Unit analysis for neutralization reaction What volume of $\pu{0.500 mol L-1}$ $\ce{HCl}$ (in $\pu{mL}$) is required to neutralize $\pu{2.02 g}$ of solid $\ce{NaOH}$ pellets?

So far I have tried regular unit analysis:

2.02 g NaOH * 1mol NaOH/40g NaOH * 1mol HCl/1mol NaOH * 500g HCl/1mol HCl
= 25.25 g HCl --> 25.5 mL HCl

This answer doesn't make sense in the context of this neutralization reaction. Is there a type of unit analysis that is used specifically for determining the volume required for a reaction?

There is no special unit analysis.
It's a recommended practice to solve the problem algebraically first using proper notations for physical quantities, and plug the numeral values at the end minding the units — this way you reduce the chance of making the erroneous calculations and keep track of all units; as a bonus, you simplify handling significant figures.

The majority of problem solving in chemistry begins with writing down the balanced chemical reaction:

$$\ce{HCl + NaOH → NaCl + H2O}$$

So, to find the volume of hydrochloric acid $V(\ce{HCl})$, you may use its molar concentration $c(\ce{HCl})$ and the amount $n(\ce{HCl})$:

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{c(\ce{HCl})}$$

Unknown amount $n(\ce{HCl})$ can be found from the reaction's stoichiometry:

$$n(\ce{HCl}) = n(\ce{NaOH}) = \frac{m(\ce{NaOH})}{M(\ce{NaOH})}$$

where $m(\ce{NaOH})$ and $M(\ce{NaOH})$ are mass and molecular mass of sodium hydroxide, respectively.
Putting everything together and plugging in the appropriate numerical values, you arrive at the volume of approx. $\pu{100 mL}$:

$$
\begin{align}
V(\ce{HCl}) &= \frac{m(\ce{NaOH})}{M(\ce{NaOH})\cdot c(\ce{HCl})} \\
  &= \frac{\pu{2.02 g}}{\pu{40.00 g mol-1}\cdot \pu{0.500 mol L-1}} \\
  &= \pu{0.101 L}
\end{align}
$$


Dalton's law for a gas mixture If $\pu{200 mL}$ of $\ce{N2}$ at $\ce{25 ^\circ C}$ and a pressure of $\pu{250 torr}$ are mixed with $\pu{350 mL}$ of $\ce{O2}$ at $\pu{25 ^\circ C}$ and a pressure of $\pu{300 torr}$, so that the resulting volume is $\pu{300 mL}$, what would be the final pressure in $\pu{torr}$ of the mixture at $\pu{25 ^\circ C}$?

I understand that to find final pressure, I need to add the partial pressure of $\ce{N2}$ and $\ce{O2}$. I'm not sure how to find the partial pressure using Dalton's law. To find partial pressure of $\ce{N2}$ first:

$$p(\ce{N2}) = X(\ce{N2}) \cdot \pu{0.3289 atm}$$

where $X$ is the mole fraction. How do I find it?

There is no need to use ideal gas law in $pV = nRT$ form or find molar fraction $X$. The process is isothermal ($T_1 = T_2 = \pu{25 ^\circ C}$, hence $p_1V_1 = p_2V_2$), so it's a matter of finding the sum of partial pressures of each gaseous component $i$ in the system:

$$p = \sum_i{p_{2i}} = \sum_i{\frac{p_{1i}V_{1i}}{V_{2i}}}$$

Also, converting the pressure back and forth between $\pu{torr}$ and $\pu{atm}$ is counterproductive as you are explicitly asked to provide an answer in $\pu{torr}$.

   $$p = p_2(\ce{N2}) + p_2(\ce{O2}) = \frac{\pu{250 torr} \cdot \pu{200 mL}}{\pu{300 mL}} + \frac{\pu{350 torr} \cdot \pu{350 mL}}{\pu{300 mL}} = \pu{575 torr}$$


Andersen thermostat in MD with rigid bond constraints I have been writing a custom MD code for some time. I have been using the Andersen thermostat, mainly because its ease of use, and the fact that I am not concerned with gathering dynamical data, but rather free energies. I have implemented the SHAKE algorithm found in LAMMPS. However, the thermostat seems to be undercooling now, when constraints are present. For instance, if I make the set point temperature 350 K, my system will equilibrate at around 330 K.

I am calculating the number of degrees of freedom for the temperature calculation as $3N-3-$number of bond constraints. Has anyone else had a similar experience with the Andersen thermostat? Would a Langevin thermostat work better here? Why is the Andersen thermostat not playing well with the SHAKE algorithm?

Hünenberger’s Thermostat algorithms for molecular dynamics simulations (PDF available here) is a generally concise and accurate reference on the topic. On stochastical thermostats such as Langevin, it does not use the number of degrees of freedom (i.e. reduced by the number of constraints) but actually uses the number of atoms.

One would have to follow the proper derivation of the thermostat's target ensemble (from the original Andersen paper) to make sure that is correct, but it seems likely to fix your problem of having too low kinetic energy.


Which coloring agents are responsible for the color in red tea and yellow tea? Have this question in a biochemistry course, can't seem to find it anywhere. Anyone happen to know an alkaloid or something else involved? Thanks a lot!

I believe that it is the different polyphenols (Flavonoids are a type of polyphenol) in tea that give it its colors. Black tea has thearubigins that give it a red-brown color. The yellow color in tea comes from theaflavins. The color of the tea also depends on the acidity of the water used, because thearubigins are very weak acids, and the anions they produce have strong colors. If the water you use is alkaline, the color will be stronger. If it is acidic, the color will be lighter.


Why do we only consider relative motion of hydrogenic atoms? In my lectures on quantum mechanics, it was said that the Hamiltonian for a hydrogenic atom could be split into relative motion and translational motion, as follows:

$$\hat{H} = - \frac{\hbar^2}{2m}\nabla^2_{cm} - \frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$$

and that one only need consider the relative motion:

$$\hat{H} = - \frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$$

This statement is reiterated in Atkins' Physical Chemistry (10th Edition), as well as his Molecular Quantum Mechanics (4th Edition), and various other online resources, but I am yet to find an explanation as to why.

Why can we, and why do we, neglect the translational motion of the atom?

We can do that because the variables have separated. The Hamiltonian of relative motion depends on the relative coordinates. The translational part depends on the coordinates of the center of mass. The two never mix. They don't interact; they don't even care about each other.

Separation of variables in various guises is an immensely powerful technique of solving PDE. You will encounter it time and again, including this very chapter of whatever textbook you are using, just a few lines below, when you'll start to actually solve the equation for hydrogen atom by splitting it into radial and angular parts.

So it goes.


Why does ionization increase from Li to Be? If ionization energy decreases from $\ce{N}$ to $\ce{O}$ due to the pairing of electrons (causing electric repulsion and greater potential energy) in the $\mathrm{2p}$ orbital in the $\ce{O}$ atom, then why doesn’t ionization energy decrease from $\ce{Li}$ to $\ce{Be}$ as electrons are paired in the $\mathrm{2s}$ orbital?

The s-orbitals have a much greater degree of penetration than the p-orbitals, so they experience the nuclear charge more strongly.  This electrostatic attraction is sufficiently greater than the electrostatic repulsion between the two electrons that the ionisation energy increases.

Because the p-orbitals do not penetrate through the s-orbitals to the nucleus as much there is less electrostatic attraction.  By placing a second electron in a p-orbital the repulsion is now more significant compared to the attractive force, and the ionisation energy is less than expected.

However, each new element also increases the nuclear charge, which in turn increases the attraction that each electron has to the nucleus.  When the next p-orbital becomes doubly occupied this stronger attraction once again dominates, increasing the ionisation energy.


Is trisulfuric acid hypothetical? In the preparation of sulfuric acid through contact process,

Hot sulfur trioxide passes through the heat exchanger and is dissolved in concentrated $\ce{H2SO4}$ in the absorption tower to form oleum:
$$\ce{H2SO4(l) + SO3(g) → H2S2O7(l)}$$
Oleum is reacted with water to form concentrated$\ce{H2SO4}$.
$$\ce{H2S2O7(l) + H2O(l) → 2 H2SO4(l)}$$

The sulfuric acid obtained is $\ce{96-98}$% pure. That means $\ce{2-4}$% is impure and contains byproducts and other oxoacids of sulfur including trisulfuric acid. If the sulfur trioxide to be adsorbed on $\ce{H2SO4}$ is in excess, it can form trisulfuric acid.
$$\ce{H2SO4(l) + 2SO3(g) → H2S3O10(l)}$$
But the amount of trisulfuric acid is negligible and can be omitted. But is it an actual compound? Can it be isolated? Can it exist in room temperature?
Pubchem identify it as a compound and also has a unique ID. It also has an unique CAS number.

It has $\ce{S=O}$ bonds and $\ce{S-OH}$ bonds (characteristics of oxoacid of sulfur) but still it is not mentioned in that list (link above). Why?

It seems that this compound has not yet been characterized, although traces of it could be found in mixtures of $\ce{H2SO4}$ and $\ce{SO3}$. The most recent paper I could find is this one from 2012, showing that salts of the respective anion exist and theoretical calculations imply that the acid is stable with respect to decomposition into $\ce{H2S2O7}$ and $\ce{SO3}$. The conditions under which it may be stable are however unknown.


Order of solubility of sodium halides and potassium iodide Order the following ionic compounds in decreasing order according to their ease to dissolve in water.
a) NaBr; b) KI; c) NaF; d) NaCl.

d > a > b > c
a > b > c > d
b > c > d > a
c > d > a > b

So, according to what I researched the solubility $s$ for each one of these compounds in water (at $\pu{20 °C})$ is:
$$
\begin{array}{lc}
\hline
\text{Compound} & s/\pu{g L-1}\\
\hline
\ce{NaBr} & 905 \\
\ce{KI} & 1400 \\
\ce{NaF} & 40.4 \\
\ce{NaCl} & 359 \\
\hline
\end{array}
$$
Therefore, my answer would be, from more soluble to least soluble:
b) KI > a) NaBr > d) NaCl > c) NaF
But this answer is not among the list. I think option 4 might be "correct", but it's kind of backwards.
Are the given answers wrong or am I doing something wrong?

The order b-a-d-c is correct, if the solubility is in gramm per liter. If it is in mole per liter, the order is a>b>c>d, as the saturated solutions of NaBr is the highest ($8.78~ mol L^{-1}$), then KI ($8.48~ mol L^{-1}$), then NaCl ($6.19~ mol L^{-1}$), and NaF ($0.95~ mol L^{-1}$)


Carbocation Rearrangement - Till which neighbouring carbon is it possible? In the above reaction, clearly the OH will be protonated followed H2O leaving, seeing that it's a good leaving group.

