text:
Will this SN1 reaction result in an equilibrium situation?
I'm trying to prove the rate law of an $S_N1$ reaction by measuring the rate of it, when I vary the concentrations of my reagents. However, I believe the reaction of the experiment I have designed will result in an equilibrium. The reaction is as follows:
$$\ce{t-BuOH + HCl -> t-BuCl + H2O}$$
I chose this reaction based on the theory outlined in this website, http://www.masterorganicchemistry.com/2015/02/27/making-alkyl-halides-from-alcohols/
Pretty simple. However, will the $S_N1$ solvolysis reaction, in the opposite direction occur and thus hinder my attempt to measure the rate of the reaction?

Thinking about the SN1 reaction generally, we know that the rate is first order, being proportional to the rate of decomposition of the C-LG bond of the reagent, with the identity/concentration of the nucleophile being completely irrelevant. Your issue, therefore, becomes one simply of the relative leaving group ability of water vs chloride.
Quantitatively, you could think about this on an energy diagram (below). For both the forward and reverse reactions, there is an intermediate carbocation (formation of this is rate limiting), once this carbocation forms there is then a fast recombination of cation with a nucleophile to form a 'stable' species. So, in essence, absolutely the reaction can go in both directions, but in practice often goes in one direction (actually microscopic reversibility dictates that this is almost always the case, but the energetics of one direction are such that reactions aren't realistically reversible).

With respect to your actual example, you might like to take a look in Clayden/Warren, Organic Chemistry. The actual example you're interested in is discussed, along with a discussion of relative leaving group/nucleophile ability.

Ref: Clayden/Warren, Organic Chemistry.

As a general side point, the website you cited isn't great. There are many errors and general mistakes. You'd be far better looking in a textbook (theres a meta post with good suggestions if you're interested).

Thank you very much for your informative and very helpful answer! I do use McMurry's Organic Chemistry textbook, but unfortunately the synthesis of alkyl halides from alcohols was not covered enough in that book. I think I shall definitely use Clayden/Warren, Organic Chemistry textbook, in future.


text:
Measuring lead content in pewter
I've bought pewter at the thrift store and I want to try casting metal. This pewter may contain toxic metals. A ready-made testing kit I've found cost €25 for each sample.
In comparison pewter of know origin (98% tin/2% antimony) costs €40 per kg,
pewter from the thrift store is €4 per kg. If possible I want to test the pewter before buying.
Is there an easy way to measure the lead content of a pewter sample?
I'm not a chemist but do have a basic understanding of chemistry.
I've calculated the density of the sample to be: $8.55~g~cm^{3} ± 0.5~g~cm^{3}$.

How did you calculate the density?

Does it look shiny like silver or dull like lead?  Also is it soft enough to bend very easily?  Pewter will not resonate much when tapped but nickel silver will.

I measured the weight and water displacement. But very roughly, that's why I gave it a large margin of error.

It bend very easy. I didn't do a ring test. But I can melt it on an electric stove.

That's ok it's mostly tin and or lead.  Not sure how best to find lead percent.  Tin has 2 forms with different densities.

@JosephHirsch I think it's SN80PB20. It does not seem to be eutectic. Thanks for your time and comments.

You are likely looking for trace levels, so I do not suggest any at home kit. At best, is it colorimetric and sensitive to interferences.
If you really want to know, find an environmental lab and ask them to run ICP-MS to determine the lead content. Atomic absorption spectroscopy would be the second best method.

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text:
Friend threw mercury in a fire, is this dangerous?
My friend broke a mercury thermometer and threw the mercury in the fire. Is this dangerous? and what should she do about the fireplace?

He is not asking for medical advice.  He is asking about the dangers of mercury and how he should treat his fireplace going forward.

This is a dangerous situation. DO NOT use the fireplace.  Get the help of professionals trained in hazardous material cleanup.

Mercury has a high vapour pressure and can be poisonous, as a heavy metal, if it get into the body , such as via under the finger nails; supposedly. The usual way to remove mercury is to cover the area with lots sulphur powder, this converts it to the sulphide which has a low vapour pressure.

Note that I generally do not use a lot of boldface in posts. Please recognise that it is required here.
Immediately get out of the room! Keep children and pregnant women away! Ventilate that room very well for a long amount of time! And never let your friend handle a mercury thermometer again!
Liquid mercury is not as dangerous as some people make it to be; a number of mercury compounds (e.g. dimethyl mercury) are much more toxic. However, that is because of the way most people would be exposed to mercury typically. If the thermometer breaks, there is a puddle of mercury on the floor and barring yourself doing anything stupid it will take a while for the mercury to fully evaporate. It’s inhaling mercury vapour that is most dangerous with respect to liquid mercury.
Mercury’s boiling point, according to Wikipedia, is some $350~\mathrm{^\circ C}$. A fire in a fireplace is typically hotter. Thus, by pouring mercury into the fire, your friend has chosen the most effective way to generate mercury vapour quickly from liquid mercury. Unless the fireplace is a closed oven, this vapour will now gladly move around the room with air currents, thus being easily breathed in and absorbed by the body.
Of all the ways to handle mercury, your friend chose the most dangerous and the most stupid!

I would start some sort of remediation immediately,  do not use the fireplace for starters. It's definitely not good.  You can look at information for mercury vapor lamps,  the fire was about the worst place to put it. Keep pregnant people away. See for example there are some CDC documents about this search for "mercury vapor exposure" and take a look.

I think dangerous is the wrong word, so no. However, your friend did an incredibly stupid thing as mercury is a neurotoxin that can cause sterility in men, so hazardous? Yes.
It has a boiling point above that which fire burns so it's already dispersed in the air and house and not much you can do beyond remove the fireplace and flue pipes. We live in the real world though and you probably can't afford a proper hazmat remediation and are better off not doing anything other than perhaps let the fire burn for a time while you are not home.
People in the 50's played with mercury and survived so it's not likely anyone will die from the mercury however mercury accumulates in the body's fat cells and is not excreted, so any exposure stays with you for life.

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What determines the high degree of selectivity in free radical brominations?
In my organic chemistry class, we were asked what determines the relatively high degree of selectivity in the reaction of a bromine radical with primary, secondary, and tertiary C–H bonds.
I said it is the overall free energy of the reaction because according to Hammond's postulate, endothermic (endergonic) reactions have transition states closely resembling the product side, making the difference between primary, secondary, and tertiary radicals more important in the transition state.
Would this logic be wrong?

Related: Why is it possible to form bromoalkanes but not iodoalkanes from free radical substitution? and What is the major product on chlorination of 2-methylbutane?

It has to do with the difference in bond strengths between primary, secondary and tertiary $\ce{C-H}$ bonds.  These two earlier answers may prove helpful: Monobromination vs. Dibromination and Peroxide effect in reagents apart from HBr and CHCl₃

Very related: Why are free radicals so reactive?

First, let’s take a look at the radical chain reaction that leads to the substitution in question:
$$\begin{align}\ce{Br^. + H-R &-> HBr + R^.}\tag{1}\\[0.4em]
\ce{R^. + Br-Br &-> R-Br + Br^.}\tag{2}\end{align}$$
Obviously, the selectivity is determined in the first step as long as no radical recombinations take place. If we break down the process in the first step into sub-processes, we get:

cleavage of the $\ce{C-H}$ bond;
formation of the $\ce{H-Br}$ bond.

Again, the second is the same no matter which radical is formed; therefore, it must be the cleavage of the $\ce{C-H}$ bond that leads to selectivity. Indeed, you can imagine radicals as being electron-deficient compounds and therefore behave similar to cations. Being similar to cations means that you can assume hyperconjugation to be a viable stabilisation mechanism for radicals; therefore, the energy lost by homolytically cleaving a tertiary $\ce{C-H}$ bond is slightly less (a less unstable product) than that lost cleaving a primary $\ce{C-H}$ bond. Exploiting this energy difference is the basis of selectivity.
This selectivity is higher if bromine is used than if chlorine is used. The reason is precisely the Hammond postulate you cited: the reaction with bromine is predicted to be more endothermic and thus the transition state more product-sided. A layman’s explanation would be that bromine can approach slowly and try out whether it feels happy about reacting with this hydrogen — think of playing Jenga and carefully testing if a brick is moveable or not. Chlorine on the other hand, is faster to react and thus has a less product-sided transition state.

There is a second factor that reduces selectivity. Assume 2-methylbutane; this compound conveniently has primary, secondary and tertiary hydrogens. Probabilities of bromine radicals to react with each of these hydrogens have been calculated; since I don’t have actual numbers on me right now, I am going to ballpark the following: assume, that a secondary hydrogen has a five times as high probability of reacting with bromine when compared to a primary one and that the tertiary hydrogen has a 15 times as high probability. Naïvely, we could assume that the reaction yields products with a ratio of $1:5:15$ — but we need to remember how many hydrogens of a given type there is.
The possible products are 1-bromo-3-methylbutane, 1-bromo-2-methylbutane (both primary attacks), 2-bromo-3-methylbutane (secondary attack) and 2-bromo-2-methylbutane (tertiary attack). For each of these, the following number of hydrogens can react to form the product in question:

1-bromo-3-methylbutane: 3
1-bromo-2-methylbutane: 6
2-bromo-3-methylbutane: 2
2-bromo-2-methylbutane: 1

Thus, to accurately calculate the ratios, we need to multiply the preference factor with the number of hydrogens, resulting in a product ratio of $3:6:10:15$ in the same order as presented — a lower fraction of tertiary bromide than presumed.

By way of comparison, see my answer to a similar question for the numbers for the chlorination of 2-methylbutane.

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What is the meaning of a decimal in a subscript in a chemical formulas?
What is the meaning of a decimal in a subscript in a chemical formulas?
I came across the fractional subscripts in a publication,[1] for example:

We introduce oxygen into a ferromagnetic metallic glass to form a $\ce{Co_{28.6}Fe_{12.4}Ta_{4.3}B_{8.7}O46}$
   magnetic semiconductor with a Curie temperature above $\pu{600 K}$. "

How do I interpret these numbers?
References:

Liu, W.; Zhang, H.; Shi, J.; Wang, Z.; Song, C.; Wang, X.; Lu, S.; Zhou, X.; Gu, L.; Louzguine-luzgin, D. V.; Chen, M.; Yao, K.; Chen, N. A room-temperature magnetic semiconductor from a ferromagnetic metallic glass. Nat. Commun. 2016, 7, 13497 DOI: 10.1038/ncomms13497.

also, from the paper "In this way, a single metal oxide Co28.6Fe12.4Ta4.3B8.7O46 (in atomic percentage) is produced "

Permeakra - you are assuming the atons are covalently bonded to eachother. It is possible to have atoms randomly dispersed within a material. A perfect example are clays which usually have non-stoichiometric subscripts in their chemical formulas.  It's not 0.5 atoms, it would be 0.5% molarity for that atom within a sample.

