How to proceed when percentage "exchange" is given in a reaction? If $3.7~\mathrm{mol}$ of $\ce{Cu}$ reacts with an excess of $\ce{Cl2}$ and the exchange is $45\,\%$, how many moles of $\ce{CuCl2}$ can at maximum be created?

So I start with write the reaction formula
$$\ce{Cu + Cl2 -> CuCl2}$$
and then balance it.
$$\ce{2Cu + 2Cl2 -> 2CuCl2}$$
I know that I have $3.7\ \mathrm{mol}$ of $\ce{Cu}$, which is $235.1202\ \mathrm{g}$ of $\ce{Cu}$.
I assume that $\ce{Cu}$ is limiting reagent and $\ce{Cl2}$ is an excess reagent.

This is where I get stuck, I don’t know how I’m supposed to proceed with the only information that the exchange is $45\,\%$.

We have this equation of reaction:

$$\ce{Cu + Cl2 -> CuCl2}$$

Which means that $1~\mathrm{mol}$ of $\ce{Cu}$ reacts with $1~\mathrm{mol}$ of $\ce{Cl2}$ to form $1~\mathrm{mol}$ of $\ce{CuCl2}$.

Therefore, we would expect $3.7~\mathrm{mol}$ of $\ce{Cu}$ to produce $3.7~\mathrm{mol}$ of $\ce{CuCl2}$ if there is excess $\ce{Cl2}$, which is stated in the question.

However, the exchange is only $45\%$, which means that only $45\%$ of $\ce{Cu}$ is converted, making we can only create $3.7\times45\%=1.665~\mathrm{mol}$ of $\ce{CuCl2}$.


Why is adenine aromatic? I don't think it follows the $4n + 2$ rule. The double bonds give 8 π electrons. Then the lone pairs on the $\ce{NH}$ and $\ce{NH2}$ groups are delocalized (are they?) so they must contribute 4 π electrons. Thus in total there are 12 π electrons, which doesn't follow Hückel's rule. So, why is it aromatic?

The $4n+2$ rule applies only for a conjugated ring.  If you assume that the amino group contributes a pair of pendant  $\pi$ electrons into the ring system then perforce you do not have a conjugated ring.  And even if you don't assume such a delocalization from the amino group, the cross-link within the ring system is still a stretch for the $4n+2$ rule.


Acid etching of float glass While we have done some etching of glass with hydrofluoric acid alone, to achieve a matte white finish on float glass, I am curious to know what formula and procedure is required to create a more transparent and smooth finish ?

(Lead crystal polishing is achieved after diamond grinding, dipping in a mixture of $\ce{HF}$ and $\ce{H2SO4}$, and alternating with rinsing.)

Can anyone advise on the possible advantages of mixing another acid with the $\ce{HF}$?

Alkaline etching of glass tends to be less rapid than HF and leaves a smoother surface. This may not be what you do for the first etch on fresh glass, but it may smooth out the etch somewhat after an HF etch. The smoothing may also be caused by silicate ions that form in the alkaline etch fluid, so perhaps a solution of potassium (or sodium) silicate (ortho or meta) would be gentler. A polysilicate might be too slow.


Deriving goldman equation So I've been trying to follow the derivation of the Goldman-Hodgkin-Katz equation on Wikipedia (https://en.wikipedia.org/w/index.php?title=Goldman_equation ) - I can follow it through after it sets up the initial model.

However, I'm having trouble using the Einstein-Stokes relationship to derive the initial model in the first place.

There is the flux from diffusion down the chemical gradient, $-D_{A}\frac{d[A]}{dx}$. The negative sign is there because I'm imagining the ions being "lost" from the side I'm considering.

Then we have the electrical field pushing ions the other way. This won't depend upon the rate of change of concentration because the electric field is uniform and so there's effectively a constant force on each of the particles. The flow will instead depend upon the constant velocity that these particles will be pushed at. So the flow will be $[A]v_d$ - the moles per unit volume being multiplied by the speed in the direction towards the membrane will give me the number of moles per unit area passing through the membrane per second.

So all I need to do is work out $v_d$ I'm guessing. So I use the electrical mobility equation, $D_A=\frac{\mu_{q}k_BT}{q}$.

$q$ will be the charge, so $nzF$ (where $z$ is the valance, $nz$ gives the moles of electrons and multiplying by $F$ gives the charge).

$\mu_q$ will be $\frac{v_d}{E} = \frac{v_d L}{V_m}$

$k_B = \frac{R}{N_A}$ where $N_A$ is Avogadro's number.

So $D_A=v_d \frac{L}{V_m} \frac{RT}{zF} \frac{1}{N_A n}$, thus $D_A \frac{V_m}{L} \frac{zF}{RT} N_A n=v_d$

Putting this back in to get the ion flux, $D_A \frac{V_m}{L} \frac{zF}{RT} N_A n [A]$.

Adding my two fluxes together to get the total flux, I get: $j_a = -D_A(\frac{d[A]}{dx} - \frac{V_m}{L} \frac{zF}{RT} N_A n [A] )$ - this is not quite what the Wikipedia article has - a stray $n N_A$ has crept in, somehow!

Please help me by pointing out where I'm making a mistake!

The mistake is in the charge $q$, that you have put in:

$
D_A=\frac{\mu_{q}k_BT}{q}
$

$q$ is simply the particle charge (and not the whole charge, as you've unfortunately considered) and should then be expressed as:

$q=\frac{zF}{N_{A}}$

with $F$ the Faraday constant (charge / mole ), ${N_{A}}$ the Avogadro's number (number of particle / mole) and $z$ the charge valence.

${N_{A}}$ will finally cancel out when you divide Boltzmann's constant $k_B$ by the the charge $q$, in the definition of diffusion coefficient. That will allow you to recover the Goldman equation found on Wikipedia.

By checking all that through an always useful dimension analysis for $D_A$ and using the defintions you have given in your question for ($\mu_q$ = $\frac{v_d}{E} = \frac{v_d L}{V_m}$ and $k_B = \frac{R}{N_A}$):

$
\frac{\mu_{q}k_BT}{q}=\frac{m/s\,\times m}{V}\times \frac{J/(mol\times K)\,\times K}{{N_A}\times C/mol}\times{N_A}=\frac{m^2}{s}
$

you find the correct units for a diffusion coefficient.


Multiplication symbol in SMARTS There is a SMARTS pattern that was published which looks like the following[1]

[#6 × 4]

and there are actually spaces around the × symbols between the 6 and 4.

Question is, should I change this multiplication notation to a lower case "x", or really use some sort of ASCII symbol?  I didn't see a multiplication symbol anywhere in the Daylight SMARTS description.

References:

Hou, T. J.; Xia, K.; Zhang, W.; Xu, X. J. ADME Evaluation in Drug Discovery. 4. Prediction of Aqueous Solubility Based on Atom Contribution Approach. J. Chem. Inf. Comput. Sci. 2004, 44 (1), 266–275. PDF

The question was answered on the Open Babel mailing list

Suffice to say, there's no multiplication symbol in SMILES or SMARTS - only standard ASCII characters. It's likely a lower-case 'x' that's been mangled by a journal.


Does carbonation permanently increase the acidity of soda water? If you carbonate water, and then expose the resulting soda water to the atmosphere and wait until it goes completely flat, will the (former) soda water end up more acidic in final equilibrium than water that was never carbonated and is also exposed to the atmosphere?
(Assume that both the water that was and was not carbonated are exposed to the exact same atmosphere, so that (a) they end up at the same temperature, and (b) the tiny addition to the level of atmospheric CO2 released during the carbonation process itself doesn't have any differential impact between the two waters.)
I understand that the soda water will be significantly more acidic than the flat water while it's still bubbly, because of the carbonic acid formed by the water and the injected pressurized CO2. I also believe that both waters will end up slightly acidic (although not as acidic as the soda water while it's still bubbly) because of the (modest) partial pressure of atmospheric CO2 in contact with both waters. My question is whether the high concentration of carbonic acid introduced into the soda water during the carbonation process permanently changes the equilibrium acidity of the water.
I'm also curious whether the answer would be different depending on whether you start with pure distilled water vs. "realistic" tap water, where the presence of impurities might somehow change the story.
My strong physical intuition is that the flattened (former) soda water will eventually "forget" about the carbonation and both waters will end up at the same equilibrium acidity, but some chemist friends disagree with me.

Both water samples ( assuming pure water ) would theoretically end up slightly acidic with $\mathrm{pH} \approx 5.6$, but the carbonated one after long time. Tap water would have its baseline $\mathrm{pH}$ determined by its $\ce{CO2(aq)/HCO3-(aq)}$ $\mathrm{pH}$ buffer.
This is valid for an ideal case, as side processes - especially for tap water -  will affect the result. As examples come to mind microbial/algae consumption of carbon dioxide, forming compounds affecting $\mathrm{pH}$, or precipitation of carbonates.


Alternative definition of the mole? 1 amu (atomic mass unit) is defined as the one twelfth of the mass of a carbon-12 atom. The mole is defined as the number of carbon-12 atom in 12 g of carbon-12. In other words,

$$\require{cancel} 1 \textbf{ mole} = \dfrac{12 \textbf{ g}}{m(^{12}C)} = \dfrac{\cancel{12} \textbf{ g}}{\cancel{12} \textbf{ amu}} = \dfrac{\textbf{ g}}{\textbf{ amu}} $$

I believe that the 12 g were chosen so that the mole could exacly express the ratio between the g and amu.

Consider now this alternative definition : the amu is defined as the exact mass of a proton and the mole is exact ratio between the g and the amu. Wouldn't it be simpler this way?

What are the advantages of one definition over the other? Is there any reason we would want to keep the former? Why scientists decided to define the mole like that and not pick a simpler definition like the one I proposed?

One of the reasons to take Carbon-12 could be that it's hard to get your hands on one mole of pure hydrogen (and to weigh it). Carbon-12 is a (room temperature) solid which is relatively easy to produce, keep pure, store and weigh.


Is 1,2-dihydronaphthalene aromatic? According to Hückel's rule a compound is aromatic if it has conjugation throughout the compound and the number of conjugated electrons is $4n+2$. This compound does not satisfy both:

The benzene ring itself is aromatic, though. Is the compound considered aromatic because of this?

Hückel's rules are a very concise set of very strict rules. Most aromatic compounds do not comply with these.

  Hückel (4n + 2) rule
  Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0 – 5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system. Examples of systems that obey the Hückel rule include:

  cyclopropenyl cation (m = 3, n = 0)
  cyclopentadienyl anion (m = 5, n = 1)
  benzene (m = 6, n = 1)

  Systems containing 4n π-electrons (such as cyclobutadiene and the cyclopentadienyl cation) are 'antiaromatic'.
  From IUPAC Gold Book, DOI: 10.1351/goldbook.H02867

Based on this it is actually quite difficult to assess aromaticity to more complex molecules. Aromatic molecules that do not comply with Hückel's rules are for example naphthalene, thiophene, or pyridine, among many more. A more suitable definition, which also makes quite clear how fuzzy the whole subject is:

  aromatic

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1HNMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

  From IUPAC Gold Book, DOI: 10.1351/goldbook.A00441

While we tend to say a molecule is aromatic, that is hardly ever the case. Think about substituted benzene molecules; while the phenyl ring retains aromatic character, the side chains are not necessarily aromatic. When describing the molecule it is better to refrain from absolutes.