Now there can be two possibilities either a alkene is formed as is, or the carbocation rearranges and ends up opening the ring, thus increasing the final stability of molecule, followed by alkene formation. (5 carbon ring is more stable)

But as per some texts, the Carbocation only rearranges until the neighbouring carbon and then alkene is formed without ring opening. Is that correct, if yes, why?

Consider that in the situation you present, it would be much easier to lose the proton to form the alkene immediately, than to migrate the positive charge until a position where ring-opening is viable.

Not to say it is impossible, but the required 'steps' to get to that point are very improbable to the point that it doesn't happen.


Doubt on process to calculate pKa I have to solve an exercise that I've been trying for a long time without success.

Information that I have:

  pH @ certain temperature of an aqueous dissolution of an acid
  molar concentration of the acid

Unknown:

  approximate $\mathrm{p}K_\mathrm{a}$ (is not meant to be calculated with calculator I suppose that is the reason they say "approximate")

So far I tried to express the chemical equation and its two steps to finally get $\ce{H3O+}$ and an oxid.
I've calculated backwards the Hydronium concentration $[\ce{H3O+}]$ with the $\mathrm{pH}$ formula. $$ \mathrm{pH} = -\log([\ce{H3O+}])$$.
I've written down the formula to calculate $K_\mathrm{a}$ from where I expected to be able to calculate $\mathrm{p}K_\mathrm{a}$.

I don't know how to extract now the concentrations of every component on the $K_\mathrm{a}$ formula in order to be able to calculate that. Moreover, what am I suppose to do with the temperature?

I hope you can give me guidance in the process to solve such exercises.

Say your acid is monobasic: suppose its formula is $\ce{HA}$.
So, it dissociates according to following equation:
$$\ce{HA + H2O <=> H3O+ + A-}$$
Suppose its molar concentration (given) is $c$, and it dissociate $\alpha$ amount in water (assume total volume is $\pu{1.0 L}$) at given temperature at equilibrium. Thus, equilibrium concentrations of $\ce{HA, H3O+}$, and $\ce{A-}$ are $(c-\alpha)$, $\alpha$, and $\alpha$, respectively.

$$\therefore \; K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}=\frac{\alpha \cdot \alpha}{c-\alpha}$$

Since $\alpha = [\ce{H3O+}]$, $-\log \alpha = \mathrm{pH}$ (which is given) and you know $c$, which is also given, you can estimate the $\mathrm{p}K_\mathrm{a}$ (and hence $K_\mathrm{a}$ if you needed).

Late edition:

Take (-$\log$) on both side of above equation. Thus:

$$\mathrm{p}K_\mathrm{a} = -2\log \alpha + \log{(c-\alpha)}=2(\mathrm{pH})+ \log{(c-\alpha)}$$
Most of the time (in cases of $c\gt\gt\alpha$),  $\mathrm{p}K_\mathrm{a} \approx 2(\mathrm{pH})$


Why does LiAlH4 reduce the double bond of cinnamaldehyde, whereas NaBH4 does not? I read an answer here about why $\ce{LiAlH_4}$ does not reduce conjugated double bonds, and that $\ce{NaBH_4}$ does. But in the case of cinnamaldehyde, $\ce{LiAlH_4}$ reduces the conjugated double bond and $\ce{NaBH_4}$ does not. Why is it so?

Citation for $\ce{LiAlH_4}$: Quora

Citation for $\ce{NaBH_4}$:

Image Source

Cinnamaldehyde (1) does not undergo 1,4-reduction but rather 1,2-reduction with LiAlH4. The product of the reduction is 3-phenylpropan-1-ol (2). If conjugate addition were to occur, enolate 3 would be produced (stable to the reductant; enolates can be prepared to protect ketones from this reductant) and aldehyde 4 would be expected, as long as the aqueous workup were to destroy the reductant before any enolate 3 could be protonated.

All of this aside, the answer was provided experimentally in 1954.1 Reduction occurs 1,2 to afford the aluminate salt of cinnamyl alcohol 5. This species undergoes hydroalumination of the double bond by a presumed intramolecular process to give alkyl aluminate 6. This cyclic aluminate was deuterated (D2O) at the benzylic position to provide 9. In addition, trapping of 6 with benzophenone afforded diol 7 in >80% yield! Reduction of cinnamaldehyde with LiAlD4 and aqueous workup would be expected to give dideuterio 3-phenylpropan-1-ol (8).
Sodium borohydride reduces cinnamaldehyde via 1,2-reduction in hydroxylic solvent to cinnamyl alcohol in 97% yield.2 The intermediate alkoxyborane (Cf.; structure 5) either is incapable of reacting with the double as the aluminate does or the alkoxyborane rapidly exchanges with solvent.

1) Dornow, A., Winter, G., Vissering, W. Chemische Berichte, 1954, 87, 629.
2) Chaikin, S.W., Brown, W.G., J. Am. Chem. Soc., 1949, 71, 122.


Normal Phase TLC using ethyl acetate as eluent Compounds 1-3 were analyzed by normal phase TLC (shown) using ethyl acetate as the eluent. Assign spots on the TLC plate to compounds 1-3.
  1: 2-acetoxybenzoic acid 2: 2-hydroxybenzoic acid 3: benzoic acid

I asked a question about TLC and I received some mixed answers, so I'd like to see if my understanding of TLC is fine.

We're using normal phase TLC so lower Rf compounds will be more polar. $0.47$ is the most polar compound, $0.70$ is the least polar compound.

Ethyl acetate is polar. Does the eluent polarity just determine where the compounds will run on the TLC?

TLC is only about polarity.
Since all three compounds share the benzoic acid functionality, we only have to consider the rest.
Then we have three compounds: $\ce{R-OCOCH3}$, $\ce{R-OH}$ and $\ce{R-H}$.

In general the polarity is:

alkane < ester < alcohol

That is what we have here also: 3 < 1 < 2.

The one with the lowest polarity will have the highest R$_f$.
To show that it works, I have run them on a silica TLC plate with AcOEt:

AcSA: acetylsalicylic acid; SA: salicylic acid; BA: benzoic acid.


What utility does the Tau bond model of orbital overlap have? In his book on molecular orbital theory, Molecular Orbitals and Organic Chemical Reactions, Ian Fleming notes that Pauling formulated an early alternative model to Huckel theory for explaining the bonding of simple conjugated polyenes. Evidently, this came to be called $\tau$-bonding. Fleming describes it as a modification to, or offshoot from, the hybridization model, in which orbitals similar to sp3 hybridized orbitals are combined. Fleming says that the $\tau$-bond model makes the extent of the conjugation less obvious, but that the model

  [...] might have some virtues, not present in the Huckel model, especially in trying to explain some aspects of stereochemistry.

I don't have access to the relevant primary literature, and the treatment of the subject in Fleming's book amounts to one paragraph with an accompanying diagram. A search online was relatively fruitless.

My questions are:

Is $\tau$-bonding just the historical precursor to the bent/banana-bond model, or does it have some unique independent significance that makes it directly relevant to modern chemists?
What are the specific "virtues" that Fleming might have been referring to, and what, if any, advantages does the $\tau$-bond model have over Huckel theory and/or other approaches grounded in MO theory?

Tau-bonds provide an alternative valid description of electron density in alkenes and alkynes. Even in the modern chemistry
A good example is the conformational preferences in propene (lower energy of the "eclipsed" conformation). You can find more information here: http://pubs.rsc.org/en/Content/ArticleLanding/2011/OB/c1ob05393k#!divAbstract


Why does strength of bond matter if melting only affects the the intermolecular of the molecules So as I've been told, when a substance melts, the actual bonds of the substance aren't broken, only the IMFs (inter-molecular forces). So why is it that metal groups decrease in melting points going down when they should be increasing because more shells exist thus ions in the lattice will be larger so more IMFs therefore it takes more energy to overcome. Instead, I'm being told that since the metallic bonds get weaker due to less attraction between electron sea and the positive ion, melting point decreases.

Could someone please help me figure out what I'm missing here?

Depending whether you have a molecular, metallic or ionic compound, the independently moving particles in the liquid state are molecules, atoms and ions, respectively.
So for a molecular solid, you have to break the intermolecular forces to turn it into a liquid (or make them non-persistent so that the interaction partners can change over time).
For a metallic solid, you have to break the metal bonds intermittently so that the atoms can move in the liquid.
For an ionic solid, you have to break the ionic interactions intermittently so that the ions can move in the liquid.
Finally, for a network covalent solid, you would have to break covalent bonds to liquify it. For this reason, there is no liquid state of diamonds.

So why is it that metal groups decrease in melting points going down [...]

According to Jim Clark:

The strength of a metallic bond depends on three things:

The number of electrons that become delocalized from the metal
The charge of the cation (metal).
The size of the cation.


How to get the SMILES of all compounds on PubChem? I would like to download all the compounds from ChEMBL and PubChem. For ChEMBL this is easy using their webUI. Nevertheless, for PubChem, it isn't clear how to download all the compounds on the database including their SMILES representations.

I am also not sure how pubchem has 33,000,000 compounds where ChEMBL has "only"  2,000,000.

The best way to download bulk data from PubChem is actually FTP, as documented in their documentation.

For example, if you want the unfiltered SMILES of every CID in PubChem, the URL is ftp://ftp.ncbi.nlm.nih.gov/pubchem/Compound/Extras/CID-SMILES.gz

You can also download subsets using the PubChem Structure Download service

And as mentioned above, there are Python and other tools to access both PubChem and ChemSpider through their documented web APIs.


What is the difference between volatility and evaporation? I've been trying to find the difference between volatility and evaporation and I can see almost everywhere that the volatility is the tendency of a substance to vaporize. So Q1: what makes the volatility special? I understand that the fuels are generally volatile, and I understand that they vaporize faster than water, for example. But Q2: is this the only difference?

I was thinking that volatility also makes the vapor have a different chemical composition, or did I get it wrong?

For instance, the water (a non-volatile liquid) vapor is still $\ce{H2O}$, right? Q3: Will the vapor of benzene $\ce{C6H6}$ still be $\ce{C6H6}$ or it brakes into other compounds?

Evaporate is a phase change from liquid to gas without changing the molecule itself (or the atoms is the liquid is atomic).

Volatility is essentially the tendency of a liquid to evaporate. So a highly volatile liquid would show a great tendency to evaporate.