@EframGoldberg and for that matter, the atoms in the clays are by no means inserted randomly. They are located in some specific positions, even if not all of them are occupied.

@Permeakra obviously placement can't violate physical laws, but if a unit cell requires a cation, then cationic species can be inserted randomly into that position. I quote from USGS "Other Group IA or Group IIA cations may also be randomly distributed over water-molecule sites" or "In K(UO,AsO,) .3H,O, one potassium ion and three H,O molecules are distributed randomly over four water-molecule sites" In fact when modeling clay structures, random distributions are often used.

In the case of complex glasses such as $\ce{Co_{28.6}Fe_{12.4}Ta_{4.3}B_{8.7}O46}$, Instead of normalizing the subscripts to have integer subscripts they normalized them by molar percent such that the sum of subscripts is $100\%$  ($28.6+12.4+4.3+8.7+46=100$)
This is fairly common practice in materials science. For many complicated materials such as this, it's hard to gain useful meaning from the chemical formula if the subscripts are normalized to integers so they are normalized to 100. People reading this formula now can compare it to similar materials they have seen and future materials are also more comparable (e.g. if they substituted the tantalum with niobium).

Many materials, like clays or alloys are not stoichiometric meaning you would not expect whole integers in the subscripts. If for example they weighed out 1 g of material and performed elemental analysis for each element, then the subscripts you see would be the %ratio of that element.
In these materials, the atoms are not necessarily covalently bonded to eachother and this allows for the non whole number integers.

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Alternative ways to write a chemical formula and nomenclature: lead(IV) oxide, Pb(NO3)2, iron (II) oxide, hydrogen oxide

Each of the following formulas or chemical names contains an error. Correct each example.

This is the solution in the answer sheet (left side of the equal sign is the question, and the right side is the answer)

lead(III) oxide = lead (IV) oxide

Pb(NO$_3$)$_3$ =  Pb(NO$_3$)$_2$

iron oxide = iron (II) oxide

dihydrogen oxide = hydrogen oxide

To take one of those examples, I know that lead exists both as lead (II) and lead (IV). Would it be wrong to correct the first name to lead (II) oxide rather than lead (IV) oxide?
Can I make similar cases for $\ce{Pb(NO3)4}$, iron (III) oxide, and dihydrogen monoxide

Yes, your question is inherently unclear. Except for one, all of your compounds could be wrong in more ways than one. Here’s how:

Obviously, lead(III) is not a stable oxidation state of lead; these are lead(II) and lead(IV). However, there are at three different, very common lead oxides (and potentially even more):

lead(IV) oxide $\ce{PbO2}$ (black)
lead(II,IV) oxide, also called the lead spinell, $\ce{Pb3O4}$ (orange)
lead(II) oxide $\ce{PbO}$

Due to their colours, they were commonly referred to as German flag in qualitative inorganic analysis in Munich. Without additional information, it is not possible to determine which lead oxide is meant.
Only one lead nitrate is known, to the best of my knowledge; thus $\ce{Pb(NO3)4}$ would be invalid. I do not know why, though, since both nitrate and lead(IV) should be oxidising agents and thus in principle able to combine.
The same problem with lead, only worse:

iron(II) oxide $\ce{FeO}$ (black)
iron(III) oxide $\ce{Fe2O3}$ (red)
Wikipedia lists six different iron(II,III) oxides, ranging from $\ce{Fe3O4}$ (the iron spinell) to $\ce{Fe25O32}$.

Hydrogen oxide, the answer sheet’s solution, should actually be marked unclear. Dihydrogen oxide clearly refers to water $\ce{H2O}$, but water is not the only hydrogen oxide. The difference here when compared to the previous answers is that it is the oxidation state of oxygen which now varies: $\ce{H2O2}$ might be called dihydrogen dioxide. And in the same vein in which other $\ce{A2B2}$ compounds lose their numerical designators, hydrogen oxide might as well refer to the $1:1$ compound $\ce{H2O2}$. I doubt that was the answer sheet’s intention, though; since dihydrogen oxide is unambiguous.

I'd suspect lead(IV) nitrate to be prone to decomposition to PbO2 and nitrogen oxides.

@orthocresol That actually makes a hell lot of sense.

I would say dihydrogen monoxide is unambiguous; I would debate about dihydrogen oxide.


text:
Extraction of monosodium glutamate (MSG) from kelp

What is the extraction process of monosodium glutamate (MSG) from kelp?
How could one do this at home?
Is there a specific kelp species to use?

Monosodium glutamate was originally discovered by Kikunae Ikeda using a simple aqueous (hot water, namely his soup) extraction of kelp that was later refined and patented.  That particular species was Laminaria japonica. This along with a more detailed description of the original process and the history of MSG processing is given here.
The sulfuric acid treatment and precipitation process described therein certainly cannot be simply and safely done at home. The main purpose of that is to convert other compounds (e.g. glutamine) into glutamate to give higher yields of pure MSG for commercial usage. As stated in this Wikipedia article:

"MSG is freely soluble in water",

so, as Ikeda enjoyed the water extract, I would suggest trying that first.  If you need a more concentrated extract, simply let the water dry.  I wouldn't use more than mild heat, maybe an oven on very low setting with a fan, to prevent degradation of your product.
An interesting side note, stated in both sources above, is that Ikeda's discovery of MSG was also recognized as the discovery of a new flavor at that time (in addition to sweet, salty, sour and bitter).

It might be worth including an outline of the possible processes used rather than just links to their descriptions.


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Why is the iron(II) on the heme molecule depicted as having 6 electrons in its 5 d non- and anti-bonding orbitals?
Approximating the $\ce{Fe}$ complex as a square base pyramid ($D_\mathrm{4h}$ symmetry) with 5 $\ce{N}$ ligands, My electron count of the system gives me 16 electrons – 10 from the $2\times 5$ $\ce{N}$ electron pairs and 6 from the iron(II). Filling the molecular orbitals of the valence shells, I ended up with $\ce{4 e-}$ in the d non-bonding orbitals – for there to be 6 in the d orbitals, I am under the impression that there must be 18 electrons, as suggested by the 18 electron rule (which I am aware is a loose guideline).
Looking at depictions of the orbital diagram however, such as this one:

and this one:

I've seen that there are consistently 6 electrons in these orbitals — which would be consistent with iron(0) but not iron(II) as I understand it. Can anyone help me make sense of this situation?
If I'm being a little unclear about my approximation, just look at the $\ce{Fe}$ molecular orbitals on the lefthand side of the images and subtract 2 electrons

Related: http://chemistry.stackexchange.com/questions/49940/why-how-is-blood-red-colours-of-hemoglobin/49942#49942

I've seen that there are consistently 6 electrons in these orbitals — which would be consistent with iron(0) but not iron(II) as I understand it.

Iron is in the 8th group (or group VIIIb in older terminology). Thus, in its ground state it has eight electrons; two of which are in the 4s orbital. However, practically always once you oxidise a transition metal and enter coordination chemistry, the orbital energies shift relatively to each other and it becomes more stable to have all the energies in the d-orbitals. Thus, the typical configuration for iron(II) is not $[\ce{Ar}]\,\mathrm{3d^4\,4s^2}$ but $[\ce{Ar}]\,\mathrm{3d^6\,4s^0}$.

I am under the impression that there must be 18 electrons, as suggested by the 18 electron rule (which I am aware is a loose guideline).

I think, you’re not aware of the looseness of the 18-electron-rule guideline. Most of the time it does nothing except for stating a maximum that can occur. However, there are still many complexes that violate it; e.g. anything octahedral with zinc(II) (22 electrons), anything with copper(II) (21), octahedral nickel(II) (20) etc. It is much more valuable when going to neutral carbonyl complexes; e.g. iron typically forms pentacarbonyliron $\ce{[Fe(CO)5]}$ due to the 18 electron rule. But in the context of standard cationic complexes. you can safely ignore it.

Where are the 2 electrons coming from to fill the d orbitals since 2 have been lost in the transition form iron(0) to iron(II)?  Also, I don't mean that there must be 18 electrons in the sense that this is needed from a stability standpoint but rather that for there to be 6 electrons in the d orbitals is indicative of 18 electrons in the overall system.
I guess my question is, considering the orbital diagram for D4h, it seems as though with 16 electrons we should only observe 4 electrons in these d orbitals, but these diagrams depict 6.

Regarding the first part of your answer, I wrote a little bit on it before: In crystal field theory, why do all the metal electrons enter the d orbitals and not the s orbital?

@jovialplutonium Huh? Iron(0) has eight electrons in the ground state ($[\ce{Ar}],\mathrm{3d^6,4s^2}$. Two are lost which we can assume to be the s-electrons leaving the d-electrons. If the coordination sphere is pentacoordinate, there will only be 16 electrons in total.

@orthocresol--Your response pretty much sums up my confusion. We have (by my perhaps erroneous count) 16 electrons in the system--looking at your MO diagram from your link (and keeping in mind that it is describing an octahedron, not a square-base pyramid [D4h]) and simply filling the orbitals up to 16 electrons, we should have only 4 electrons in the t2g non-bonding orbital, right? But when looking at diagrams of heme, I've seen that there are 6 electrons in these orbitals--which is indicative of 18 electrons, not 16.

@Jan I misread your answer a bit, but my question still stands. If you look at the MO diagram from the link provided by orthocresol, and fill it in with 16 electrons, it seems as though we should only observe 4 electrons in the non-bonding t2g orbitals, not 6. The orbital diagram is for an octahedron but I believe that the change in geometry wouldn't necessarily change this fact.

@jovialplutonium You need to remember to remove one ligand-centred orbital (i.e. one of the lower ones) since you only have a pentacoordinated metal centre. Thus, for pentacoordinate 16-electron systems, the central metal has six electrons in its $\mathrm{t_{2g}}$ orbitals.

@Jan That makes sense--I'll have to work through that, then.
Thanks!

@jovialplutonium Yes, exactly that. If you have five ligands, you have five lower-energy ligand-based orbitals, for 10 electrons. The remaining six belong to iron.

Can either of you recommend any good resources about these kinds of diagrams for a variety of different point groups?


