How to calculate the equilibration time of two connected chambers of salt solution, initially at different concentrations? I am trying to build a system of two connected chambers, one of which is also connected to a cuvette, so I can perform a spectrometry experiment with different concentrations of $\ce{KNO3}$ solution. In its simplest form, the cuvette and the first chamber would hold solution at the same concentration, while the second chamber would hold solution of higher concentration. I would perform one experiment, open a valve between the chambers, and wait for the concentration to equilibrate before performing the next experiment.

My advisor suggested this system and said that the equilibration time would likely be on the order of minutes.

I used this equation to calculate equilibration time between two chambers:

$$
t = \frac{\Delta m \cdot  l_{v}}{A_{v} \cdot D \cdot 10^6 \cdot \Delta c}
$$

where $t$ = equilibration time in s
$\Delta m$ = the difference in mass of $\ce{KNO3}$ between the chambers in g
$l_{v}$ = the length of the valve in m
$A_{v}$ = the orifice area of the valve in m^2
$D$ = the mutual diffusion constant of $\ce{KNO3}$ and $\ce{H2O}$, which is on the order of 10^-9 m^2/s
$\Delta c$ = the difference in concentration between the chambers in g/cm^3
the factor of 10^6 is to convert between g/cm^3 and g/m^3

For a cylindrical valve 1 cm long and 2 cm in radius (improbably short and wide) and chamber volumes 95 cm^3 and 400 cm^3, this gives an equilibration time of 4.7 days!

This is much more than I was expecting. I can reduce the sizes of the chambers to decrease the time demand, but I would prefer to keep them as large as I can, to increase the final concentration that can be achieved in the cuvette.

Is there something obvious that I'm missing in this calculation? Is there a better way to do it?

Edit:

The equation above was derived by equating the mass flux $j_{m}$ with its definition from Fick's first law (using $c$ as mass/volume rather than amount/volume):

$$
j_{m}=\frac{\Delta m}{A_{v}\cdot \Delta t}=D\cdot\frac{\Delta c}{\Delta x}
$$

I know that strictly the RHS should contain $\frac{\delta c}{\delta x}$ rather than $\frac{\Delta c}{\Delta x}$. Does this approximation make much difference? For now, I am also making the fairly large assumptions that the volumes of solution in each chamber do not change, and that the density of the solution is independent of concentration. I think I can include these considerations later on.

This is an actual system that I need to build (well, ask our machine shop to build for me) and I need to have a good idea of the best valve sizes to use as they have a significant effect on the overall structure.

In the future please specify where your equation came from, justify its use and specify all of your parameters.

When only a concentration gradient is present, the diffusion coefficient, $D$, is typically described by Fick's first law of diffusion. To determine $D$ in these terms we must know the flux, $j$, and the gradient, $d c / dx $.  $c=$ concentration of our species divided by the volume and $x$ is the distance this occurs over.

Flux, $j$, is the (flow quantity/time)/area or (flow rate)/area. Fluxes depending on potential gradients are molecular, but bulk flow flux is called convective. Here we have molecular flux present and the units of this are kmol/(m$^2 \cdot$ s). (It's true that diffusion creates its own convection but there is no bulk flow so we ignore this diffusion-induced convection.) Properly our flux is the product of a transport property by the gradient of the driving force, $ j = (D) ($Gradient of driving force$)$. Note that $D$ has units of m$^2$/s. For example, $D$ is $10^{-5}$, $10^{-9}$, $10^{-10}$ m$^2$/s for some gases, liquids and solids respectively at STP.

Now let's think back to basic physics. Say we have a feather at $h$ height here on Earth and we drop it. It will experience an acceleration due to gravity, $g$. Remember that gravity has units m/s$^2$. We can estimate the time it takes to travel distance $h$ crudely by dimensional analysis: knowing that $g$ has units of m/s$^2$ and $h$ has units of m then time is simply $\sqrt{\frac{h}{g}}$. In fact the equation that describes this is $h = \frac{1}{2}gt^2$ and so we actually needed to multiply our original answer by $\sqrt{2}$. In the same way we can crudely estimate the time by dimensional analysis on the diffusion coefficient, $D$ with units m$^2$/s and going a distance with units of m. So we need to take the distance$^2$ divided by $D$ to get seconds. Say ( $0.01$m )$^2$ divided by $10^{-9}$ m$^2$/s gives $10^5$ seconds which is about 1.1 days. This could easily be 4 days if  we needed to multiply by 3.7 which is certainly possible if your orifice is small. It is calculating these other things which turns out to be tricky. Although it's obvious the single easiest thing you can do is switch to a gas when considering just diffusion because a gas would take on the order of 10 seconds because $0.01^2/(10^{-5}) = 10$. You can also change the time by introducing another flux, such as by stirring the solution.

A better answer requires differential equations. Are you sure you want to venture down that road?


Can sodium borohydride reduce hemiacetals? In the synthesis of Panacene by Canesi et al. [1], a step involves the installation of an $\ce{-OH}$ group on a double bond, using the oxymercuration-demercuration procedure. However, the product happens to be a hemiacetal, as shown in the picture. With a 78% yield, it seems that the hemiacetal is not reactive towards the added sodium borohydride. Is this true in general, that hemiacetals will not be reduced to diols by $\ce{NaBH_4}$? What about $\ce{LiAlH_4}$?

References

Cyrille Sabot, Didier Bérard, Sylvain Canesi, "Expeditious Total Syntheses of Natural Allenic Products via Aromatic Ring Umpolung," Organic Letters 2008, 10(20), 4629-4632 (DOI: 10.1021/ol801921d).

The major purpose of $\ce{NaBH4}$ in the second step of given reaction sequence is the demercuration, which stops at the hemiacetal as the applied condition is the limit (at $\pu{0 ^\circ C}$ for $\pu{1 min}$).
In general under usual conditions (at $\pu{25 ^\circ C}$ in ethanol for for an hour or two), $\ce{NaBH4}$ reacts with hemiacetals and reduce them to corresponding diols. For instance, $\ce{NaBH4}$ reacts with glucose, which is a hemiacetal in aqueous solutions, to give sorbitol, the corresponding diol (Ref.1). The reaction has been done in water and at room temperature ($\pu{20-25 ^\circ C}$) for $\pu{1-2 h}$. Thus, running the given reduction in $\pu{0 ^\circ C}$ for shorter time make sense.
Also, although lithium aluminum hydride ($\ce{LiAlH4}$) is too vigorous a reagent for reduction, under carefully controlled conditions, it can be used to convert lactone to lactol (cyclic hemiacetal such as monosaccaride) without reacting further to give the corresponding diol. However, excess $\ce{LiAlH4}$ would give the corresponding diol after reacting further with lactol (Ref.2,3):

References:

M. Abdek-Akher, J. K. Hamilton, F. Smith, "The Reduction of Sugars with Sodium Borohydride," J. Am. Chem. Soc. 1951, 73(10), 4691–4692 (DOI: https://doi.org/10.1021/ja01154a061).
2.Glen E. Arth, "Reduction of Lactones to Hydroxyaldehydes with Lithium Aluminum Hydride," J. Am. Chem. Soc. 1953, 75(10), 2413–2415 (DOI: https://doi.org/10.1021/ja01106a040).
Robert A. W. Johnstone, "Chapter 1.11: Reduction of Carboxylic Acids to Aldehydes by Metal Hydrides," In Comprehensive Organic Synthesis - Selectivity, Strategy and Efficiency in Modern Organic Chemistry, Volumes 8: Reduction; Ian Fleming, Volume Editor; Barry M. Trost, Ian Fleming, Editors-in-Chiefs; Pergamon Press: Oxford, United Kingdom, 1991, pp. 259-282 (ISBN: 978-0-08-040599-1).


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

From the last page of this paper:

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

Edit: Article title is Autoignition Temperature for Mixtures of Flammable Liquids at Elevated Pressures by Brandes, Hirsh, and Stolz

1) From this, it seems that pressure and fuel concentration effects are related. Is the critical fuel concentration related to the equivalence ratio and the ability to reach this equivalence ratio based on the saturation pressure of the fuel?

2) I do not get the relation between the above explanation and the last sentence regarding re-increase of AIT for longer chain molecules?

3) I'm not sure why larger volumes leads to lower AIT as stated in ASTM E659 Appendix X2.

1) One could argue that yes. All other things being equal, larger molecules will clearly have lower vapor pressures at the same temperature so it is reasonable that there is a limit at which the fuel-air mixture is too lean to support autoignition.

2) After reading the paper I can only conclude that they are referring to the above idea that autoignition temperatures increase for larger compounds due to the need to increase the vapor pressure to reach an appropriate fuel concentration that supports autoignition.

3) Larger volumes have lower surface-area-to-volume ratios. The lower surface area means a the walls of a vessel play less of a role in the combustion chemistry leading to a lower AIT. These surface effects are detailed by Babrauskas with references to those who have observed this.

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


How does PCC not oxidize aldehyde? I am wondering how does PCC actually work in not oxidizing aldehyde. Like the reaction mechanism I saw online, it seemed as if not having a acidic solvent is what made aldehyde not tranform to carboxylic acid when PCC is present. So the why can't you just use chromium containing compounds without the acidic solvent and then you will be able to isolate aldehyde.

The "trick" is in the enviroment, not in the PCC itself.

Alcohols tend to be oxidized to carboxylic acids when they are in aqueous enviroment.

PCC was invented as a workaround to this problem: it works as a reactant in anhydrous enviroment, hence stopping the reaction at the aldheyde stage.

An alternative way to see PCC is to follow its preparation: if you add $\ce{CrO3, HCl}$ and pyridine, PCC is formed.
If you add $\ce{CrO3}$, acid and water, chromic acid is formed.

So, you see that the reagents are similar: the point is in finding at which stage the difference is.

Mechanistically speaking, the difference occurs as follows: in aqueous enviroment, water can form a geminal diol intermediate by attacking the aldheyde. This step adds the second oxygen to the aldheyde that, with further reaction with chromic acid, yelds a carboxylic acid.

When water is not present, this step doesn't occur, and the reaction stops.


How to distinguish diastereomers of unsaturated ketones by NMR? I am synthesising a natural product that contains an α-methylated α,β-unsaturated ketone (see image below). The natural product’s stereochemistry is an (E) configured double bond. I selected this double bond as one of the retrosynthetic cuts of the overall molecule and I want to be sure that when I synthesise it that I get the correct stereochemistry. Is there any easy way to distinguish the (E) from the (Z) isomer?

Generic structure of an (E)-α-methylenone.