Within this framework it is possible to identify molecular entities that posses aromatic character and refer to these. Strictly speaking the answer that 1,2-dihydronaphthalene is aromatic is therefore incorrect. Better is to say that because of the phenyl subunit, 1,2-dihydronaphthalene possesses some aromatic character. In this particular case you could even argue that because of the additional double bond, the  main reactivity of this molecule is more similar to alkenes than it is to aromatic compounds.
Aromatic character is sometimes like a unicorn: it's hard to describe in unambiguous terms, but when you see it, you'd probably know what it is.

There is another definition of relevance in the IUPAC Gold Book, which actually gives a clearer, more systematic aproach for determining aromatic properties, and can be seen as an extension of the above:

  aromaticity
  The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.
  From IUPAC Gold Book, DOI: 10.1351/goldbook.A00442

One of the key elements is the thermodynamic stability and the tendency of retaining the structural type during the course of a reaction. This certainly reflects the benzene-like aspect of the earlier definition.
The last part comes directly from the original (4n + 2) criterion, but significantly weakens it, as it is not reduced to the very symmetry restricted number Hückel derived from the hydrocarbon cycles.
For simple compounds like thiophene it is easy to construct an approximate MO diagram (if you for further simplification ignore the σ-sub-structure). And the same can be done more complex molecules like 1,2-dihydronaphthalene.
In general I am not a big fan of oversimplifying complicated situations; questions that are asking for absolute statements are always oversimplified. In case of aromaticity it is tempting to reduce it to a simple electron-count, which therefore still prevails in textbooks although horribly incomplete; but it is also dangerous, because it neglects a lot of the underlying chemistry.

Aromaticity is a beast; fuzzy definitions, or criteria, the ongoing debate among scientists only show that you need to be open minded and accept that many things are not as easy as it seems at first glance.
For the time being you can help yourself by identifying molecular entities, or moieties that are derived from (simpler) aromatic compounds and check whether a conjugated π-system was destroyed while creating them. As one of the key features you can be almost certain, that aromaticity is retained in all but the most extreme cases.


Cycloaddition vs Electrocyclic Pathway Is there a molecular orbital/Woodward-Hoffmann explanation as to why Diels-Alder reactants don't commonly undergo electrocyclization instead of cycloaddition?

The Woodward-Hoffmann rules only tell you whether a reaction is thermally allowed or not, they do not give any indication of the relative rates of multiple allowed reactions.

I guess you're asking why the 4π component in a [4+2]-cycloaddition doesnt undergo a conrotatory 4π electrocyclic ring-closure? This is plausible, but I suppose even if it did happen the cyclobutene product is highly strained so the equilibrium would lie in favour of the ring-opened diene anyway.


Electron flow: How does osmium tetroxide react with alkenes? This is a question that has bothered me for some time. I have always thought that the part of one molecule will only attack part of another molecule if it has a higher positive charge. So with this in mind, why does the alkene double bond attack the $\ce{O}$ atom of $\ce{OsO4}$. Isn’t the $\ce{O}$ atom negatively charged?

I have a suggestion that could explain the attacK of electron pair of alkene pi bond to the oxygen of osmium tetroxide : The electron pair of alkene pi bond repels the electron density of the oxygen when the osmium tetroxide approaches the alkene, creating a temporary δ+ on the oxygen that is closer to the pi bond of alkene. The electron-rich alkene then attacks the electron-poor oxygen.

Dimitrios Kariotoglou


Any substance too poisonous to measure an LD50? A recent question on poisons was answered in part by bringing up the concept of $\pu{LD_{50}}$ and animal testing and so on; none of that was new to me, but it did bring to mind a question I don’t know the answer to:

Has there ever been a substance that scientists have actively tried to determine $\pu{LD_{50}}$ for, and could not because they were unable to determine a dosage low enough to kill “just” half of the subjects?

Please note that this question is not restricted, strictly, to any technical definition of “poison.” Rather, answers must meet these requirements:

There must be research (considering that I am looking for something that might be a negative result and those do not get published as much as they should be, I won’t demand peer-reviewed publication, but the goal should have been to attempt peer-review publication if it had worked out—effectively, I mean serious work), and
that research must use the term “$\pu{LD_{50}}$” to characterize the substance.

Basically, I don’t want to get into debates about what is, or is not, a poison here. If a researcher is willing to call something $\pu{LD_{50}}$, then I am willing to accept it as a “poison” for the purposes of this question.

I also don’t want speculation, or for a user here to characterize something as $\pu{LD_{50}}$ when the underlying research doesn’t call it that. It is not enough to say, for example, “well I’m sure even a single atom of antimatter inside your body would be pretty bad,” you need to cite a particular researcher who has performed experiments with the goal of determining what they themselves called $\pu{LD_{50}}$ for that substance.

I suspect the answer is no, but I have no idea how to research something like this.

Of course no. Botulotoxin is probably the strongest known, and still its $\rm LD_{50}$ is counted in nanograms per kilogram, which is pretty manageable. Sure, working with such tiny amounts requires some special measures, but still, it is way greater than one molecule. You can divide it again, and again, and again.

So it goes.


Book reference for electronic occupation of TM Is there a book I can reference that has a table like this? I can't reference an online source on my thesis and all the pictures I have found online like this don't make their reference explicit.

Thanks!

Element Name and Symbol + Atomic Number + Electron Configuration

NIST Standard Reference Database 111 provides an up-to date reference for the electron configuration of neutral atoms in the ground state [1] in both tabulated and periodic table formats.

Common Oxidation States

An overview of elements' oxidation numbers is provided in practically every inorganic chemistry textbook. Arguably, the most extensive one is given throughout Greenwood and Earnshaw's Chemistry of the Elements [2].

More compressed information is provided in CRC handbook of chemistry and physics (Section 4. Properties of the Elements and Inorganic Compounds) [3, pp. 4-1 — 4-92].
There is also a periodic table at the end which includes oxidation numbers of the elements in the upper right corner of each cell:

References

Kramida, A., Ralchenko, Yu., Reader, J., and NIST ASD Team (2018). NIST Atomic Spectra Database (ver. 5.6.1), [Online]. Available: https://physics.nist.gov/asd. National Institute of Standards and Technology, Gaithersburg, MD. DOI: https://doi.org/10.18434/T4W30F
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3


Isomerization reaction involving carboxylic acid derivatives I have tried searching for what type of organic reaction this is in my textbooks and ended up with nothing. Thus, I could not fully understand and refer to the mechanisms that the solution has explained.
So, please tell me the types of organic reactions (i.e. name the reactions involved for me) that are involved in this question. If possible you can describe how the mechanism of this reaction. Thank You!

tl;dr A slightly anthropomorphic approach with some basis in general organic chemistry:

Step 1: The protonation of the carbonyl (An acid-base reaction)

Here, we have three options – we can protonate the nitrogen (deactivated by the resonance with oxygen), the alcohol group ($\mathrm{sp^3}$) or the carbonyl group ($\mathrm{sp^2}$). As you can see this is basically an acid-base reaction and therefore the strongest base (having highest electron density) would react the fastest and this means since the carbonyl has the highest electron density here the carbonyl is protonated.
Step 2: The intermolecular attack ($\ce{O}$ is very sneaky)

Well, we began with $\ce{HCl}$. Time for the $\ce{Cl-}$ to make an appearance? Not really. As you can see, we have an alcohol group in the compound which has electrons to donate. The carbon on the (protonated) carbonyl group is highly electropositive (It was positive enough when there was no positive charge on the oxygen and now it just got more electropositive). Now, the oxygen with the lone pair makes an appearance. It attacks the carbonyl carbon and at the same time the π electrons move to the previously protonated oxygen making it neutral.
Step 3: An intermolecular proton transfer (Another acid-base reaction?)

Now, as you may have learnt, oxygen is not the most electropositive element. If possible, it doesn't stay positive. Nitrogen comes to the rescue. Another acid base reaction takes place where the newly active amine group deprotonates the oxygen forming $\ce{>NH2+}$.
Step 4: The final rearrangement (We're done here(?), you can figure it out)

We still need the final isomer and we also need to remove the positive charges on the compound before we end. Therefore, the ring breaks and the electrons move to the nitrogen removing the positive charge on it. This also means a carbocation is formed in the carbon where the alcoholic group is attached which stabilises it using resonance to form a $\ce{C=OH+}$ bond. (try drawing this step).
But we aren't done.
Step 5: The true final step (A small deprotonation)

Nobody needs any positive oxygens, do we? Therefore the oxygen is deprotonated. (Note: this is in equilibrium since medium is acidic). All $\ce{H+}$ used has been released which also makes this a catalytic reaction.
There might be one final question in the back of your mind (for the curious).
Why not go back to the starting compound?
This is because as @user55119 very well pointed out in a comment to the other answer. My additions in italic.

In the absence of acid, heating amino ester (a) would give the starting material, the hydroxyamide because esters are more labile (easily broken down or displaced) than amides. The reason this reaction can occur is that the amino group is protonated in acid which shifts the equilibrium.

Welcome to the realms of organic chemistry.


Preference of locants in organic nomenclature Why can't we write the compound 5-ethyl-4-methyloctane as 4-ethyl-5-methyloctane?

As pointed out by Loong, the name you prefer is the correct name: 4-ethyl-5-methyloctane.

Generally we try to do the following:

Assign the smallest locant possible to the carbon bearing the highest priority functional gorup.
Assign locants to generate the lowest possible sum of the locants.

Often it is straightforward. In this case, both assignments of generate locants of 4 and 5, which sum to 9. Both place an alkyl substituent at the lower locant.

In this case, the locants are assigned so that the alphabetically first of the two substituents of equivalent priorty receives the smallest locant.

  Correct - 4-ethyl-5-methyloxctane: ethyl is alphabetically before methyl and should get a smaller locant

  Incorrent - 5-ethyl-4-methyloctane: ethyl is alphabetically before methyl and should get a smaller locant


About how much carbon monoxide is produced in these two different reations? The U.S. Department of Agriculture produces a gas cartridge to poison subterranean rodents.  The product is provided as a cardboard tube sealed at one end and then loaded with a mixture of 144.6 grams of the following ingredients:

53.0% Sodium Nitrate
28.0% Charcoal
19.0% Inert Ingredients

Once filled, the cardboard tube is sealed at the other end with a thin cardboard cap which can be punctured and a fuse inserted when ready to be used.  The gas cartridge's advertised function once lit is to generate a large volume of carbon monoxide that kills the rodent within a few minutes in its burrow.  My experience has been it is very effective when used according to the instructions.

The USDA gas cartridge is getting harder to obtain anymore and a commercial producer sells a similar product which is utilized in the same applications.  But it is smaller and uses a somewhat different formulation.  It does not advertise how the rodent is poisoned, so I don't know if the resulting gas from combustion is primarily CO or something else.  My experience has been that it doesn't seem to be as effective as the USDA gas cartridge, but I don't know if it is because of its different formulation or just its smaller size.  The 56.7 gram mixture of ingredients in this commercial gas cartridge are as follows:

50.0% Sodium Nitrate
38.0% Sulfur
9.0% Charcoal
3.0% Inert Ingredients

The question is how much carbon monoxide is generated by each of these cartridges when their ingredients are ignited and what are the quantities of other resulting gases?  Just straight theory here is good enough for me.  I know there are probably variables such as how well the ingredients are ground and mixed and what are used as the inert ingredients in each.  But lets just leave those variables out of the equation.