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Hydrogen bond of ozone
Why can't $\ce{O3}$ form hydrogen bonds with water? It has lone pairs of electrons. I do not know the answer.

It can.$\mathstrut$

Sorry I meant O2

It can as well.

Then edit question with what you really want

As far as I know, I would predict it does form H-bonds with water. Please provide an authoritative (scholarly) reference claiming it doesn't. The only information I found on the web was the claim that it decomposes slower in water than in the gaseous state. To me, this implies H-bonding is occurring.


text:
Why are condensation reactions endergonic?
My lecture notes state that

Condensation reactions, like all biosynthetic reactions, are endergonic.

I am unsure as to why this would be the case. I thought bond formation was an exothermic reaction, so you would get a negative enthalpy change from the bond being formed. I appreciate that you wold have a decrease in entropy when going from two small molecules to one larger molecule, but would this really counter-balance the negative enthalpy change every time? Perhaps the small molecule produced in the biosynthetic reaction, e.g. $\ce{H2O}$ for condensation reactions, is important here?
In a nutshell, I am just confused as to what ensures that all biosynthetic reactions (and condensation reactions) are always endergonic?

One of the most simple biosynthetic reactions that generates ‘oligomers’ from ‘monomers’ (between inverted commas because that description is not fully correct here) is the ester formation used to synthesise lipids. It can be broken down to:
$$\ce{R-O-H + H'-O'-C(=O)-R' -> R-O-C(=O)-R' + H'-O'-H}$$
With each important bond highlighted and the oxygen and hydrogen atoms made distinguishable by primes. A simplistic way of deriving standard reaction enthalpies is to use Hess’ law and add up the bond dissociation enthalpies of bonds that are broken and subtract that from the bond dissociation enthalpies of bonds that are formed. Doing that in this case gives:

Broken:

$\ce{C-O'}$
$\ce{O-H}$

Formed:

$\ce{C-O}$
$\ce{O'-H}$

Since the bonds being formed and those being broken are identical, at first approximation $\Delta_\mathrm{r} H^0 \approx 0$, meaning the reaction depends entirely on the entropic term.
On the entropic side of things, you are creating one huge, dissolved entity out of two not-as-huge dissolved entities. Making an exact call here is difficult, as it depends on a number of things, most notably how large the hydration sphere is for all three of these compounds. If each of them has a hydration sphere of approximately the same diameter, the product side will have more entropy since less water molecules are needed for an identical hydration sphere (akin to the hydrophobic effect). The opposite effect is also possible, though.
I would like to challenge the ‘all condensations are …’ It seems to simplistic to me. Many condensation reactions are probably endergonic but likewise I expect a few to be overall exergonic. The cell just takes no chances and uses an $\ce{ATP}$-dependent pathway in all cases to ensure the outcome it desires.

That all condensation reactions are endergonic is certainly not true. And as can be seen by the Helmholtz equation $\Delta G = \Delta H - T\cdot \Delta S$, this also depends on the employed temperature. One example is cupper(II)sulfate penta hydrate that spontaneously releases water upon heating, so there you have an exergonic condensation reaction. Forcing T higher will always favor a condensation reaction as typically $\Delta S_\text{condensation} > 0$ due to release of water.
That all (or most, anyway) biological condensation reactions are endergonic is hopefully the case, as otherwise biology would be unable to control the conversion by providing the energy to enable it. If there is an enzyme that converts A to B in an exergonic fashion, all of A will be converted to B eventually (one could argue that the activation barrier is high so that A to B conversion is prevented, but even when the activation barrier is too high for the process being spontaneous the cell would not be able to withdraw the enzyme so the enzyme still catalyzes the reaction in an uncontrolled fashion). But typically the cell will employ pyrophosphates (e.g. ATP) hydrolysis to drive and control those reactions. These will generally be endergonic, so that the cell can convert the endergonic process into an exergonic one by ATP hydrolysis and thereby control the process.
So, indeed my guess would be that most enzymes catalyze endergonic processes. However, there are also exergonic processes like the hydrogen peroxide hydrolysis catalyzed by the catalase enzyme. In that case, there is no need of the cell for A, so it gets converted to B as quickly as possible (and with equilibrium far on the side of B).
So my bet would be that most condensation reactions in the cell are ATP driven, and are thus endergonic.

$\text{exergonic}\ne\text{no activation energy}$. And the dehydration of copper(II) sulphate pentahydrate would not be a condensation in the strict sense. Your argument for $\Delta S$ is also not correct.

I know that exergonic $\neq$ no act en what has that to do with my answer?

Because most processes in the cell are not spontaneous because they require activation; not because they are endergonic.

Any elementary step requires activation, its only a question how much. But yeah sure its not the question if it happens, but to what degree. E.g. catalase will convert all $\ce{H_2O_2}$ to $\ce{H_2O}$ and thats fine with the cell. However if e.g. peptide formation happens spontaneously in the cell, and all amino acids would convert completely to 'the product side of the reaction' it would be a big problem for the cell. So I guess that it is a design principle of the cell to employ endergonic reactions to prevent all monomers to completely react to the condensation product.

So its not a kinetic but a thermodynamic argument...


text:
Why is the production of aramid fibers expensive?
When comparing knitted fibers for the production of clothing, aramid based fibers trade at many times the price of classics synthetic fibers per $\mathrm{m}^2$.
Despite the limited supply of aramid fibers, what are the core drivers in production costs when compared to classical synthetic fibers?


text:
How do I find the ground state term symbol for transition metal complexes?
I have a problem where I'm trying to match metal complexes with the proper ground state term symbol.  I've tried doing a Google search and looking for a process, but haven't had any luck.  I decided to try the following for complexes that are octahedral ($O_\mathrm{h}$ symmetry):

Determine the d-electron count for the oxidation state of the metal atom
Look at the corresponding Tanabe-Sugano diagram to obtain the appropriate ground state term symbol

Is that considered even somewhat correct?
The other problem I'm having is that I there are certain configurations (e.g. $\ce{Cu^2+}$ has a $\mathrm{d^9}$ configuration) for which there are no Tanabe-Sugano diagrams (at least none that I can find) and so my process doesn't work.  Can someone please point me in the right direction?  Thank you!

The purpose of Tanabe-Sugano diagrams are to help assign the transitions in the UV-vis spectrum, and hence calculate the value of $B$ (the Racah parameter) and $\Delta$ (the octahedral splitting). They are useless for $\mathrm{d^9}$ configurations; Wikipedia explains it quite well. In fact, I think that would work very well as a separate question ;)

My inorganic professor told me that, while not incorrect to use the Tanabe-Sugano diagram to determine the ground state term symbol, it was not the way he intended for the ground state term symbol of the complex to be determined.

To find the ground state term symbol, you should be using symmetry and group theory arguments, you shouldn't have to resort to searching Tanabe-Sugano diagrams to get the answer.
We'll start with octahedral complexes (the general idea can be extended quite easily to tetrahedral or square planar complexes). As requested in the question, I will only cover ground-state term symbols, but this procedure can also be extended easily to excited configurations.
In an octahedral complex, the d orbitals are split into a $\mathrm{t_{2g}}$ set and an $\mathrm{e_g}$ set. You will need the direct product table for the $O_\mathrm{h}$ point group, which you can find here (I will reproduce it here for convenience). You simply need to tack on the g subscripts on everything (since $\mathrm{g \otimes g = g}$, there will never be any ungerade states involved).
$$\begin{array}{c|ccccc} \hline
 & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\ \hline
\mathrm{A_1} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\
\mathrm{A_2} & & \mathrm{A_1} & \mathrm{E} & \mathrm{T_2} & \mathrm{T_1} \\
\mathrm{E} & & & \mathrm{A_1 + [A_2] + E} & \mathrm{T_1 + T_2} & \mathrm{T_1 + T_2} \\
\mathrm{T_1} & & & & \mathrm{A_1 + E + [T_1] + T_2} & \mathrm{A_2 + E + T_1 + T_2} \\
\mathrm{T_2} & & & & & \mathrm{A_1 + E + [T_1] + T_2} \\ \hline
\end{array}$$
I also want to make the disclaimer that there is not necessarily a one-to-one correspondence between electronic configurations and term symbols. Therefore, the configuration $(\mathrm{d}_{xy})^2(\mathrm{d}_{z^2})^1$ does not necessarily lead to one specific term symbol. A term symbol may often be a linear combination of multiple electronic configurations, and vice versa. Nevertheless, I will use the traditional "orbital diagrams" to illustrate points as necessary.

$\mathrm{d^1}: {}^2T_{\mathrm{2g}}$

The ground state is a $\mathrm{(t_{2g})^1}$ configuration, which corresponds very simply to a $T_\mathrm{2g}$ state. The spin of a single electron is $S = 1/2$, corresponding to a multiplicity of $2S + 1 = 2$.

$\text{High-spin }\mathrm{d^6}: {}^5T_{\mathrm{2g}}$

The high-spin configuration of $\mathrm{(t_{2g})^4(e_g)^2}$ can be considered to be the sum of a $\mathrm{(t_{2g})^3(e_g)^2}$ $\mathrm{d^5}$ configuration (in black), and a $\mathrm{(t_{2g})^1}$ $\mathrm{d^1}$ configuration (in blue).
The five spin-up electrons are symmetrically distributed amongst the five d-orbitals, and as a consequence, these five electrons together have a spherically symmetric electron density. You can prove this quite easily by looking up the wavefunctions of the five d-orbitals, and evaluating the function:
$$\rho(\mathrm{d^5}) = |\mathrm{d}_{xy}|^2 + |\mathrm{d}_{xz}|^2 + |\mathrm{d}_{yz}|^2 + |\mathrm{d}_{x^2-y^2}|^2 + |\mathrm{d}_{z^2}|^2$$
In fact, it is probably much easier to add up the complex spherical harmonics (add up the square modulus of the five functions with $l = 2$). The radial part $R(r)$ can be factored out (and ignored) since it only depends on the quantum numbers $n$ and $l$ (and is therefore constant for all five 3d orbitals).
Whichever way you do it, you would find that the resulting electron density $\rho$ would only be dependent on $r$, and no longer on $\theta$ and $\phi$. Such spherically symmetric spatial components transform as the totally symmetric irreducible representation, which in our case is $\mathrm{A_{1g}}$. (This is always the case, but if you need further evidence, it is provided in the tables of descent in symmetry.) Since $\mathrm{A_{1g}} \otimes \Gamma = \Gamma$ (where $\Gamma$ can be any generic irrep), we can safely discard the $\mathrm{A_{1g}}$ component, and focus on the blue component...
...which we already found in the previous section to be $\mathrm{T_{2g}}$. Therefore, the spatial component of a high-spin $\mathrm{d^6}$ configuration is also $\mathrm{T_{2g}}$.
The spin, however, is different. This time, there are four unpaired electrons, such that $S = 2$ and $2S + 1 = 5$.