A measure of volatility is the vapor pressure. There is a very related term partial pressure. So if we think of air as being nitrogen and oxygen, then there is a partial pressure for nitrogen and a partial pressure for oxygen, and the sum of the two partial pressure is the actual pressure (the absolute pressure).

To make things a bit more confusing chemists think of absolute pressure. So because we are in earth's atmosphere there is always 1 atmosphere of gas pressure. So when you pump a car tire up to 30 lbs/sq inch, there is already 14.7 lbs /sq inch air pressure due to the atmosphere. So the absolute pressure inside the tire is 44.7 lbs/sq inch.


What happens to acids/bases left out overnight? I conducted a titration experiment a few days back and was unable to neutralize the solution in time, so I had to leave the solutions of HCl and NaOH out overnight in order to use them the next day. Also, one of my partners blew into the buret to get a drop of NaOH out. At the end of the experiment, we found that we required 6-7 mL more base to neutralize the HCl solution; another group who was forced to leave their solutions overnight used 3 mL more even than we did.

My hypothesis is that CO2 from the air formed carbonic acid in both solutions, mainly due to the instance where my partner blew into the buret, and thus the acidity of both solutions required that we use more base than we would have originally. As far as I've researched, this seems to be the case. Just to be sure, though: is this a correct assumption? If not, what could be the cause of this discrepancy?

First of all, blowing into the buret seems like a less-than-safe thing to do. $\ce{NaOH}$ could easily fly out into his/her mouth and nose.

I would expect that your acid and base solutions were already at the equilibrium concentration of $\ce{CO_2}$ when you got them, since they were exposed to air. I think that dissolving large amount of $\ce{CO2}$ into solution at atmospheric pressure is no easy task. I just don't think small amounts of $\ce{CO2}$ would account for the difference.

When you say "left them out" do you mean uncovered? If so, there's probably a fair amount of evaporation happening. The analyte solution could be affected depending on your experiment, and the concentration of your titrant would go up, which means you'd actually need less of it to reach the end-point.


Second electron affinity of sulfur and oxygen Why is the second electron affinity of oxygen greater than that of sulfur?
I think it should be like- second electron gain enthalpy of oxygen should be greater than the sulfur but electron affinity should be less than sulfur.

The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.

It is the tendency of an atom to gain electron. It is numerically equal to the negative of electron gain enthalpy.

Although fluorine has the highest electronegativity, chlorine has the highest electron affinity and this is because the considerable repulsion in the tightly packed $2p$ subshell of fluorine.

While electron gain enthalpy is enthalpy change when an isolated gaseous atom attain an electron to form a monovalent gaseous anion.

Chlorine ($\ce{Cl}$) has more negative electron gain enthalpy than fluorine the reason is same  as above.

Source:

Wikipedia and chemistry reader


Why is cyclopentadiene anion is aromatic but cycloheptatrienyl anion is not? I just can't seem to figure this out. The cyclopentadienyl anion (1) is aromatic, but the cycloheptatrienyl anion (2) is not, though its cation (3) is. Why?

That happens because of Huckel's rule of aromaticity. The cycloheptatrienyl anion has 8 electrons which translates as 4n electrons, not the 4n+2 as implied by Huckel. If you are about to construct the MOs for both cycles, you'll find that putting 4n electrons will rise in a diradical molecule, not stable as it already sounds. A quick trick for doing that is using Frost's circle. All you need to do is inscribe the correspondent polygon in a circle and make sure that you have a vertex touching the circle as low as possible. The vertices will give you the qualitative energy of the MOs.


Measuring pH using pH paper makes solution change color I have an unknown substance dissolved in water, and it's milky white. When I put in a pH paper, the paper shows that it is slightly basic at pH ~8. However, after putting in the pH paper, the solution changed color from milky white to a very faint blue. What possible compound could this be?

What about the volume of the analyzed mixture? I doubt that anyone can guess what the substance is, but my guess about the color is that the indicators on the pH paper might have been partially dissolved in the mixture. And, since the color of litmus paper at a pH = 8 is blue, it could be a possible explanation for the color, especially if you are dealing with a small volume of liquid.


Gas solubility in water Good morning,
I know that Henry's law describes the gas solubility in water with partial pressure of a specific gas.
For example the oxygen concentration in water, under atmospheric pressure, is

$\mathrm{C = P_{O_2} * k}$, where $\mathrm{P_{O_2}}$ is $\mathrm{0.21}$ bar.

But if I have a steam-water mixure, how I can calculate the solubility (for example in a deaerator)?
What data I need to calculate that solubility?
Thanks

You need to know the oxygen content (partial pressure) of the steam-water mixture with which you are purging the system.

The water being purged will come to equilibrium with the purge gas based on Henry's law, just as in your example of atmospheric conditions, except that the partial pressure of oxygen in the steam should be much lower.


Can artificial sea water be prepared by mixing sea salt with water? I am looking for ways to easily prepare artificial sea water. The chemical composition doesn't have to exactly match that of natural sea water. On-line I can find recipes to prepare sea water. But I was wondering, can I just buy natural sea salt and dissolve it in fresh water (Until I get the required salinity ).

The commercial shrimp farms in my region, that use natural sea water cause enormous environmental damage. That is the primary motivation behind my interest in artificial sea water. I want to explore if artificial sea water can be a eco-friendly sustainable alternative.

My intention is not to breed shrimp, but just to grow the shrimp that I get from hatcheries. And yes, I am planning to get the help of people with relevant expertise in this. I just want to gain some prior knowledge and experience before I go for their help.

Some of my present concerns are similar to those of oceanarium though on a much much smaller scale :-) Like them I also want to prepare sea water, which I can control to my exact needs. Locally we get these "raw sea salt" used for industrial purposes, which are claimed to be very similar to natural salts found in sea. They do not contain iodine, caking agents or any other additives etc. I am guessing probably the sea salts that oceanarium imports might be same as these (probably of much better quality).

Yes, it is possible though not easy to make artificial seawater by mixing sea salt and fresh water. An interesting example of this happens at the oceanarium in Lisbon, Portugal, which is one of the largest in Europe. After considerable investigation, it was found that the seawater present in massive quantities just outside the building could not be economically filtered, purified and adjusted to attend the surprisingly demanding and delicate needs of the marine organisms to be displayed. The chemical composition of seawater is actually quite complex, varying significantly with the location and time of collection. As with all life, marine life can be very sensitive to small changes in concentration of compounds, not to mention possible biological incompatibilities between the micro- and macrobiota present in seawater from different regions. The solution chosen for the Lisbon Oceanarium was to import some 16.5 metric tons of sea salt monthly from the Red Sea in Israel and mix it with local fresh water in order to replace the tank water. [1] [2] (Couldn't find a reference in English)

Having said that, you really should consult people experienced in setting up seawater aquariums before trying anything yourself. Seawater aquariums have significantly tougher upkeep and maintenance requirements compared to fresh water aquariums; you can't just mix any old cooking sea salt into water, it has to be in the right concentration, and replacement of the tank water has to be more gradual and frequent, not to mention that the salty water will corrode the actual aquarium more quickly. A good aquarium supplies store should be able to better inform you.

Edit: Now I realize you may actually be interested in cultivating shrimp rather than putting them on display. That's a whole different matter, many many times more difficult. You really need to contact some experts in marine cultivation.


Is it possible to protect molten magnesium with Butane I am currently experimenting with simple methods of metal casting at home. It is conventional wisdom that magnesium is better left to the professionals due to the unfortunate arrangement of its combustion and melting temperatures. I dont have access to any inert gasses or vacuum furnaces; but like a challenge.

The first thing that occurred to me was to use CO2 cylinders you can buy at the supermarket to protect my casting area. Luckily I didnt try that experimentally, since it is a known bad idea. Same applies to water vapor. Molten magnesium will happily steal the oxygen from either, leaving carbon resp. hydrogen gas behind. And a lot of heat of course.

Which brought me to the following counter-intuitive idea: if magnesium cares so little for carbon or hydrogen, could you protect your magnesium from burning with hydrocarbon gas? If I douse my casting mould in butane, then pour my molten magnesium in, I am likely to set the butane ablaze, sure. But I can deal with that; as long as my magnesium stays put.

Is this the dumbest idea ever? Or might it actually work?

To make my question more concrete: what is the known chemistry of magnesium at 700c and pure hydrocarbons? Note that there will also be a lot of silicon dioxide present, which can form a thermite with magnesium, but normally shouldn't under these circumstances. But perhaps the hydrocarbons could catalyze that reaction?

if it might pull the oxygen out of sand , try using magnesium oxide, surely it won't pull the oxygen from its own self. Or using burnt lime, surely it won't pull oxygen away from calcium.
It might react with butane to form magnesium hydride and carbide ....which would be fun to play with . Also magnesium easily reacts with nitrogen to make the nitride...which is an excellent way to make ammonia


How to produce fumes in chemistry lab? In chemistry lab we perform experiments to find the unknown solution. I just want to produce some fumes in lab. We are only allowed to use few common dilute solution like $\ce{NH_4Cl}$, $\ce{NH_4OH},$ $\ce{HCl}$, and few more which are generally allowed to students.

Is it possible to form fumes in lab?

Aqueous ammonia and hydrochloric acid are solutions of gases.

In a fume hood place two beakers with aliquots of these two solutions (one in each beaker, don't mix) side to side.

Did I mention the fume hood?

You'll see the formation of solid ammonium chloride from the reaction of the two gases.

$$\ce{HCl(g) + NH3(g) -> NH4Cl(s)}$$

Don't forget to clean up the mess!


Bond strength of pi acceptors and metals with filled d(pi) orbitals I was reading in The Organometallic Chemistry of the Transition Metals (Robert Crabtree) that lewis acids like BF3 often accept electrons from d(pi) orbitals. Is this generally true for all pi acceptor ligands? That is, do metals with filled d(pi) orbitals form thermodynamically stable bonds with pi acceptor ligands? I would imagine that d(pi) electrons would only be available for donation if said metal was in a low oxidation state. Correct me if I'm wrong.

If this post merits two different question threads, I'd be happy to separate them.

TL;DR Yes, you need low oxidation states unless there is a significant $\ce{M\bond{->}L}$ σ contribution to the bonding.

There are a few kinds of π acceptor ligands to distinguish between. These classifications are quite arbitrary and the line is not always clear, but their behaviour is sufficiently different and it is hopefully an instructive endeavour.