About the best that can done is simply to look at the ratio of carbon in the two cartridges.

    $\dfrac{56.7*0.09}{144.6*0.28} = 12.6$%

The ratio of carbon dioxide to carbon monoxide formed on burning would depend on particle size, how well mixed an so on. The second formulation with sulfur is akin to black gun powder. The sulfur burns first at a lower temperature to ignite the charcoal. The sulfur dioxide would be toxic as well, but not as fast acting as carbon monoxide.


Relationship between a temperature increase of an exothermic reaction, Keq, and kf (forward rate constant) I'm currently stuck on problem #30 for the Chemistry Olympiad 2015 local exam which reads:

For a reversible exothermic reaction, what is the effect of increasing temperature on the equilibrium constant ($K_\mathrm{eq}$) and on the forward rate constant ($k_\mathrm{f}$)?
(A) $K_\mathrm{eq}$ and $k_\mathrm{f}$ both increase
(B) $K_\mathrm{eq}$ and $k_\mathrm{f}$ both decrease
(C) $K_\mathrm{eq}$ increases and $k_\mathrm{f}$ decreases
(D) $K_\mathrm{eq}$ decreases and $k_\mathrm{f}$ increases

The answer for this problem is D.
My approach:
To find the relationship between $K_\mathrm{eq}$and $k_\mathrm{f}$, I first used this theoretical reaction of
$$\ce{A -> B + \mathrm{heat}}$$
The forward rate can be defined as
$\mathrm{rate(f)} = k_\mathrm{f}[\ce{A}]$
and the backwards rate as
$\mathrm{rate(b)} = k_\mathrm{b}[\ce{B}]$
Since we're finding the relationship between these rates with $K_\mathrm{eq}$, rate(f) = rate(b) because the reaction is at equilibrium. This allows the equation
$k_\mathrm{f}\ce{[A]} = k_\mathrm{b}\ce{[B]} $
to be set up. Dividing both sides by $k_\mathrm{b}$ and [A] gives us
$k_\mathrm{f}/k_\mathrm{b} = \ce{[B]/[A]}$
And
$K_\mathrm{eq} = \mathrm{ products/reactants }= \ce{[B]/[A]} = k_\mathrm{f}/k_\mathrm{b}$
Due to the fact that the temperature is raised with an exothermic reaction, the reaction is reactant favored and more A will be formed. Since
$K_\mathrm{eq} = \mathrm{ products/reactants }= \ce{[B]/[A]}$
$K_\mathrm{eq}$will decrease. However, why doesn't $k_\mathrm{f}$ decrease? Since
$K_\mathrm{eq}= k_\mathrm{f}/k_\mathrm{b}$,
how could $k_\mathrm{f}$ increase while $K_\mathrm{eq}$ decrease if they're equal to each other in that equation? Shouldn't $k_\mathrm{b}$increase or $k_\mathrm{f}$ decrease to "=" to the decreased value of $K_\mathrm{eq}$?

You know that both the forwards $k_f$ and reverse $k_r$ rate constants will increase with temperature, simply because there is more energy present and so it takes less energy to surmount the activation barrier, both from the product and reactant sides.

The question is then which is changed more by changing the temperature. The equilibrium constant is as you write $K_Q=k_f/k_r$ and now using the Arrhenius equation we get
$$K_Q=\frac{A_f\exp(-\Delta E/(RT))}{A_r\exp(-(\Delta E+E_0)/(RT))}$$

where the A's are pre-exponential factors (and are constant vs temperature) and $\Delta E$ the activation barrier for the forwards reaction and $E_0$ the extra energy for the back reaction and  is positive since the reaction is exothermic. (If you sketch the barriers hopefully this should be clear).

Simplifying gives
$$K_Q=\frac{A_f}{A_r}\exp(E_0/(RT))$$

As T  is increased $K_Q$ becomes smaller. Thus the answer is (D).


Determine surface area of heat exchange tube A heat exchanger is used to cool down a stream of $\pu{20 kg s-1}$ of water from $\pu{80 °C}$ to $\pu{60 °C}$. The cooling water enters at a rate of $\pu{12 kg s-1}$ and $\ce{20 °C}$. The overall heat transfer coefficient is $\pu{2 kW m-2 K-1}$ and the heat capacity of water is $\pu{4.183 kJ kg-1 K-1}$

Calculate the surface area.

I am stuck on what the last step of the calculations should be.
I began with calculating the outlet temperature for the cooling stream using that $q$ is constant, I set IN = OUT:
$$m_hC_pΔT_\mathrm{in} = m_cC_pΔT_\mathrm{out}\label{eq:1}\tag{1}$$
From $\eqref{eq:1}$ I got that $T_\mathrm{2out} = \pu{53 °C}$
Then I calculated the logarithmic temperature difference, which I got to be
$ΔT_\mathrm{ln} = \pu{-33.075 °C}$.
After this I am confused as to how I can get the surface area of the tube.
Edit
I did a heat balance, IN - OUT = 0:
$$(m_hC_pΔT_\mathrm{in} + m_cC_pΔT_\mathrm{in}) - (m_hC_pΔT_\mathrm{out} + m_cC_pΔT_\mathrm{out}) = 0$$
simplified into:
$$m_hC_pΔT_\mathrm{in} - m_hC_pΔT_\mathrm{out} - UAΔT_\mathrm{ln} = 0$$
However when I try to solve for A I get a negative value, which isn't correct.

The outlet temperature  should be 53.3.  The log-mean temperaturr difference should be positive.  Calculate the heat load and divide by the log-mean temperature difference and by the heat transfer coefficient to get the area.

ADDENDUM

This addendum is for the benefit of @MaxW (the downvoter).

The heat load $\dot{Q}$ of the heat exchanger is given by:  $$\dot{Q}=\dot{m}_HC(T_{H,in}-T_{H,out})=20(4.183)(80-60)=1673\ kW$$This heat load represents the rate of heat transfer from the hot fluid to the cold fluid.
The exit temperature of the cold stream is determined from $$\dot{m}_CC(T_{C,out}-T_{C,in})=12(4.183)(T_{C,out}-20)=\dot{Q}=1673$$

Solving for the exit temperature of the cold stream gives  $$T_{C,out}=53.33\ C$$
3.  The log-mean temperature difference is given by:
$$(\Delta T)_{LM}=\frac{(T_{H,in}-T_{C,out})-(T_{H,out}-T_{C,in})}{\ln{[(T_{H,in}-T_{C,out})/(T_{H,out}-T_{C,in})]}}=\frac{26.67-40}{\ln{[26.67/40]}}=32.88\ C$$

The heat transfer area A is determined from the equation:  $$U_oA(\Delta T)_{LM}=2.0A(32.88)=\dot{Q}=1673$$Solving this equation for the heat transfer area A gives:
$$A=25.4\ m^2$$

Questions?


Transition from ground state to excited state I don't understand how to correctly find the term symbol of an excited state. I know that the first two permitted transitions for the Ne atom are 16.8eV and 19.8eV and I want to find the electronic configuration and the term symbol.

I assume that from the $2p$ orbital I'll have a transition to the $3s$ orbital (first excited state) and to the $4s$ orbital (second excited state). The selection rule $\Delta l=\pm1$ is obeyed. But what about the total angular momentum $j$? To find $j$ I have to know something about the spin $s$. Is it something about singlet and triplet states?

First you need to have the ground state electron configuration, for Neon this is 1s$^2$2p$^6$ or what ever it is for your atom. This will give you the term symbol for the ground state. The excited state has one electron promoted, say p to a d orbital, then 1s$^2$2p$^5$3d$^1$.

A term symbol has the form $^{2S+1}L_J$, where S is the total spin ($2S+1$ is the multiplicity) , L the orbital angular momentum and J the total angular momentum.

Because angular momentum is added vectorially it is necessary to follow some rules to get all the values. Each type of angular momentum forms a series of values (the Clebsch-Gordon series). If total spin angular momentum is S and for each electron the spin quantum number is s1 and s2 then the series is $S= |s1+s2|..|s1-s2|$ with steps of 1 between values. Note that all values are positive and there may only be one value in the series. The same type of series applies to L and J total spin and orbital angular momentum.

A p$^5$ can be considered as the same as p$1$ as only 1 electron is unpaired. Thus using the series if there are two electrons each with spin quantum numbers s1, s2 = 1/2 each the the total spin for two electrons is S = 0 or 1 and the spin multiplicity is 1 (a singlet) or 3 a triplet. The series is |s1+s2| = 1, and |s1-s2| =0.

The L values is calculated with l1=0 (s orbital) and l2=1 (p orbital) thus L only has 1 value = 1. Thus far the possible term symbols are $^1P$ and $^3P$.

The total angular momentum  J is found in the same way as S and L values were using the Clebsch-Gordon series, $$J= |L+S|...|L-S| $$ Again this means find the maximum value, then the minimum make it positive, and separate any other values by 1. Sometimes there is only 1 value so be careful. There are $2J+1$ degenerate J levels, these are split in an external magnetic field into separate states.

This for the triplet state $S=1$ and so $J=|1+1|..|1-1|$, or 2, 1 and 0. (note three values from the series). Thus the term symbols are $^3P_2$ , $^3P_1$, $^3P_0$. But there is also a singlet term so the J values for this are $J=|1+0|..|1-0| = 1$ (one term only). Thus the last term symbol is $^1P_1$. Although this looks rather complicated it becomes quite quick with a bit of practice.

Once you have all the term symbols you can then apply selection rules to see which transitions are allowed.

Finally if the configuration contains, say, p$^2$ then one has also to consider the Pauli principle and it is more tricky. The way to do this is to list all s1, s2, l1, l2 values and remove any with the same set of quantum numbers. Finally when you have the term symbol Hund's rules are used to determine which levels and states are lowest in energy. (Many text books work through the p$^2$ case, See for example Steinfeld, 'Molecules & radiation')


Relation between the virial coefficients and van der Waals constants I don't know where to begin with this question.

A function $f(x)$ can be expressed as a special power series called the Taylor expansion. The common expansion is $(1-x)^{-1}=1+x+x^2$ Note that the higher terms become neglibible small and usually are not included in expansion. Find the correlation between the virial coefficients and $a$ and $b$ in the corresponding van der Waals equation of state. Start with the equation shown below and compare it with the virial equation of state.
$$ p=\frac{RT\left(1-\frac{b}{V_{\mathrm m}}\right)^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2}$$

The virial equation of state is

$$ p=\frac{nRT}{V}\left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+\ldots\right)$$

I get that there are $A$, $B$, $C$… coefficients, and that the $A$ coefficient should be 1.
The best I have is that:
$$\begin{align}
A &= 1 \\
B &= -\left(\frac{b}{V_{\mathrm m}}\right)^{-1} \\
C &= -\frac{a}{V_{\mathrm m}^2}
\end{align}$$
If I were to relate that to the terms in the Taylor expansion, then $p=(1-x)^{-1}$ and then $A$, $B$, $C$… are the terms on the right side of the equation.
Is this the correct approach? Can the virial coefficients have a negative value?