$\text{Low-spin }\mathrm{d^6}: {}^1\!A_\mathrm{1g}$

In fact, there is nearly no group theory here to be discussed. This is an example of a closed-shell configuration: all electrons are paired up and there is no unequal population of degenerate orbitals. Such configurations always lead to totally symmetric term symbols, i.e. a spatial part of $A_\mathrm{1g}$. Since there are no unpaired electrons, $S = 0$ and $2S + 1 = 1$.

$\mathrm{d^9}: {}^2\!E_\mathrm{g}$

This is a good point in time to introduce what is known as the hole formalism. The idea is that a $\mathrm{d^9}$ configuration is the same as a $\mathrm{d^{10}}$ configuration, minus one electron. A fully filled d subshell is spherically symmetric, meaning that the $\mathrm{d^{10}}$ configuration has a $A_{\mathrm{1g}}$ term symbol. Therefore, if the $\mathrm{d^{9}}$ configuration transforms as $\Gamma_{(9)}$, and the "missing" $\mathrm{d^{1}}$ transforms as $\Gamma_{(1)}$, then:
$$\Gamma_{(9)} \otimes \Gamma_{(1)} = \mathrm{A_{1g}}$$
There is a result in group theory that states that the totally symmetric irrep ($\mathrm{A_{1g}}$ in our case) only occurs along the main diagonal of the direct product table. In other words, the above equation can only be fulfilled if
$$\Gamma_{(9)} = \Gamma_{(1)}$$
Therefore, our task is simply to find what the term symbol of the "missing" electron is. The missing electron is known as a hole.
In our case, in order to minimise the energy of the $\mathrm{d^9}$ electron configuration, we would want to put the hole in the higher-energy $\mathrm{e_g}$ orbital. As a result, our task becomes to find the term symbol for a $\mathrm{(e_g)^1}$ hole configuration, which is simply $E_\mathrm{g}$.
The total spin of a $\mathrm{d^9}$ configuration is easily seen to be $S = 1/2$, leading to a doublet state again.

$\mathrm{d^2}: {}^3T_{\mathrm{1g}}$

The ground state is a $\mathrm{(t_{2g})^2}$ configuration. Clearly, the green configuration is more favourable than the red one (Hund's first rule). The question then becomes how you obtain the possible term symbols for a $\mathrm{(t_{2g})^2}$ configuration, and which of them correspond to the green configuration and which to the red. The different possibilities are given by the direct product
$$\mathrm{T_{2g} \otimes T_{2g} = A_{1g} + E_g + [T_{1g}] + T_{2g}}$$
The square brackets are used to indicate spatial states that are antisymmetrised, i.e. are antisymmetric under the permutation of two electrons. Due to the spin-statistics theorem, the total electronic wavefunction, which is a product of spin and spatial parts, must be antisymmetric. To satisfy this, you have two possible choices:
$$\begin{array}{cc|c} \hline
\text{Spatial} & \text{Spin} & \text{Overall} \\ \hline
\color{green}{\text{AS } (\mathrm{T_{1g}})} & \color{green}{\text{S (“triplet")}} & \color{green}{\text{AS}} \\
\color{red}{\text{S } (\mathrm{A_{1g}, E_g, T_{2g}})} & \color{red}{\text{AS (“singlet")}} & \color{red}{\text{AS}} \\ \hline
\end{array}$$
The antisymmetric spatial wavefunction is more stable than the symmetric spatial wavefunctions, due to exchange energy. Consequently, the ground-state term symbol has a $T_{\mathrm{1g}}$ component.
As can be seen from the above table, this necessitates that the spins of the electrons are parallel, i.e. $S = 1$ and $2S + 1 = 3$. We can therefore associate the $T_\mathrm{1g}$ state with the green electronic configuration, with two spin-up electrons in different orbitals.

$\mathrm{d^3}: {}^4\!A_\mathrm{2g}$

The $\mathrm{(t_{2g})^3}$ configuration here is simply a spherically symmetric $\mathrm{d^5}$ configuration, minus an $\mathrm{(e_g)^2}$ hole configuration. To find the symmetry of the two holes, we take
$$\mathrm{E_g \otimes E_g = A_{1g} + [A_{2g}] + E_g}$$
Since the two holes are in different orbitals, their spatial component has to be antisymmetrised. Therefore, it is really the $A_\mathrm{2g}$ state that we want here.
As for the multiplicity, $S = 3/2$ gives $2S + 1 = 4$.

Now that all the ground work has been done,
we can whizz through the remaining electronic configurations - how exciting! It's a good idea to try working these out yourself before looking at the answer, to see if you truly understand the concepts above. I'm also not going to comment on the spin multiplicities any more - hopefully you can look at the diagram, calculate $S$, and work out $2S + 1$ yourself.

$\text{High-spin }\mathrm{d^4}: {}^5\!E_\mathrm{g}$

Using the hole formalism, you simply subtract one $\mathrm{e_g}$ electron from a spherically symmetric $\mathrm{d^5}$ configuration, which straightforwardly gives the term symbol above.

$\text{Low-spin }\mathrm{d^4}: {}^3T_\mathrm{1g}$

The black $\mathrm{d^3}$ configuration has symmetry $\mathrm{A_{2g}}$, and the lone blue d-electron has symmetry $\mathrm{T_{2g}}$. The direct product gives the symmetry for the low-spin $\mathrm{d^4}$ case:
$$\mathrm{A_{1g} \otimes T_{2g} = T_{1g}}$$

$\text{High-spin }\mathrm{d^5}: {}^6\!A_\mathrm{1g}$

We already established earlier that this is spherically symmetric and therefore transforms as the totally symmetric irrep $\mathrm{A_{1g}}$.

$\text{Low-spin }\mathrm{d^5}: {}^2T_\mathrm{2g}$

This is best treated as a $\mathrm{t_{2g}}$ hole in a totally symmetric low-spin $\mathrm{d^6}$ configuration, which gives rise to a $T_\mathrm{2g}$ state.

$\text{High-spin }\mathrm{d^7}: {}^4T_\mathrm{1g}$

Treating this as a sum of a totally symmetric $\mathrm{d^5}$ configuration and a $\mathrm{d^2}$ configuration, we can tell that this has the same spatial symmetry as the original $\mathrm{d^2}$ configuration, i.e. it is a $T_\mathrm{1g}$ state.

$\text{Low-spin }\mathrm{d^7}: {}^2\!E_\mathrm{g}$

Noticing that this is the sum of the totally symmetric low-spin $\mathrm{d^6}$ and one $\mathrm{e_g}$ electron makes this easy.

$\mathrm{d^8}: {}^3\!A_\mathrm{2g}$

This is, again, a sum of the totally symmetric low-spin $\mathrm{d^6}$ configuration and the $\mathrm{d^3}$ configuration, which has $\mathrm{A_{2g}}$ symmetry.
For extra practice, why not try doing this using the hole formalism instead? You should reach the same answer!

Hi there. I tried setting up the spherical harmonic equations to prove $A_1g$ is spherically symmetric, but I am still ending up with theta terms. Is this set up correctly? http://i.imgur.com/DsCagX5.png

@Blaisem No. 1) One of your equations for the spherical harmonics isn't correct, specifically $Y^0$. Where are you getting them from? I used the ones from Wikipedia and they give me the correct answer (theta-independence). Or you might have made a typo, idk. 2) You forgot to take the square modulus. So for example $|Y^{+2}|^2 = (1/16)(15/2\pi)\sin^4\theta$ is the quantity you should be adding. If you can't get the final answer, ask a new question and I'll be happy to type it up. (Ping me here also.)

$Y^0$ was a typo on the 3. I missed the square exponent, but I still am not canceling the theta terms. I've got an exam tomorrow morning so don't have time to type up the work, but I will submit a question and respond to your other helpful response to my other question during the week. Will delete this comment and ping you here again then. Thanks for your time.

No problem, good luck on your exam.


text:
Ring contraction in pinacol rearrangement of cyclohexane-1,2-diol
I came across this question in three different textbooks.
The question was: What is the major product obtained on dehydration (treatment with concentrated $\ce{H2SO4}$) of cyclohexane-1,2-diol.
I figured it would undergo a pinacole-pinacolone rearrangement. I expected the product to be cyclohexanone but to my surprise the answer was 1-cyclopentylmethanal.
What can be the driving factor for the ring to preferentially contract? Shouldn't the intermediate carbocation be stable in a six membered ring rather than a five membered ring?

Your question is essentially why the C-C bond migrates instead of the C-H bond, right? You may want to check out this answer.

@orthocresol I actually went through the answers and the links therein. However, in the cases discussed where the ring degraded, there was not an option for hydride shift as the alpha carbon containing hydroxyl group was lacking an H atom. But in my case, when there is a H shift possible, why does it not happen?

Why do you think the bond angles in a 6-member ring are $120^{\circ}$?

@Zhe A mistake on my part. I forgot that the six membered ring would be puckered and not planar. but anyways, there needs to be a driving force for the ring to degrade and I'm interested in knowing that reason.

It could be equilibrium driven. You immediately form the aldehyde after ring contraction.

Related: Pinacol pinacolone rearrangement

Background

You don't specify whether we are starting with cis- or trans-cyclohexane-1,2-diol, but it doesn't matter since both isomers will interconvert under the reaction conditions.  Therefore we will need to consider the reaction products from both isomers.
These reactions run in strong acid are never clean.  I'd bet that some cyclohexanone was also formed, but that it is the minor product.  Here is a link (see section 2. Pinacol Rearrangement, example #3) showing that 1,2-dimethylcyclohexan-1,2-diol produces >90% of the corresponding cyclopentyl analogue and around 6% of the cylcohexanone analogue.
Predominant formation of the cyclopentyl carbaldehyde indicates that our reaction must be kinetically driven. 5-membered rings have ~6 kcal/m more strain than 6-membered rings.  Therefore, if the reaction were thermodynamically controlled the 6-membered ring would predominate.
In the case of cyclohexane-1,2-diol, if a protonated hydroxyl group departed, a seconday carbocation would form.  Secondary hydrocarbon carbocations are not particularly stable, so it is likely that rather than generate a secondary carbocation there is some migrating group migration at the same time the protonated hydroxyl is departing. Such migrations have a strong preference for occurring by way of an anti-periplanar geometry.