Firstly you have ligands which are mostly σ donors, but are also capable of acting as π acceptors, albeit weak ones. Two common examples are the cyanide anion and alkyl phosphines. Because the $\ce{M-L}$ bond is mostly donation from ligand to metal, you don't necessarily need particularly low oxidation states for these complexes to form: so $\ce{[Fe(CN)6]^3-}$ is OK despite having a relatively high oxidation state of iron, $\ce{Fe(III)}$.

Next up you have ligands which are poor σ donors and pretty good π acceptors. Carbon monoxide falls into this category: it is not really all that good a σ donor and backbonding is often required for a stable $\ce{M-L}$ bond to be formed (for some exceptions, google "non-classical carbonyl complexes"). Effective backbonding, in turn, requires high-energy d electrons on the metal. So, this usually precludes higher oxidation states. Many metal carbonyls have the metal in a negative oxidation state or zero, and $\ce{[Fe(CO)6]^3+}$ most certainly doesn't exist.

Lastly you have weird ligands such as $\ce{BF3}$ which are both σ and π acceptors. In the case of $\ce{BF3}$, it is a σ acceptor via the empty p orbital on boron, and a π acceptor via the $\ce{B-F}$ σ* orbitals. These are pretty obscure - I have never heard of them before - and are called Z-ligands (Wikipedia). Obviously any bond that forms must involve significant contribution of electron density from the metal. It turns out that as of 2011 there were no examples of unsupported $\ce{M-BF3}$ bonding,1 but there are quite a few examples of similar molecules $\ce{EX3}$ where $\ce{E}$ is a heavier Group 13 element.

In all these cases the metal fragment is very electron-rich and is again in a low oxidation state (zero or negative). I am sure the electron density on the metal required for these to form are even stricter than for carbonyls; otherwise we would undoubtedly have heard about $\ce{[M_pZ_q]}$ complexes, since there are already lots of $\ce{[M_p(CO)_q]}$ carbonyls known.

Reference

Amgoune, A.; Bourissou, D. σ-acceptor, Z-type ligands for transition metals. Chem. Commun. 2011, 47 (3), 859–871. DOI: 10.1039/C0CC04109B.


Why do other sugars melt whereas sucrose decomposes? Sucrose decomposes at 186 °C. It doesn’t melt but when it reaches decomposition temperature, sucrose decomposes into fructose and glucose.

After cooling and hardening it is no longer sucrose. Am I right?

You're mostly right.  Sucrose is a disaccharide composed of glucose and fructose.

It is a crystalline material.  Heating sucrose results in a complex thermal process that involves both melting and decomposition.  Depending how fast you heat it, you will observe different things.  Basically, sucrose will start to decompose when it starts to melt.  If there is some water present during the heating process, then a reaction known as sucrose inversion will occur and the sucrose will break down into its components, one molecule of glucose and one molecule of fructose.

However if dry sucrose is heated to its melting \ decomposition point, then more complex things happen.  Significant amounts of glucose are initially detected.  Continued heating leads to "caramelization".  Caramelization is a condensation process where water is eliminated from sucrose; the initial loss of water is followed by isomerization and polymerization.  Over 4,000 products are thought to be formed during caramelization.  When this complex mixture is cooled, it is no longer sucrose.


Can we force reactions with positive delta G? Do these reactions always have to be paired with another energetically favorable reaction with a greater delta G magnitude, or is there a way to force reactions with a positive delta G to happen?

The change of the Free / Gibbs enthalpy $\Delta_{\text{R}}{}G$ consists of a contribution by reaction enthalphy $\Delta_{\text{R}}H$, and a contribution by reaction entropy, $\Delta_{\text{R}}S$:

$$ \Delta_{\text{R}}{}G = \Delta_{\text{R}}H - T \cdot \Delta_{\text{R}}S $$

Hence, the answer to your question is no.  By variation of temperature $T$, for example, a reaction previously endergonic ($\Delta_{\text{R}}{}G > 0$) may become exergonic ($\Delta_{\text{R}}{}G < 0$), as summarized here.


What are the products of the thermal decomposition of potassium cyanate with carbon at elevated temperature? What are the products of the thermal decomposition of potassium cyanate ($\ce{KOCN}$) with carbon ($\ce{C}$) at elevated temps ($\ce{300^\circ C - 1300^\circ C}$)?

I guess some of the nitrogen is lost and forms something. I'm trying to find out the full equation, but a google search brings up nothing concerning reduction of potassium cyanate with carbon.

[...] some of the nitrogen is lost and forms something [...]

Carbon is a good and cheap reductant, often used in the processing of oxidic ores - except when metal carbides are formed. Carbon dioxide a reasonable product in these reactions.

$\ce{2KOCN + C ->[\Delta] 2 KCN + CO2}$

Let's agree that potassium cyanate is reduced to potassium cyanide, while carbon dioxide is released.


Effect of stereoisomerism and racemization on solubility I was on the Wikipedia page of Tartaric Acid, and I noticed that its solubility in water varies depending on whether the solution is racemic, meso, or optically active:
$$\begin{array}{|c|c|} \hline
 & \text{solubility (kg/L)} \\ \hline
\text{D or L (active)} & 1.33 \\ \hline
\text{DL (racemic)} & 0.21 \\ \hline
\text{meso} & 1.25 \\ \hline
\end{array}$$

After some pondering, I can see why the D-solution and the meso solution may have different solubilities; I would guess it is because of varying degrees of intramolecular H-bonding that can occur between the -OH and the -COOH groups. I would be grateful if anyone could formally justify this.

What surprised me, however, was the difference in solubility of the D-solution versus the racemic mixture. I would imagine that due to symmetry, every individual molecule of the acid in a racemic mixture would have an equal chance of being ionized/solvated as any other, as well as equal to the chances of molecules in a pure D-solution. Ergo, their solubilities should be identical. However, it appears my intuition is wrong here.

What is the reason, then, for these differences in solubility between the three types of solutions?

Indeed a racemic mixture can be considered - at least respect to collective properties - as a "third stereoisomer", so much that it deserves the special status of raceme.

Not only the 1:1 ratio of the two enantiomers results in optical inactivity, but its physicochemical properties differ from those of the two separated compounds.

As for the solid state, any single stereoisomer behave as usual. This means that the presence of a second compound lowers the melting point.

However, a stereoisomer can have a higher (lower) affinity and bound better to its specular companion as compared to itself. This could be due to a better filling of the space or the formation of H bonds at opposite sites of the molecules, for instance.

In this case, the addition of the L form to a pure D sample initially lowers the melting point. Nothing special here.

However, because of lattice symmetry, once the ratio becomes 1:1 a more stable structure is achieved and the sample has now a melting point higher than the two stereoisomers alone.

This is right the case for tartaric acid,

Now consider the Delta G for the solubilisation process. As correctly suggested in a comment, the equilibrium involves not only the solvated molecules but the solid lattice as well.

Standard thermodynamic consideration tell us that as the latter gets more stable, solubility at a given T gets lower

according to the experimental data that you have posted.

(Note that for both enantiomers and for all their mixtures the entropic terms are surely comparable and primarily dictated by the disruption of the solid lattice, especially so given the specular structure of the molecules).


Why does calcined feldspar crumble? For context, I am a potter and am interested in collecting and processing all materials that I use by myself. One of the processes which I use is calcination of certain minerals like feldspar, so that I can turn them into fine powder.
For reference, most feldspars are mixtures of the three chemicals: $\ce{KAlSi3O8,\ NaAlSi3O8,}$ and $\ce{CaAl2Si2O8}$. I usually collect pieces of granite which have a high feldspar content. I then put the granite into my electric (oxidation) kiln and fire it above $\pu{1000 ^\circ C}$. After it has been calcined, the granite falls apart very easily, with the exception of the pieces of quartz which remain quite hard. I am then able to crush the feldspar up into a fine powder.
I gather that the heat does something to the feldspar which makes it crumble easily but chemically speaking I don't know what this involves. Does the chemical formula of the feldspar change? Does it just break down in some way?

You're replicating the natural weathering process of granite.
Granite is a rock comprising the minerals feldspar, quartz, and mica.
There is a continuous solid-solution ternary diagram here that illustrates the various feldspars as a function of calcium, potassium, and sodium content: the end members are K-spar (potassium variety), albite (sodium variety), and anorthite (calcium variety).
The weathering process is complex and involves so-called incongruent dissolution, where one component in the solid state reacts to form one or more nother solid state components. This is why you see granitic feldspar weathering faster than the quartz or mica components out in nature.
You are reacting various minerals (the exact composition of your feldspars and micas would be determined by XRD or a similar method) with oxygen, carbon dioxide, and water from the atmosphere as well as any waters of hydration that may be in your micas, which are sheet silicates from a general class of silicates known as phyllosilicates.
Back to your query about crumbling: you're dehydrating the micas (muscovite on the light-colored end, biotite on the dark-colored end) if they are hydrated at all, but mostly you're making various clays from the feldspars via the incongruent dissolution pathway. Those new minerals you're making are the crumbly ones, mostly clays and maybe some brittle mica scraps. But it's mostly clay.
Generally speaking, carbon dioxide and water form an aqueous solution of carbonic acid species which reacts in the presence of oxygen and a feldspar to form a clay and aqueous calcium and carbonate ions. There are many, many complex reactions.
Here is an excellent write-up about anorthite hydrolysis, specifically regarding the region around Egersund, Norway. In it, the author does some modeling using PHREEQC and gives specific examples of some of the weathering reactions. Here is one of the reactions that takes a mixed albanitic-anorthitic feldspar to kaolinite:
$$\ce{2NaCaAl3Si5O16 +
      8CO2 +
      9H2O =
      2Ca^2+ +
      2Na^+ +
      6HCO3- +
      4SiO2 +
      3~Al2Si2O5(OH)4 +
      2CO2}$$
High temperature speeds things up, and also will dictate what products you get. Same for water and carbon dioxide, of course.
In summary: you are making mostly crumbly clay from granite by speeding-up the chemical weathering process of (mostly) the feldspar component(s). The particular composition of the clay varies with the entire system composition.
This is a very complex sub-field of study in aqueous geochemistry. An excellent reference (which I used in my grad school days and beyond) is Stumm and Morgan.


Can hydrofluoric acid etch glass at a pH of 7.0? Will it still etch glass when very dilute? Will the reaction still proceed slowly? At what pH will the reaction stop completely?

From Wikipedia on hydrogen fluoride: "Although hydrofluoric acid is regarded as a weak acid, it is very corrosive, even attacking glass when hydrated."
On hydrofuoric acid: It etches glass "by reacting with silicon dioxide to form gaseous or water-soluble silicon fluorides. "

It seems fishtank glass can be etched even when the water is kept at a pH of 7.0.
"You might try a razor scraper, to see if that will remove any of it. If not, there is a very high chance they are due to etching. Etching is caused by the action of water, especially acidic water, over a long period of time. "

In $\ce{SiO2}$ etching, there are two active fluorine-containing species, $\ce{HF}$ and $\ce{HF2-}$.