You have to bring both equations into the same form and then compare the coefficients. You seem not to have used the tip with the Taylor expansion and I guess that's why you got an incorrect result.

As you wrote: The virial equation of state is

\begin{align}
p &=\frac{nRT}{V} \left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+ \ldots \right) \\
  &= \frac{RT}{V_{\mathrm{m}}} \left(1+ B\frac{1}{V_{\mathrm{m}}}+ C\frac{1}{V_{\mathrm{m}}^2}+ \ldots \right)
\end{align}

and the van der Waals equation is

\begin{align}
 p &=\frac{RT(1-\frac{b}{V_{\mathrm m}})^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2} \ .
\end{align}

Now, you have to bring the van der Waals equation into the same form as the virial equation by factoring out $\frac{RT}{V_{\mathrm{m}}}$

\begin{align}
 p &= \frac{RT}{V_{\mathrm{m}}}\left( \left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} - \frac{a}{R T V_{\mathrm{m}}} \right) \ ,
\end{align}

using the Taylor expansion $\left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} = 1 + \frac{b}{V_{\mathrm{m}}} + \frac{b^2}{V_{\mathrm{m}}^2} + \ldots$ and ordering the terms according to their power in $\frac{1}{V_{\mathrm{m}}}$

\begin{align}
 p &= \frac{RT}{V_{\mathrm{m}}}\left( 1 + \left(b - \frac{a}{R T} \right) \frac{1}{V_{\mathrm{m}}} + b^{2} \frac{1}{V_{\mathrm{m}}^{2}} + \ldots \right) \ ,
\end{align}

so that you can compare the coefficients in both equations to get the result

\begin{align}
 B &= b - \frac{a}{R T} \\
 C &= b^2 \ .
\end{align}


Base used in Claisen Condensation Here's what I read about which base to use in Claisen Condensation reaction (two esters or ester + carbonyl compound forming a beta keto-ester or beta-diketone):

  The base used must not interfere with the reaction by
  undergoing nucleophilic substitution or addition with a carbonyl
  carbon. For this reason, the conjugate sodium alkoxide base of the
  alcohol formed (e.g. sodium ethoxide if ethanol is formed) is often
  used, since the alkoxide is regenerated. In mixed Claisen
  condensations, a non-nucleophilic base such as lithium
  diisopropylamide, or LDA, may be used, since only one compound is
  enolizable. LDA is not commonly used in the classic Claisen
  or Dieckmann condensations due to enolization of
  the electrophilic ester.

Here's what I didn't understand, particularly -

Why is the conjugate sodium alkoxide base of alcohol generated, used? The paragraph talks about regeneration of alkoxide ion; but I didn't really understand that part. (Seems like something related to chemical equilibrium?)
They said that Lithium Diisopropylamide is NOT generally used due to enolisation of electrophilic ester. What do they mean by it? Why can't we use LDA? Which bases can we use instead?

Could someone please answer the above two questions, with proper explanation? Thanks a lot!

Consider that there may be other reactions also going on, not just acid-base chemistry. So, for instance in the Claisen reaction of ethyl acetate, if methoxide was used instead of ethoxide then you may get methyl acetate formed as well.


Where can I find (citable) FTIR spectra of CuCl and CuCl2? I can't seem to find any published/citable infrared (FTIR) spectra in the 4000-400 cm-1 range of CuCl and CuCl2. Does anyone know of a source available, either free or paid?

If your institution holds a subscription to Elsevier's database Reaxys, than you may find, based on chemical name, molecular formula, or CAS number (the later for example from a catalogue of a supplier of chemicals) links to primary literature in the section of spectroscopic properties. Reaxys tries to cover both organic materials (formerly deposited in the Beilstein database) and inorganic / metalorganic materials (formerly deposited in the Gmelin database).  Otherwise, Scifinder from the ACS equally points towards primary literatures, too.

Beside the databases behind a paywall, freely accessible databases like the NIST webbook may be worth to consult; the entry about $\ce{CuCl2}$ here
mentions in the references already one with IR data of gaseous $\ce{CuCl2}$ (here).


What happens when 3-(2-chlorophenyl)propanenitrile is treated with sodium amide? I'm having some trouble figuring out what happens in the above reaction. Based on the strongly basic conditions I think that a benzyne mechanism should be operative, but I'm not sure what else should happen, or whether the nitrile is involved in the reaction. What is the product and how is it obtained?

You are correct that a benzyne is formed as an intermediate in this reaction. However, the nitrile is not an innocent bystander. Recall that protons α to a nitrile group are acidic: this is because the negative charge can be delocalised onto nitrogen. The pKa of a typical nitrile is around 30, lower than that of ammonia, which is roughly 40 (see: Evans pKa table). So, sodium amide is capable of deprotonating α to the nitrile.

This anion can now act as a nucleophile in the benzyne mechanism which you are no doubt familiar with. Note that the intramolecular attack of this nucleophile is likely to outcompete intermolecular attack of the amide ion. Also, since the nucleophile must attack intramolecularly, there is no question about which triply bonded carbon it attacks; it can only reach the position ortho to itself:

This specific reaction to form a benzocyclobutene is described in, for example, J. Med. Chem. 1966, 9 (5), 656–659.


Order of Basicity of Nitrogen Trihalides I was asked to arrange nitrogen tri-halides in order of increasing basicity. On looking up the answer online, it turned out to be the following: $\ce{NF3}$ < $\ce{NCl3}$ < $\ce{NBr3}$ < $\ce{NI3}$
The reason given was that more electronegative halides attract negative charge & hence reduce its availability.

I thought that the order should be $\ce{NF3}$ > $\ce{NI3}$ > $\ce{NBr3}$ > $\ce{NCl3}$ because:

all the halides except fluorine have vacant d-orbitals. So, back-bonding should reduce the basicity of all but $\ce{NF3}$.

$\ce{NCl3}$ shows more effective back-bonding as compared to $\ce{NBr3}$ which shows more effective back-bonding than $\ce{NI3}$. Better the back-bonding, lesser is the availability of the lone pair on the nitrogen atom.

It seems to imply that electronegativity is a more dominant factor as compared to back-bonding. But when we compare the acidity of boron tri-halides, we consider back-bonding to be the dominant factor. Same goes for Ipso effect.
So, what is wrong with my assumptions?
PS: I got the order from Yahoo! answers(which is pretty unreliable BTW)
PSS: I often come across questions of this sort which ask me to compare acidity & basicity of two or more compounds(both organic & inorganic). I think of looking at the pKb/pKa values of the chemicals can help me to reliably get the order in some cases. But I couldn't find any source/compilation of those values. Do you know any source which lists the pKb/pKa values of chemicals? If so, please add it to your answer. It may prove to be really useful for me & other people.

In case of Boron, boron has vacant orbitals which leads to pi back bonding from halides to boron and that concept is used to explain the basicity of Boron. However Nitrogen doesn't have vacant orbitals, hence that pi bonding is not possible from halide to Nitrogen. Consequently electronegativity is used for explanation for basicity of Nitrogen Halides.


Calculate force on atoms given the energy of configuration, trajectory of simulation and position of atoms in each frame Background

I am using a neural network to calculate the potential energy of atoms in a configuration and then adding energy of all atoms to compare it with the true energy of the configuration(label) to update the weights.

I am doing so for 10000 configurations of 100 atoms each for liquid Argon at 100K.

Note I don't have the actual energy of each atom and only the energy of the configuration.

Problem

I need to calculate the forces on each atom to see if the sum of forces for all the atoms is zero in X,Y,Z axis. In case it is not I want to find out how off is it from zero.

How do I calculate the force on atoms using their position and potential energy for a given trajectory. I know I need to do $\frac{dE}{dr}$ but what should be my E and r. Also since I don,t a formula for E(r), what should my $dE$ and $dr$ be respectively?

tl,dr: Forces are equivalent to the negative of the gradient, use small, arbitrary $\mathrm{d}r$ for each atom

The problem, as you mention, with many neural network methods in chemistry at the moment, is that the $E = f(x,y,z)$ is a black box and thus analytical gradients (and thus forces) are often not available.

First, as you clearly know, $F = -\nabla E(x,y,z)$ so you need to solve $\frac{\mathrm{d}E}{\mathrm{d}r}$ for your neural network. (I indicate this for anyone reading who doesn't realize that the force will inherently be the negative of the gradient of the energy function.

So what you need is a finite difference method. This is, of course, a common problem (e.g., on Math Stack Exchange or Wikipedia):

$$ \frac{\partial E}{\partial x} \approx \frac{f(x+\epsilon)-f(x)}{\epsilon}$$

Picking a good step $\epsilon$ is a bit tricky, but I'd probably pick something like 0.01 Å or 0.001 Å, as a start. In your case, it sounds like you want Cartesian gradients, so you'll take your $f(x,y,z)$ and perform the partial derivatives for all three axes (i.e., the initial point $(x,y,z)$ plus 3 extra energy evaluations per atom.

That is, you'd loop through each atom, move it a bit in the x, y, and z directions and evaluate the system to get the gradient/force on that atom. You'll, therefore, have $3N$ energy evaluations to get all the forces on all the atoms.

You can perform better (and more computationally expensive) finite difference methods if you choose.


How are point group character tables typeset correctly? I want to know how to typeset character tables of point groups correctly. Consider the following two extremes using the $C_{\text{2v}}$ (or is it $\text{C}_{\text{2v}}$?) point group as an example.

Firstly, one where all the symbols (except for the symbols denoting vertical) are cursive (your default LaTeX math mode behaviour):

Second, a table where all the symbols are in roman font.

Which is correct? Or is it a mixture?

Notations and conventions used for the description of symmetry in rigid molecules are established in Notations and conventions in molecular spectroscopy: Part 2. Symmetry notation (IUPAC Recommendations 1997):

  Symbols for symmetry operators are printed italic, with sub- and super-scripts that are upright except for the variables $k$ and $n$ which are replaced by numbers for a specific operator.

For example:
$E$ – identity operator
$\sigma$ – reflection operator for reflection across a general plane
$\sigma_\mathrm{h}$ – reflection operator for reflection across a horizontal plane
$\sigma_\mathrm{v}$ – reflection operator for reflection across a vertical plane
$C_n{}^k$ – $n$-fold rotation operator for $k$ successive rotations through an angle of $2\pi/n$ about an $n$-fold rotation axis, where $n = 2, 3, \ldots$; $k = 1, 2, \ldots, (n-1)$
$S_n{}^k$ – $n$-fold rotation-reflection operator for $k$ successive rotation-reflections about an $n$-fold rotation-reflection axis for a rotation through an angle of $2\pi/n$ followed
by a reflection in a plane perpendicular to the axis, where $n = 2, 3, \ldots$; $k = 1, 2, \ldots, (n-1)$
$i$ – inversion operator through the centre of symmetry

Likewise,

  Symbols for symmetry groups are printed italic with upright subscripts, following common usage.

For example:
$C_n$, $S_{2n}$, $D_n$, $D_{n\mathrm{h}}$, $D_{n\mathrm{d}}$, $C_{n\mathrm{v}}$, $C_{n\mathrm{h}}$, $T$, $T_\mathrm{h}$, $T_\mathrm{d}$, $O$, $O_\mathrm{h}$, $I_\mathrm{h}$, $C_{\infty\mathrm{v}}$, $D_{\infty\mathrm{h}}$

However,

  Symbols for irreducible representations of point groups, which are usually called symmetry species in spectroscopy, are printed upright.