Solution
In the figure below I've drawn the 3 isomers of cyclohexane-1,2-diol. The isomer on the left is the axial, axial trans isomer, the isomer in the middle is the equatorial, equatorial trans isomer, and then on the right is the axial, equatorial cis isomer.  I've also labeled the hydroxyls (A and B) for clarity.  In an equilibrium mixture of these isomers we would expect the equatorial, equatorial isomer to predominate because both hydroxyl groups are in the sterically less encumbered equatorial position and both hydroxyls can hydrogen bond with each other in this conformation.

Examining the isomer on the left, neither the $\ce{C-H}$ bond nor the $\ce{C-C}$ bond is aligned anti-periplanar to either hydroxyl A or B, consequently this isomer should be relatively unreactive.
Examining the middle isomer only $\ce{C-C}$ bonds are aligned anti-periplanar to the hydroxyl groups.  Rearrangement of this isomer can only produce the cyclopentyl derivative.
Examining the isomer on the right we see that a $\ce{C-H}$ bond is aligned anti-periplanar to hydroxyl A, while a $\ce{C-C}$ bond is aligned anti-periplanar to hydroxyl B. Hence, this isomer can produce both the cyclopently carbaldehyde and cyclohexanone.
We see that there are 3 pathways leading to the 5-membered ring product and just one leading to the 6-membered ring product. This suggests that the 5-membered ring product should predominate.  If we further consider that, as explained above, the middle isomer should predominate at equilibrium, then our product mixture should be even further enriched in the 5-membered ring product just as you reported.

I hadn’t seen new answers of your’s in a while which I felt was a shame. Nice to ‘see’ you’re ‘back’ (if you ever ‘left’ which I can’t verify). =)

@Jan  Hi Jan...There just don't seem to be as many interesting organic questions these days. Then when one does come along, few people seem really interested in the question or answer, so I've kinda become more of an observer.  I enjoy your answers, you seem to have a passion for chemistry.

a really good explanation Ron. I never considered the cis-trans isomers.

@ron two hydroxyl groups in diequitorial position have converted to axial equatorial positions.Is the interconversion possible ?

@ChakravarthyKalyan Yes, as noted in my first bullet in the "Background" section of my answer, the cis and trans isomers interconvert under the conditions of the pinacol reaction.

Absolutely magnificent!;-)

This is perhaps one of the most beautiful explanations in organic chemistry I have seen recently...

@ron You are absolutely a magnificent explainer!!!!

@ron: Do you really want an equilibrium arrow between the cis and trans diols? And yes, I enjoy your answers and I also bemoan the dearth of good organic questions.

@user55119  Yes, protonation and loss of water from the diol is likely reversible.  It surely depends on the specific substituents and reaction conditions, but generalizing, it seems that all steps are potentially reversible.  See example 3 in the "Pinacol Rearrangement" section where it is claimed that the cis and trans isomers of the dimethyl analogue interconvert.

@ron: At some acid strength it may occur but certainly more difficult than in the dimethyl (tertiary) analog. I'm happy if you are happy. ;)

@user55119 Yes, I agree.


text:
How to calculate atmospheric composition
Okay, so I'm a novice in chemistry and am trying to figure something out.
Say that you want to turn carbon dioxide into oxygen using photosynthesis, given an infinite supply of water. Say that the atmosphere's volume is 100km^3 and has a pressure of 5kPa. The current composition of the atmosphere is 90% Carbon Dioxide, 5% Argon, 5% Nitrogen. You want to get the Oxygen in the atmosphere to the point that it's at 20% of the atmospheric composition. How would you calculate the atmospheric composition after you do that?
A step by step guide with just formulas would be fine, you don't necessarily have to solve this exact problem; I just need to figure out how you go about solving it. I'm only asking because when I tried doing the math and putting it back into a percent via # mols or their mass, I thought that in the end it should still have 5% Argon and 5% Nitrogen in the atmospheric composition since the chemical equation is only using the Carbon Dioxide in the air to make Oxygen.
Keep in mind when explaining it to me that I've only just completed my first semester of college chemistry, so make sure you tell me what each variable is!

As stated by @iad22agp, there is some important missing information to this very interesting, and potentially someday very relevant, idea.  What is the proximity to a sun, other energy sources, is the planet geologically active, what are the dominant minerals, and very importantly does it have a magnetic field.  Mars for example has bad-news answers to most of these questions.  I'm not bringing up these things to shoot you down but to encourage you to give these complications some consideration, which will also help others to come up with more plausible scenarios to answer your question.

A really interesting problem, but I don't think you have enough information to solve it. At first glance, it seems like you would get one oxygen molecule in the atmosphere for every $\ce{CO2}$ molecule consumed. If we assume we are only making carbohydrates, then the process would be $\ce{CO2 + H2O -> [CH2O] + O2}$. That would give a $70:20:5:5$ ratio of $\ce{CO2:O2:N2:Ar}$. However, plants will do more than just produce carbohydrates into their structures. They also produce fats, proteins, etc. Some of the $\ce{N2}$ would become incorporated. In addition, $\ce{CO2}$ is appreciably soluble in water (more so at lower temperatures). I think both factors would reduce the $\ce{CO2}$ and $\ce{N2}$ levels in the final mix.

\ce{} command embraced with $ is very convenient for LaTeX chemical formatting


text:
Any way to guess what these crystals are?
Is there any way to guess from these crystals what minerals formed them?  They came from ordinary tap water.
I was cleaning out a room humidifier which hasn't been used since last year when I noticed lots of little crystals in the bottom of it.  Most of them are square, which would suggest halite (NaCl) but some of them appear to have a pyramidal, or inverted pyramidal, shape.  This picture was taken with a cell phone and the largest crystals in the picture are probably 1/8".    Soaking them in warm water for an hour dissolved them.  The color - a pale brownish-yellow in the picture is accurate.

Sodium chloride is a possible answer, it forms crystals with such a shape in some conditions, I've seen it personally (the crystals are actually cubic, it's the way they grow that gives them this form). The color may come from some impurities. However, I have no idea how you can have so much solid residue in your tap water.

The key here is the source - a humidifier. This device is used to evaporate water, and is regularly refilled (and in the case of the one I own), and less regularly cleaned. This build up could be over many refills of the humidifier (and thus many liters of water).

@Ben Norris - exactly -  these were the result of last season's multiple refillings -  so maybe 120-150 refillings without a cleaning.  I'm unclear on the process oof the crystal formation since where they form is mostly underwater during the season, but because of the evaporation the water has a very high concentration of whatever this is so I assume it precipitates out -  but why it doesn't dissolve again at the next refilling surprises me.

You are right that multiple refill-evaporation cycles cause the building-up of the solid residue. But I would suspect a limescale-like residue, not something water soluble. I guess your tap water have an unusual mineral composition.


text:
Burning different isotope combinations of hydrogen and oxygen, how much does the energy vary?
In this comment I suggested that there would be a (very) small difference in energy released from burning a stoichiometric mixture of hydrogen and oxygen, depending on the isotopes, due to a slight difference in bond energy. The differences can be simply related to different masses of the nuclei, or to other quantum mechanical effects due to the large differences in nuclear spin and moments.
I'd like to repeat the calculation in this answer just to see the sensitivity to the isotopic shifts.
Approximately how large are these small differences in bond energies? Are the differences mostly due to the mass difference or due to the other nuclear properties?
Isotopes in question: $\ce{{}^1H, {}^2H, {}^3H$,  ${}^{16}O$, ${}^{17}O$, ${}^{18}O}$
Either the bond energies themselves, or just a link to a source where I can look it up myself would be fine.

In the Born-Oppenheimer Approximation the nuclei are infinitely heavy point particles that move in the effective field of the electrons. Since the mass is considered to be infinitely heavy, within the BO approximation there is no change in the potential energy as a function of the internuclear distance. Effects of the finite nuclear mass can be incorporated and are referred to as adiabatic corrections to the BO energies. These corrections are very small (the difference in the adiabatic corrections for H$_2$ and D$_2$ is 3 cm$^{-1}$ or 0.0085 kcal/mol [Pachucki and Komasa, J. Chem. Phys. 141, 223103 (2014)]). Finite nuclear size effects are much smaller than that and can only be observed in high-precision spectroscopy of the simplest molecules.
A more important effect of the different mass for the different isotopes is the effect on the vibrational energy structure. When solving the Schrodinger Equation for the nuclear potential, one obtains the rovibrational energies (and wave functions). As you probably know, the lowest energy solution is not located at the zero of the potential energy curve, but has an offset which is typically referred to as the zero point energy. The zero point energy is roughly equal to halve of the harmonic frequency of the potential and since the harmonic frequency scales as $\omega_e\sim \sqrt{\frac{1}{\mu}}$, with $\mu$ the reduced mass of the system, the dissociation energy (i.e. bond energy) with respect to the ground state of the potential becomes larger with increasing $\mu$ as the ZPE becomes smaller.
For H$_2$, $\omega_e=4401$ cm$^{-1}$ using the scaling with $\mu$, we find for D$_2$ a value for $\omega_e(\text{D}_2)=\omega_e(\text{H}_2)/\sqrt(2)=3112$ cm$^{-1}$ and thus a difference of 1289 cm$^{-1}$. The difference in the dissociation energy $D_0$ is half of the difference in the harmonic frequency, so $\Delta D_0=644.5$ cm$^{-1}$. In good agreement with the experimental result of 630 cm$^{-1}$ (Liu et al. J. Chem. Phys. 130, 174306 (2009) and Sprecher et al., J. Chem. Phys. 132, 154301 (2010)).
When looking at the energy released in a reaction, you should also consider the change in $D_0$ for the products of course.

Thank you for the very helpful answer! I hadn't even through of  the vibration/rotation energies. Thanks also for the references. At say 20C the molecules with have some distribution of excited (rotational/vibrational) states. In addition to the shift in zero-point energy, the level spacing may also shift in some cases at least. But that doesn't actually matter so much because that would just slightly shift the population fractions so that the average energy would still be $k_B T$?

Yes, the different mass also reduces the rotational spacing and therefore the peak of the distribution might be at a different rotational level, but overall it will still be a Boltzmann distribution.

Let me add to Paul's excellent answer with a simple thermodynamic calculation.