The rate of etching at 25 degrees C in Angstroms per second is:

$$9.66[\ce{HF2-}] + 2.50 [\ce{HF}] -0.14$$

according to A Study of the Dissolution of SiO2 in Acidic Fluoride Solutions J. Electrochem. Soc. vol. 118, pp. 1772-1775.

The relavent equilibrium constants at 25 degrees C and ionic strength of 1M are

$$\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 0.0013M $$

$$\frac{[\ce{HF}][\ce{F-}]}{[\ce{HF2-}]} = 0.104M $$

Because of the -0.14 term in the rate equation, only at enormously high F- concentration would there be etching at pH 7 (such as in this situation).

This study was concerning high (~1M) concentrations of fluorine species, so it might not be completely relevant for low concentrations or very long time periods.


Computational Efficiency of Periodic Boundary Conditions From what I understand, in a molecular dynamics simulation the number of molecules included is limited by the computational power available.

In order to simulate macroscopic properties, one needs to ensure that the particles experience forces as if they were in the bulk, so the system is replicated in all directions through the application of a periodic boundary condition.

How is the creation of this larger, periodic system any more computationally efficient than just simulating a non-periodic system of similar size?

The application of periodic boundary condition does not mean that the system is explicitly replicated. You only deal with the particles within the unit cell (the box that you replicate), but you let them interact with the replicated ones outside of this unit cell.

Consider a simple 1-dimensional case with three atoms in the unit cell: $\ce{A-B-C}$, and lets assume we only need to treat nearest-neighbor-interactions. So the "monomer" has only 2 interactions: AB and BC.

One option to simulate the macroscopic system would be to explicitly deal with $\ce{A-B-C-A-B-C-A-B-C-A-B-C}$ (or even longer), but this requires a lot of atoms with accordingly many interactions between them. And even then both ends of the chain would be missing one neighbor.

Using periodic boundary conditions, one explicitly treats $\ce{A-B-C}$ only. But this time we will allow $\ce{C}$ to interact with $\ce{A}$, since the periodic boundary conditions implicitly tell us, there is another $\ce{C}$ to the left. Similar, there is an implicit $\ce{A}$ to the right of $\ce{C}$. Compared to the monomer from above, we have again 3 atoms, but now there are 3 interactions to be considered: AB, BC and AC.


Troubles with Gibbs free energy The book I'm reading writes that for a system at constant pressure and temperature it holds:
$$dG=-TdS_{tot}$$
where $dG=dH-TdS$.
So if you want to see if a certain process is spontaneus, you just have to look at the sign of $dG$, since it depends on the sign of $dS_{tot}$.
But I have a lot of doubts about this formula, indeed read the following scenario.
You consider a certain amount of liquid inside a piston at a fixed pressure given by the weight of the piston. Let's assume the liquid is changing its phase, so also its temperature stays constant at the value $T$. Out there is a heat reservoir at the temperature $T_{res}$, which touches the piston. Since the temperature can't be a discontinous function, you assume that there is a piece of metal which connects the reservoir and the piston. You have:
$$dS_{tot}=dS_{res}+dS+dS_{metal}$$
If you consider a stationary transform, the state of the piece of metal doesn't change, then $dS_{metal}=0$.
$$dS_{tot}=\frac{\delta Q}{T_{r}}-\frac{\delta Q}{T}$$
Since pressure is constant:
$$dS_{tot}=dH[\frac{1}{T_{r}}-\frac{1}{T}] $$
Note that if $T_{r}=T$ then $dS_{tot}=0$ since there are no irreversibility in the universe (no finite differences of temperature), otherwise it must be $dS_{tot}>0$. I would stop at this point but let's introduce the Gibbs free energy.
Definig the Gibbs free energy as:
$$G=H-TS$$
$$dG=dH-dTS-TdS$$
$$dH=dG+dTS+TdS$$
For constant temperature:
$$dH=dG+TdS$$
You finally get:
$$dS_{tot}=(dG+TdS)[\frac{1}{T_{r}}-\frac{1}{T}] $$
Which is not the same thing that writes my book. Also note that the temperature and the pressure of the system are constant, so I have the same hypothesis of the book but I get different results. Someone could clarify this? I'd really appreciate.

Your error occurs when you write that

$$dS_{tot}=\frac{\delta Q}{T_{r}}-\frac{\delta Q}{T}$$

since this assumes that

$$\delta Q=-TdS$$

which implies reversibility and therefore that $dG=0$ at constant T and p.

The more general statement is that

$$dS_{tot}=\frac{\delta Q}{T_{r}}+dS$$

which means that

$$dS_{tot}=-\frac{dH}{T_{r}}+dS$$

(since positive heat here means exothermic)

Then

$$T_{r}dS_{tot}=-dG+(T_{r}-T)dS$$

or

$$dG=-T_{r}dS_{tot}+(T_{r}-T)dS$$

Now if you let $T=T_{r}$ (thermal equilibrium with surroundings) then

$$dG=-TdS_{tot}$$


Benzene to phenol in a single step My teacher told me we can convert benzene to phenol within a single step using $\ce{V2O5}$/ $\ce{O2}$ at 300 ⁰C (not 500 ⁰C). Later I searched for that reaction but I couldn't find it in all over Google except in Allen's Chemistry handbook.

Does that reaction really occur and if it occurs then what is the mechanism of that reaction?

The reaction of benzene over $\ce{V2O5/ PtAu}$ catalyst at lower temperatures, can convert benzene  to phenol with some success.

See this book here Direct hydroxylation of benzene

[The original question before editing was: What is the Ratta Maar Reaction?] Someone has played a prank with you with the named reaction. "Ratta maar" is a slang for "rote memorization" in Hindi/Urdu, which implies keep on memorizing without understanding and perhaps this is what your teacher wanted too. I assume this Allen's Chemistry Handbook is being taught or used in India.


Can a clay brick waste in a soil be a source of minerals for plants? Clay brick is made by heating them so their clay structure changes. What i understand is ultimately all materials have one or another base minerals in it particularly here this brick still it remains in some other form. So after process of heat, those minerals or nutrients can remain in such condition a plant can absorb? Can a brick dust be a good source of nutrients for plants to grow?

In general clay is not a good source of nutrients. Clay is a class of hydrated alumina-silicates containing small amounts of metals. The alumina and silica are not plant nutrients, but some of the metals can be, in particular: potassium, magnesium, calcium, and iron.

Of these, only potassium is needed in large amounts by plants, and conversely, is fairly uncommon in clays. Calcium and iron are much more common, but are needed by plants in very small quantities.

For any mineral nutrient in any case I can think of, other sources would be better than clay.

Therefore, in the specific case where there was a mineral deficiency in the soil, and you didn't have another option available, clay could provide that nutrient.

However, clay can help plants in another manner. This article describes the ability of clay soil to attract and hold positive ions (metals that can be absorbed by plants are positive ions) due to the high surface area and negative surface charge of the clay particles. Since leaching can be a problem in nutrient-poor soils, having a soil component that "holds on" to those nutrients can be helpful.

Whether or not brick dust would be helpful to plant growth would depend on the specific case, in particular:

The type of plant(s)
The composition of the rest of the soil
The type of clay used to make the brick
The climate conditions (amount of rainfall, etc.)


Could you make a helium atom from hydrogen gas? Hydrogen gas is $\ce{H2}$, and contains 2 electrons and 2 protons. My question is, with sufficient pressure, could you fire two neutrons at the molecule and have them form a single Helium atom?

The hydrogen nuclei are spaced out by about $\pu{70pm}$, if I guessed/remembered the number correctly. To generate a helium nucleus, this distance has to be reduced to almost $\pu{0pm}$ against the repulsion of like charges. So if you just bombarded a hydrogen molecule with neutrons from one side, nothing would happen since the two would not fuse.

It may work if you manage to shoot both neutrons into different hydrogen atoms in such a manner that both hydrogens are displaced towards the other. But this type of collision can be considered more of a thought experiment than anything practical. (Even just hitting a single molecule with two neutrons at the same time is hard enough.)


Reading number of outer shell electrons and other properties from periodic table? How can one read from the periodic table the number of outer shell electrons that an atom has, to predict how these atoms will make bonds with other atoms? For example to see that hydrogen ($\ce{H}$) has 1 electron free, while carbon ($\ce{C}$) has 4 and oxygen ($\ce{O}$) has 2? This allows us to infer that carbon can make four bonds with 4 different hydrogen atoms. It's not clear what in the periodic table would tell us this; it is neither in the group nor in the period of an atom, it seems.

Related to this: can someone summarize concisely all the properties (like number of free outer shell electrons) that one can read off about each element from its position in the periodic table? I was having a hard time finding a concise summary of these.

Start with a periodic table that shows the electron configurations. You might try the Dynamic Periodic Table; as you mouse over an element, its electron configuration is presented.

Then go to Electron Configurations in the Periodic Table, Electronic Structure of Atoms or Block (periodic table) to see how shells and orbitals are added. It's pretty straightforward until the transition metals, and things get much more complicated for the lanthanides and actinides, where a new shell may start before an inner shell is complete.

The general rule is: an orbital is more "stable" when filled, or half-filled, so fluorine, for example, can readily accept one more electron to finish its outer 2s2 2p5 shell and it will then have the configuration of neon, 2s2 2p6. Many periodic tables don't present the inner shells, using instead the shorthand of [Ne] 3s2 3p5 for chlorine, rather than the full 1s2 2s2 2p6 3s2 3p5.

In fact, you're not alone in trying to understand the electron structure; see the current debate on the position of Lw in the table: New data on synthetic element trigger rethink of periodic table.

BTW, note there are different ways a shell can be completed. Carbon will share its four outer electrons with four hydrogen atoms; this is called a covalent bond. Chlorine outright "steals" an electron from sodium; this is an ionic bond. See Covalent Lewis Dot Structures and Ionic Bonding and Electron Transfer for more info.


Choise of solvent for extraction In the synthesis of (S)-Diphenyl(pyrrolidine-2yl)methanol, I need to extract the product using an organic solvent and water. Which organic solvent is the most suitable in this case?