According to Noack and Jones, unsaturated ketones with an α substituent — even if it is only methyl — generally adopt an s-trans conformation due to steric reasons. Only if the α substituent is hydrogen can an equilibrium between s-cis and s-trans be observed.[1] Therefore, we should better draw the structure this way:
$\hspace{30ex}$
$\hspace{28ex}$s-trans configuration of α-methylenones
I already explicitly added the unsaturated hydrogen atom to the image. As we can see, it is either rather close to the keto group or pointing away from it. If the alkene’s configuration is (E) and the hydrogen is thus pointing ‘inward’ it experiences a larger deshielding due to the close carbonyl group. Therefore, its signal in 1H-NMR is shifted to lower field when compared to the (Z)-configured double bond.
You can also compare experimental shifts. Unfortunately I didn’t find a pair of identical molecules differing only in the double bond orientation, and not even the solvent is the same, but I think the data still shows a clear trend, namely that for (E) enones the chemical shift is in the high alkene range while for the (Z) enones it is in the low range:
$\hspace{33ex}$
(E)-3-methylhex-3-en-2-one. 1H-NMR ($500~\mathrm{MHz}$, $\ce{CDCl3}$) $\delta =$ $\mathbf{6.59}$ (tq, $J = 7.2,1.3~\mathrm{Hz}$, 1 H), $2.28$ (s, 3 H), $2.24$ (app. qnt, $J = 7.4~\mathrm{Hz}$, 2 H), $1.73$ (s, 3 H), $1.05$ (t, $J = 7.6~\mathrm{Hz}$, 3 H).[2]
$\hspace{33ex}$
(Z)-3-methylnona-3,8-dien-2-one. 1H-NMR ($300~\mathrm{MHz}$, $\ce{C6D6}$) $\delta =$ $5.71$ (ddt, $J = 17.1, 10.5, 6.6~\mathrm{Hz}$, 1H), $\mathbf{5.38}$ (tt, $J = 7.5,  1.5~\mathrm{Hz}$, 1 H), $4.92$–$5.02$ (m, 2 H), $2.29$ (q, $J = 7.5~\mathrm{Hz}$, 2 H), $1.92$ (q, $J = 7.2~\mathrm{Hz}$, 2 H), $1.85$ (s, 3 H), $1.60$ (t, $J = 2.7~\mathrm{Hz}$, 3 H), $1.35$ (qnt, $J = 7.5~\mathrm{Hz}$, 2 H).[3]

References:
[1]: K. Noack, R. N. Jones, Can. J. Chem. 1961, 39, 2225. DOI: 10.1139/v61-294.
[2]: W. F. Austin, Y. Zhang, R. L. Danheiser, Org. Lett. 2005, 7, 3905. DOI: 10.1021/ol051307b.
[3]: L. Jiao, C. Yuan, Z.-X. Yu, J. Am. Chem. Soc. 2008, 130, 4421. DOI: 10.1021/ja7100449.


Are there Exchange integrals in Semiempirical methods? One of the major features of semiempirical methods is that they significantly reduce the cost of evaluating 2-electron integrals.

In general, they follow the formula for HF, but, introduce many simplifying assumptions. In HF for same spin electrons you need to calculate both coulomb and exchange integrals.

When applying the simplification in, for instance, CNDO and INDO, that $(\mu \nu | \lambda \sigma ) = \delta_{\mu \nu} \delta_{\lambda \sigma}(\mu \mu | \lambda \lambda )$, this seems to de facto rule out calculating an exchange integral.

Is there a scenario where the exchange integral is computed in semiempirical methods (for simplicity, in the case of CNDO/INDO methods)?

Textbooks do not discuss the exchange integral when going over semiempirical, so the answer must be obvious... I just don't know if it is obvious that they cannot be performed, or if they can, and I am missing something.

Semiempirical quantum chemistry models do carry out proper Hartree-Fock calculations, including both Hartree (electron-electron electrostatics) and Fock exchange contributions to the total electronic energy. As noted in the question, the tensor of 4-center Coulomb integrals is heavily sparsified, approximated, and parameterized to simplify and accelerate semiempirical calculations. Even in the most approximate models (CNDO), the exchange interaction between molecular orbitals does not vanish in general. For example, the exchange integral between a pair of molecular orbitals $i$ and $j$ with real atomic-orbital coefficients $\phi_{i \mu}$ is simplified by the zero-differential-overlap (ZDO) approximation stated in the question as

$(ij|ji) = \sum_{\mu,\nu,\lambda,\sigma} \phi_{i \mu} \phi_{j \nu} \phi_{j \lambda} \phi_{i \sigma} (\mu \nu | \lambda \sigma) = \sum_{\mu,\lambda} \phi_{i \mu} \phi_{j \mu} \phi_{j \lambda} \phi_{i \lambda} (\mu \mu | \lambda \lambda)$.

This answer might be confusing precisely what the original question means by "exchange integral". For example, no explicit integrals are involved in calculating the Coulomb matrix elements used by most semiempirical models, and there may be some class of Coulomb matrix elements in the atomic-orbital basis that are considered to be "exchange-like" that are indeed neglected in most semiempirical models (I'm not familiar with all variants of this nomenclature). The neglect of diatomic differential overlap (NDDO) approximation that eventually superseded ZDO does allow for nonzero $(\mu \nu | \lambda \sigma)$ values when $\mu$ and $\nu$ (similarly, $\lambda$ and $\sigma$) are different atomic orbitals on the same atom.


Determining the rate equation from a reaction where one reactant is in excess Lets say
rate = k[A][B]n
(A is in excess, lets say at a concentration of 50mM)

My data includes the initial concentration of B, and the initial rate of reaction at that concentration

I can rewrite the rate equation as as:
 rate = kapp[B]n

I then take the natural logarithm of this equation and plot a graph of ln(rate) against ln[B], the gradient will give me the value of n.

Am I right in thinking that the y intercept will give a value of kapp? And if it does, to determine the true rate constant do I just divide kapp by 50? If not, how do you determine the true rate constant?

There is no real need to invoke a "$k^{app}$".

Let the rate be $r$, so:

$$r=k[\ce{A}][\ce{B}]^n$$

Since as $\ce{A}$ is in large excess:

$$[\ce{A}]=C(A)\gg[\ce{B}] \implies r=kC(A)[\ce{B}]^n$$

So $[\ce{A}]$ doesn't vary perceptibly during the rate measurement interval.

Plot the value of $\ln r$ versus $\ln [\ce{B}]$ in the knowledge that:

$$\ln r=\ln k + \ln C(A) + n\ln [\ce{B}]$$

So the gradient of that graph is $n$.

And the intercept of that graph is $\ln k + \ln C(A)$. Since as $C(A)$ is known, $\ln k$, and thus $k$, is also known.


Why is fluorine the most electronegative atom? It seems related to the atomic size but hydrogen has a smaller atomic size than fluorine. Why is fluorine the most electronegative atom?

Fluorine is the most electronegative element because the definition of electronegativity makes it so. The electronengativity scales are defined based on experimentally determined properties of the elements. Fluorine has appropriate values for all of the common scales to ensure it has the highest electronegativity.

The Pauling scale, which is the first electronegativity scale proposed, relates the difference in electronegativity $\Delta\chi_{A,B}$ between two atoms $(A\ \& \ B)$ is related to the bond dissociation energies $(E_d)$ of the diatomic species $AA,\ BB,\ \& \ AB$:

$$\Delta \chi_{A,B}=|\chi_A - \chi_B| = (\text{eV})^{-1/2}\sqrt{E_d(AB)-[E_d(AA)+E_d(BB)]/2}$$

Since only differences can be determined, one element (hydrogen) was chosen to have an arbitrarily set value. On the Pauling Scale Hydrogen has a value of $2.20$. Since the $\ce{H-F}$ bond is much stronger $(E_d=586\ \text{kJ/mol})$ than the average of the  $\ce{H-H}$ bond $(E_d=436\ \text{kJ/mol})$ and the $\ce{F-F}$ bond $(E_d=157\ \text{kJ/mol})$. Data from here. In general, since fluorine forms stronger bonds to all other elements than it does to itself, fluorine will have the highest electronegativity.

The Mulliken scale relates electronegativity to two other properties: electron affinity $E_{ea}$ and ionization energy $E_i$:

$$\chi_A = \frac{E_{ea}(A)+E_i(A)}{2}$$

Fluorine has a high electron affinity (easy to gain an electron) and a high ionization energy (hard to lose an electron).

There are other scales, including the Allred-Rochow scale, based on effective nuclear charge and covalent radius, the Sanderson scale based on atomic size, and the Allen scale based on the average energy of the valence electrons.


When we heat a borax crystal, why does it first expand and then contract? In the borax bead test, we take a crystal of borax and then heat it.

  A small loop is made in the end of a platinum or nichrome wire (as used in the flame test) and heated in a Bunsen flame until red hot. It is then dipped into powdered borax, and the adhering solid is held in the hottest part of the flame where it swells up as it loses its water of crystallization and then shrinks, forming a colourless, transparent glass-like bead (a mixture of sodium metaborate and boric anhydride)

Whenever I've done the experiment, I've noticed that the process looks like this:

It heats up in a second, in this time there are no visible changes
It starts expanding/blooming. Bubbles appear, and little pseudopodia(closest work I can think of) of borax flail about. This takes approx five seconds. At the end of this, the borax looks like whipped cream that's been...well.. whipped .
The thing suddenly contracts into a tiny bead.

Now, as mentioned in the quote, basically borax loses its water of crystallization, and then decomposes into $\ce{NaBO2 +B2O3}$, the small bead.

The flailing about is probably due to steam escaping.

But, the reaction proceeds in such an interesting manner, I can't help but think that there's something else going on as well.

What are the details of the mechanism of the borax bead preparation reaction?

I've never seen this reaction, but it sounds quite entertaining. And I learned a couple of new words in looking at the literature to explain this phenomenon.

Intumescence: The property of a material to swell when heated. Borax is intumescent.
"Calcination is a thermal treatment process in presence of air applied to ores and other solid materials to bring about a thermal decomposition, phase transition, or removal of a volatile fraction. The calcination process normally takes place at temperatures below the melting point of the product materials."

The mechanism for the calcination of borax quite complex and is covered in some detail in this article. Borax decahydrate (BDH), $\ce{Na2B4O7.10H2O}$ is easily dehydrated to Borax pentahydrate (BPH), $\ce{Na2B4O7.5H2O}$ without much visible change. The actual form of the borate ion in both BDH and BPH is $\ce{B4O5(OH)4^2-}$ with 2 of the water molecules actually incorporated as hydroxyl groups in the ion. (The other 2 hydroxyl groups come for the water hydrogens and 2 oxygens from the $\ce{B4O7^2-}$.) That means that the last two water molecules can only be removed by molecular decomposition. So in BPH, you actually have only 3 "free" (i.e., easy to remove) water molecules.

So here's what happens. The borax decahydrate dehydrates smoothly with minimal density changes (no swelling) until you get to borax pentahydrate. At this point, the BPH dissolves  in its own water of dehydration and the "blossoming" you see is from the solution erupting through the partially dehydrated crystal surface as it's heated. The density of the borax changes dramatically as it goes from the pentahydrate (about $0.85\ \mathrm{g/cm^3}$) to the dihydrate  (about $0.14\ \mathrm{g/cm^3}$) which is why it ends up looking "whipped".

To do the final transformation to a tiny bead, the remaining 2 waters of hydration (those tied up as hydroxide ions) are driven off as the $\ce{B4O5(OH)4^2-}$ decomposes to give metaborate, $\ce{BO2^2-}$ and boric anhydride, $\ce{B2O3}$. $$\ce{B4O5(OH)4^2- -> 2 BO2 + B2O3 + 2H2O}$$

Fascinating.


What chemical elements does a nuclear explosion release into its mushroom cloud? I'm interested in the mushroom cloud produced by a nuclear explosion: What is it made of?