Hydrogen oxidation.
$$\ce{H2 + 0.5 O2 -> H2O(g)}$$
This reaction has a standard free energy change of about -229 kJ/mol.  Using $\Delta G^\circ = -RT \ln K$, I get the equilibrium constant for this reaction of $K=1.2\times10^{40}$.
Deuterium isotope exchange between water and hydrogen.
A 1976 paper gives the value for this reaction.
$$\ce{H2O(g) + HD -> HDO(g) + H2}$$
The value given is $K = 3.5$, which by the standards of isotope exchange reactions is very large. Using $\Delta G^\circ = - RT \ln K$, we can calculate that the standard free energy change of this reaction is $\Delta G ^\circ =$ -3.1 kJ/mol.
Add the above reactions.
$$\ce{HD + 0.5 O2 -> HDO(g)}$$
We can add the free energies of the reactions to get that $\Delta G = $-231 kJ / mol.

Thus, burning $\ce{HD}$ gas provides about 1.4% more free energy than burning $\ce{H2}$ gas. And this is by far the largest difference that we would observe for any common stable isotope.  For nearly any isotopic exchange reaction that involves other isotopes, such as $\ce{^{13}C}$ or $\ce{^{18}O}$, the relevant equilibrium constants will be $0.9 \lt K \lt 1.1$ or so.  Only deuterium vs. protium (i.e. $\ce{D}$ or $\ce{^2H}$ vs $\ce{^1H}$) would often have values outside that range.  Thus for isotopes like $\ce{^{13}C}$ or $\ce{^{18}O}$, the difference will be closer to 0.1%, or, as isotopic geochemists like to say, 1 ‰ (i.e. one per mille).

That you for working the problem through. I'm really surprised the difference is so large, but it makes sense it would be mostly from the large H/D mass difference. Does this have implications for the question I've linked to? Would exposure of  typical water to HD gas lead to an exothermic 'reaction' or 'exchange'? - not sure what to call it. if there's a relevant answer maybe post there.

This calculation here is a thermodynamic calculation, but the question you are asking now is about kinetics.  "Is an exchange reaction energetically favorable" is different than "will an exchange reaction occur".  Similarly, when you visit the Grand Canyon, "is me falling all the way to the bottom energetically favorable" has a (hopefully) different answer than "will I spontaneously fall all the way to the bottom".

Thank's for the visuals. I'll ask a new question once my vertigo subsides. :)


text:
Why refine tar sands oil?
Why refine tar sands oil into petroleum and the rest of the oil processing byproducts rather than burning it directly in the furnaces of private residencies? At least in the snowbelt areas where gas is not readily available.

Tar sand oil is extremely heavy and would produce a lot of soot. It also contains a lot of sulfur, so the smoke would contain a lot of sulfur dioxide. This means, that burning it would require exhaust purifications. This in turn would require costly equipment that is not practical for individual household.
Furthermore, heavier fractions of oils have hard time burning. They have to be heated significantly before a flame can form.
Consequently, tar oils must be refined to be of practical use. This includes three key processes:

hydrocracking, which breaks down larger molecules into smaller ones, that produce less soot and burn easier.
hydrothreatment, to remove sulfur, metals (yes, heavier oil fractions contain measurable amount of metals, vanadium and nickel in particular, that are known cancerogenes)
distillation, separating lighter fractions from the heavier ones. Typically fractions C5-C20 are used for internal combustion engines, some С21+ are usable in industrial-grade furnaces, and the heaviest and most troublesome fractions are used as asphalt.

So, propane or propane-butane mixture is a lot more practical for small furnaces. It is very easy to refine, produce no smoke and contains no sulfur, so burning it is pretty easy. Furthermore, in can be liquefied under pressure, but it is still a gas at most common conditions. So equipment to burn it is extremely simple and exhausts are virtually harmless in most cases. Propane is reasonably cheap in most setups to use.

ok, thanks. I further distilled the question here http://chemistry.stackexchange.com/questions/64476/whats-the-smallest-practical-application-for-a-tar-sand-burning-furnace


text:
How do you determine the hybridisation state of a coordinate complex?
If an exercise asks for the hybridisation state of a coordinate complex (ion in solution) consisting of a central cobalt atom surrounded by $\ce{NH_3}$ ligands, and neither coordination number or complex ion is given, how do you decide which configuration will occur?
The answer is that in a solution this will form a $\ce{[Co(NH_3)_6]^{3+}}$ complex ion. I know that this means the hybridisation is $\mathrm{d^2sp^3}$ but can't figure out how to get there since cobalt is found in different oxidation states.

Again, I feel a bit like a broken record. You should not use hybridisation to describe transition metal complexes.
Instead, you should discuss transition metal complexes using molecular orbital theory; a simplified version of it such as the crystal field theory may also suffice for simple complexes. Below, I have attached the orbital scheme of an octahedral coordination complex of a random transition metal not including π effects.

Figure 1: molecular orbital scheme of an octahedral $\ce{[ML6]}$ complex. Originally taken from Professor Klüfers’ coordination chemistry course’s web scriptum (in German).
Forcing hybridisation onto this system is very bad and very wrong; look at the huge energy differences between the 3d, 4s and 4p orbitals. Hybridisation would also lead us to assume six equal bonds, which is not the case: the six $\ce{M\bond{->}L}$ bonds transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ which leads to three different energies. And finally, hybridisation is not able to explain crystal field split, and why $\mathrm{e_g}$ — the orbitals that should take part in hybridisation and thus form bonds — are raised in energy (they are $\mathrm{e_g^*}$ in MO terms).
In most cases, coordination complexes would default to octahedral or derived. The main catch case is low-spin $\mathrm{d^8}$ in which case square-planar should be assumed. This is relatively independent of the type and number of ligands and most exceptions are due to sterics. Thus, octahedral complex is a safe first guess. Note that the only difference between two complexes of the same metal and ligands but with different metal oxidation states is typically whether a Jahn-Teller distortion is relevant and how strong it is.

You'd perhaps be surprised to know that in Indian curriculum we are supposed to use hybridization even in CFT. Like in hexaaquacobalt(III), after splitting into t2g and eg, cobalt's six electrons fill the t2g and eg is empty. We are then supposed to use the two empty d orbitals of eg, plus 4s and 4p to give it a d2sp3 hybridization, which are then filled with electrons from aqua ligands. Sounds pretty horrible to you, I expect ;)

I wonder why it's still like this @TRC


text:
Boiling point elevation for a mixture of ethanol and water

Properties such as boiling point, freezing point and vapour pressure
of a pure solvent change when solute molecules are added to get
homogeneous solution. These are called colligative properties.
Applications of colligative properties are very useful in day-today
life. One of its examples is the use of ethylene glycol and water
mixture as anti-freezing liquid in the radiator of automobiles. A
solution M is prepared by mixing ethanol and water. The mole fraction
of ethanol in the mixture is 0.1.
Given:
Boiling point elevation constant of water =0.52 K kg/mol Boiling point
elevation constant of ethanol = 1.2 k kg/mol
Boiling point of water =373 K
Boiling point of ethanol =351.5 K
The boiling point of this solution is?

Now,
I am aware that for a solution of a non-volatile solute in a solvent
The elevation in boiling point is given as
$\Delta\mathrm{T}=im\mathrm{K}_{b}.$
In this case,which is the solute and which the solvent?
The material that I am studying from considers ethanol to be the solute and water,the solvent.And thus,calculates the elevation in boiling point of water.
Therefore,the new boiling point is higher than that of pure water
My problem with this is,since the boiling point of a mixture of ethanol and water,for all compositions,lies between that of water(maximum) and ethanol(minimum), how can a mixture of water and ethanol of 0.1 molarity (by ethanol) have a higher boiling point than that of water?

I heard someone say "non-volatile". That's the  key.

@IvanNeretin Indeed,the formula I stated appeared with regards to elevation of boiling point when a non-volatile solute is added to a solvent.
With regards to a binary mixture of volatile liquids,I did not encounter any such formula .Only ones relating to changes in vapour pressure( Raoult's law).
But , we know the boiling point does change for a liquid-liquid mixture (from b.p of lower boiling liquid to b.p of higher boiling liquid).
Is there then a law/formula which relates this change with composition?

Your best bet is to use the approximate expressions for vapor pressures of both compounds, equate their sum to atmospheric pressure, and solve for temperature.

Are you sure that is the right equation to answer the question? It only holds true for non-volatile compounds. You should be able to use Raoult's Law. Perhaps this helps http://www.chemguide.co.uk/physical/phaseeqia/idealpd.html


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How fast do electron's quantum spin flip?
In a recent lecture it was taught that carbenes with their electrons in a triplet configuration do not undergo concerted reactions as one of the electrons must first undergo a spin flip; on what timescale does this occur?

It's briefly mentioned in the last paragraph of this answer.


text:
Schweizer's reagent from/with CuO?
While attempting to make Schweizer's reagent in household conditions, using $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route (all at least 95% pure, either technical or lab grade purity), I've stumbled upon a problem. While trying to precipitate $\ce{Cu(OH)2}$, the temperature of the solutions rose to about 50 Celsius, and instead of expected blue precipitate, I got an almost black (with a hint of violet) one instead, which I assumed is $\ce{CuO}$. However, I was able to dissolve some of the precipitate in $\ce{NH4OH}$ slightly (a suspension was made from about 10 g of precipitate in 200 ml ammonia), resulting in dark blue solution with only traces of suspended black particles ($\ce{CuO}$?), appearing very like the expected Schweizer's reagent itself.
The resulting substance was able to dissolve regular cotton & cellulose samples overnight, however the dissolution was quite slow and unimpressive, with some of the sample material remaining almost intact; a chalk-coated paper was quite resistant to it (although it also dissolved somewhat eventually).
I don't have prior experience with Schweizer's reagent, so I don't know if the rate of dissolution I got here was the proper, expected behaviour here or a sign of dud.
Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?
As side questions:

was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?
how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route) in dissolving various celluloses in practice?

Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?

Well in general this is true, but because the fresh precipitate can be partially hydrated and amorphous, which does allow the ammonium hydroxide to dissolve it.

was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?

Your precipitate was likely a hydrated copper oxide. Heat causes the copper hydroxide to lose its water.  To prevent $\ce{CuO}$ from forming, you want to keep the solution cool, perhaps with an ice bath. The easiest way to verify this is by visual observation. Copper oxide is black and copper hydroxide is blue, there is no false positive/negative to this for your preparation.

how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route) in dissolving various celluloses in practice?