Ethyl acetate: I think that it is miscible with water and it can form an emulsion, for this reason I would not choose it.
Hexane: it is an apolar solvent so maybe the product is not very soluble in it.
Ethanol: I think I cannot use it because it is miscible with water.
Diethylether: it is an apolar solvent so maybe the product is not very soluble in it.
Dichloromethane: I think that this is the best choice because it is not miscible with water and the product should be soluble in it.

Is my reasoning correct?

The perspective about ethyl acetate should be revisited.  EtOAc is slightly soluble in water; not infinite (as e.g. ethanol), nor this little as hexane(s).  It is greener than halogenated solvents such as dichloromethane, and the boiling point higher than the for dichloromethane or diethyl ether*) is an advantage during summer.  Thus, EtOAc is an often used polar solvent of extraction.
Equally recall the partition coefficient -- multiple extractions with a small volume of organic are more efficient than one with a large volume as long as after shaking the organic and the aqueous layer separate well from each other. Typically, for the isolation of your raw-product later to be purified by crystallization or chromatography, three extractions just «good enough» may suffice for you.  Except you intend a Craig-Extraction to separate similar compounds from each other (as here), a precursor of the countercurrent chromatography.
The same logic of using multiple small volumes instead of one large is applied when washing the organic phase, e.g. by brine, and back extraction.  Indeed, the presence of salts in the aqueous phase may break an emulsion during an extraction; prior to application, it should be tested on small (e.g., test tube) scale ahead though.
*) Compared to diethyl ether, consider MTBE ether as alternative: it is much less prone to the formation of peroxydes, has a higher boiling point, and is cheaper (per L) than the former, too.


What is the bonding picture in Peroxomonosulfate? In $\ce{SO5^{2-}}$, can somebody tell me which orbitals are used to form the pi bonds in this anion.
Please tell the hybridization.
And the lewis structure.

Well I tried and the answers I got was sp3 & the pi bonds are formed using d orbitals.I just need to know whether I'm correct

The peroxomonosulfate is a hypercoordinated molecule, in which the sulfur appears to exceed Lewis' rule of two, better known as the octet rule. In former times it was believed that d orbitals would contribute to bonding. This is not the case.
Bonding is better represented by

three equivalent $\ce{S-O}$ single bonds
sulfur having a formal charge of +2
these three oxygens having a formal charge of -1
one $\ce{S-O}$ single bond that leads to the peroxyl oxygen
this oxygen has a formal charge of zero
one $\ce{O-O}$ single bond
and the terminal oxygen having a formal charge of -1

This bonding picture is much more in accordance with quantum chemical calculations and an NBO analysis on the DF-BP86/def2-SVP level of theory. The involvement of d orbitals is minimal (<1%).

It is noteworthy, that this molecule pretty much has the same bonding around the sulphur as in sulphate. It can therefore similarly be described. Have a look at my rant against hypervalency or the bonding picture of the $\ce{PO4^{3-}}$ anion, which is isoelectronic to sulphate.


Equation for collision frequency of molecules/ions in aqueous solution? There are equations that describe collision frequency in the gas phase, but I am wondering if this collision theory has been extended to describe the collisions of molecules and/or ions that are dissolved in water. I would imagine this equation would incorporate the aqueous diffusion coefficients of the species in some way and also not have the gas constant (R) in the expression.

I believe this page from the University of Malta goes into at least some of what you are looking for. In particular, it describes how the diffusion coefficients influence the rate of reaction in solution and discuss limiting cases, such as diffusion controlled and activation controlled, for a generic aqueous system.


Wurtz reaction on acidic compounds Why can Na not be used for the Wurtz reaction of compounds having an acidic group like alcohol or carboxylic acid?

With alkanols, sodium will form alkoxides (and hydrogen gas).
$$\ce{2CH3OH + 2Na -> 2 CH3ONa +H2 ^}$$
With alkanoic acids, sodium will form alkanoates (and hydrogen gas).
$$\ce{2CH3COOH + 2Na -> 2 CH3COONa +H2 ^}$$

If your starting material contains both an $\ce{OH}$ group and a bromo substituent, the alkoxide formed upon the reaction with sodium may attack the carbon bearing the $\ce{Br}$ atom in a substitution reaction to yield a cyclic ether.


What is the molecular structure of the fragment ion m/z 161 of rosmarinic acid? Attached is an ESI-MS/MS spectrum of product ion scan of rosmarinic acid (m/z 359.10) (source). I'm trying to figure out the structure of the fragment m/z 161. It's not clear to me what the structure should be. Does a triple bond form perhaps?

The source you link to says that the spectrum was acquired in negative mode.

  The CID experiment on $\ce{[M - H]-}$ ion at m/z 359.08 identified as rosmarinic acid gave the two main constituents of rosmarinic acid, namely caffeic acid at m/z, 179.0 and the 2-hydroxy derivative of hydrocaffeic acid at m/z
  197.0 as illustrated in Figure 3. Similar pattern of fragmentation of rosmarinic acid during CID analysis has been reported by several authors (17, 25, 26) in analyzing extracts of Lamiaceae spices.

This means that the parent ion is $\ce{(C18H16O8 - H)-}$, and thus the (hydrated) fragment in question is $\ce{C9H7O4-}$.

The first question in any CID mechanism is whether the fragment arises via charge migration or via charge retention.  I think in this case a charge retention mechanism is pretty likely, because phenolic hydroxyl groups are acidic enough to ionize readily via proton loss.

The next question is where the loss of water comes from.  It's worth noting that the mass spectrum of even just plain old catechol, $\ce{C6H6O2}$, shows a loss of water in negative-mode CID.  That is, the parent ion $\ce{(C6H6O2 - H)-}$ loses water somehow to form $\ce{(C6H3O)-}$.  I see no reason why the same thing couldn't happen to the rosmarinic acid fragment in question.

It's impossible to say for sure without doing a lot of esoteric spectroscopic characterization of gas-phase ions, or least the synthesis and fragmentation of a series of partially deuterated or $\ce{^{18}O}$-labeled rosmarinic acid isotopologues.  Neither of those is an easy task (far from it), so it's nearly impossible to say for sure what the structure of the fragment is.

But think a perfectly reasonable proposition is that the ion is the same as whatever the structure of the catechol dehydration fragment is, perhaps a deprotonated hydroxybenzyne or maybe an epoxybenzene anion?  Loss of aromaticity in CID fragmentation not unusual.  In fact, a previous Q&A here on Chemistry StackExchange provides yet another example of this.

High-energy gas-phase ion chemistry is weird.


Why does the Michaelis-Menten constant remain constant in the event of non-competitive enzyme inhibition? A while back, I "answered" the question: Why does the Michaelis-Menten constant decrease in the presence of non-competitive inhibitor?

At the time of posting my answer, both the OP and I were under the impression that non-competitive inhibition reduces the Michaelis-Menten constant ($K_\mathrm{M}$).

However, the Wikipedia article on "Enzyme inhibitors" states that $K_\mathrm{M}$ remains constant in the event of non-competitive inhibition. (second emphasis mine)

  In non-competitive inhibition, the binding of the inhibitor to the enzyme reduces its activity but does not affect the binding of substrate. As a result, the extent of inhibition depends only on the concentration of the inhibitor. $V_{max}$ will decrease due to the inability for the reaction to proceed as efficiently, but $K_m$ will remain the same as the actual binding of the substrate, by definition, will still function properly.

But this goes against what I thought earlier (see my answer to the linked question). Wikipedia defines $K_\mathrm{M}$ as:

  The Michaelis constant $K_\mathrm{M}$ is the substrate concentration at which the reaction rate is half of $V_\mathrm{max}$. Here, $V_\mathrm{max}$ represents the maximum rate achieved by the system, at saturating substrate concentration.

Wikipedia attributes the invariance of $K_\mathrm{M}$ in non-competitive inhibition to the fact that the actual binding of substrate to an active site is unaffected. Now I agree that the actual binding of the substrate is unaffected here, but how does that have any effect on $K_\mathrm{M}$?

When $V_\mathrm{max}$ decreases during non-competitive inhibition (the only point I and Wikipedia concur on), shouldn't the $K_\mathrm{M}$ decrease correspondingly (based on the definition of $K_\mathrm{M}$)? Is the Wikipedia article incorrect in this regard?

This is a question that is relatively easy to answer mathematically.

To keep the algebra a little simpler, let me use a simplified Michaelis-Menten mechanism, where we assume that the enzyme-substrate complex (ES) and the free enzyme are in equilibrium.  (We could use the less restrictive pseudo-steady state hypothesis on (ES) and obtain the same result for non-competitive inhibition, at the expense of more complicated math.)

$$\ce{E + S <=>[K_M] (ES) ->[k3] E + P}$$

Non-competitive inhibition means that the inhibitor doesn't care at all whether the substrate is bound to the enzyme.  Either on (ES) or on (E), it will bind with affinity $\mathrm K_{\mathrm i}$.  We assume that (ESI) does not react, at all, to form products.  So the only other two reactions we need are:

\begin{align}
\ce{E + I &<=>[K_i] (EI)}\\
\ce{(ES) + I &<=>[K_i] (ESI)}
\end{align}

Now to get the rate equation, we first perform a mass balance on enzyme:

$E_0 = E + ES + EI + ESI$

We use the equilibria we've assumed above to get:

\begin{align}
(ES) &= \frac{(E)(S)}{K_M}\\
(EI) &= \frac{(E)(I)}{K_i}\\
(ESI) &= \frac{(E)(S)(I)}{K_M K_i}
\end{align}

Thus

\begin{align}
E_0 &= E\Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)\Big)\\
E &= \frac{E_0}{\Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)}
\end{align}

The rate equation we are after is just the formation of product, $\frac{dP}{dt}=k_3 (ES)$.

$$k_3 (ES) = k_3 \frac{(E)(S)}{K_M}$$

Now we know $(E)$ from our enzyme mass balance, so

\begin{align}
\frac{\mathrm{d}P}{\mathrm{d}t} &= k_3 E_0 \frac{(S)}{K_M \Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)}\\
&=k_3 E_0 \frac{S}{K_M + S + \frac{I}{K_i}\left(K_M + (S)\right)}\\
&=k_3 E_0 \frac{S}{(K_M + S)\cdot (1 +\frac{(I)}{K_i})}\\
&=\frac{k_3 E_0}{(1 +\frac{I}{K_i})} \frac{S}{(K_M + S)}
\end{align}

The only thing different about this expression is that the "$V_{max}$" part of the expression, which would have been $k_3 E_0$ for a simple reaction with no inhibitor, is now $\frac{k_3 E_0}{1 + I/K_i}$.  In other words, $V_{max}$ is the only parameter changed by the inhibition mechanism we have assumed.