I know that the uranium fission reaction is the following (from this web page)

$$\ce{{}^{235}U + {}^1_0n -> {}^{95}Sr + {}^{139}Xe +2~{}^1_0n + 180~MeV}$$

I know that these elements are not stable and decay, so what is left over?

According to The Preparatory Manual of Explosives, the mushroom cloud is produced by the partial vacuum left after initial expansive shock wave. That implies that it is primarily the condensation of water that makes it visible.  These clouds form in high explosive as well as in nuclear detonations near the ground. See https://books.google.com/books?id=J55D3HcgPuoC&pg=PA98#v=onepage&q&f=false.


Interference between 2 electrons If 2 electrons undergo destructive interference (as they show wave nature)  will they disappear or will they have no wave nature left?

The mathematics of quantum mechanics predicts an unevenness of the observation, similar to wave equations in that there are points where high values of electron density/reaction and low (~zero) values occur. If you have places (~50% of the places) where "interference" has yielded zero electron density/reaction, then the other 50% of the places will have twice the expected (average) electron density/reaction.

The question should include the entire observation, the whole experiment. For example, it is easy to imagine a photon experiment where diffraction causes interference effects upon a photographic film: photons abandon their even spread by being diffracted, and don't go to certain places on the film, but do go (in twice the expected density) to other places. Two photons don't just cancel out in some places - they "cancel out" and also "add up" in other places. Oh, heck, just think of a sine wave (square its amplitude) plopped down on the film and where the high peaks are, the film turns dark and where the wave bottom is, nothing happens. Two photons, two sine waves; there has to be a slight difference in path so that when you superimpose them, they add up more in one place and subtract more in another.

That thought experiment was reasonably easy because we think of photons as sort of fuzzy, wavy, intangible things. We usually consider electrons to be touchable, therefore firm on some sort of basis, so how can they interfere and cancel out of existence? Well, the wave nature of electrons means that they have a spread in space which has some of the characteristics of a wave, and they can, under the right conditions, exhibit this effect strongly. Getting back to the photographic film experiment, electrons can be made to exhibit the same effects as diffraction does to photons. Where it becomes more difficult to understand is that while the photon (say, ~500 millimicrons wavelength) is much fuzzier than a silver ion (~0.25 millimicron diameter in AgCl), you would expect an electron to be, if anything, a bit smaller than an atom, since, after all, electrons (many of them!) will fit into an atom or ion.

So you can see that a photon with a spread of 500 millimicrons will smash down over close to a million Ag+ ions, but can only activate one, but you might expect a little electron to be like a bullet and just hit one silver ion. How could you miss? Consider the bullet to be wobbly (wavy) enough to hit either one Ag+ or its neighbor. The faster it goes, the wobblier (wavy) it gets. And these waves interfere so that some paths just don't happen, and others are travelled twice as frequently - this doesn't mean that two electrons cancel out of existence or form a bi-electron.

At least that's the way I look at it without the comfort and brevity of mathematical equations. But  words only poorly describe the situation and it takes forever to try to explain.


What properties of an element determines the maximum number of bonds it can make? From my conjecture, I think it's a mix between valence electron and the principal energy state, but I'm not sure.

For example, C, N, O and F can only make a maximum of four bonds, as they can only have up to sp3 hybrid orbitals.

For Si, P, S and Cl, they can make a maximum of nine bonds, as they can have sp3d5 hybrid orbitals.

That's what I think, but I'm probably incorrect in some aspect.

Considering the number of other atoms a central atom can be bonded to, "hybridization [is] irrelevant" and "the coordination number ... is not restricted by the number of
hybrid orbitals that the atom can form", quoting On the role of d orbitals in sulfur hexafluoride J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593.

Examples such as $\ce{CLi6}$ and $\ce{TeF8^{2-}}$ are cited and emphasis is placed upon the ionic character of the bonding by the cited reference.

In my opinion, the main limiting factor is how many atoms can simultaneously be close to the central atom, this increasing with increasing radius of the central atom and with decreasing radius of the surrounding atoms.


Work done in expanding a gas reversibly and irreversibly So, my chemistry teacher gave the class following $P_{external}$ versus $Volume$ diagrams for reversible and irreversible expansion of a gas which are as follows.

(Reversible expansion)

(Irreversible expansion)
Then he gave us following results to memorize:
$$\mid \left(W_{reversible}\right)_{expansion}\mid \gt\mid \left(W_{irreversible}\right)_{expansion}\mid$$
Where W is for work.
According to him, these results are always true.
But consider the following graph

If I'm not wrong, this can also be considered an irreversible expansion.
But as we can see clearly,
$$\mid \left(W_{reversible}\right)_{expansion}\mid \lt\mid \left(W_{irreversible}\right)_{expansion}\mid$$
This thought is really bothering me. Where did I go wrong?

I will call your processes (1), (2), and (3). I assume (1) is reversible isothermal expansion, (2) is irreversible isothermal expansion, and (3) is unspecified.
The key idea is that the inequality $w_\text{rev} \geq w_\text{irrev}$ can only be applied to reversible and irreversible analogues of the same process. (1) and (2) satisfy this requirement because (1) is the reversible analogue of (2), but (1) is not the reversible analogue of (3).
Physically, (1) is like having a pile of sand on top of a container (containing an ideal gas) whose lid is a piston, and then slowly blowing away the sand. (2) is the same scenario, except you sweep off the sand in one grand gesture. But (3) is a different scenario entirely.
If we refer to the first law, $\Delta U = q + w$, it's clear that, given an initial and a final state, we can make $w$ as large as we want as long as we have a commensurate value of $q$. Just because processes (1), (2) and (3) all have the same initial and final states does not mean that it makes sense to apply the inequality $w_\text{rev} \geq w_\text{irrev}$ to any two of them.


Why is the ionization energy for Hydrogen non-zero? There are no other electrons to collide, repel and kick Hydrogen's single electron to a distant nucleus. And that a single electron is tightly attracted to the nucleus by the electrostatic energy between them. So it seems to me, that Hydrogen does not require ionization energy at all.

But when I checked on the sources - the Hydrogen's ionization energy is relatively high.

Even Hydrogen makes $NaH$, which just doesn't stay for long and has extreme need for becoming $Na$ metal and Hydrogen gas.

So in this case why can't we say that Hydrogen does not require ionization energy?

The ionization energy (IE) of an atom or molecule describes the minimum amount of energy required to remove an electron (to infinity) from the atom or molecule in the gaseous state.

With ionization energy, an electron is not "kicked out" by other electrons, but rather it is "the energy required for the electron to 'climb out' and leave the atom."

Since the electron "is drawn inwards by positive electrostatic potential," it would make sense to infer that the more "tightly drawn" the electron is to the nucleus, the more energy it would require to "climb out."

More of this can be found on the wikipedia page.


Phenolphthalein indicator spill onto skin I was using a phenolphthalein indicator to test the acidity of some 0.1 M HCl. I was using the pipette to draw the phenolphthalein indicator out of the tiny jar but it wasn't working, so I thought that it might not have much left in it, so I tilted it at an angle to try and get the remaining indicator. However it turns out the pipette wasn't working and it was full and a couple of millilitres spilt out, some spilt onto my right index finger, but most of it spilt onto the bench. I immediately rinsed my finger under running water for about 7-10 seconds. Just wondering if it'll have any negative effects. I'm pretty sure it was 99.9% methanol and 0.08% phenolphthalein measured in weight.

Didn't you receive the SDS for phenolphtalein (or its methanolic/ethanolic solutions)?
Weren't you given instructions on how to handle it?
Didn't you wear suitable gloves in addition to your safety goggles?

Back in the days, phenolphthalein was apparently used as a laxative. This is no longer recommended!

However, the current GHS classification is as follows:

  Germ cell mutagenicity, Category 2; H341
  Carcinogenicity, Category 1B; H350
  Reproductive toxicity, Category 2; H361f

(cited from the GESTIS database)

To my knowledge, phenolphtalein is on the REACH Candidate list of SVHC (substances of very high concern) since 2011.

My suggestion is:

Contact a physician.
Report the exposition to your supervisors.


Thermodynamic vs kinetic reaction control with radical substitution Doing a revision for a chemistry class test tomorrow, I stumbled upon thermodynamic and kinetic reaction control in the context of radical substitution.

Here, my chemistry teacher said, that thermodynamic reaction control would basically be about favoring stabilized (originally instable) transition states, e.g. by increasing temperatures. According to him, this would "thermodynamically prefer" the substitution of tertiary hydrogen atoms relative to primary ones.

However, he also told us that kinetic reaction control would, instead, prefer the substitution of primary hydrogen atoms. More precisely, the fastest reaction (which would result in the substitution of primary hydrogen atoms due to less steric hindrance) would be favored under certain conditions.

The question is, under which conditions a certain type of reaction control would be prefered.

Thermodynamic control typically implies reversibility of all steps. Say you had three reactants $\ce{A, B}$ and $\ce{C}$ which could react either to product $\ce{A-B-C}$ or to product $\ce{A-C-B}$ (equation $(1)$). Then thermodynamic control would assume that the ratio of products depends only on the difference in Gibbs free energy between them, signified by the equilibrium constant (equation $(2)$).

$$\ce{A-C-B <=>[$\Delta G_1$] A + B + C <=>[$\Delta G_2$] A-B-C}\tag{1}$$

$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = K = - \frac{\Delta\left(\Delta G\right )}{RT}\tag{2}$$

On the other hand, kinetic control implies that the reactions are basically irreversible and that it depends on the transitions states and their energies which product will be formed preferentially. This is often expressed as a difference in rate constants $k$ as can be seen in equation $(4)$ for the reaction $(3)$.

$$\ce{A-C-B <-[$k_1$] A + B + C ->[$k_2$] A-B-C}\tag{3}$$

$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = \frac{k_1}{k_2}\tag{4}$$

One (simple) method of achieving thermodynamic control is to heat the reaction to oblivion, until one can assume that all transition states are sufficiently populated and the energy differences of products can take effect. Kinetic control, on the other hand, is often achieved by cooling down and higher dilution, since both allow the difference of transition states to take greater effect.

To apply this to the radical halogenation of hydrocarbons, Hammond’s postulate must be invoked, which basically states that the lower-energy intermediate (or product) will be reached by a lower-energy transition state. Since radicals are electron-deficient species, a tertiary radical is more stable than a secondary one, which in turn is more stable than a primary one. Thus, kinetic control implies that the lower-lying transition state determines the dominating product which would be the one leading to a tertiary radical.

A secondary factor under kinetic control is the number of hydrogens that can react. Of course, given a molecule such as isobutane, there are much more primary hydrogens ($9$) than there are tertiary ones ($1$). Yet, the reaction rate of the tertiary hydrogen is still much faster. Kinetic control still favours the tertiary halide.

Under thermodynamic control, the heats of formation of the different products must be examied. But conincidentally, these lead to the same conclusion. Since the $\ce{C-Br}$ bond is polarised towards bromine, the molecule is more stable if the ipso-carbon is tertiary. However, since we are no longer comparing rate constants but Gibbs free energy values, the ratio will be a different one.

Thus, both control types would favour the same product — a tertiary halide — albeit to different extents. This is the standard case! Only rarely, such as in certain Diels-Alder reactions does the main product differ when changing from thermodynamic to kinetic control.


