it is oxygen that is behaving oddly.  One equivalent of oxygen in a peroxide would be 16 grams instead of 8 grams.


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How do electronegativity and lone pairs affect bond angles?
How do bond angles vary in molecules with a lone pair and central atom of different electronegativity, but in the same period so that electronegativity matters more than orbital size? Let's assume that the molecules I am comparing have the same substituents and one lone pair.
Will the molecule with the more electronegative atom in the middle have the lone pair electrons closer to the nucleus, so they take up more bonding space and the angles for the other substituents will be smaller?

see Why is the bond angle H-P-H smaller than H-N-H?

@ron: I think the OP is comparing NH3 with H2O.

@AbelFriedman  Would the arguments used in the above-referenced question\answer, especially those relating to electronegativity, change when discussing groups vs. periods?

This seems really challenging to answer, since O vs. N involves different number of valence electrons and thus bonding patterns (e.g., comparing a lone pair to a bonding pair)

The electronegativity of the central atom does not really influence bond angles in any meaningful way. The size of atoms and the orbitals used and useable are much more important.
Sometimes, students may be taught concepts such as electronegativity or ‘lone pairs require space’ to explain why $\ce{H2O}$ has a smaller bond angle ($105^\circ$) than $\ce{NH3}\ (107^\circ)$ or similar. However, careful analysis of the molecular orbitals reveals that sterics are a much more important reason than ‘lone pairs requiring more space and thus making tetrahedral $\mathrm{sp^3}$ contract’. Indeed, as soon as a lone pair exists on a central atom, there is a tendency to have the lone pair occupy an orbital with an s-contribution as high as possible and to have bonding orbitals with a maximised p-contribution. p-orbitals, having a direction, can form stronger σ bonds due to better overlap.
This is why the bond angle drops quickly when comparing second-period to third-period central elements to close to $90^\circ$ but then hardly changes down the group. On the other hand, the reduced bond angles observed from ammonia to water are mainly due to more steric stress on nitrogen — which needs to accomodate an additional hydrogen and thus needs to space them out more.
Indeed, if you compare the bond angles of different $\ce{OX2}$ compounds, the only general trend is ‘the larger the second element’s atoms, the greater the bond angle.’ See the table below (all values taken from Wikipedia).
\begin{array}{ccc}\hline
\ce{X} & \angle(\ce{X-O-X}) & \angle(\ce{X-N-X})\\ \hline
\ce{H} & 104.5^\circ & 107.8^\circ \\
\ce{F} & 103^\circ & 102.5^\circ \\
\ce{Cl} & 110.9^\circ & -/- \\
\ce{Br} & 112^\circ & -/- \\ \hline \end{array}
The drop in bond angles from hydrogen to fluorine is sufficiently explained by the much longer bond lengths of the $\ce{OF2}/\ce{NF3}$ species when compared to the corresponding hydrogen species due to fluorine using p-orbitals for bonding where hydrogen uses its 1s orbital.
While the bond lengths also increase when going from fluorine to chlorine and bromine, the angle also does. Note especially $\ce{OBr2}$ whose bond angle is larger than the tetrahedral bond angle. Any argument involving electronegativity should assume a smaller angle here since the electronegativity of the partners drops from fluorine to bromine.