The apparent $K_M$ of the non-competitively inhibited reaction is still $K_M$, because if you plug in $S = K_M$, you get a rate that is half of the maximum (of the inhibited reaction).

  Wikipedia attributes the invariance of  $K_M$ in non-competitive inhibition to the fact that the actual binding of substrate to an active site is unaffected. Now I agree that the actual binding of the substrate is unaffected here...but how does that have any effect on $K_M$?

You can think of $K_M$ as an apparent equilibrium constant for the binding of substrate to free enzyme.  So if something doesn't affect substrate binding, it probably won't affect $K_M$.

  So, when $V_{max}$ decreases during non-competitive inhibition (the only point I and Wikipedia concur on), shouldn't the $K_M$ decrease correspondingly (based on the definition of $K_M$? Is the Wikipedia article incorrect in this regard?

No, Wikipedia is correct.  Enzymologists have adopted very confusing terminology for no good reason, but non-competitive inhibition is not the same uncompetitive inhibition.  In uncompetitive inhibition, both apparent $K_M$ and apparent $V_{max}$ are changed.

If I gave you two vials, A and B, of the same enzyme, and one of them had been pre-mixed with a high concentration of a non-competitive inhibitor, and I asked you "what's the $K_M$ you measure for each of these two different enzyme preparations?,"   then you'd get the same number for each vial.  If 5 mM of substrate is the concentration at which enzyme from vial A reaches half of its maximum rate, then you'll find that 5 mM of substrate is also the concentration where enzyme from vial B reaches half of its maximum rate.


Proper Resonance Structures Would this be proper resonance structures? I know there's a lone pair on the top carbon. Having a carboanion isn't favorable as opposed to a negative formal charge on the more electronegative carbon so I flipped a bonding pair to oxygen's valence. I then proceeded to move the pi electrons inside the ring as well. This seems to satisfy all octet requirements.

For the second resonance diagram I saw the negative charge on the oxygen and decided to take one of its lone pairs and make it a bonding pair so that oxygen has 0 formal charge. This in turn necessitates pushing around a few pi electron pairs inside the ring. Same thing for the third one; I tried giving nitrogen a positive formal charge.

Yes, they are all valid resonance structures.  In the first series, there is a third resonance structure you could draw with the negative charge alpha to the carbonyl carbon. In the second series, the phenoxide anion, you can draw two more structures.  One with the negative charge in the other ortho position and one with the negative charge in the para position.  This significant resonance stabilization of the phenoxide anion explains why phenol is such a strong acid.  In your third series there is also a third structure with the positive charge situated alpha to the nitrogen.


After a unitary transformation, is Koopmans' theorem still valid? Koopmans' theorem says that the energy of the HOMO of the Hartree-Fock orbitals is equal to the first ionization energy of whatever system is being studied. This is only approximate because it assumes no changes in the orbitals in the ionic state, no relativistic effects, and no electron correlation. Ignoring those approximations, there is another feature of this which I don't understand.

The wavefunction as determined using HF can be unitarily transformed so that the total energy and wavefunction are preserved, but the "orbitals" themselves are not. Under many of these transformations, I assume that the orbital energies change. Is this true? If it is true, I assume Koopmans' theorem is still valid, but is now more complicated so that the ionization energy is some combination of orbital energies. Can anyone expand on this and let me know if my thinking is correct?

Also, what is it that is so special about the Hartree-Fock orbitals that Koopmans' Theorem has such a simple interpretation, while, if what I say above is true, it seems to be quite convoluted for other sets of orbitals?

As a note, I asked a question which more or less contained this question, but was a bit broader. The question I'm asking right now is the question I am really more interested in having answered.

You can find that other question here: Observability of Orbitals and Orbital Energies

Disclaimer: I like Tyberius' answer, but I'd like to go a little further.
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$

The beauty of Koopmanns' theorem is its simplicity. Quoting from the gold book:

  Koopmans' theorem

  Directly relates experimental ionization potentials with energy levels of molecular orbitals. The theorem states that the ionization potential required to remove an electron from the orbital $\Psi_i$ is given by the negative value of the energy of the orbital, $−\varepsilon_i$, as calculated within the Hartree–Fock approximation. The theorem is not applied to localized molecular orbitals, which are not eigenfunctions of the effective hamiltonian.

While Koopmans' text (in German: Physica 1934, 1, 104-113.) is a bit hard to comprehend, due to the ancient language I never learned, it basically boils down to two important facts.

There is a unitary transformation of the Lagrange multipliers which diagonalises them. These form a characteristic set of elementary wave functions, matrix, which in turn have a characteristic eigenvalue each. If all eigenvalues are different, then all elementary wave functions are specified through the Hamiltonian. (If there are equal eigenvalues, a freely chosen unitary transformation can be performed.)

In other words: There is one set of canonical orbitals that diagonalises the Fock matrix; each canonical orbital has an eigenvalue. For non-degenerate systems those orbitals are specified through the Hamiltonian.
This choice of functions has physical meaning, as the eigenvalue (disregarding a small correction) can be equated to the ionisation energies of the corresponding electron.

The second statement implies that there is no relaxation of the orbitals when removing an electron. This is oftentimes referred to as "frozen MO" approximation. Koopmanns himself appreciates that there should be a contraction of orbitals when removing an electron, however, he doesn't go into detail.

Nowadays we explain Koopmans' theorem in some simpler terms, where we remove the $k$th electron:

\begin{align}
&& E_N &= \sum_{i=1}^N H_{ii}
        + \frac12\sum_{i=1}^N\sum_{j=1}^N (J_{ij} - K_{ij})
        + V_{\mathrm{nuc}} \tag1\\
&& E_{N-1}^k &=\sum_{i=1}^{N-1} H_{ii}
        + \frac12\sum_{i=1}^{N-1}\sum_{j=1}^{N-1} (J_{ij} - K_{ij})
        + V_{\mathrm{nuc}} \tag2\\\hline
&& E_N - E_{N-1}^k &=  H_{kk}
        + \frac12\sum_{i=1}^N (J_{ik}-K_{ik})
        + \frac12\sum_{i=1}^N (J_{kj}-K_{kj}) \tag{$1-2$}\\
\therefore&&
E_N - E_{N-1}^k &=  H_{kk} + \sum_{i=1}^N (J_{ki}-K_{ki})\\
\therefore&&
E_N - E_{N-1}^k &= \varepsilon_k\\[2ex]
\text{with}&&
H_{ii} &= \langle \phi_i(\mathbf{x}_1)|
          \mathbf{H}^\mathrm{c} | \phi_i(\mathbf{x}_1)\rangle\\
&&
J_{ij} &= \langle \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) | r_{12}^{-1}
          | \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) \rangle\\
&&
K_{ij} &= \langle \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) | r_{12}^{-1}
          | \phi_j(\mathbf{x}_1) \phi_i(\mathbf{x}_2) \rangle\\
\end{align}

As you can see, Koopmans' theorem is not limited to the HOMO, but can be used for any occupied MO.

There is a very important point to be always be kept in mind when working with any Hartree-Fock based approaches:
Even though the equation
$$\op{F}_i\phi_i = \varepsilon_i\phi_i
     \tag{3}\label{fock-pseudo}$$
suggests an eigenvalue problem, it is not. Remember the definition of the Fock operator and the operators contained
\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}
As you can see, the "one-electron" Fock operator depends on the solution of all "one-electron" Fock operators (ref. Szabó-Ostlund p. 115). The Hamilton operator is not the sum of all Fock operators, and the total HF energy is not the sum of all orbital energies. As a result of that, the canonical orbitals are actually unique solutions.
The Fock operator is associated with the whole wave function, and while unitary transformations will keep the wave function and its energy equivalent, everything else will fall apart.
Another important consideration is the fact that the Fock operator is only well defined for occupied MO. You sometimes find the statement
$$E_{N+1}^{l>N} - E_N = \varepsilon_l,$$
which can not be applied in the same way; it is very basis set dependent.

From all that above it is obvious, that Koopmans' theorem only works for systems where a single determinant approximation is reasonable. It also explains, why it only works for HF; although there are generalisations for DFT.

Is Koopmans' theorem still valid after a unitary transformation?

No. As stated above, the Fock operator is associated with the $N$-electron wave function. After a unitary transformation and then removal of an electron, the energy of the $N-1$-electron wave function is not conserved.
As an illustrative example:
\begin{align}
&& \sum_{i=1}^{N} H_{ii} &= \sum_{i=1}^{N} H_{ii}'\\[2ex]
\text{with}&&
H'_{ii} &= \langle \phi'_i(\mathbf{x}_1)|
          \mathbf{H}^\mathrm{c} | \phi'_i(\mathbf{x}_1)\rangle\\
\text{and}&&
|\phi_i\rangle &\color{red}{\neq} |\phi'_i\rangle\\[2ex]
&& \sum_{i=1,i\neq k}^{N-1} H_{ii} + H_{kk}&=
   \sum_{i=1,i\neq k}^{N-1} H_{ii}' + H_{kk}'\\
\text{for } H_{kk} = H_{kk}':
&& \sum_{i=1,i\neq k}^{N-1} H_{ii} &=
   \sum_{i=1,i\neq k}^{N-1} H_{ii}'&&
\implies|\phi_i\rangle \color{red}{=} |\phi'_i\rangle\\
\end{align}

I have indicated the contradiction in red. Therefore it follows:
$$|\phi_i\rangle \neq |\phi'_i\rangle
  \implies H_{kk} \neq H_{kk}'
  \implies
  \sum_{i=1,i\neq k}^{N-1} H_{ii} \neq \sum_{i=1,i\neq k}^{N-1} H_{ii}'
$$

You can follow that trough for the other terms and will see that the expectation value of energy of the $N-1$-electron wave function needs to be different from the expectation value of energy of the $N-1$-electron wave function after unitary transformation.


Why must the leaving group in E1cb be poor? Typically the leaving group for E1cb is poor (like -OH or -OR) but why must this be the case? The substrate appears in the rate equation so surely a good leaving group would be beneficial?

There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another.  For example, in some elimination reactions the E1 and E2 pathways can operate in competition with one another.  An activation energy is associated with each of these 3 reaction pathways.  Whichever pathway has the lowest activation energy will be the major pathway followed.  By changing solvent, reaction temperature, relative strength of the nucleophile, relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.

  Typically the leaving group for E1cb is poor (like -OH or -OR) but why
  must this be the case?