How to visualize the rearrangement of a pentacyclic triterpene alcohol? I was looking into the exam paper of last year when I found this:

Propose arrow-pushing mechanism for the reaction below.

I suppose it's a Wagner–Meerwein rearrangement. (Sorry for the dirty mechanism graph.)

Is the answer correct? How would I explain the stereochemical configuration of the reaction? And why doesn't the elimination (the leaving of proton) happen at carbon A, as the cation intermediate is just as stable as the others?
I tried drawing chair conformation graph of the reaction. I marked the possibly important groups in red for convenience.

You are correct about the beautiful chain of eight Wagner-Meerwein rearrangements. The stereochemistry of each of these is given by the fact that this arrangement is suprafacial and proceeds under retention, thus a group on the top of the molecule will stay above the ring plane and a group below will stay below. During the rearrangement process, since each intermediate cation is planar, the rings A to D (from where the hydroxy group was to where the double bond will be) will undergo a ring flip each, preserving the general trans-decalin structure for each pair of rings.

Once the final rearrangement has occurred, we arrive at that position in the molecule which features a cis-decalin system: note how in your drawing of the starting material the final ring E is not aligned with all the others but pointing downward. Assuming that final proton, which is eliminated, would actually migrate, we would generate the final carbocation at the cis-decalin junction. The methyl group between the eliminating proton and carbon A is in the wrong orientation to migrate; the carbon belonging to ring E has a much greater tendency because it is better aligned (anti to the proton). That migration would lead to a 6,5-spiro compound and a trisubstituted double bond. That isn’t bad in itself, but the pathway asked for in the question leads to a trans-decalin system by elimination, which features much less strain than a spiro compound. Additionally, a tetrasubstituted double bond is also more stable.

To arrive at the all-trans-polcyclic system, the proton on the junction of rings D and E must eliminate which it is happy to do. Ring D flips, which now turns the cis-decalin system of DE into a trans-decalin system, ring strain is released and the reaction terminates.


IUPAC name for branched alkane I'm trying to apply IUPAC rules to name the compound below.

TLDR: Why is this 3-ethyl-2,3,4-trimethylpentane and not 2,3-dimethyl-3-(1-methylethyl)-pentane?

Is this the correct source to use?
According to above, p. 60:

  P-15.1.7.1.4
  The longest chain is numbered from one end to the other by arabic numbers, the direction being chosen so as to give the lower locants to the substituent groups (side chains) [see P-14.4(f)]. The lower set of locants is defined as the set that, when compared term by term with other locant sets cited in order of increasing magnitude, has the lower term at the first point of difference (see P-14.3.5).

The longest chain possible here is a pentane chain, and ChemSketch says that this is 3-ethyl-2,3,4-trimethylpentane.

However, if I want to select the lowest set of locants, wouldn't 2,3-dimethyl-3-(1-methylethyl)-pentane be preferred, as its ordered set (2,3,3) is lower than (2,3,3,4)?
Also, is 1-methylethyl preferable to prop-2-yl?

That doesn't affect 3., as it would be 2,3-dimethyl-3-(prop-2-yl)-pentane.

Am I correct in assuming that prop-2-yl is preferable to isopropyl?

Otherwise, that would make it 2,3-dimethyl-3-isopropylpentane.

Answers to numbered questions:

1.  Yes (with the errata applied)
3.  Yes, however before the lowest locant set rule (P-45.2.2), maximum number of substituents as prefixes rule (P-45.2.1) is to be used. (See e.g. Alkane nomenclature - choosing the parent chain - maximize or minimize substituents?)
4–5. No, “prop-2-yl” is an incorrect spelling of propan-2-yl (P-29.3.2.2), which is preferred to both isopropyl and 1-methylethyl (P-57.1.1.1).


Why do Cu⁺ ions spontaneously form copper metal and Cu²⁺ ions in solution? The standard electrode potentials for three reactions involving copper and copper ions are:

  $$
\begin{align}
\ce{Cu^2+(aq) + e- &-> Cu+(aq)} &\quad  E &= \pu{+0.15 V} \\
\ce{Cu^2+(aq) + 2e- &-> Cu(s)} &\quad   E &= \pu{+0.34 V} \\
\ce{Cu+(aq) + e- &-> Cu(s)} &\quad   E &= \pu{+0.52 V} \\
\end{align}
$$

  Which statement is correct?

  A. $\ce{Cu^2+}$ ions are a better oxidizing agent than $\ce{Cu+}$ ions.
  B. Copper metal is a better reducing agent than $\ce{Cu+}$ ions.
  C. $\ce{Cu+}$ ions will spontaneously form copper metal and $\ce{Cu^2+}$ ions in solution.
  D. Copper metal can be spontaneously oxidized by $\ce{Cu^2+}$ ions to form $\ce{Cu+}$ ions.

Why is the answer C, and not B? I thought that because copper is more likely to get oxidised than $\ce{Cu+},$ it is a better reducing agent and thus B is the answer. But I cannot explain why C is the correct answer. Why is this?

Any help would be greatly appreciated.

Answer:
Both reactions involving Cu+ have the lowest and highest Standard Electrode Potentials respectively. Thus, Cu+ has to be the strongest oxidizing/reducing agents, and therefore it undergoes disproportionation.

Because the formation of Cu has a higher SEP than Cu+, it is more likely to get reduced and therefore is a better oxidizing agent - NOT a better reducing agent - than Cu+ (which is more likely to get oxidized as it has a higher SEP).


Hybrid functional calculations using different approach In hybrid functional such as B3LYP, PBE0, HSE etc, some part of HF exchange and/or correlation is added to some part of DFT X/C part in order to get better results. My question is if I am adding 1/4 th of HF exchange part that means I already calculated HF exchange part and HF is undoubtedly better than DFT, then why don't we just use the full HF solution that we already have in our hand?

In hybrid functional such as B3LYP, PBE0, HSE etc, some part of HF
  exchange and/or correlation is added to some part of DFT X/C part in
  order to get better results.

First, there is no correlation in HF, only exchange, so you can not add part of the HF correlation into DFT, as you said. In principle, of course, if you want to add not only exchange from the wave function theory (WFT) methods but also some correlation, you could use other that the HF sources, but the conventional hybrid DFT corresponds to mixing in just (a part of) the HF exchange.

Now, you very question of

  why don't we just use the full HF solution that we already have in our
  hand?

is addressed in almost any textbook on DFT. I suggest that you read, for instance, Section 6.6 Hybrid Functionals in Koch & Holthausen, specifically, pages 78-80. The authors provide a relatively simple argument for for the well-known failure of the exact exchange/density functional
correlation combination approach.

In short, the source of the problem is that the separation of an exchange-correlation entity (would it be energy, or hole, or something else) into exchange and correlation contributions is artificial. Only the whole exchange-correlation entity has a physical meaning, while the above mentioned individual contribution are introduced just for convenience. Usually, it means that exchange and correlation contributions has to be chosen consistently so that their sum, an exchange-correlation entity, has some desired properties.

For instance, the exchange-correlation hole in molecular systems is usually relatively localized, so that the exchange hole and the correlation hole must both be either localized or delocalized (in a very special way so that combined they result in a localized hole). In LDA- and GGA-DFT both the exchange and the correlation hole are localized, but when we mix HF exchange and LDA/GGA-DFT correlation in the most straightforward way, i.e.,
$$
E_{\mathrm{XC}} = E_{\mathrm{X}}^{\mathrm{HF}} + E_{\mathrm{C}}^{\mathrm{DFT}} \, ,
$$
then, I quote:

  We combine the exact, delocalized exchange hole with a localized model
  hole for correlation. Because the cancellation between the two
  individual holes cannot take place, the resulting total hole has the wrong characteristics.

1) Koch, W., & Holthausen, M. C. (2001). A Chemist's Guide to Density Functional Theory, Second Edition, Wiley-VCH Verlag GmbH.


How to debug a failed Gaussian IRC calculation? During IRC calculation I found the following error message. Anybody can help me?

IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC

Error in corrector energy =          -0.0000002157

 Magnitude of corrector gradient =     0.0003105753

 Magnitude of analytic gradient =      0.0003306121

 Magnitude of difference =             0.0001786220

 Angle between gradients (degrees)=   32.1573

 Pt  9 Step number  20 out of a maximum of  20

 CORRECTOR INTEGRATION CONVERGENCE:

   Recorrection delta-x convergence threshold:    0.010000

   Delta-x Convergence NOT Met
 Maximum number of corrector steps exceded.

 Error termination via Lnk1e in /home/adhikary/g09/l123.exe at Sun Aug 21 05:49:50 2016.

According to the output, it seems like OP is using Gaussian 09 software package. And according to the error message, the problem is that at 9th step of IRC procedure, Gaussian could not optimize the geometry in 20 steps (which is the default number). I suggest first to increase the number of geometry optimization iterations for each and every IRC step by using the MaxCycle=N option as follows:

IRC=(...,MaxCycle=100)


Work done on a container, resulting in pressure change Why does pumping gas into a rigid container (causing the pressure and temperature to increase), not mean that work has been done on the container?

Work is given by the formula :  W = -P ΔV

If I go off the formula alone, ΔV=0 (as the container is rigid) and hence W=0

But if I ignore the formula and take the definition of work as "the energy it takes to move an object against a force", wouldn't pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container e.g. like pumping a bike tyre, the gas is not going to go into the container unless I pump it in?

For reference:

Question from textbook: A rigid container of constant volume is used to store
compressed gas. When gas is pumped into the container,
the pressure of the gas inside the container is increased and
the temperature of the container also increases. Which
statement is true of the work done on the container?

(a) The work is equal to the increase in the pressure
inside the container.

(b) The work is equal to the increase in the temperature inside the container.

(c) The work is equal to the sum of the pressure and temperature increases.

(d) There is no work done on the container.

Answer from textbook:  (d) the container does not expand, so as there is no
change in volume of the container, no work is done on the
container.

Yes, the explanation

  the container does not expand, so as there is no change in volume of the container, no work is done on the container.

is right as it takes into account that the container is rigid, so its volume can not be changed. Thus, there is no work done by the gas on the container.

Also, if you say

  " pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container "

Yes, there is work done on the gas, but it is by the piston(or the pump).

You should see how the laws of Thermodynamics make clear about the boundary between system and its surroundings, so work has to be done by "something" on "something".
Also, In the question, it is clearly asked "Which statement is true of the work done on the container?" So, as the container's volume is unaltered, there is no work on it.


Will this SN1 reaction result in an equilibrium situation? I'm trying to prove the rate law of an $S_N1$ reaction by measuring the rate of it, when I vary the concentrations of my reagents. However, I believe the reaction of the experiment I have designed will result in an equilibrium. The reaction is as follows:

$$\ce{t-BuOH + HCl -> t-BuCl + H2O}$$

I chose this reaction based on the theory outlined in this website, http://www.masterorganicchemistry.com/2015/02/27/making-alkyl-halides-from-alcohols/

Pretty simple. However, will the $S_N1$ solvolysis reaction, in the opposite direction occur and thus hinder my attempt to measure the rate of the reaction?

