ICHO Synthesis Problem 2018 I've several doubts about this ICHO synthesis problem, which was a part of the preparatory problem package of 2018 (so yes, it's not a part of an ongoing contest)

Before sharing the answer given, I'd like to share my thoughts and efforts towards the problem -

The first step is clearly an oxidation step, performed in presence of the peroxy acid above the arrow; which is subsequently reduced to the corresponding carboxylic acid. The oxidation product should be an amine oxide, but the question is, which one? (We have three nitrogens to play with)
For the sake of continuation, let's assume A is what they expect it to be (check the image below). Now, I'd expect more than just one hydrogen of the ring to be replaced by deuterium in the base catalysed tautomerism step. What's the next step (acid catalysed) for? Also, in basic medium (1) the initial ester would undergo saponification to acid.
B to C is a reducing step. Pd/C in presence of formic acid is known to be a reducing agent, but how do we know it reduces the amine oxide back to amine? (It'd be great if some sources could be provided for this, I couldn't find any)

C to the final product is pretty direct, the carboxylic acid group converts to acyl halide which undergoes nucleophilic addition by the ammonia derviative.

Thanks a lot.

P.S. Only a sound explanation for the A to B conversion remains - rest has been taken care of by the first posted answer. Please make additions/include relevant stuff for other conversions if you find any.

You have asked quite a number of questions in your post so let's tackle it one by one.

Why does the oxidation take place at that particular nitrogen atom?

To understand this, we first need to acknowledge that peroxyacids $\ce {RCO3H}$ are electrophilic oxidising agents. Specifically, the oxygen atom attached to the $\ce {H}$ is the electrophilic atom. From the frontier MO perspective, the reactive LUMO of the peroxyacid is the $\ce {O-O \sigma^*}$. This implies that the most nucleophilic nitrogen atom is likely to be favourably oxidised.

In fact, the oxidation of pyridine, which is a chemical species containing a nucleophilic $\ce {N}$ atom, to form pyridine $\ce {N}$-oxide using peroxyacids is well-known. It is documented by Clayden, Warren & Greeves (2012) on p. 730 as a method of increasing the reactivity of pyridine towards both electrophilic and nucleophilic aromatic substitution reactions. Notice that the bottom $\ce {N}$-containing ring actually resembles a pyridine fused with another benzene ring. However, note that the lone pair on $\ce {N}$ is not delocalised into the ring(s). Thus, there is no significant decrease in nucleophilicity of the bottom $\ce {N}$ in this compound relative to pyridine's $\ce {N}$ atom, suggesting that oxidation of the peroxyacid can very likely take place there.

Of course, we must be thorough in our analysis and also take a look at the relative nucleophilicities of the other two $\ce {N}$ atoms. Immediately, we will notice that for the $\ce {N}$ atom that appears to be making 3 single bonds has a lone pair that is heavily delocalised, in both the benzene ring below and the 5-membered ring it is in. Thus, we can eliminate the possibility of the the oxidation taking place at that nitrogen atom.

For the final top-most $\ce {N}$ atom, the lone pair may appear to be of the same nature of that of pyridine, i.e. held in an $\ce {sp^2}$ hybrid orbital. However, note that the presence of the adjacent $\ce {N}$ atom may result in inductive withdrawal of electron density from that $\ce {N}$ atom, decreasing its nucleophilicity.

Ultimately, I believe this first part is trying to test students if they are able to relate the structure and reactivity of pyridine towards oxidation by peroxyacids to the bottom pyridine-like ring. If students can make the connection, then they would instinctively know that the oxidation has taken place at that $\ce {N}$ atom.

From A to B what happened?

Well... There are a number of things which happened here. The most obvious being the first step: Basic hydrolysis occurs, with the leaving of $\ce {EtO^-}$. The next step is an electrophilic aromatic substitution, with the acidic deuterium ions as the electrophile. For pyridine $\ce {N}$-oxide, the 2- and 4-positions are activated towards EAS reactions. This can be shown simply by drawing the resonance structures of the compound with an $\ce {N=O}$ double bond. Similarly, the EAS takes place at the 2-position, relative to the $\ce {N}$-oxide group. It is difficult to explain why it doesn't take place at the 4-position as well. In general, for this type of structure elucidation questions, we would need to predict structures by looking at the resultant structures as well, especially if they are given. In this case, we see that only the 2-position has deuterium substitution taking place and we should simply leave it as that. Of course, if anybody has a good explanation for why that is the case, please propose it. Also, the other positions on the other rings may not be activated enough for the substitution to take place. Thus, we do not see substitutions taking place at those positions. Again, I am not able to provide a rigorous explanation for why that is the case.

Finally, the third step with $\ce {H2O}$ effectively makes the pH of the medium neutral again and also changes the carboxylic acid group from $\ce {COOD}$ to $\ce {COOH}$. This is simply due to a concentration effect. The high concentration of $\ce {H2O}$ simply pushes the equilibrium towards species containing $\ce {COOH}$.

What about from B to C?

From Wikipedia, it seems that $\ce {B}$ to $\ce {C}$ is simply a reduction step. The ammonium formate decomposes to give $\ce {NH3}$, $\ce {H2}$ and $\ce {CO2}$ in the presence of $\ce {Pd/C}$. The hydrogen gas evolved then gets adsorbed onto the palladium catalyst, where it can reduce the $\ce {N}$-oxide to give back the original $\ce {N}$-heterocycle.

And the last step involving conversion of C to the final product is as you have proposed.

Reference

Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry (2nd ed.). Oxford University Press: New York, 2012.