The E1 mechanism involves ejecting a leaving group in its first step, while the E1cb mechanism involves removing a proton in its first step. Let's consider how changing the leaving group can shift an elimination reaction towards one pathway or the other.  To a first approximation, changing a leaving group will not affect how hard or easy it is to remove the proton. So it is reasonable to assume that the activation energy for the E1cb process doesn't change as we vary our leaving group.  If we use a better leaving group (make our leaving group more stable) that means that we have made ejecting the leaving group a lower energy pathway and the E1 process will become more favorable relative to the other elimination mechanisms.  If we change our leaving group to one that is an extremely poor leaving group (make our leaving group less stable), then ejecting it becomes a higher energy process and the E1 reaction becomes less competitive with the other reaction pathways. Saying this last sentence differently, when we use a poor leaving group we raise the activation energy for the E1 mechanism.  Since the rate of the E1cb process is not affected by the leaving group, its rate remains unchanged.  Consequently, a poor leaving group will disfavor the E1 process making the E1cb process more competitive. If the leaving group is bad enough, we can disfavor the E1 process so much that we wind up pushing our reaction all the way over to the E1cb side.


definition of oxidizing agent For the reaction $\ce{SF4 + F2 -> SF6}$, since oxidation state of $\ce{S}$ increased from +4 in $\ce{SF4}$ to +6 in $\ce{SF6}$, can $\ce{F2}$ be referred to as the oxidizing agent?

Oxidation means increase in Oxidation state or loss in electrons.

Example: $\ce{Fe^2+ -> Fe^3+ + e-}$

Oxidizing agent is a substance that oxidizes other substances and reduces itself.

In your example, $\ce{F2}$ will act as oxidizing agent as in the reactant side its oxidation state is $0$ but in the product side its oxidation state is $-1$, which means that its oxidation state in decreased, i.e. $\ce{F2}$ is reduced.


Why is there a comparative lack of variety in positive counterions in drugs? Pharmaceutical salts are important in the process of drug development. Using different chemical species to neutralise the parent drug can produce a diverse series of compounds, and this process is traditionally used to improve drug solubility and drug dissolution rates.

For negatively charged drugs which are paired with cationic counterions, there seems to be a general lack of variety in the counterion choice, which is usually a metal ion such as sodium. Other counterions such as ammonium cations (e.g choline and respective cations of benzathine, diethanolamine, diethylamine, meglumine, piperazine, procaine) or silver have not been used over the past 25 years.

On the other hand, there seem to be many more anionic counterions, most notably chloride, but also mesylate (e.g imatinib mesylate), besylate (amlodipine besylate), hydrocortisone succinate, and so on.

How can such a trend be explained in drug development process or chemistry of substances thereof? Could it simply be because there are fewer negatively charged drugs, or are there other explanations?

Here is what I managed to find:

In various researches, cationic counterions were associated with a high prevalence of unexpected incompatibilities in drug development processes. Owing to their low solubilities, (typically  <0.1% ) cationic counterions are infrequently used in many pharmaceutical preparations, however may be preferred to provide prolonged slow absorption while achieving effective plasma concentrations for some therapeutic uses, which is the case for benzathine and procaine penicillin G intramuscular injections.1 Furthermore cationic counterions, to a greater extent are susceptible to salting. Salting out results when highly hydrated inorganic ions (e.g $\ce{K+}$, $\ce{Na+}$) deprive organic ions and molecules of adequate water molecules to remain dissolved.

In addition another important factor influencing pharmaceutical salt selection is compatibility with excipients (diluents, binders, disintergrants, glidants, preservatives etc). Most amine derived counterions are incompatible with excipients. For example, the counterion of tromethamine (a primary amine) should not be used with reducing sugars as excipients because of a possible Maillard reaction, consequently limiting the available entities in the cationic counterions' pool .2

The trend can also be explained as a consequence of the changes in research techniques employed by the pharmaceutical industry. It can also be attributed to small absolute numbers of approved drug products containing salts formed from acidic entities.3

Overally, anionic counterions are favoured in drug development due to the wide availability of  weakly  basic pharmaceutically  active principles (approximately 75% of drugs are estimated to be weak bases) 4, consequently leading to correspondingly low availability of cationic counterions for acidic drugs which comprise only about 20%, with remaining percentage being neutral.

References

Newton D, W. Drug Incompatibility chemistry (American Journal of Health-System Pharmacy) Available from: http://www.medscape.com/viewarticle/590261.
Wu, Y., Levons, J., Narang, A. S., Raghavan, K., & Rao, V. M. (2011). Reactive impurities in excipients: Profiling, identification and mitigation of drug-excipient incompatibility. AAPS PharmSciTech, 12(4), 1248-1263.
Paulekuhn, G. S., Dressman, J. B., & Saal, C. (2007). Trends in active pharmaceutical ingredient salt selection based on analysis of the orange book database. Journal of Medicinal Chemistry, 50(26), 6665-6672.
Manallack, D. T. (2007). The pKa Distribution of Drugs: Application to Drug Discovery. Perspectives in Medicinal Chemistry 1( )25–38.


At what temperature can molecules no longer form? // hyper simplistic terminology in use for the question - apologies.

Molecules are formed by atoms bonding together, and those bonds have an equivalent energy.

At what temperature does the thermal energy of an atom exceed the bond energy, preventing it from forming any molecule?

I think an equivalent question is "what is the temperature needed to spontaneously dissociate any molecule into its component atoms?"

The bond will break if you can put more than its dissociation energy into it; $\ce{CO}$ has a huge dissociation energy of $\pu{1080 kJ/mol}$, $\ce{I2}$ less than most and is only $\pu{151 kJ/mol}$. This energy can be easily obtained using a laser of the correct wavelength. (You will need to convert units to electron volts to get wavelength needed. $\pu{1 kJ/mol} = \pu{1.036E-2 eV}$, $\pu{1eV} = \pu{8065 cm^-1}$).

However, when we excite using a laser we only use only a small range of energy, using heat at a given temperature is different as temperature is an average measure of a distribution of energies given by the Boltzmann expression $n_i \approx \exp(-E_i/k_\mathrm{B}T)$ where $n_i$ is the number of molecules with energy $E_i$ at temperature $T$, $k_\mathrm{B}$ is the Boltzmann constant. If  dealing with molar units replace $k_\mathrm{B}$ by $R$, the gas constant.

Thus, even at room temperature ($\pu{300 K}$),  there are, in principle, some molecules of $\ce{CO}$ that will be decomposed  even though the dissociation energy is so large; the chance of being decomposed is $\exp(-1080000/(300\times8.314) \approx 10^{-190}$. It is utterly minute even for $6\cdot 10^{23}$ molecules in a mole, but for iodine $\approx 10^{-27}$, still minute so these molecules are stable at room temperature. At $\pu{1000 K}$, 1 in $\approx 10^8$ $\ce{I2}$ molecule decompose and $10^{-4}$ at $\pu{200 K}$ so this is appreciable over a period of time.

(Strictly we should integrate from $T$ to infinity to get the exact answer but the exponential decays so quickly just one value gives us a good idea)


Xylan vs. Teflon for non-stick food pans? I am looking at non-stick pans and some are Xylan® others, and most, are Teflon™.

What are the chemical differences?
What is better for cooking?
What is better for health safety?

I do know that Teflon can be consumed without a problem, as it is never degraded, but at very high temperatures the fumes are dangerous.

Both Xylan and Teflon are chemically polytetrafluoroethylene. But a polymer just by name can mean a lot of different things if you consider the structure/arrangement, additives and impurities. These are highly process-dependent and, guess what, we do not know their chemistry for sure because they are industrially protected. For actual matters, both Xylan and Teflon are unsafe above 260 ⁰C (500 ⁰F), when they start to break down to toxic fluorine polymer fumes.


Spontaneous Reaction and Gibbs Free Energy I know how to derive the expression for the equilibrium constant starting from $\mathrm dG=0$,
$$\ln(K)=-\frac{\Delta G ^\circ}{RT}$$

However, if we are not at equilibrium, $\mathrm dG$ is not necessarily zero. In this case:
$$\frac{\partial G}{\partial \epsilon} = \sum \mu_i \nu_i = \Delta G ^\circ + RT \ln \left( \prod a_i ^{\nu_i} \right)$$

Where $\nu_i$ are the stoichiometric coefficients, $\mu_i$ are the chemical potentials, $\Delta G^\circ$ is the Gibbs free energy of reaction at standard state and $a_i$ are the activities.

Hence,
$$\frac{\partial G}{\partial \epsilon} = RT \ln \left(\frac{Q}{K} \right)$$

Where $Q=\prod a_i ^{\nu_i}$ is the reaction quotient.

So the question is, if we want to kown if a reaction is spontaneous, do we look at the ratio $Q/K$? If this is correct, if $Q > K$ then $\frac{\partial G}{\partial \epsilon} > 0$ and the reaction would be non-spontaneous, whereas if $Q < K$ then $\frac{\partial G}{\partial \epsilon} < 0$ and the reaction is spontaneous.

If so, if we start with reactants only, $Q=0$ (since the activites of the products are zero because there are none) and the reaction would be spontaneous, and we would always get at least a little bit of products.

However, various pages in internet say that if $\Delta G ^\circ < 0$ then the reaction is spontaneous, and it is not if $\Delta G ^\circ>0$.

So how do you know, then, if you start with only reactants, if a reaction is going to happen?

I think you are confusing two related concepts here. Spontaneity of a reaction is determined by the net change in free energy, $\Delta G$ - notice the capital Delta.

Which direction you will move along a reaction coordinate is determined by $dG$ - note the little $d$.

$\Delta G$ is the overall change in free energy when comparing products to reactants. It tells you whether the reactants are "uphill" or "downhill" of the products.

$dG$ is the slope of the free energy surface at your current position on the reaction coordinate. In the equation you gave, that's given by Q/K. K basically tells you what the equilibrium ratio between reactants and products is, and Q tells you what the current ratio is. The ratio of Q/K is an indication of how far to the left or right you are of equilibrium.

This picture might help explain it a little better:

Image the free energy surface as a gravitational potential energy surface (we can get away with this because free energy is the thermodynamic potential for systems at constant T & P). The ball represents the reacting system, and the x-axis is a reaction coordinate.

$dG$ (the differential) tells us the slope at a given point - it tells us where the system is going to go right now. $\Delta G$ tells us the overall change - it tells us where the system is going to eventually end up.

When you look up standard state $\Delta G$ values for a reaction, you are determining whether an overall reaction is spontaneous or not - you are figuring out which side of the equilibrium has more stuff - reactants or products.

When you look at Q, you are figuring out which direction things need to go to get back to equilibrium, based on where you currently are.