Thinking about the SN1 reaction generally, we know that the rate is first order, being proportional to the rate of decomposition of the C-LG bond of the reagent, with the identity/concentration of the nucleophile being completely irrelevant. Your issue, therefore, becomes one simply of the relative leaving group ability of water vs chloride.

Quantitatively, you could think about this on an energy diagram (below). For both the forward and reverse reactions, there is an intermediate carbocation (formation of this is rate limiting), once this carbocation forms there is then a fast recombination of cation with a nucleophile to form a 'stable' species. So, in essence, absolutely the reaction can go in both directions, but in practice often goes in one direction (actually microscopic reversibility dictates that this is almost always the case, but the energetics of one direction are such that reactions aren't realistically reversible).

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

  Ref: Clayden/Warren, Organic Chemistry.

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


Haloalkanes and E1, E2 reactions In an E1 reaction, I understand that the base reacts with the hydrogen of the carbon which is next to the positive charged carbon of the cation (of an haloalkane), because this hydrogen is supposed to be acidic. Why is that? And again, in an E2 reaction, why is the hydrogen that is attacked by the base, acidic? I am not sure how the polarity of the halogen affects the situtation.

After the first step of the reäction (i.e., loss of the halogen), you have the carbocation.  You could draw a resonance structure of this: push the electrons from the C–H bond into making a new C–C bond, leaving you with an alkene and a nearby proton (note the conservation of charge, moving the + from a carbon to the hydrogen).  This is like the reverse of an addition of a hydrohalic acid to an alkene, whereïn you'd have the double bond attack a proton, then have the halogen attack the carbocation.  Either way, you've drawn a very reasonable* resonance structure showing an alkene and H+.  That H+ is pretty much the definition of a Brønsted acid, so that would be a hand-waving explanation of how to see this hydrogen as being particularly acidic.

For the E2 example, we can't use the same method because we can't show this resonance structure (well, we kind of can if you want to get into a really tedious discussion that would be highly debated).  You'd probably want to use the argument about halogens being inductively electron withdrawing, which pulls electron density away from nearby bonds.  This results in less electron density existing in the C–H bond, making the bond longer, and slightly increasing the positive charge on the hydrogen.  Since Brønsted acid theory defines the acid as being H+, the more we can show a partial positive charge on the hydrogen, the more Brønsted-acidic we can argue it to be.

* Note that I mentioned drawing a "reasonable" resonance structure.  That's important, because we could technically show a resonance structure of benzene as having a carbon with a negative charge and a hydrogen with a positive charge, but that wouldn't be an argument for that hydrogen being particularly acidic, as it's a horrible resonance structure to draw.

It certainly is more complicated than this, but I feel like this highlights some of the most important effects without oversimplifying to the point of making anything explicitly wrong or creating a bunch of exceptions.


Flory–Schulz type distribution with stoichiometric imbalance In a typical step-growth polymerization, the Anderson–Schulz–Flory distribution is the probability mass function that describes the number fraction or weight fraction of polymers of length $x$ at a given extent of reaction $p$.  The extent of reaction, $p$, is a value between 0 and 1, and $x$ is the degree of polymerization for any given oligomer or polymer present in the system.

$$\begin{align}
\text{mass fraction} &= p^{x-1} \cdot (p-1)^2 \\
\text{weight fraction} &= x \cdot p^{x-1} \cdot (p-1)^2
\end{align}$$

A nice derivation of these equations can be found in this video.  Figure 1 illustrates a plot of the number fraction distribution as a function of the degree of polymerization for several extent of reactions.  Figure 2 illustrates the plot of the weight fraction distribution as a function of the degree of polymerization for several extent of reactions.

Note that high extents of reaction in step-growth polymerizations result in very broad distributions, and are also required for high molecular weight polymers.
Carothers' equation describes how stoichiometric imbalances necessarily limit the possible degree of polymerization:

$$x_\text{average} = \frac{1+r}{1 + r - 2rp}$$

where $x_\text{average}$ is the average degree of polymerization, $p$ is the extent of reaction, and $r$ is the mole fraction of the reacting functional groups. $r$ is always a number between 0 and 1.  The excess functional group is always taken to be in the denominator. When $r$ is equal to 1, stoichiometric balance exists, and the equation reduces to:

$$x_\text{average} = \frac{1}{1-p}$$

For example, if I react a diol and a diacid under step-growth conditions:

My goal is to purposely limit the degree of polymerization of a chemical reaction by introducing a stoichiometric imbalance in order to form short chain oligomers.  I'd like to predict/calculate the statistical distribution of molecular weights for a given extent of reaction, p and stoichiometric imbalance, r.

I'm having trouble convincing myself that the Flory–Schulz distribution is appropriate for modeling step-growth systems that contain stoichiometric imbalances. Is there a straightforward way to incorporate stoichiometric imbalance into these distribution functions? Any help would be appreciated.

I came up with a way to solve my own problem as I was writing my question, but I feel as though it is a bit hand-wavy.  I would appreciate any input regarding the validity of this solution.

The following is a workaround to force the Flory distribution to have a maximum value that is consistent with the average degree of polymerization due to stoichiometric imbalance via Carothers' equation.  Carothers' equation models how stoichiometric functional group imbalances limit the average degree of polymerization for the step-growth mechanism.  This average degree of polymerization $(\text{DP}_\text{average})$ can be calculated very simply:

$$\begin{align}
\text{DP}_\text{average} &= \frac{r+1}{r+1-2rp} \\
\end{align}$$

When there are no stoichiometric imbalances, and the Flory distribution is valid:

$$\text{DP}_\text{average} = \frac{1}{1-p}\\
f(\text{DP}_\text{average}) = x \cdot p^{x-1} \cdot (p-1)^2$$

Let $p_\text{eff}$ be the effective extent of reaction, which is related to $r$ and $p$ via:

$$\frac{1}{1-p_\text{eff}} = \text{DP}_\text{average} = \frac{1+r}{1+r-2rp}\\
p_\text{eff} = 1 - \frac{1+r-2rp}{1+r}$$

This $p_\text{eff}$ can be used to calculate the ideal Flory distribution according to

$$F(\text{DP}_\text{average}) = x \cdot p_\text{eff}^{x-1} \cdot (p_\text{eff}-1)^2$$

The function $F(\text{DP}_\text{average})$ will force the molecular weight distribution to be properly centered about the  $\text{DP}_\text{average}$ that is consistent with Carothers' equation.  I'm not sure if this is chemically sensible, or accurately describes a real-ish polymer system.  Comments/alternative answers would be appreciated.


Which ideal gas law does a constant volume thermometer work on? I searches online for my question and found mixed answers. I think the Ans. Should be Gay Lussac's Law as we are dealing with a case of constant volume but variable pressure and temperature. In many books the answer is given to be Charles Law. Would appreciate if someone gave me the ans. Thanks.

Gay-Lussac's Pressure Temperature Law.

The law simply states

  The pressure of a gas of fixed mass and fixed volume, is directly proportional to the gas's absolute temperature.

As telling temperature rises, the gas in the thermometer has its pressure increased in accordance with the law. Measuring this pressure gives us the temperature as per the ideal gas equation, or by using Gay-Lussac's law directly, taking the pressure ratio against a set base temperature and pressure.


Is this relation in case of a mixture of ideal gases valid? I came across the following relation in a book (written as a fact, with no proof), for a mixture of ideal gases both at the same pressure and temperature.  The final volume of the mixture of gases after mixing is:
$$ V_\text{mix} = \frac {M_1V_1 + M_2V_2}{M_1+M_2} $$

where $M_1$, $M_2$ are the molar masses of the individual gases and $V_1$, $V_2$ are the respective volumes.  I could not figure out its proof.  Is this relation valid at all?
Please do not confuse with a similar looking relation,with $ M $ corresponding to molarity, in case of a solution mixture.

This relation is not valid if the gases are assumed to be ideal. see Amagat's law of partial volumes
Moreover, if the gases are non ideal, the relations between the volumes of the combining gases and the final resultant volume is very complex and the given relation is too simple and includes too less variables to even begin describing the actual non-ideal gas mixing.
It, however unlikely, might represent a certain special case that can only be judged from the context wherein the aforementioned equation was taken from.


Buffer Power of the amphiprotic substance The mass balance for a generic amphiprotic substance $[\ce{AH^-}]$ this:

$$
[\ce{HA^-}]=C \cdot \dfrac{K_{a1} [\ce{H_3O^+}]}{K_{a1}\cdot [\ce{H_3O^+}] + [\ce{H_3O^+}]^2 + K_{a1} \cdot K_{a2}} \quad Eq.1
$$

where:
 C = initial concentration of the amphitrofic substance

The Eq.1 derivative  is:

$$
\dfrac{d[\ce{HA^-}]}{d[\ce{H_3O^+}]} = -\dfrac{C \cdot [\ce{H_3O^+}] \cdot K_{a1} \cdot (2 [\ce{H_3O^+}] + K_{a1}}{(K_{a1}\cdot [\ce{H_3O^+}] + [H_3O^+]^2 + K_{a1} \cdot K_{a2})^2} + \dfrac{C \cdot \cdot K_{a1}}{K_{a1}\cdot [\ce{H_3O^+}] + [\ce{H_3O^+}]^2 + K_{a1} \cdot K_{a2}} \quad Eq.2
$$

The zero of this derivative is found at $[\ce{H_3O^+}] = \sqrt{K_{a1} \cdot K_{a2}}$, where is the maximum of $[\ce{AH^-}]$.

The buffer power (BP) can be defined $BP = \dfrac{dC_B}{dp \ce{H}} = -\dfrac{dC_A}{dp \ce{H}} = - 2.3 \cdot [\ce{H_3O^+}] \cdot \dfrac{d[\ce{HA^-}]}{d[\ce{H_3O^+}]} \quad Eq.3$

For example:

If $K_{a1} = 6.2  \times 10^{-8}$ and $K_{a2} = 4.8 \times 10^{-13}$ and C=0.1, the maximum oncentration of the specie $\ce{AH^-}$ is found at $[\ce{H_3O^+}] = \sqrt{6.2  \times 10^{-8} \cdot 4.8 \times 10^{-13}}  = 1.725 \times 10^{-10} \quad pH=9.76$

The curve that represents the buffer power, plotting the Eq.3, is:

In this way I can say that at maximum concentration of the amphiprotic substance the buffer power is zero. Is this statement true?

As the curve that represents the buffer capacity does not simply rely on one fraction of the whole system but on all species of the acid-base system it is not enough to only consider the derivative of $c(\ce{HA-})$.

You also have to take care of the other deprotonated species in your system. For the case of a diprotic acid this means that $c(\ce{A^2-})$ is also important albeit not much as you can see in the following figure.

Figure 1: Buffer capacities for a diprotic acid with different amounts of assumptions taken into account – denoted by their plot labels. The lower graphs are zoomed versions of the upper graphs.