For example:
$\mathrm{A}$, $\mathrm{B}$, $\mathrm{E}$, $\mathrm{A}_1$, $\mathrm{A}_2$

The example table given in the question should thus be typeset as follows:


Is an iron (II) sulphate solution supposed to turn cloudy when silver nitrate is added? $\ce{AgNO3}$ is not suppose to cause a precipitation when mixed with $\ce{FeSO4}$, but the acidified solution turns slightly cloudy when $\ce{AgNO3}$ is added.

Is this suppose to happen? solutions are all prepared fresh. Any reason as to why? This also happens when silver nitrate is added to magnesium sulphate.

Picture is as attached, the left one has the $\ce{AgNO3}$ added (labelled Ag) while the one on the right has $\ce{Ba(NO3)2}$ added to show the comparison

According to this resource for solubility rules, $\ce{BaSO4}$ is insoluble and $\ce{Ag2SO4}$ is slightly insoluble, which would explain the the cloudiness of each solution, as well as why the $\ce{BaSO4}$ solution is cloudier than the $\ce{Ag2SO4}$ solution.

According to this table, $\ce{BaSO4}$ has a $K_{\mathrm{sp}}$ of $1.1 \times 10^{-10}$. Comparably, $\ce{Ag2SO4}$ has a $K_\mathrm{sp}$ of $1.2 \times 10^{-5}$, which is considerably higher than $\ce{BaSO4}$.

If you are not aware the $K_{\mathrm{sp}}$ is a mathematical measurement describing the solubility of a salt at a certain temperature (usually $25^{\mathrm{o}} \mathrm{C}$), and is found by multiplying the concentration of the salt's dissolved ions, each raised to the power of their respective stoichiometric coefficients. The higher the $K_{\mathrm{sp}}$, the more soluble a salt is. For example, the $K_{\mathrm{sp}}$ for $\ce{Ag2SO4}$ would be given by the expression: $K_{\mathrm{sp}}=[\ce{Ag+}]^2\cdot[\ce{SO4^{2-}}]$.


N in Nernst Equation My teacher told me that in the Nernst equation, N was the number of electrons transferred. For example, $n = 2$ in $\ce{Ni^{2+} + Zn -> Zn^{2+} + Ni}$.

However, say all of the coefficients were doubled: $\ce{2Ni^{2+} + 2Zn -> 2Zn^{2+} + 2Ni}$. Would $n = 2$ or $n = 4$ in that case?

The Nernst equation is as follows,
\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{nF}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}.
\end{align*}
where $c$ is the concentration of the species specified (usually the chemical activity, $a$, is used, but the equations follow the same principle). In the first case, $n=2$. So,

\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{2F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}.
\end{align*}

Suppose we double the coefficients in the equation. Now, $n=4$. However, the concentrations of all the species have also doubled. This requires the reaction quotient (the stuff inside the natural logarithm) to be squared. Therefore,
\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)^2}.
\end{align*}

The laws of logarithms means that we can move the power of 2 to be a multiplication of the logarithm,

\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{2RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)},\\\\
    &=E_\text{red}^{0}-\frac{RT}{2F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)},
\end{align*}
which is the exact equation we had previously, thus, nothing has "changed".

So to answer your equation. Yes, $n=4$, but the equation does not change, as the reaction quotient alters accordingly, also.


If radium has such a long half-life, how can radon possibly be a threat to us? If the probability is so low that a radium atom will decay into radon at any given time (the half-life is over 1600 years), then there will be a low amount of radon produced, granted it will be produced for a long time. If this is so, how can radon possibly be dangerous to us?

There are two factors here:

Radon is a gas and can thus enter the body very easily, through inhalation.
$\ce{^222Rn}$ and its first 4 decay products are extremely radioactive, with halflives of miliseconds to a few days, releasing alpha and beta radiation.

When the two are combined, we have a problem. Even if the quantity is low, inhaling radon places an extremely radioactive substance at very close proximity to the cells in one's lungs. Normally, alpha and beta radiation are relatively harmless as they cannot penetrate tissue very well, but now they have direct access. All it takes is one mutation in the wrong place to cause lung cancer.

To give you an idea of what a difference these make, the volumetric activity which is one guideline for needing to mitigate radon accumulation in a house is around 800 $\mathrm{Bq/m^3}$, which corresponds to only 141 fg of radon per cubic metre of air. This isn't like a uranium mine or something, but at this level, a certain increased risk of cancer is expected. Apparently some study found a 50% increased risk at a much lower level ~150 $\mathrm{Bq/m^3}$, too.

N.B. a becquerel (Bq) is a reciprocal second—1 Bq means one atom of the radioactive substance disintegrates every second.


Odd/Even swap rule for chiral compounds in stereochemistry I had this question asked in my exam, where I was supposed to find if this pair is a diastereomer:

Now, I have learnt this rule where: Making odd number of swaps in a molecule means they both are enantiomers, and making even number of swaps means they are identical.
So according to this rule, I made even no. of swaps in the molecule to the right, which means it is identical to the inital molecule that was on the right. Now this newly formed molecule superimposes the molecule to the left perfectly. Hence, in my opinion the two molecules in the picture above should be identical. But the answer given in my answer key is that they are diastereomers, and not identical.
Is there any flaw in my justification?

While this rule is certainly not infallible, it is often used in organic compounds, especially for Fischer projections.

The odd/even rule implies that no change of configuration occurs if an even number of swaps are made at a single chiral center and change of configuration occurs if an odd number of swaps are made at a single chiral center.

Here, you have made 3 swaps at both ends which means you changed the configuration of 2 out of 4 chiral centers. Hence, you can make the inference that 2 chiral centers (C-2 and C-3) are the same, while another 2 (C-1 and C-4) are different. These are hence diastereomers.

I am assuming that you thought you made 3 + 3 = 6 swaps, so the compound should be identical. But the odd/even swap rule is only applicable to a single chiral center, hence the error.

Finally as pointed out already, this rule doesn't apply anywhere where the central atom doesn't have a tetrahedral geometry. As counterexamples, coordination complexes often have non-tetrahedral geometries:

octahedral: $\ce{[Co(NH3)6]^3+}$
square planar: $\ce{[PtCl4]^2-}$
dodecahedral: $\ce{[Ce(NO3)6]^2-}$

Note that these specific examples are optically inactive but by changing the ligands for different ones, it is possible to create optically active versions with the given geometries.
However, since organic chemistry mostly involves tetrahedral carbon as the chiral center, the rules certainly do apply there.


Number of stereoisomers of tetramethylspirononane The above compound – 2,3,7,8-tetramethylspiro[4.4]nonane – clearly exhibits both optical and geometrical isomerism. The question is, how many stereoisomers does it have in all?

As the number of possibilities are many, I couldn't really visualise the molecule's all possible stereoisomers properly. Could someone please help me count the total number of stereoisomers?

I figured that the carbon common to both rings is sp3 hybridised (tetrahedral geometry), and the two rings don't lie in the same plane.

The compound 2,3,7,8-tetramethylspiro[4.4]nonane has five potential stereogenic or pseudoasymmetric centres: the four C atoms with the methyl substituents and the spiro atom. In total there are $2^5=32$ possible configurations. However, many of these combinations are actually identical compounds.

A simple and obviously exhaustive (but also exhausting) approach is to draw all 32 configurations and name them. (This approach may also be used to check the results if someone finds a more elegant solution.) The resulting names for the 32 combinations are:

(2R,3R,7R,8R)-2,3,7,8-tetramethylspiro[4.4]nonane (2 times)

(2R,3R,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (8 times)

(2R,3R,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,5r,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,5s,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (8 times)

(2S,3S,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (2 times)

Therefore, there are seven different combinations.


Predicting the product of bromination of acetanilide I figured the answer would be option (B) as the group attached to the benzene ring is ortho- and  para- directing.

I thought $\ce{FeBr3}$ would be used in electrophilic aromatic substitution.

However, the model answer to the question is (D).

Is this correct? If so, what makes it possible?

This is a classic $\alpha$-bromination of a carbonyl. You can achieve this with an electrophilic source of bromine like in the reaction conditions provided or something like N-bromosuccinimide.

The reason you're not getting the EAS product is because that's a much higher energy pathway since you need to temporarily break aromaticity. Given more time, heat, and bromine, you can probably get the ortho- and para- products, but you will still $\alpha$-brominate on the amide first.


Can potassium carbonate be used to calibrate hygrometers? Context:
In the process of trying to calibrate a number of hygrometers, with Sensirion MEMS electronic sensors, I came across the method of using wet salt: an airtight container will stabilize at a predictable relative humidity over time, given a particular salt and temperature.
For a list of salts and their RH values at particular temperatures see this page on the website of Vernier, a maker of STEM education equipment. The various salts suggested for this method repeatably produce a range of low to high relative humidities, also allowing for two-point calibration.
I require my hygrometers be calibrated close to 40% RH. With a predictable 43% RH that makes potassium carbonate the ideal candidate. However, when I wettened potassium carbonate with water it underwent an moderate exothermic reaction. Googling what was going on, I read "potassium carbonate dissociates completely in water into potassium (K+) and carbonate ions (CO32-)."
Which makes me wonder:
Can potassium carbonate be used at all for the wet salt calibration method, or does its swift dissociation render it useless for this application?

Note that all salts, used in humidity calibration, dissociate to their respective hydrated cations and anions when dissolved.
From the solution crystallizes the sesquihydrate $\ce{K2CO3·\frac 32 H2O}$ ("potash hydrate"). Heating this solid above 200 °C gives the anhydrous salt (Wikipedia).
It seems you have anhydrous carbonate, that hydrates itself exothermically. You can still use it, but you need to prepare rather the saturated solution with excess of carbonate, that would integrate the crystal water into itself.
Or, let the prepared saturated solution partially evaporate and crystalize.

What is important is maintained equilibrium.
$\text{Air 43% humidity }\ce{ <=> K2CO3} \text{ saturated solution } \ce{<=> K2CO3 . 3/2 H2O}$
Anhydrous $\ce{K2CO3}$, wetted by water, rather just absorbs water into its crystall lattice. If you add too little water, no solution would left, similarly what would happen to the plaster of Paris.
You need the carbonate in hydrate form, together with its saturated solution. Take some water, add there carbonate to dissolve as much of as it can, and then some. Let is settle.
If you had the hydrate, you could just moisten it. But you do not have it.
There is the third option - moistening  it enough so some liquid phase always remains. For $\pu{100 g }\ce{K2CO3}$, you need about $\ce{20 mL}$ of water just for the crystal water, so it becomes the part of the solid.
The carbonate solubility is $\pu{110.3 g/100 mL}$, so I would carefully mix some $$\pu{100 g}\  \ce{K2CO3} + \pu{50 mL}\ \ce{H2O}$$
or recalculated to  more suitable amounts.


Activating effects of amino and hydroxyl groups in different pH How do the activating effects of substituents on an aromatic ring depend on the medium? Explain using the two compounds shown below.

I know that different groups on benzene ring can either deactivate (e.g. $\ce{-NO2}$) or activate (e.g. $\ce{-OH}$) the benzene ring towards attack by electrophiles. How does this activating or deactivating effect change in acidic and basic media?