This is not easy to answer and depends greatly on the concentration of your reagents, temperature, pH and so on. I've seen videos on YouTube however showing about 2g of cotton balls dissolving in 500mL solution as a reference


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What are complete active space methods and how are such spaces defined for molecules?
I am new to computational chemistry and would like to know about the active space and inactive space terminologies in MCSCF type calculations.
For understanding this I have four systems:

$\ce{H2O}$
$\ce{Be}$
$\ce{Be(OH)2}$
$\ce{NO3-}$

For above atom, molecules and ion what would be the CAS space configuration (which is typically written as eight digits) separately for inactive and active space. If you could give explanation then it will be useful.

What program are you using?

@Fl.pf.  Trying with DALTON

Multiconfigurational self-consistent field (MCSCF) calculations can be further categorized. What you are referring to is a complete active space SCF (CASSCF) calculation.
In CASSCF, the 1-particle orbitals are partitioned into 3 groups, called spaces, that must be continuous and non-overlapping:

core or inactive orbitals; these must always hold two electrons
active orbitals; these are allowed to be partially occupied
virtual or external orbitals; these must always hold zero electrons

What "complete active space" means is that within the active orbitals, a full configuration interaction (FCI) calculations is performed, so electrons may occupy orbitals in any configuration as long as spin is conserved. The core/inactive orbitals are identical to occupied orbitals and the virtual/external orbitals are identical to unoccupied orbitals from Hartree-Fock (HF), by definition only; starting orbitals for any MCSCF theory don't necessarily need to be HF orbitals. You will see the notation CAS(n,m), where n is the number of active electrons and m is the number of active orbitals. Again, this is for a given total spin.
Here is a rather crude depiction of what a starting active space before optimization would look like; here, there are 8 (active) electrons in 8 (active) orbitals to give a CAS(8,8):

There is also restricted active space and generalized active space (RASSCF and GASSCF). For RASSCF, there are additional spaces, 1 below and 1 above the CAS, where electrons can only be doubly excited from the lower space to the upper space. By definition, RASSCF is more general than CASSCF. The Molcas manual has a good explanation. For GASSCF, all restrictions are lifted and an arbitrary number of active spaces can be chosen with arbitrary restrictions on excitations between them. I will defer mostly to the primary literature.
I'm not sure what is meant by "typically written as eight digits". If you're referring to the strings that are printed to the screen by programs, like 222ab0ba00a or 21102110 (those are made-up examples), they represent the orbital occupations (and potentially electron spins) of a single determinant that comprises a configuration state function (CSF). The length is entirely dependent on how large the chosen active space is in terms of number of orbitals, and all the numbers should add up to how many electrons have been chosen for the active space.
Choosing the active space for an atom or molecule is done so that the static or nondynamical correlation is qualitatively recovered in the wavefunction. In FCI, all possible electron correlation is captured within a given basis. As the excitation level is truncated from all possible excitations to all possible 1 through 6-fold excitations (CISDTQ56) to all possible single and double excitations (CISD) and so on, we make the artificial distinction between static and dynamic or dynamical correlation. Crudely put, static correlation is a result of near-degeneracies in the wavefunction, requiring a linear combination of Slater determinants for a proper qualitative representation. Dynamic correlation is a result of electrons interacting with each other instantaneously and explicitly, rather than in an average way such as the mean field SCF approaches. There are more detailed explanations here and here. Again, in the FCI limit, or even as you approach something like CISDTQ or CCSDTQ for simple systems, these are the same thing.
Because an FCI calculation is being performed within a subspace, and the computational cost of FCI in general grows factorially, this subspace is by necessity kept small. If no static correlation is present in a system, then it's rather pointless to perform any MCSCF calculations, and it's better to do a dynamical correlation calculation (perturbation theory, truncated CI, coupled cluster) on top of a single-configurational wavefunction from HF.
With regards to these specific cases,

there is no static correlation in the water molecule, so any energy lowering from MCSCF would actually be caused by dynamic correlation. Compare correlation energies (post-HF energy corrections) between a singlet CAS(2,6) calculation (just to pick an active space that will run quickly) with an MP2 calculation.
For the beryllium atom, it is well-known that the ground-state wavefunction from CASSCF should be ~90% $\mathrm{(1s)^{2}(2s)^{2}}$ and ~10% $\mathrm{(1s)^{2}(2p)^{2}}$ divided evenly over the 3 $p$-orbitals (they are degenerate in the absence of any fields), so a CAS(2,2) will qualitatively capture this effect.
My guess for $\ce{Be(OH)2}$ is that this effect is only qualitatively changed by the presence of two hydroxide groups breaking spatial symmetry, which will remove the $p$-orbital degeneracy and require more orbitals in the active space.
I am not familiar with nitrate, but looking at the Lewis structure, there are 3 resonance structures which clearly contribute equally, with 2 electrons moving between 3 bonds. Without looking at orbitals, I would try a CAS(2,3) and a CAS(2,6) to start.

I have neglected to discuss any of the subtleties of MCSCF calculations which are numerous and not always obvious. The one I most blatantly ignored above is I assume the starting orbitals are in the correct order. Unless you reorder the orbitals prior to starting MCSCF iterations, CAS(2,2) will pick the last 2 occupied orbitals and the first 2 unoccupied orbitals. These might not be the ones you want! Look at your starting orbitals! The other one I left out is technical, which is how the MO and CI coefficients are optimized during CASSCF iterations leading to final CSFs, and single-state versus state-averaged or multi-state MCSCF.
Image source: Chemgapedia

P.S. If you need help with DALTON or basically any other program, come to our chat and I can help.

Choosing the active space for an atom or molecule is done so that the static or nondynamical correlation is qualitatively recovered in the wavefunction. Could you explain it a  bit more pls.

I've edited the answer at that location, and I highly suggest looking at those two added links. Let me know if anything is still unclear.

@pentavalentcarbon that's interesting. So the Be example, does this mean that the electrons either occupy a 2s or a 2p orbital (90:10 % of the time) or are they in orbitals which are a mixture of those orbitals, and therefore higher in energy than if it would be 2s?

@DSVA If you need a physical picture, you can think of it as a mixture of orbitals, but it leads to a lowering of the energy since the use of multiple Slater determinants allows for more flexibility in the wavefunction. If it was only a single determinant, like HF/DFT, and you did only LCAO-MO with 2s/2p, that would raise the energy. The "occupation percentage" picture is better suited for dynamical correlation but still not very appropriate.

@DSVA To my understanding, the 'simultaneous occupation' picture is roughly analogous to that of traditional resonance, like the pi bond structure of benzene. There's not a transient 'flip-flop' among configurations like with, e.g., tautomerization -- the orbitals just are a simultaneous mixture of 2s/2p. (penta, correct me if I'm wrong here...)

@hBy2Py That is correct (the superposition picture). I try not to read into it too much though, it's easy to get too philosophical.


text:
Why do increasing concentrations of zinc chloride not linearly decrease the freezing point?
I did an experiment on how different concentrations of $\ce{ZnCl2}$ affect the freezing point of demi water. My results were that till a concentration of about $10\ \mathrm{mol\,dm^{-3}}$ the freezing point decreased (more or less in a linear manner). After that at concentrations of $15~\mathrm{mol\,dm^{-3}}$ the freezing point started to increase again (it again approached $0~\mathrm{^\circ C}$).
The graph below shows the freezing point depression of different $\ce{ZnCl2}$ solutions. The $x$-axis is the concentration of $\ce{ZnCl2}$ in $\mathrm{mol\,dm^{-3}}$, the $y$-axis is the freezing point depression in kelvin.

Why does the freezing point depression initially decrease, and then increase again after a certain concentration has been reached?

Removing the smooth curves and showing the actual data points might really help you one day.

You've reached the other side of the eutectic.
See, that graph of lowering freezing temperature is always a part of a bigger picture.

Once you are past that minimum, it is no longer freezing point that you are measuring. Instead, it is the point where the solution becomes saturated and $\ce{ZnCl2*6H2O}$ (or whatever hydrate is stable at those conditions) starts crystallizing.
You'd see a similar pattern with pretty much any soluble salt. Zinc chloride, if anything, is good in that it dissolves immensely well, so the graph is wide, and its features can be seen more clearly.

For the freezing point depression to keep increasing with increasing solute the $\ce{Zn^{2+}}$ and $\ce{Cl^-}$ ions need to be free of interactions with each other. In concentrated solutions this isn't true for two reasons.
First there aren't enough water molecules. Each ions actually has severance layers of water molecules in a hydration cluster.
Second the cations and anions attract in concentrated solutions. So the activity of the ions doesn't increase linearly with concentration.


text:
How can a substance have a higher boiling point but lower melting point than another substance?
Looking at 1,3-Butadiene and Isoprene, I'm wondering: why does Isoprene have a higher boiling point, but a lower melting point than 1,3-Butadiene?
I'm thinking this might be because the Methyl group on C(2) in Isoprene might cause it's lattice to be less tightly packed, thus requiring less energy to be split (-> lower melting point). As both substances can't form any ion-ion, nor dipole-dipole, nor hydrogen bonds, Isoprene probably has a higher boiling point than 1,3-Butadiene because it has a larger surface, enabling it to develop stronger London (VdW) forces.
Can anyone confirm/ disprove this?
Thanks a lot in advance!


text:
Why are there hydroxide concentrations in a solution of a strong acid?
For example HBr + H2O -> H3O+ + Br-
There is clearly no OH- here, but according to the formula [OH-] = Kc/[H+] indicates that there are OH-. Why is this?

The formula is correct; note that water contains about $10^{-7}\,\mathrm{M}$ hydroxide, but with the $\ce{HBr}$ added (effectively $\mathrm{pH}=0$) the $\ce{OH-}$ concentration is indeed reduced--now to ~$10^{-14}\,\mathrm{M}$ -- but not to zero. If you had $18$ grams of this solution (with about $6 \cdot 10^{23}$ water molecules), you would still have about $10^9$ hydroxide ions present. Not enough to do a lot of chemistry with, however.

Thanks, Zhe. My computer does not allow the browser add-in.


text:
Chemical Reaction ammonium hydroxide + aluminum chloride (3NH4OH + AlCl3 -> 3NH4Cl + Al(OH)3)
What is the color of the precipitate of Al(OH)3?

Can you clarify what your issue is? If it is as stated, please change the title of your question as it doesn't have anything to do with the actual question.