As you have come to the mass balance for $c(\ce{HA-})$ I don't have to show you how to come to the mass balance for $c(\ce{A^2-})$:
$$c(\ce{A^2-}) = \frac{c_0(\ce{H2A})~k_{a1}~k_{a2}}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2}}$$

This means that you have to derive the following equation:
$$C_B=\frac{c_0(\ce{H2A})~\left(c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}\right)}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}}$$
or if you want to include he self-ionization of water it has to be
$$C_B=\frac{k_W}{c(\ce{H3O+})}-c(\ce{H3O+})+\frac{c_0(\ce{H2A})~\left(c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}\right)}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}}$$

Let's derive (with the help of D. D. van Slyke, J. Biol. Chem. 1922, (52), 525-570 and Mathematica (or by hand if you've got time for that))
$$\beta=-\frac{\mathrm{dC_B}}{\mathrm{dpH}}=-\ln(10)~c(\ce{H3O+})~\left(\frac{\mathrm{d C_B}}{\mathrm{d}c(\ce{H3O+})}\right)$$
$$\frac{\mathrm{d C_B}}{\mathrm{d}c(\ce{H3O+})}=-\frac{k_W}{x}-1-\frac{c_0(\ce{H2A})~k_{a1}~c(\ce{H3O+})~(2~c(\ce{H3O+})+k_{a1})}{(c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2})^2}+\frac{c_0(\ce{H2A})~k_{a1}}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2}}$$

Using this final equation you should come up with plots like mine and see that the buffer capacity (for $c_0(\ce{H2A})=0.1~\ce{mol/L}$) at $\ce{pH}=\frac{1}{2}\left(\ce{pK_{a1}}+\ce{pK_{a2}}\right)$ is not zero albeit not being very high.

complete: $\beta=0.0014~\ce{mol/L}$
complete w/o water: $\beta=0.0013~\ce{mol/L}$
w/o $c(\ce{A^2-})$: $\beta=0.0001~\ce{mol/L}$
w/o $c(\ce{A^2-})$ and water: $\beta=1.1 \cdot 10^{-10}~\ce{mol/L}$

So ... is it correct to say that at maximum concentration of the amphiprotic substance the buffer capacity is zero? No but for your to acidic constants it's nearly zero.

But is it always zero at $c_\text{max}(\ce{HA^-})$?

Let's look at those systems a little bit closer. Imagine a few systems and their corresponding buffer capacity curves with $pK_{a1}=4$ and $pK_{a2}$ values between 4 and 10 like in the following figure.

If you create a plot with the buffer capacity over $\frac{1}{2}\left(pK_{a1}+pK_{a2}\right)$ you would end up with something like this

What you can see now is that the buffer capacity at $c_\text{max}(\ce{HA-})$ depends clearly on the distance between both $pK_a$ values. If they are close together the buffer capacity at $c_\text{max}(\ce{HA-})$ is high and if $\Delta pK_a \approx 3$ the buffer capacity at $c_\text{max}(\ce{HA-})$ is nearly zero but never exactly zero.


How to simulate a Lorentzian broadened spectrum in MATLAB? I'm trying to simulate the 2nu2 band of HCN in MATLAB. I generate a stick spectrum from the data I have (taken from the HITRAN database) and then use conv() to apply my line shape function. However, when I plot the data, the broadened spectrum does not line up with the stick spectrum at all. I cannot figure out what it is that I'm doing wrong or what I need to change to align the spectra. Does anyone know what is I need to fix or a resource I could use to solve this issue?
This is the spectrum I've simulated:

and the function I use to generate the spectrum is
function f = func()
HCN2nu2 = importdata('I:/HCN 2nu2.txt');
HCN = HCN2nu2(1:111,1:2);
AX = HCN2nu2(1:111,2);
v = [];
s = [];
L = [];
q = 1;
p = HCN(q);
increase = false;
for i = 6200:0.01:6700
    for x = 1:111
        if i == round(HCN(x,1), 2)
            z = AX(x);
            increase = true;
            break
        else
            z = 0;
        end
    end
    v = [v; i];
    s = [s; z];
    if increase == true
        q = q + 1;
        increase = false;
        if q > 111
            q = 111;
        end
    end
    t = (1/(1 + ((i - p)/0.5)^2));
    L = [L; t];
end
hold on
plot(v,s)
k = conv(L,s, 'same');
plot(v,k)
ylim([0 8e-21]);
hold off
end

Any help would be greatly appreciated
Link to the data file used to generate this spectrum:
https://github.com/tkh97/HCN-2nu2.git

Update : In the given data by the OP, the x scale is not evenly spaced so the discrete convolution approach will not work to generate a spectrum until and unless your data is evenly spaced! The trick mentioned by EdV is a very nice shortcut. Here, if you really want to play around with convolution via MATLAB, then read below.
Solution: You can satisfy yourself as follows:

Make sure you have a even spacing of the x-axis. Your uneven axis threw me off, and it took a while to discover why it is not working. I wish you mentioned it earlier.
Your sticks are basically scaled delta functions as Prof. Ed said.
Generate a vector for a single Lorentzian peak, whose center is zero, which is also evenly spaced. The x-axis of the Lorentzian would be symmetric [-length (wavenumber/2):Sampling rate:length(wavenumber/2)-Sampling rate].
Perform convolution of your spectrum and this Lorentzian.

Let us start with a Dirac delta centered at point $x.$ I don't know how to draw a displace Dirac delta in MATLAB. I manually made it in an Excel file. If you convolute it with a zero centered Lorentzian, then you will get a Lorentzian centered at $x.$ This is called the sifting property.
Basically, all you have to ensure is that your Lorentzian is centered at zero. First, play with convolution using single stick. Once you correct the code, so that the line position does not change, apply it on the entire spectrum.

A great book is Bracewell's Fourier Transform And Its Applications. It is all there. From MIT 2.14 / 2.140 Analysis and Design of Feedback Control Systems, Spring 2007 handout:

1.1 The “Sifting” Property of the Impulse
When an impulse appears in a product within an integrand, it has the property of ”sifting” out the value of the integrand at the point of its occurrence:
$$\int_{-\infty}^\infty f(t)δ(t - a)\,\mathrm dt = f(a)\tag{5}$$
This is easily seen by noting that $δ(t − a)$ is zero except at $t = a,$ and for its infinitesimal duration $f(t)$ may be considered a constant and taken outside the integral, so that
$$\int_{-\infty}^\infty f(t)δ(t - a)\,\mathrm dt = f(a)\int_{-\infty}^\infty δ(t - a)\,\mathrm dt = f(a)\tag{6}$$
from the unit area property.


When water undergoes electrolysis where in the beaker does the water molecule break apart? When I was thinking about it I remembered reading two things, one was that certain gases "form" on certain electrodes. The other was something like, such and such ion "travels" towards such and such electrode.
Does the molecule break up on the electrodes?
Or perhaps it breaks up at a random spot between?

The water molecules form long networks of hydrogens bonds; let's imagine one such H-bonded chain that stretches from the anode to the cathode. Reduction will occur at the cathode, so the nearest water molecule (on the end of our chain) will be deprotonated and that proton will be reduced to $H\cdot$ radical, which can combine with a nearby $H\cdot$ to form the reduction product, molecular hydrogen.

Now, the water molecule that was deprotonated can re-protonate itself with the hydrogen of its neighbor (the second $H_2O$ in our hypothetical chain), likely the very hydrogen that they were sharing as an H-bond. This leaves the second water molecule deprotonated, so it fully claims the proton the third molecule was offering in their H-bond.

The process continues, with the negative charge effectively traveling the length of the chain toward the anode as each water molecule in turn briefly becomes hydroxide. This "proton hopping" behavior is known as the Grotthuss Mechanism. This animation can give you a sense of how this phenomenon happens, but note that the animation shows an acidic mechanism (extra proton, positive charge traveling) rather than the basic mechanism I described (a missing proton, with the negative charge traveling). Both are likely at play during the electrolysis of water. (Actually, the two mechanisms probably cannot be distinguished due to the simultaneous nature of the redox process, which I'm about to describe).

At the anode, hydroxide will be oxidized to $O_2$ through a few steps that aren't necessarily relevant to your question. The big idea is that the hydroxide will be created from a water molecule at the anode end of the chain of H-bonded molecules, and it will probably happen nearly simultaneously as a water molecule at the cathode end is deprotonated to form the proton reactant. Most probably both ends of our H-bonded chain of water molecules will react at more or less the same time.

If you like metaphors, think of it as a line of people each holding a basketball. Everybody passes their basketball to the right, and received a basketball from the person on their left... Except the people standing on the two ends. One will throw their basketball into the hoop, and the other will be torn to shreds and evolve molecular oxygen.


Should binding constants be unitless when deriving fractional occupancy equations from reactions? It is known that binding (aka association) constants are in fact unitless, as has been discussed here already. However, I'm not a chemist and am confused about when one should or should not use units when working with association constants. One source says:

  $K_\text{eq}$ for a reaction with unequal numbers of reactants and products is always given with units, even in published papers.

But why is that? Why not always use unitless values? Is there something inherently wrong with never using units for binding constants?

Consider this example. Let $\ce{A}$ bind to $\ce{X}$ as an n-mer (e.g. $\ce{A}$ can be a transcriptional activator binding to gene promoter). This results in active state, $\ce{X_{A}}$ (e.g. a state leading to gene transcription):

$$\ce{X + nA ->[k_{\text{on}}][k_{\text{off}}] X_{A}}$$

Association constant: $K_\text{A} = \frac{k_\text{on}}{k_\text{off}} $

Assuming equilibrium and law of mass action:

$$K_\text{A} \cdot \text{X} \cdot \mathrm{A^n} = \mathrm{X_A}$$

Now, the fractional occupancy (active states to all states ratio) is:

$$y=\mathrm{\frac{X_A}{X+X_A}} = \frac{K_\text{A}\cdot \mathrm{A^n}}{1 + K_\text{A}\cdot \mathrm{A^n}}$$

As $y$ must be unitless (i.e. it can be interpreted as a probability), for this particular equation, $K_\text{A}$ must have units of $\mathrm{M^{-n}}$ (in general, $\text{concentration}^{-n}$) for the equation to work out, correct (assuming concentration of $\ce{A}$ has units $\mathrm{M}$)? So in this particular case, using unitless $K_\text{A}$ seems wrong, but is it really, or is there something I'm missing?

Now let's extend this equation so that it includes a half-saturation constant $h$, i.e. concentration of $\ce{A}$ required for $y=0.5$ (50% activation). If I'm doing this correctly, we get:

$$y=\frac{K_\text{A}\cdot \mathrm{A^n}}{K_\text{A}\cdot h^n + K_\text{A}\cdot \mathrm{A^n}} = \frac{\mathrm{A^n}}{h^n + \mathrm{A^n}}$$

Note that this is equivalent to the Hill equation for an activator. This generalized equation, unlike the previous one, works just fine regardless of whether or not $K_\text{A}$ is unitless.

Is my understanding of this correct, and does the choice of unitless vs non-unitless binding constant indeed depend on the formulation of a specific equation?

The equilibrium constant does not need to be unitless, because it is depending on its definition. See goldbook:

  Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always unitless, as it is defined differently (goldbook)

  Standard Equilibrium Constant $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{\mathcal{R}T}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $\mathcal{R}$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

In your example both would work if you straighten your definitions, involving concentrations
\begin{aligned}
K_A \cdot c(\ce{X}) \cdot c^n(\ce{A}) &= c(\ce{X_{A}}),\\
y=\frac{c(\ce{X_A})}{c(\ce{X})+c(\ce{X_{A}})} &= \frac{K_A\cdot c^n(\ce{A})}{1 + K_A\cdot c^n(\ce{A})},
\end{aligned}

or activities
\begin{aligned}
K^\circ_A \cdot a(\ce{X}) \cdot a^n(\ce{A}) &= a(\ce{X_{A}}),\\
y=\frac{a(\ce{X_A})}{a(\ce{X})+a(\ce{X_{A}})} &= \frac{K^\circ_A\cdot a^n(\ce{A})}{1 + K^\circ_A\cdot a^n(\ce{A})}.
\end{aligned}

In both cases $y$ remains unitless.