Both the amino and hydroxyl groups activate benzene rings, as the lone pairs on N or O are in conjugation with the benzene π system.

However, in acidic media, the amino group is protonated to form an ammonium cation. Now, the lone pair on nitrogen is no longer available: it has been used to make a new N–H σ bond.

Consequently, the $\ce{-NH3+}$ group no longer acts as an electron donor via resonance. Instead, it is an electron-withdrawing group via the inductive effect (neutral amines are, too, but the resonance donation far outweighs the inductive withdrawal). For further details, you may wish to read this question: Is the ammonium substituent (-NH3+) really meta-directing in electrophilic substitution?

Phenols, on the other hand, can be deprotonated in basic media to form phenolates. Perhaps unsurprisingly, this has the opposite effect from before: the negatively charged substituent $\ce{-O^-}$ is even more electron-donating than the neutral hydroxyl group.

It's tempting to think that this is because "there are more lone pairs on oxygen", but that's not the truth. In order to have maximal overlap between a lone pair and a π system, they must be parallel to each other: on this oxygen, there can only be one lone pair which is parallel to the benzene π system. So, whether a substituent has two or three lone pairs doesn't make all that much of a difference.

The explanation is, instead, to do with the availability of the lone pair. We could think of the neutral phenol as being the same as the phenolate, but with an extra proton which is pulling one lone pair on oxygen towards it. This stabilises the lone pairs, and therefore makes them "less willing" - in a sense - to be delocalised into the benzene ring. (Technically speaking, the lower energy of the lone pair makes for a poorer energy match with the benzene ring LUMO.) In the deprotonated phenolate, though, there are three lone pairs around oxygen which repel each other. This means that they are higher in energy, and are more delocalised into the benzene ring.

As a result, phenolate reacts with very poor electrophiles such as $\ce{CO2}$ (Kolbe–Schmitt reaction) quite well.


R32 Space Group (Crystallography) Why R32 Space Group is so named? I mean, if R stands for Rhombohedral and 3 for Threefold Axis, what 2 stands for?

I don't see any Twofold Axis in the cell:

http://img.chem.ucl.ac.uk/sgp/large/155az1.htm

EDIT:

What does it means 1/2, 2/3 and 1/6 on the arrows?

R32 have two cells.

This one: http://img.chem.ucl.ac.uk/sgp/large/155bz1.htm

And this other: http://img.chem.ucl.ac.uk/sgp/large/155az1.htm

Are they equivalent?

EDIT:

I have traced some lines to mark the screw-axes and the twofold-axis, but i can't see any point that follow the simmetry operator and the element of simmetry. Maybe i don't understand how it works.

Which is the height of the green line? And, which are the points of the screw-axis of the yellow line?

The 2-fold axes are parallel to the "paper plane ", i.e. perpendicular to the three-fold. They are depicted as arrows, the half-arrows indicate screw-axes. The numbers next to it show the "height" of the axis, i.e. where the plane parallel to the paper-plane cuts the unit cell: 1/2 would be in the middle, 1/3 would be a third from the bottom/top, etc. (obviously one only gives relative numbers here, as the absolute unit cell length varies depending on the real-life crystal you are looking at).

Also have a look here for the symbols: http://img.chem.ucl.ac.uk/sgp/misc/guide.htm

Otherwise, the two links you provided are equivalent, as you can see from the pictogram in the lower lefthand corner - the first link is in the "tri-/hexagonal", the second link in the "rhombohedral" setting. The symmetry for both is obviously the same (otherwise they would not be the same space group), it is just a different way of depicting the spacegroup by changing the point of origin.

See here how they relate to each other: https://www.uwgb.edu/dutchs/symmetry/rhombo-lattice.htm

This one should also be quite enlightening - but I am not quite sure whether it might be behind a paywall: http://it.iucr.org/Ab/ch7o1v0001/sgtable7o1o155/
(Alternatively, have a look at the hardcopy of the International Tables of Crystallography, Volume A and look up spacegroup 155 - any decent crystallography lab or library should have it).


Why doesn't the atomic number change by two in the process of production of Americium? Looking at the bomardment of uranium-238 this is what I see.

First, I see a uranium-238 atom with 92 protons bombarded with a neutron. The neutron then undergoes beta decay into a proton. So we then get Neptunium as the neutron decayed into a proton and hence, atomic and mass number increase by one. An electron is also released from the atom.

That makes sense.

Looking at the second equation of the bombardment of Plutonium-239.

Plutonium-239 with 94 protons gets bombardment with 2 neutrons. These neutrons undergo beta decay to produce protons. Thus, the atomic and mass number must increase by two. The mass number, indeed, does increase by two. The atomic number does not. An electron is also released? Not sure why aswell.

What might be confusing is that it's actually two reactions. Two neutrons are captured followed by a single beta decay:

$$\ce{^239_94Pu ->[2(n,\gamma)]^241_94Pu->[\beta^-]^241_95Am}$$

The first reaction leaves the new nucleus in an excited state, so a gamma ray is emitted. The second reaction is actually quite slow with a halflife of ~14 years. It's a beta minus decay so that's why an electron is emitted.


Defining general acidic strength The $K_\mathrm{a}$ or $K_\mathrm{b}$ values provide an idea of how likely a specific step is to happen.
However, how do you specify one acid just being stronger than another, as in the case of $\ce{H2CO3}$ $(\mathrm{p}K_\mathrm{a1} = 3.6,$ but $\mathrm{p}K_\mathrm{a2} = 10.32)$ and phenol $(\mathrm{p}K_\mathrm{a} = 9.95)?$ How do you compare them?

According to the Brønsted-Lowry acid base theory, an acid is a compound that releases a $\ce{H+}$ ion to give a conjugate base. For most acids, this reaction exists in a dynamic equilibrium. The equilibrium constant of this reaction is what defines $K_\mathrm{a}$. Generally more stable the product, the more of the forward reaction takes place.
Using this last statement, when comparing two compounds, we can say that if the conjugate base of one compound is more stable, it would mean that the value of its $K_\mathrm{a}$ would be greater. This implies its $\mathrm{p}K_\mathrm{a}$ would be less, meaning that the compound would be more acidic than the other.
For example, taking the case of two compounds, $\ce{HCOOH}$ and $\ce{C6H5-OH}$: Here formate ion would have resonance via oxygen whereas phenoxide ion would have resonance via carbon. This makes phenol less acidic than formic acid.
$\mathrm{p}K_\mathrm{a} \ {\text{ of phenol} = 10.0}$
$\mathrm{p}K_\mathrm{a} \ {\text{ of formic acid} = 3.75}$
Now, if we consider polyprotic acids, we can only consider one hydrogen at a time. Successive $K_\mathrm{a}$s decrease in magnitudes of very large orders.
For example, $\ce{H3PO4}$ has $\mathrm{p}K_\mathrm{a1} = 2.12$ whereas $\mathrm{p}K_\mathrm{a2} = 7.21$ and $\mathrm{p}K_\mathrm{a3} = 12.68$.
Hope this answered your question.


Why is the solution of ammonium bifluoride more acidic than aqueous ammonium fluoride? At low concentrations of aqueous $\ce{NH4HF2}$ $(w = 1–2\,\%)$ $\mathrm{pH}\approx 3.$
But aqueous $\ce{NH4F}$ $(w = 20–40\,\%)$ has $\mathrm{pH}\approx 7.$
Yet the equilibria of
$$
\begin{align}
\ce{HF &-> H+ + F-} &\quad K &= \pu{6.85E-6 M} \tag{1} \\
\ce{HF + F- &-> HF2-} &\quad K &= \pu{3.963 M^-1} \tag{2}
\end{align}
$$
I understand that $\ce{NH4F}$ comes from a weak acid and a weak base. So, the $\mathrm{pH}$ is around neutral. I don't understand how this equilibria makes $\ce{NH4HF2}$ more acidic. Does the second equilibria make the $\ce{HF2-}$ a stronger acid? I'm having a hard time understanding this.

This question and the one referenced by Mithoron address NH$_4$HF$_2$ as an entity in itself which must be examined as a whole.
It is easier conceptually to rewrite the formula for the compound as NH$_4$F.HF. Consider adding NH$_4$F to water: you get a pH near 7. In a separate container, add HF to water (~0.1 M); you get a pH ~1 https://ehs.princeton.edu/book/export/html/197.
Combine the solutions and the pH will settle out somewhere between 1 and 7, because of the common ion effect: F$^-$ from the NH$_4$F inhibits the ionization of HF, but the solution is still acidic. The approximate pH can be calculated from the ionization constant of HF and the concentrations of NH$_4$F and HF.
The concept of the HF$_2^-$ moiety dominating the discussion for a solution is probably making the analysis murkier than it needs to be. It's just a hydrogen-bonded complex that happens to retain some semblance of existence in the solid. It is fairly stable: mp 126$^0$ C, bp 240$^0$ C (with decomposition). https://en.wikipedia.org/wiki/Ammonium_bifluoride


Is the rate determining step the step with the largest Ea? I've seen some controversy on this question while doing a brief search. For example, this SE answer quotes Wikipedia and says that the RDS is the step with the largest $E_a$. However, this UC Davis page says that

  In the potential energy profile, the rate-determining step is the reaction step with the highest energy of transition state.

My textbooks don't explicitly mention whether the RDS is the step with the largest $E_a$ or the step with the highest energy transition state. When I was doing a past exam, one of the questions was on this topic, and the correct answer assumed that the RDS is the step with the highest energy transition state. However, I've always thought that the RDS is the slowest step, and the slowest step has the highest $E_a$. So which viewpoint is actually correct?

Yes, the rate determining step is the largest energy difference between any starting material or intermediate on a potential energy diagram and any transition state that comes after it. That transition state will then be the rate-determining step of a given reaction.

The transition state with highest absolute energy may not necessarily correspond to the rate determining step, because if it is transforming from another high energy state, then little additional energy is required to reach this (low $E_a$).


Which is the most intuitive to perform Dynamic Monte Carlo calculations? I'm trying to find a (free is possible) code to perform Dynamic Monte Carlo calculations over chemistry systems. Which one is the most intuitive one?

If you want to perform Kinetic Monte Carlo for a chemical reaction system, you can use ZACROS. I am not sure what your system is? You should follow some good text book like "Understanding Molecular Simulation" to identify the type of your problem and appropriate software for that. There are plenty of freely available code on the internet but they all are for some specific purpose, So basically it is your responsibility to choose the best available software that will suit your need.


Abstraction of a proton in cycloheptatriene So, whenever there is abstraction of H+ by an acid-base reaction, I see that the proton on the sp3-hybridized carbon is taken. Why is this if you take into account that vinylic hydrogens are more acidic?

Also, on a sidenote, what is the hybridization state of a carbon in a C=C double bond where the vinylic proton was removed by a base?

The pKa for a vinylic proton is ~ 43, the pKa for propene is ~ 40, and the pKa for 1,4-pentadiene is ~ 35.  Consequently it's not the vinylic protons that are the most acidic, it's the allylic ones.  Now the question whether tropylium anion is aromatic/antiaromatic is complicated.


Degeneracy of second excited state of H-? This is a question presented in a IIT-JEE 2015 paper I exam. It says,

  . Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H atom is 9, while the degeneracy of the second excited state of H- is-

I saw a solution which says it's 3.