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

I'm quite sure that it's white.  Maalox is essentially an aluminum hydroxide solution and it's white.  There is nothing structurally or electronically to suggest that it would be any other color.


text:
Why is the H−C−H bond angle in сhloromethane larger than 109.5°?
Symmetrical tetrahedral molecules (like $\ce{CH4}$) have a bond angle of $109.5$. Those with lone pairs in place of one atom (like $\ce{NH3}$) have bond angles less than $109.5$.
The standard explanation for why is that the lone pairs are "larger" than the hydrogens, creating repulsive forces that push the hydrogens down, decreasing the bond angle.
However, this explanation fails when confronted with $\ce{CH3Cl}$. According to the above logic, the $\ce{Cl}$, which is much bigger than the hydrogens, should repel the hydrogens just like a lone pair, making the $\ce{H-C-H}$ bond angle smaller than $109.5$. However, the $\ce{H-C-H}$ bond angle is actually $110.5$ (see here, I could not find a more authoritative source).
Why is this (or is my source wrong)?

The reason is Bent's rule. You might want to look this up in a textbook, or alternatively read this question or a little further going question.

Chlorine is more electronegative than carbon which is further more electronegative than hydrogen and thus chlorine pulls the bond pair towards itself, polarizing the bond. This decreases the electronic repulsion between the $\ce {C-Cl}$ bond pair and the $\ce {C-H}$ bond pair, decreasing the $\ce {H-C-Cl}$ bond angle. Consequently, the $\ce {H-C-H}$ bond angle increases to 110.5°.

I was offered the standard explanation that lone pairs are larger back in high school but in the first year of uni that was overturned by a pretty simple counterexample: the heavier homologues $\ce{H2S, PH3}$. While $\ce{SiH4}$ remains tetrahedral and has a similar orbital structure to $\ce{CH4}$, the bond angles in $\ce{H2S}$ and $\ce{PH3}$ are very close to $90^\circ$ as are the heavier homologues $\ce{H2Se}$, $\ce{H2Te}$, $\ce{AsH3}$ and $\ce{SbH3}$. This provides the first clue: it is not actually the size of the lone pairs but that of the central atom that matters. If oxygen and nitrogen weren’t so small, they also would have bond angles much closer to $90^\circ$. The advantage of a $90^\circ$ bond angle is that pure p orbitals can be used to form the bond. Not only are these directional and thus point towards the bonding partner, they also have a higher energy which typically means their energy is closer to the hydrogen’s orbital’s and thus it allows for a greater stabilisation. The problem with the small sizes of oxygen and nitrogen are the coulombic interactions between the two $\ce{X-H}$ bonds which force these bonds further apart until the perfect trade-off point is reached.
When increasing the bond angle from an ideal $90^\circ$ to an ideal $109.5^\circ$, what you can imagine on a molecular level is an increased s orbital contribution from the central atom to the bonding orbitals. In turn, this means that (one of) the lone pair(s) will have a lesser s contribution which raises that lone pair’s energy.
This explanation has now served us well for $\ce{XH_n}$ molecules, where the hydrogens and lone pairs total 4, but your actual question concerned $\ce{CH2Cl2}$ where we are dealing with two different substituents on carbon. Naïvely, we might expect all four bonds to be equal but they shouldn’t be because chlorine and hydrogen have very different electronic and bonding properties. There will be two bonds to a more electron-withdrawing group and two to a less-electron withdrawing group—it’s easy to see from electronegativities that the former is chlorine and the latter is hydrogen.
Just by noticing we have two different bond energies per se (deriving from the different partner atoms), we can start to consider: what if we don’t use identical orbitals from carbon? Instead of using four $\mathrm{sp^3}$ with each having a $25~\%$ s contribution, we might be better off using two with, say, $20~\%$ and two with, say, $30~\%$ (ultimately, whatever we put in must add up to one full s and three full p). Comparing the two resulting pairs a priori, the one with a higher p contribution (lower s contribution) will be higher in energy, less close to the carbon nucleus, more directional and involve a smaller bond angle (as we learnt above in the $\ce{NH3/H2O}$ case). The other one will be the exact opposite. A more electron-withdrawing group can form stronger bonds with a higher-energy atomic hybrid orbital while the opposite is true for a less electron-withdrawing group. Thus, we assign the orbitals with a higher s contribution to the $\ce{C-H}$ bonds and the ones with a lower s contribution to the $\ce{C-Cl}$ bonds. The result is a $\ce{Cl-C-Cl}$ angle less than $109.5^\circ$ and a $\ce{H-C-H}$ angle larger than $109.5^\circ$.
This long-winding explanation has been thought up before and it has been deemed significant enough to acquire a name: Bent’s rule. In general, Bent’s rule says that electronegative partners bond best with atomic orbitals that have a lower s contribution; the angle discussion is a consequence of that. You may find a better description of Bent’s rule in this question.

Notes:

Bent’s rule is a generalisation of an observation. That’s important, because there will be cases where it works better and where it works worse and as far as I know you cannot use it to predict the numeric value of angle difference.
I have made it seem in this discussion as if a certain hybridisation is best which results in a certain geometry. Actually, it is the other way around: the atoms arrange in a certain lowest-energy geometry, we observe that and can calculate orbital hybridisations from said geometry.

The Cl atom is electronegative relative to the C atom which polarizes the electrons to Cl. This puts p character in the C-Cl bond at the C atom and shifts the HCH bond angle from sp3
109.5° toward sp2, toward 120°. Remember that if the C-Cl bond at the C atom were pure p the σ bonds would have to be coplanar and the bond angle would be 120°. Obviously the effect is not great. You can calculate just how far hybridization takes the bond angle by using formulas found here. https://en.wikipedia.org/wiki/Isovalent_hybridization.


text:
Which rate is faster: one step with Ea or two steps each with Ea/2?
I have been struggling with the following problem posed at a prior lab meeting. The answer seems trivial to me, so can you tell me if I'm doing this wrong? Thanks!
Consider the following two scenarios. They represent a small portion of the same catalytic cycle.

There is a reaction step of that cycle wherein $A\rightarrow P$ with an activation energy $E_{a}$.
Upon further inspection, the above reaction is actually $A\rightarrow I\rightarrow P$, where each of the two steps has an activation barrier of $E_{a}/2$ such that their sum is equal to the barrier height in the first scenario. Refer to the schematic here for a depiction if needed.

The question is: which scenario is faster? My thought process is as follows.
First, we need to define which rate we are talking about. I am going to assume the rate of production of product, $dP/dt$, is compared. I need information about pre-factors, but I'll have them all equal an arbitrary value of 1 for the time being. I have set $[A]_{t=0}=1$, $[P]_{t=0}=0$, and $[I]_{t=0}=0$. I now write out my ODEs and solve them. I've assumed $E_{a}=RT$ for demonstration purposes.
Case 1:

Case 2:

Comparison of Rates:

My Issue:
Can the question not be definitively answered? It seems like the answer (whether case 1 is faster than case 2) is that it depends on what conversion you're running to since the answer changes at a certain point. However, this was asked in the context of catalysis. I've posed this problem to multiple people and they all brought up the point that they don't usually think about time or conversion in this regard. They just want to know which one of these two cases will be faster when looking at two potential energy profiles. Isn't that too naive, or am I thinking about this all wrong? At full disclosure, this is mildly related to a prior question I posed in the comments located here.
Edit: To emphasize, if I change the pre-factors or the value of $E_{a}$, the order of which case is faster will also change. So, the answer to the question in the title is that "it depends".

You've answered your own question by plotting the rates and concentrations. The single step case is always faster to reach a given point in the reaction (say 50% complete) but at some points the instantaneous rate of the two step process is higher.

Yes, I agree. Thank you for the confirmation. Just to note for others that visit this post, that trend changes if I change the parameters. It's very easy to get the two-step case to be faster if you change the pre-factors or value of $E_{a}$. So, the end conclusion is that you can't a priori say which one is faster without further information.

Instead of looking at $\mathrm{d}[\ce{P}]/\mathrm{d}t$ against time $t$, a better idea might be to plot $[\ce{P}]_t$ against $t$, which you already did. By eyeballing the graphs I'm fairly sure that $[\ce{P}]_t$ in case 2 is always below $[\ce{P}]_t$ in case 1.
In org chem, people probably aren't too interested in what the absolute rate of the reaction at time $t$ is, they are probably more interested in how long the reaction will take to run. Because of the maths, the reaction will never hit 100%, so you can't use that as a benchmark, but you can use e.g. 95% complete (i.e. $[\ce{P}]_t = 0.95$) as a benchmark.

Thanks for the reply. That does seem like a more appropriate metric one would care about. You are right that Case 2 will always be better in that regard. I'm going to try this for a variety of pre-factors to see how that impacts things. Thanks.


text:
Water and Ethanol Vapor Phase diagram for specific range
I am looking for a Vapor Phase diagram for Water and Ethanol where Ethanol would be from 0 to 5%.
The diagrams I have found are for Ethanol concentrations of 0% to 100% and this doesn't allow me the accuracy for the range I have in my process.
If anyone could point me to a table of values that the 0 to 100% diagrams are generated from I can generate a diagram for the range I am looking for.

The Dortmund Data Bank might be a resource to start searching.
They do seem to have a page for vapour-liquid equilibrium data for the ethanol/water mixture.
On a first view, however, the data seem to be for less than atmospheric pressures only and are compiled from Ind. Eng. Chem., 1942, 34, 1501–1504.


text:
How do I determine stereochemistry of the product
When pure (L) lactic acid is esterified by racemic-2 butanol then what is the nature of the stereoisomers so obtained i.e. whether they are diastereomers or enantiomers.
I studied the mechanism of esterification but I didn't find anything regarding the stereo isomerism that can affect the formation of the product.Any ideas.Thanks.

You can start by identifying all stereocenters in reactants and products. You should assume the reactants are racemic so you will get all combinations in the product.

@Zhe in the question it is given that a racemic mixture of 2 butanol is used as the reactant but I do not get how it might affect the resulting stoichiometry of the product.thanks.

@aniline  You are correct, esterification will not affect the stereocenters.  So keeping in mind that the stereocenters are preserved, what products result?

@ron Yes that would mean that L configuration would remain unchanged and two products with different configuration on other chiral carbon would result in diastereomers.

@aniline  Nice going!

The only complication that is not conveyed in this question is that there may be some level of diastereoselectivity. It's probably not very high given these reagents, but for extreme cases, you might be missing one diastereomer altogether because it's so much easy to make the other one.

The reaction doesn't change the stereo configuration for either reactant, so we can assume that the product is enantiomerically pure with respect to the carbon attached to the hydroxyl group (in orange below), and a racemic mixture with respect to the tertiary carbon (in blue).

Since there are multiple stereocenters, you can either have both the same (same molecule), both different (enantiomers), or have some different and some the same (diastereomer).  The products would be diastereomers.  In particular they would be epimers, which differ in only one stereocenter.