In experimental chemistry it is much easier to observe concentrations instead of activities. Therefore to fist named constant is more often in use. A straightforward use of the thermodynamical equilibrium constant might sometimes prove a little tricky.

For reasonable dilutions one can simply substitute activity with a unitless concentration as given through
$$a(\ce{Y})=\gamma\frac{c(\ce{Y})}{c^\circ},$$
with $\gamma\approx1$ for $c(\ce{Y})\to0$ and $c^\circ=1\:\mathrm{mol/L}$.


Why does H2O electrolysis produce O2 on the anode instead of H2O2 If the oxidation of $\ce{H2O}$ produces $\ce{OH-}$ ions and the anode separates electrons from them leaving neutral $\ce{OH}$ molecules, why don't they bind to each other covalently?

It does, as an intermediate product. The peroxide formed is on the surface of the electrode which changes it's chemical behavior.

Electrolysis is a four-electron process. In the first step, an electron is removed from a water molecule, producing a neutral OH adsorbed to the surface of the electrode. The second step removes another electron from the adsorbed OH, producing a neutral oxygen atom adsorbed to the electrode. Third, another electron is removed from a water molecule leaving an adsorbed OOH. Then the fourth electron comes from the OOH to produce O2.

My understanding is that the interactions with the electrode partially stabilize the unstable intermediate states. This is the reason some electrodes work better than others as catalysts.

This paper has more than you'll ever need to know: https://doi.org/10.1002/cctc.201000126


Can a six member ring expand to achieve octet completion to stabilize a carbocation? (Probably unnecessary background: I came up with this carbocation while predicting the major product of $\ce{NaNO2/HCl}$ with 1-(methylamine)cyclohexan-1-ol)

Will the following carbocation rearrangement occur?

I have opinions both for and against it.

For: The ring expansion leads to octet completion of the empty p orbital of the carbocation. That stabilizes the carbocation.

Against: expansion from six member ring to seven member ring is uncommon. In school, I was taught that - for rings less than size ten - the cyclohexane ring is the most stable ("strain-free"). So, rearrangement shouldn't occur.

A Scholar search gives me no results for the proposed expansion. I know the search probably won't get me anywhere because I don't know the right keywords, and hence I'm stuck. So, can a six member ring expand to achieve octet completion to stabilize a carbocation? Also, does the expansion lead to the major product?

The reaction which we are talking about here is called Tiffeneau–Demjanov rearrangement.

According to Wikipedia,

  the Tiffeneau–Demjanov rearrangement (TDR) is the chemical reaction of a 1-aminomethyl-cycloalkanol with nitrous acid to form an enlarged cycloketone. The Tiffeneau–Demjanov ring expansion, Tiffeneau–Demjanov rearrangement, or TDR, provides an easy way to increase amino-substituted cycloalkanes and cycloalkanols in size by one carbon. Ring sizes from cyclopropane through cyclooctane are able to undergo Tiffeneau–Demjanov ring expansion with some degree of success. Yields decrease as initial ring size increases, and the ideal use of TDR is for synthesis of five, six, and seven membered rings. A principal synthetic application of Tiffeneau–Demjanov ring expansion is to bicyclic or polycyclic systems.

According to The Encyclopedia of Reagents for Organic Synthesis (DOI: 10.1002/047084289X), the yield of your product will be around 40-42%:

You can check out the Wikipedia page for more details.


How to know whether to add latent heat when calculating final temperature? From the examples in my textbook, it is usually very easy to see whether you need to account for the latent heat when calculating the final temperature. For example, if you add 2 x 13 gram ice blocks to a 250 ml cup of water, it is obvious that the ice will melt, and you will have to account for the latent heat of fusion for the ice.

However, what if it isn’t so obvious whether one of the substances will undergo a phase change? For example, what if we added 15 grams if ice to 20 ml of oil? If the final temperature is below 0, then we don’t need to account for the latent heat for ice, otherwise we will have to. How do we determine whether the ice will experience a phase change, and more generally, how do we determine whether one of the substance will experience a phase change?

It is less obvious than you seem to assume (say, if water cup in your example were already at $0^\circ$C, then the ice would not melt), but still pretty obvious.

The rule of thumb is:

Assume there is no phase change.
Calculate the final temperature.
Check whether there is a phase change in any of the substances involved as they go from starting to final temperature.
If no, we are done. If yes, assume there is a phase change and go back to step 2.


How does the energy released during a bond formation typically manifest itself on atomic level? I've been reading about how energy is released when new chemical bonds are formed but it's hard to find info on how that typically manifests itself (besides that it's via heat). I read somewhere that the valence electrons will move into lower energy orbits during a bond, but I thought that always created a photon.  How and where is the heat actually "created?". When two H atoms form H2, does the new molecule start moving faster (discounted for increased mass)? What happens if there are no surrounding molecules?

As pointed out by @Paul in case of the reaction between 2 hydrogens atoms you need a third body to take away the excess energy. This third atom(M) can be another hydrogen molecule or other inert gases(eg. Ar, He, N2).

Or you can have the presence of a wall(eg. gas-phase hydrogens atoms in a container):

Another possibility is adding a reactive gas(eg.oxygen) and in this case, the mechanism is much more complicated with many steps.

All these reactions are discussed in ref.1.

Comment: when two hydrogens atoms combine 4.5eV is released. In the case of an inert third body, I haven't found any quantitative information about how much energy is transferred to the third inert gas. If all of it is transferred to the third body(eg.H2 molecule) as kinetic energy you end up with an H2 molecule moving at 20'800m/s that corresponds to a kinetic temperature of 34'800K!

Another example is the CsBr formation in the presence of a third inert body(R):

From ref.2:

  Analysis of the detailed dynamics of recombination at small impact parameters showed that the stabilization of an ion pair occurs through the collision of the third body with the two ions and that the dynamics of energy transfer in this region is almost
  completely determined by the repulsive interactions of the Xe atom with both the ions.

  In the range of impact, the main mechanism of stabilization is energy transfer energy through repulsion between the third body and both the ions or through consecutive collisions of the third body with the ions approaching each other.

Regarding how much energy is transferred to the third inert body:

  The amount of energy transferred to the third body is apparently determined by the collision configuration and by the phase of the collision of the ions at the moment of energy transfer.

References

Bennett, J. E., and D. R. Blackmore. 1971. “Rates of Gas-Phase Hydrogen-Atom Recombination at Room Temperature in the Presence of Added Gases.” Symposium on Combustion 13 (1): 51–59.
Kabanov, D. B., and L. Yu Rusin. 2012. “Mechanism of the Direct Three-Body Recombination of Atomic Ions in a Central Collision.” Russian Journal of Physical Chemistry B 6 (4): 475–85.
Stepukhovich, Aleksandr Davidovich, and Viktor Markovich Umanskii. 1969. “Kinetics and Mechanism of Three-Body Recombination of Atoms and Radicals.” Russian Chemical Reviews 38 (8): 590.


How to decide from the solubility product constants which hydroxide precipitates first? Salts of $\ce{Fe^3+}$ and $\ce{Al^3+}$ are in a beaker of water. If NaOH is added slowly, what compound would precipitate first?

For me, I think it's $\ce{Fe^3+}$ because of its lower $K_\mathrm{sp}$ value when it reacts with $\ce{NaOH}$. The two equations I wrote were

$\displaystyle\ce{Fe^3+(aq) +3NaOH(aq) -> 3Na+(aq) +Fe(OH)3(s)}$
$\displaystyle\ce{Al^3+(aq) + 3NaOH(aq) -> 3Na+(aq) + Al(OH)3(s)}$

The lower the $K_\mathrm{sp}$ value, the more insoluble the substance is (the precipitate is insoluble). So, $\ce{Fe}$ would precipitate first.
Is this correct?

If the question is about what compound would precipitate first, then there can be at most one answer.

Given equal stoichiometry  for the precipitation of $\ce{Fe(OH)3}$ and $\ce{Al(OH)3}$ and solubility product constants of $6 \cdot 10^{38}$ and $4 \cdot 10^{15}$ respectively, $\ce{Fe(OH)3}$ will be the correct answer.

$$\ce{Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} -> Fe(OH)3_{(s)}}$$

However, in practice the precipitate will probably contain less water than stoichiometric $\ce{Fe(OH)3}$, and also contain small amounts of $\ce{Al(OH)3}$


Finding the volume needed to make a new pH In the titration of 20.00 mL of 0.500 M HCl by 0.500 M NaOH, calculate the volume of 0.500 M NaOH required to reach a pH of 2.0.

I have tried to change the 20.00 mL into the amount of substance (0.01 moles), then I turned the 2.0 pH into concentration (0.01 M). I then tried to find the total volume, which came out to be 1 L, which creates 980 mL of NaOH. Plugging it back in though doesn't give 2.0 pH. So, I was wondering if anyone knew how to find the volume, or if there was a formula that I could use to solve it.

$$\ce{NaOH + HCl -> NaCl + H2O}$$

Let $x$ Litres be the amount of $\ce{NaOH}$ needed.

$$-\log[\ce{H+}] = 2$$

$$-\log\left(\frac{(0.5 \times 0.02 - x \times 0.5)}{(0.02 + x)}\right) = 2$$

$$x = \frac{49}{2550}$$

$$x \approx 0.0192 \; \text{(3 sig. fig.)}$$

Thus, $\pu{0.0192 L}$ of $\ce{NaOH}$ is needed.


What happens when an explosion can't lead to volume increase of the thermodynamic system? What if an explosion is not able to exit the container?

Take a pipe bomb for example; assuming that the pipe bomb was hypothetically made out of such a strong material that the explosion is not able to rip the material apart, what would happen?

Asking this because I was told: NO MATTER WHAT the explosion has to and WILL "exit", but what if it just can't?

I'm not an expert in thermodynamics and gas physics so there might be some errors in this calculation but here goes:

According to several sources (wikipedia included) TNT decomposes according to this equation:
$$\ce{2C7H5N3O6 -> 3N2 + 5H2O + 7CO + 7C}$$

If 100kg of TNT was placed in an indestructible $1m^3$ container containing an inert gas (we will say helium) at standard temperature and pressure here is a rough estimate of what might happen if it was detonated.