I think by 9 degenerate energy levels the question meant 9 possible orbitals for M shell. But I am not sure how's that 3 for for H-.

Please don't delete this question as I haven't found it anywhere.

I think it is important to understand that for hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy. For instance, $E_\mathrm{2s} = E_\mathrm{2p}$, $E_\mathrm{3s} = E_\mathrm{3p} = E_\mathrm{3d}$, etc. Thus,

The first excited state of hydrogen atom would be one in which either $\mathrm{2s}$ or one of the three $\mathrm{2p}$ orbitals is occupied and it will be 4-fold degenerate: $\mathrm{1s^0 2s^1}$, $\mathrm{1s^0 2p_x^1}$, $\mathrm{1s^0 2p_y^1}$, $\mathrm{1s^0 2p_z^1}$.
Analogously, the second excited state of hydrogen atom would be one in which either $\mathrm{3s}$ or one of the three $\mathrm{3p}$ or one of the five $\mathrm{3d}$ orbitals is occupied and it will be 9-fold degenerate: $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_x^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_y^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_z^1}$, plus 5 configurations with only $\mathrm{3d}$-orbitals occupied.

For hydride the situation is different: it is not a one-electron system, so different orbitals from the same shell do not have same energy anymore. For instance, $E_\mathrm{2s} < E_\mathrm{2p}$, $E_\mathrm{3s} < E_\mathrm{3p} < E_\mathrm{3d}$, etc. Thus,

The first excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another $\mathrm{2s}$ one, i.e. a non-degenerate $\mathrm{1s^1 2s^1}$ state.
The second excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another one of the three $\mathrm{2p}$ ones, i.e. a 3-fold degenerate state: $\mathrm{1s^1 2s^0 2p_x^1}$, $\mathrm{1s^1 2s^0 2p_y^1}$, $\mathrm{1s^1 2s^0 2p_z^1}$.


Na2So4(aq)+HCl(aq) (Any reaction?) I am just reading what a double  replacement  reaction is and what are the examples.

So I came up with  this  question : what is a reaction actually?

For example when we add $\ce{Fe(NO3)3}$ and NaOH to water we see $\ce{Fe(OH)3}$ forms as a precipitate. So hay! A reaction  has occurred!

But the reaction in the title doesn't form any precipitate depending on solubility rules. So does any reaction really takes place or the substances just dissociate into separate ions?

$\ce{Na2SO4}$ will dissolve and dissociate into separate ions. And there will be some protonation $\ce{H+ + SO4^{2-} -> HSO4-}$.


Can anyone please help me how do you determine kinetically the fastest product of sulphonation of $\beta$ napthol? I was asked a question in a paper where I was asked to identify the kinetically fastest product of sulphonation of $\beta$ naphthol.

Can anyone please help me how do you determine kinetically the fastest product ??

The answer that is provided was that  the ortho position to the hydroxyl group is the position where the fastest attack would take place. Because when the carbo cation is formed, the most stable one initially.

But I did not find this explanation satisfactory because ultimately the whole molecule is in complete resonance and it should not matter whichever Ortho or para position the sulphonation happens because ultimately the positive charge would end up being stabilised by OH group. Please help me.

The solution argued that this is less kinetically fast as the initial carbocation is less stable. Also there are 2 Ortho positions, but it does not specify which is the faster one.

The solution given was that Oxy Tobias acid is fastest but based on above arguments , R acid is equally likely and Schaeffer's acid is more probable ...(as there is no steric factors

Also what would the thermodynamically most stable product be ??

Electrophile can substitute 2-naphthol ring on 3 possible positions: 1-position through intermediate I, 6-position through intermediate II, and 3-position through intermediate III (See following diagram). Among these three intermediates, intermediate I is the most stable (You must visit Mechanism of formation of 2-naphthol red dye to get excellent explanation why substitution at 1-position is the preferable one over 3-position, although both are ortho to the phenolic $\ce{OH}$ group). Thus, the kinetically fastest product of sulphonation of 2-naphthol is 2-hydroxynaphthalene-1-sulfonic acid (Oxy Tobias acid). There will be no 3-hydroxynaphthalene-2-sulfonic acid (Ref.1).

The best evidence for this preference has been displayed during specific reaction using Claisen rearrangement (See above diagram in the box). After first rearrangement free phenolic $\ce{OH}$ group was allylated again and subjected to heat. For our surprise, the rearragement is proceeded to give 100% 1,1-diallyl derivative as shown. No 1,3-diallyl-2-naphthol was detected (according to NMR evidence).

Based on Ref.1, some 2-hydroxynaphthalene-6-sulfonic acid (Schaeffer's acid) has also been produced under kinetic control. Yet, longer heating time allowed to give disubstituted product (2-hydroxynaphthalene-1,6-disulfonic acid) preferably, thus thermodynamic product cannot be determined by these evidence.

Reference:

Paolo Beltrame, Giorgio Bottaccio, Paolo Carniti, Giuseppina Felicioli, “Sulfonation of 2-naphthol by sulfuric acid: rate measurements and kinetic model,” Ind. Eng. Chem. Res. 1992, 31(3), 787-791 (https://doi.org/10.1021/ie00003a022).


What does the equilibrium equation look like when calculating the solubility from Ksp? For example, given a reaction

$\ce{NaCl -> Na+ + Cl-}$

and its $Ksp$ value, I know that you can set up
$$Ksp\; =\; \left[ \ce{Na^{+}} \right]\left[ \ce{Cl^{-}} \right]$$ and solve from there.

My question is why do you not put $\ce{[NaCl]}$ in the denominator, as isn't that part of the equilibrium expression also?

The reason why $K_{sp}$ does not include $[\ce{NaCl}]$ is because it is solid:

$$\ce{NaCl(s) -> Na+ (aq) + Cl- (aq)}$$

The equilibrium constant $K$ would include $[\ce{NaCl}]$:

$$K=\dfrac{[\ce{Na+}][\ce{Cl-}]}{[\ce{NaCl}]}$$

However, the concentration of sodium chloride in solid sodium chloride is constant, thus we can write an equation for the observed rate constant that is based only on the variable concentrations. $K_{sp}$ is this observed rated constant:

$$K_{sp}=K_{obs}=K[\ce{NaCl}]=[\ce{Na+}][\ce{Cl-}]$$


Why does acidity increase going from sulfide to sulfoxide to sulfone? It is fairly well-known that the acidity of a proton alpha to a sulfide/sulfoxide/sulfone increases in that order. Some pKa values (in DMSO), courtesy of the Evans pKa table:

Traditionally this has been explained by resonance, or delocalisation of the carbon-based negative charge onto oxygen (left, below). But this necessarily invokes hypervalency and a S=O double bond. The major resonance form in a sulfoxide should be the charge-separated, octet-compliant form (right, below).

What is the proper explanation for the acidity trend above, then?

The oxidation state of sulphur and its therefore increased inductive effect should be enough to explain this effect.

Besides the number of oxygens, the principle difference between the three structures is the oxidation state of sulphur that rises from $\mathrm{-II}$ to $\pm0$ to $\mathrm{+II}$. The higher an element’s oxidation state, the stronger it draws electrons towards itself since a higher oxidation state typically correlates with less electron density.

Unfortunately, we cannot draw good comparisons. Phosphorus would suggest itself, but neither the Evans table nor the Bordwell data have enough data points for scrutinisation. Furthermore, phosphorus is less electronegative than carbon, meaning the bonds would be polarised towards carbon.

Chlorine might be an option being on the other side of sulphur in the periodic table, but to the best of my knowledge no ‘chlorone’ or ‘chloroxide’ organic compounds are known let alone their acidities.

Nitrogen in its higher oxidation state (a nitro group) has a double bond which can resonate with the negative charge so it is not a good comparison. And the elements in the fourth period and below are again too electropositive in general. If anything, we may be able to draw conclusions based on the β-acidity of an ether, the α-acidity of an acetal/ketal or the α-acidity of an — if these are even stable enough to be deprotonated.


Radioactivity of materials Suppose that a raidoactive brick of an unknown element with a half-life of one year (dangerous stuff!) is left on a regular desk for a day. After the day has passed, the brick is removed. Does the desk now exhibit any radioactivity even when artificialy induced radioactivity is considered?(Assume that the brick exhibits alpha, beta, and gamma decay products and ignore the dimensions of the brick and the realism of placing a radioactive brick on a random desk.)

No. Radioactivity means that the nucleus of a particular atom is unstable, and the nucleus will decay to a more stable state either by decomposing or emitting energy (as a photon). This instability will not make other nuclei unstable unless the decay of one of the radioactive atoms causes a fission process in one of the atoms in the desk. This is a secondary effect though and surely will not happen once the material is removed.

Another way of thinking about this is that there are materials in which people house radioactive substances. These materials do not become radioactive or else there would be no point in having them in the first place. Not just any material will work though because radioactive emission products might be transparent to many materials or are energetic enough to melt/decompose other materials. So, I would bet the desk would get destroyed, but from normal decomposition not from decay.

Finally, the reason that it seems like things become radioactive once there has been radioactive material nearby is because we usually think about the products of nuclear fission. These products will have quite long half-lives and thus the material will stay around for a long time decaying.


Calculating pH of diprotic and amphoteric solutions I am trying to understand this problem and its corresponding solution (complete solution not shown)

First off, from the above one can see that the equilibrium constants when $\ce{H2ASO4-}$ donates and receives a proton are as follows:
\begin{aligned}
\ce{H2AsO4 &<=> HAsO4^{2-} + H+} & K_2 &= 1.1 \cdot 10^{−7}\\
\ce{H2AsO4- + H+ &<=> H3AsO4}    & K_{-1} &= 1.6 \cdot 10^{2}\\
\end{aligned}

However, if this is an amphoteric substance as the solution claims, shouldn't both $K_a$ values be comparable? It seems logical that if it is able to act as both an acid and a base, the equilibrium constants should at least be similar to each other. Instead, one is several order of magnitudes larger than the other? Why can this be true and $\ce{H2ASO4-}$ still be an amphoteric substance?

Next, how did they arrive at the statement

  We approximately have the equality relation: $$\ce{[H3AsO4]} = \ce{[HAsO4^{2−}]}$$

What is the evidence that shows this to be true? Isn't the $K_a$ value of the basic reaction ($1.6 \times 10^2$) much larger than that of the acidic reaction? Why should the concentrations be equal?

Finally, let's say you are calculating the pH for the diprotic substance $\ce{H2S}$, which undergoes the following diassociations:

$$\ce{H2S <=> H+ + HS-}$$
$$\ce{HS- <=> H+ + S^{2-}}$$

To find the pH you would assume that the first reaction is dominant, and that the effects of the second reaction consuming $\ce{HS-}$ are minimal, and then calculate the pH from there — right?

However, why can't you treat $\ce{HS-}$ as an amphoteric substance (because it can both gain and lose a proton) and then calculate the pH as you would for an amphoteric substance (as seen in the image at the beginning)?

I'll address your questions in 3 parts:

(A) Why amphoteric

(B) How to solve the problem without making unexplained assumptions

(C) Why the solution in the book uses the approximate equality
    $\ce{[H3AsO4]=[HAsO4^{2-}]}$.