Using a Hess's Law cycle we can calculate the enthalpy of decomposition of TNT from formation enthalpy data: $$\Delta_{decomp}H = 63.2*2 - 285.8*5 - 110.5*7 = -2076.1~ kJmol^{-1}$$

Then we can calculate the energy released by the decomposition: $$M_r~of~TNT = 227.13~gmol^{-1}$$ $$n = \frac{100}{0.22713} = 440.3~mol$$ $$Energy~released = 440.3 * 2076.1 *1000 = 914.1~MJ$$

I'm not sure how to go about specific heat capacities for gas mixtures so I won't put a figure on the temperature rise that this will cause but I suspect it will be in the thousands of degrees due to the amount of energy and so the resultant pressure rise will also be very significant, probably enough to liquefy the gases. Also as @Klaus Warzecha hinted at in his comment these kind of temperatures and pressures will probably cause some interesting chemical reactions to take place.

http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf


why do electron pairs not take axial positions in the VSEPR structure of ClF3? when we make a Valence shell electron pair repulsion theory based structure for an molecule after calculating its hybridisation ; when we have to decide the positions of the ions we put them in such positions so that repulsion is minimized .It follows the basic rule of lone pair-lone pair repulsion> lone pair-bond pair repulsion >bond pair -bond pair repulsion . So when it comes to deciding the position of lone pairs of electrons in ClF3 why dont we put them on the axial positions so that they have an angle of 180 deg between them as compared to 120 deg that they currently have ?

Let's take a look at the electronic configuration of $\ce{ClF3}$.

For $\ce{ClF3}$ actually three isomers are possible.

It is highly unfavorable for the lone pairs to be 90° apart which rules one of the three. In order to rationalize the obervation that the isomer with both lone pairs in equatorial positions is the observed form, it is necessary to count the number of 90° bond pair - lone pair interactions. The observed isomer  has four such angles and the isomer with the lone pairs in axial positions has six which is (presumably) less favorable. The molecule might be called T-shaped based on the atomic positions only.


Calculate the formula, and moles If $\ce{1.84\times10^{-4}}$ moles of element A was completely reacted with element B to create $\ce{1.231\times 10^19}$ molecules of $\ce{(A3B2)_x}$ a hypothetical compound, what is the value of x?

I tried this:

$\pu{1 mol} = 6.02 \times 10^{23} ~\pu{molecules}$

$y~\pu{mol} = 1.231 \times 10^{19} ~\pu{molecules}$

$y = \pu{2.0449 \times 10^-5 mol}$

$\ce{(A_3B_2)_x} : 1.84\times10^{-4}+ {2\over 3}(1.84\times10^{-4})=3.067\times10^{-4}$

So it produces $3.067\times10^{-4}$ mol of the compound

and, $3.067 \times 10^{-4}~\pu{mol} \div 2.0449 \times 10^{-5}~\pu{mol} = 15$

And since 1 compound has 5 atoms, $15 \div 5 =3$

Therefore, $x$ is 3

I am not sure if I did this question correctly. Even if I did, is there a more efficient and organized way of solving this problem?

At first you said that $x = \pu{2.0449\times10^−5 mol}$ and then later said $x=3$. It is clearly wrong even without going into the maths!

That said, I believe you are confusing the stoichiometry of the compound with the number of moles of the compound produced. The $x$ in $\ce{(A3B2)_x}$ is a part of the chemical formula for that compound, and not the moles of $\ce{(A3B2)}$ produced.

For example, $\ce{AlCl3}$ usually exists as a dimer in gas phase, so we write it as $\ce{Al2Cl6}$ or simply $\ce{(AlCl3)2}$ for clarity. You should hence realize that the $x$ in $\ce{(A3B2)_x}$ represents the number of $\ce{A3B2}$ molecules that have polymerized together to form a single molecule. And the question states that $\pu{1.231\times10^19}$ of such molecules was produced.

With the conceptual clarity attained, we'll now solve the numerical. Note that the question says that all the moles of elemental A were obtained in the compound. This implies:

$$
\begin{align}
\text{moles of A in the compound} &= \text{original moles of elemental A}\\
\implies 3x\times(\text{moles of compound }\ce{A3B2})&=1.84\times10^{-4}
\end{align}
$$

Solving this, you'll get approximately $x=3$, which is the expected answer.


Nucleophiles in protic and aprotic solvent If the nucleophiles have almost same size (not a large difference) such as $\ce{C- , N- , O-}$ then how their nucleophilicty varies in protic and aprotic solvents?\

Does it depend on types of reaction such as Sn1 or Sn2 ?

From what I learned in organic I, their nucleophilicity does not change with the solvent because if the solvent is polar Protic, each nucleophile will be hindered in a similar manner (due to the solvent cage). If the solvent is polar aprotic, this hindrance does not matter. In both cases, nucleophilicity matches base strength.

I believe the solvent only affects nucleophile strength going down the periodic table. There are also polarizability effects for larger molecules but I’ve been told in general, a smaller molecule is a stronger nucleophile and larger molecules are weaker in aprotic solvents. This trend reverses for polar Protic solvents.

Also keep in mind, the strength of the nucleophile only matters for SN2 because of the backside attack being the slow step.


Most efficient way to purify household vinegar: distillation or freezing? I am aware that there are better ways to purify acetic acid out of a water-acid mixture than these, but which one out of distillation (using heat) and fractional freezing (using cold) would gain a higher concentration of acetic acid?

There is no acetic acid/water azeotrope, so distillation can produce glacial (pure) acetic acid.


Physical significance of chemical potential and fugacity I'm studying equilibria and thermodynamics and came across these two terms.
My problem is, unlike other thermodynamic properties that I can understand physically like volume, pressure, enthalpy etc. do these two quantities have physical significance or, is it just that we define fugacity (without any physical meaning) just to make sure the equation
$$\mu - \mu_0 = RT\ln\frac{p}{p_0}$$
looks the same for real gases? If yes, then what is the motivation for the name “fugacity” meaning literally “escaping tendency”?
Similarly, do we just define chemical potential randomly, since it's extremely useful in various calculations when the composition varies, or does it have some physical significance?
Also, why do we call it chemical potential?

The chemical potential is the partial derivative of the Gibbs free energy with respect to the number of moles of the specified species at constant T, P, and numbers of moles of all other species.  The equation you wrote is the what this partial derivative reduces to for a species in an ideal gas mixture, where P is the partial pressure of the species in the mixture.  For a real gas, the fugacity of the species replaces the partial pressure of the species in the same equation for the chemical potential.


Nitrogen bonding and electron affinity How does nitrogen bond with itself to form N2 if it has no affinity for electrons? If a nitrogen atom can't pull on the electrons of another nitrogen atom how can N2 have covalent bonds.

It doesn't matter that the electronegativity difference between nitrogen with itself is 0. It just means the bonds it makes with another nitrogen atom have $100~\%$ covalent character.

A typical chemical bond basically spans the spectrum of bonds with $100~\%$ ionic character, uneven ionic/covalent character, and $100~\%$ covalent character.

$\mathbf{100~\%}$ ionic character: $\ce{A^+B^-}$ or $\ce{A^- B^+}$, with no sharing of electrons at all
Uneven ionic/covalent character (i.e. polar): $\ce{{\stackrel{\delta+}{A}}-{\stackrel{\delta-}{B}}}$ or $\ce{{\stackrel{\delta-}{A}}-{\stackrel{\delta+}{B}}}$, with deviation from perfect sharing of electrons and direction of favored negative/positive end of dipole determined by electronegativity difference
$\mathbf{100~\%}$ covalent character (i.e. $100~\%$ nonpolar): $\ce{A-A}$ or $\ce{B-B}$, with perfect sharing of electrons

It just so happens that $\ce{N2}$ is one example of $100~\%$ covalent character. You can argue the same for $\ce{O2}$, and any similar homonuclear diatomic molecule.

(Granted, there are some exceptions to 100% covalent character, like in iron's metallic bonding, but anyways...)


If a good choice of a unit cell should be the one of most symmetry ,then why keep body centered tetragonal if face centered cubic exists? For any lattice a good choice of a unit cell is the one that is the most symmetric.
Now if every lattice made by a body centered tetragonal unit cell can be made by a face centered cubic unit cell ,which is more symmetric ,then why keep the body centered tetragonal unit cell as a class of Bravais lattice ?

Body-centered tetragonal is face-centered cubic only if $c/a=\sqrt2$.  If you try your transformation with a $c/a$ value greater/less than $\sqrt2$, your "cube" will have lateral edges that are longer/shorter (resp.) than the basal edge and so really remains tetragonal.
You can also read this the other way.  If you are presented with a body-centered tetragonal lattice and $c/a$ seems to be close to $\sqrt2$, maybe it's really fcc.


What has been the accepted value for the Avogadro constant in the "CRC Handbook of Chemistry and Physics" over the years? Some questions about the constant made me curious as to exactly how much the accepted value has changed over the years. I am pretty sure that I learned a rounded value of $6.023\times10^{23}$ about 1970 which would have been around the 50th edition. I know that sometime after that the accepted value rounded to 4 significant figures was $6.022\times10^{23}$. Of course since 2019 the accepted value has been fixed at exactly $6.02214076\times10^{23}$
It would seem that the CRC handbook is a reasonable source for the accepted value of the constant since it has been updated almost yearly.
Please update the table in the answer in the community wiki page with other values

I dislike offsite image links since they disappear over time. So below is Table 1 from andselisk's comment on the question.
The paper itself is behind a firewall and I can't see it.

See Table 1 in Becker, P. History and Progress in the Accurate Determination of the Avogadro Constant. Rep. Prog. Phys. 2001, 64 (12), 1945–2008. DOI: 10.1088/0034-4885/64/12/206. It covers year up to the year 2001 and is going to save a lot of time. – andselisk


Where can I find the total enthalpy and heat of dissolution in water of a given components? I don't really have a background in chemistry but I'm actually studying in aerospace engineering, sorry if this question may seem trivial.

I am interested in the decomposition process of the hydrogen peroxide ($\ce{H2O2}$) and I need to find the total enthalpies of $\ce{H2O2}$, $\ce{H2O (l)}$, $\ce{H2O (g)}$ and $\ce{O2 (g)}$ and the heat of dissolution ($\Delta H_{\text{dis}}$)of $\ce{H2O2}$ in water, as a function of the solution concentration of $\ce{H2O2}$.

Firstly, I tried to find the values of the total enthalpies, but I couldn't find them. However, I found a lot of values about the $\Delta H^0$ and $h^o_f$. I can't understand if I can retrieve the value of $H$ with the $\Delta H^0$ and/or $h^o_f$.

Secondly, I can't find a good answer to another question. Does the $H$ is a function of the pressure or not? If so, how can I find the value of $H$ at a given pressure and a given temperature?

It's very unlikely you actually need the enthalpy.  Like energy (of which it is a variation), absolute enthalpy is almost unmeasureable, normally one deals only in enthalpy changes.  In your case I would hazard a guess that you're  interested in things like dH_rxn, the heat of reaction for the decomposition of H2O2.  You can readily calculate this from the dH0 of the components.  The latter are called the "standard heats of formation" and they are combined with "Hess's Law."  Googling either of those terms should lead you to explanations of how the data you can find can be combined into the data you want. How exactly you do it will depend on what quantity you want, and the thermodynamic conditions (e.g. T and p) under which you want it, and the accuracy you need.

Yes, H is generally a function of pressure and temperature.  The starting search term there would probably be "derivatives of the enthalpy" because you would generally calculate H at one (T,p) from H0 at some other (T0,po). You'll find the derivatives can be related to material properties like the heat capacity, compressibility, and coefficient of thermal expansion.  For changes over significant temperature ranges these can be integrated, provided they are known as functions of temperature, or you are willing to use approximate models (like the ideal gas model) that have formulae.