PART (A)

An amphoteric species (molecule or ion) is one that can either gain a proton with relative ease, or lose a proton with relative ease. "Relative ease" means, for example, that it can happen in aqueous solution. If your species will gain a proton in the presence of enough $\ce{H3O+}$, and if it will lose a proton in the present of enough $\ce{OH-}$, it can certainly be called amphoteric.

Translated into equilibrium constants, if your species has a pKa between 0 and 14, and its conjugate acid (what you get when you protonate it) also has a (necessarily lower) $pK_a$ between 0 and 14, your species will act in aqueous solution as either as an acid (losing a proton in the presence of strong base) or as a base (gaining a proton in the presence of strong acid).

The $pK_a$ of (neutral) arsenic acid is between 2 and 3 (because the equilibrium constant for removing the first proton, $K_1$, is between $10^{-2}$ and $10^{-3}$). The $pK_a$ of the version that lost one proton, $\ce{H2AsO4-}$, is around 7 (from the value of $K_2$). So clearly $\ce{H2AsO4-}$ will gain a proton if the pH is below 1, say, and it will lose a proton if the pH is above 10. This justifies why it's amphoteric.

PART (B)

In a monoprotic acid like HF, all the F appears either as F$^-$ or as HF (two species). You can write down 1 equilibrium equation, a mass balance for F and a charge balance, and solve all three equations simultaneously; when you combine the three you get a quadratic equation that can be solved exactly.

But with a multiprotic acid, things get complicated fast. In this problem you have three equilibrium equations, plus the mass-balance and charge-balance equations.  That's 5 equations in 5 variables.

Fortunately it's possible to combine 4 of them easily, using only high-school algebra. They are the 3 equilibrium equations and the charge balance. This reduces the problem to two variables (the pH and the concentration of one of the arsenic ions) in two equations.

Let's write and combine the 4 equations. There are four arsenic species with varying charges; I'll call their concentrations $A_0,A_1,A_2,A_3$ -- the index denotes the charge (which is 3 minus the number of hydrogens). So $A_0=[\ce{H3AsO4}]$, $A_1=[\ce{H2AsO4-}]$, and so on.

By definition, $A_1[\ce{H^+}]/A_0=K_1$, or equivalently $A_0=A_1[\ce{H^+}]/K_1$. This relates two concentrations, $A_0$ and $A_1$. Similarly $A_1=A_2[\ce{H^+}]/K_2$ and $A_2=A_3[\ce{H^+}]/K_3$.

The important thing is that all these four concentrations are proportional and given one you can find all others if you know the pH. (We don't know the pH, but I'll address that in a moment.)

The charge balance (expressed in moles present in 1 L of solution) is simply
\begin{equation} 0\,A_0+ 1\,A_1+ 2\,A_2+ 3\,A_3+K_w/[\ce{H^+}]= [\ce{H^+}]+0.1.
\end{equation}
On the left we have the negative charges, on the right the positive ones. Notes:

The concentration of each arsenic species is multiplied by its charge, obviously.
On the left side, $K_w=10^{-14}$ is the dissociation product of water, and $K_w/[\ce{H^+}]$ represents the concentration of OH$^-$; it will turn out to be negligible, but we don't know that a priori, and including it doesn't make the problem any harder.
The 0.1 on the right comes from Na$^+$ ions; being an alkali metal, sodium does not exist in water solution except in ionic form, so we know [Na$^+$] exactly. This explains something that will be important later: The 0.10 M solution of $\ce{NaH2AsO4}$ is exactly what you obtain by dissolving 0.10 moles of NaOH in one beaker, 0.10 moles of $\ce{H3AsO4}$ in another, and mixing the two, assuming the combined volume is 1 L.

Now we express the left side of (1) in terms of just two
concentrations, $A_3$ and $\ce{[H^+]}$: $$
\begin{aligned}A_1+2\,A_2+3\,A_3&=(\ce{[H^+]}/K_1)\,A_2+2\,A_2+3\,A_3\\
&=(\ce{[H^+]}/K_1+2)\,A_2+3\,A_3\\
&=(\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)\,A_3+3\,A_3\\
&= \bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr)\,A_3
. \end{aligned} $$

The last line is the key; it allows us to relate the concentration of one arsenic
species to the pH.  With this (and moving the $K_w/\ce{[H+]}$ to the
right)
the charge balance becomes
$$ \bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr)\,A_3
=\ce{[H^+]}+0.1 -K_w/\ce{[H+]}$$
or again
$$ A_3=
=\frac{\ce{[H^+]}+0.1 -K_w/\ce{[H+]}}{\bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr).}$$

This look complicated, but it's easy to put into a spreadsheet. The last step is this: The right [H$^+$] is the one that, with $A_3$ calculated as above, obeys the mass balance, $$ A_3 + A_2 + A_1 + A_0 =0.1.$$

At this point we could declare the multiple-choice problem solved, by cheating a bit: take the various proposed values of  [H$^+$] and see which one works.  For answer (a) the value of the sum $ A_3 + A_2 + A_1 + A_0$ is 0.62; for (b) it is  0.128; for (c) it is 0.10165, and for (d) it is 0.100002.  So the best answer is (d).

What if you didn't have multiple choice? Then you'd want to draw a graph or a table, taking (say) every integer value of pH and connecting them.  Here is such a graph:

The horizontal axis is pH and the vertical one is total arsenic concentration.  The correct pH is where the vertical coordinate is exactly 0.1, and you can get a good idea of its value by eyeballing, or solve by successive approximations.

Note that this method allows you to solve with no more effort the following problem: what is the pH when you add 0.1 mol/L of $\ce{NaH2AsO4}$ (as in the original problem), plus 0.05 mol/L of $\ce{H3AsO4}$?  The only difference is that now you'll be looking for the point on the graph where the vertical coordinate is 0.1+0.05 = 0.15, instead of 0.1.

PART (C)

Why does the book say that $\ce{[H3AsO4]=[HAsO4^{2-}]}$, approximately? And is this a good approximation?

The equality comes from the claim that the "dominant reactivity" of $\ce{[H2AsO4-]}$ is $$\ce{2\,H2AsO4- \leftrightharpoons H3AsO4{}+HAsO4^{2-}}.$$ If $\ce{H2AsO4-}$ exchanges protons only with itself, and not with, say, water, then every proton lost by one $\ce{H2AsO4-}$ ion means a proton gained by another $\ce{H2AsO4-}$, and the result is that $\ce{[H3AsO4]}$ and $\ce{[HAsO4^{2-}]}$ are generated in equal amounts.

So given the claim about "dominant reactivity", the equality is true, and by the way, the equality implies that the pH is the average of $pK_1$ and $pK_2$. (This follows, by taking logs, from the analysis that I'm guessing your book gives right after the part of the solution you quoted.)

The problem with the claim is that it does not follow from the amphoteric nature of $\ce{H2AsO4^-}$ alone. It requires two additional conditions: You're starting from only $\ce{H2AsO4-}$, and in relatively high concentrations. I'll illustrate why these conditions are necessary:

(1) You have to start with only $\ce{H2AsO4-}$ (plus a counterion that doesn't participate further, like Na$^+$). If you started with some $\ce{H3AsO4}$ in addition, you wouldn't have the equality $\ce{[H3AsO4]=[HAsO4^{2-}]}$, even approximately. Or, to use the example you gave, if you start with H$_2$S the amphoteric character of HS$^-$ wont help you calculate the pH of the resulting mix. On the other hand if you started by dissolving NaHS alone, it would.

(2) You have to start with a strong solution. If the original question had specified a 0.1 mM solution instead of a 0.1 M solution, the resulting H$^+$ concentration would not be $2.5\times 10^{-5}$ anymore but $3.3\times 10^{-6}$, and the equality $\ce{[H3AsO4]=[HAsO4^{2-}]}$ would be off by a factor of 60!

I hope this helps answer your very interesting and perceptive questions.  Don't hesitate to ask for further clarification if necessary!


How do I start studying chemical nomenclature? Chemical nomenclature seems to be the way chemistry "talks", allied with the mathemathics. But where to start?... and what reasons and theories are behind? Some "names" come from latin... I know there are rules IUPAC etc, but it is so demanding and exaustive to learn all that rules at once, for organic and inorganic chemistry. Maybe there are reasons and rules that help us to name a chemical or compound... Since I am interested in metallurgy, can you give me some advice to start? To associate theory to the formula, ad strutural formula, it would be nice. The implicit "why" in everthing included is always welcome.

Chemical nomenclature is mixture of historical traits, empirical knowledge and little bit of systematics.
And as you mentioned, it is exhaustive to learn all at once.

To understand the nomenclature, you must understand the systems themselves. Because it's their structure and reactivity which is reflected in the name.

This makes an unfortunate cyclic dependency. To start studying any given chemistry branch, you must know its language. But without knowing the science behind, the language does not make sense. It's like that and you probably cannot do anything about it, just cut this circle and crunch through the nomenclature. It will start to make sense, once you get sufficiently deep. Additionally, the "why" is good indicator. Once you have enough knowledge to understand the answer why it is named so, than you can probably answer yourself.


Hydro-Bromination Product 1,4-Benzoquinone What will the products on addition reaction of 1,4-benzoquinone with HBr ?


Can molten Strontium and Barium Hydroxides be electrolysed to give the metals? Can strontium and barium hydroxides be electrolysed to give the metals? I've noticed they have relatively low melting points. What challenges would this present?

Expanding on @Oscar Lanzi comment: first, all the hydrogen (and half the oxygen) would be electrolyzed before then alkaline earth metals.  At that point, if kept liquid, then the $\ce{BaO}$ and $\ce{SrO}$ remaining would be split. That would not be easy, as the melting point of $\ce{BaO}$ is 2,196 K and that of $\ce{SrO}$ is 2,804K (though a eutectic mix of the oxides would be a bit lower).

[It's more efficient to dissolve the oxides in a halide salt. The electrolysis of $\ce{CaO}$ in $\ce{CaCl2}$ has been explored here and here.


Amine Nomenclature – When to use "N" I'm currenty wondering what the correct naming scheme for the following molecule would be:

Is this called "2,N,N-trimethylpropan-1-amine" or "2-methyl-N,N-dimethylpropan-1-amine". In other words, do you group the substituents attached to the N with the other substituents.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ‘N’ is a locant just as the ‘2’:

  P-14.3 LOCANTS

  P-14.3.1 Types of locants

  Traditional types of locants are arabic numbers, for example, 1, 2, 3; primed locants, for example, 1′, 1‴, 2″; locants including a lower case Roman letter, for example, 3a, 3b; italicized Roman letters, for example, O, N, P; Greek letters, for example, ‘α-, β-, γ-’ and compound locants, for example, ‘1(10)’ and ‘5(17)’.

Note that ‘N’ is lower than ‘2’:

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
  (…)
  Italic capital and lower-case letter locants are lower than Greek letter locants, which, in turn, are lower than numerals.

Identical simple substituents are indicated by multiplicative prefixes, such as ‘di’, ‘tri’, etc. The current IUPAC recommendations do not contain any rule that would stipulate that such substitutents should be grouped. By way of comparison, similar examples given in the Blue Book include names such as N,N,2-trimethyl-3-{4-methyl-3-[2-methyl-3-(methylamino)-3-oxopropyl]phenyl}propanamide.

Therefore, the preferred IUPAC name (PIN) for the compound given in the question is N,N,2-trimethylpropan-1-amine.