What bond\crystal has the most regular pattern (i.e. least irregularities)? Salt is a famous regularly repeating cubic crystal, but there are always natural deformities and flaws present. I have heard that graphene is an extremely uniform substance. With these in mind, what is the most regular crystal pattern, natural or man-made?

In nature you need to look for precious gamestones and then do optical selection (diamante and ruby would be good candidates). The same would be true for synthetics but of course as mentioned by Jon you can also take elements (Si, Ge). General rule: they should be hard (smaller chance that pressure will affect structure, by pressure I mean also your tools) and you need to grow them slowly.

btw it's hard to call graphen crystal pattern -- it's 2D.


How many alkanes having a molecular weight of hundred, are chiral? I successfully found 2, but the answer key says that there are 4 such compounds possible. I found:

3-methyl hexane
2,3-dimethyl pentane

We can't have a cyclic ring as the masses won't match. What else could be made folks?

You are essentially correct.  Of the nine isomers of heptane, only two are chiral.  However, if you count each enantiomer as a separate molecule, then there are "four" chiral isomers:

(R)-3-methylhexane
(S)-3-methylhexane
(R)-2,3-dimethylpentane
(S)-2,3-dimethylpentane


How to calculate the start concentration for a first order reaction? The half-life of the first order reaction, A → products, is 33.8 s. What was the original concentration, [A]O in mol/L, if after 2.5 minutes, the concentration of A is 0.0796 mol/L? Round your answer to three significant figures.

I think the formular is 1/(A)t = kt + 1/(A)0

So I tried 1/33.8s = (0.0796)

so it = 1/0.0796.

I know this is not correct, but I think I'm using the right formula.  I looked back at my notes and nothing is really making any sense, but this formula.

For a first order reaction $\ce{A ->[k]P}$, the rate constant is given by

$$k[A] = - \frac{d[A]}{dt} $$

So, at any time $t$, the concentation $[A]_t$ is

$$[A]_t = [A]_0 \cdot e^{-kt}$$

At the half-life time $t_{1/2}$, $[A]_t = \frac{1}{2}[A]_0$, isn't it?

Would you agree that we can calculate $k$ using $k = \frac{ln 2}{t_{1/2}}\ $?

Knowing $k$ and $[A]_t$, calculation of $[A]_0$ doesn't seem impossible now ;).


What is the molecular shape of Sulfur Pentafluoride Anion? $\ce{SF6}$ is an octahedral shape which makes perfect sense. $\ce{SF5^+}$ arranges 5 pairs of electrons in a trigonal bipyramidal structure. But what if you add a lone pair to this? Will the electron pairs arrange into an octahedral shape, but due to extra repulsion of lone pair, or will it have some distortion?

Will it be similar to $\ce{BrF5}$ structure?

Yes, $\ce{SF5^-}$ does infact have an octahedral shape. It's structure would be

Yes, it is similar to the structure of $\ce{BrF5}$


Determining the temperature at which 90% of the reactants in a equilibrium have reacted I have the following problem I am stuck on, specifically the last bit (iii). The first parts are fairly trivial and I have arrived at the following equation to determine the temperature:

$$T=\frac{\Delta H}{\Delta S - R \ln K}$$

However I am unsure on how to proceed. I do not have a value for $K_p$ and I can't think of how to deduce it. I have tried the following, by assuming that
$n_{\ce{Fe3O4}}=n_{\ce{CO}}$, going to the following expression

$$K = \frac{(0.9 x)^3 (0.9 x)}{(0.1x)(0.1x)}$$

but obviously that doesn't cancel.

Any help? I'm probably barking up the wrong tree with the above expression as I'm aware I'm looking for $K_p$ in terms of partial pressures but it would surely result in the same overall problem.

Thanks.

Assuming $\ce{FeO}$ and $\ce{Fe3O4}$ are solid throughout the range of temperatures we are interested in, they won't appear in the expression for the equilibrium constant $K$ (which in this case is equivalent to $K_p$). If $n$ is the total number of moles of $\ce{CO}$ that you start with,

$$K_p = K = \frac{p_{\ce{CO2}}}{p_{\ce{CO}}} = \frac{n_{\ce{CO2}}}{n_{\ce{CO}}} = \frac{0.9 n}{0.1 n} = 9$$

Plugging this into the formula previously derived, we find $T = 1388~\mathrm{K}$.


Gaussian vs. Lorentz/Cauchy distributions for artificial broadening When I am supposed to calculate an IR-, Raman-, or UV/Vis-spectrum, the calculations give me some x-y-values, e.g., Frequency-Intensity, or Energy-Oscillator Strength. Those values can either be printed as stick spectra or a sum of Gaussian-/Lorentz-distributions is being used to broaden the peaks to generate a more "natural" spectrum.

I've been asked a couple of times which broadening function to choose when, but actually never had a good answer/given explanation. The only thing that I wrote down quite some time ago, was that for IR- or Raman-spectra of solids a Gaussian broadening and for every other type of IR- or Raman-spectra a Lorentz broadening should be used.

Google searches for "gaussian or lorentz broadening" and "line broadening" did yield results that try to explain it, but I didn't find most of them really helpful.

So the question is: When do I choose the Gaussian or the Lorentz-functions for artificial broadening of IR-, Raman-, or UV-Vis stick spectra and why?

I think that any of these functions, Gaussian, Lorenzian, Voight, will do, they do not differ by much but the Lorenzian has a slightly wider base and in overlapping spectra this may increase background too much. I assume that you will convolute your stick spectrum to get your 'real' spectrum, in this case you can zero pad the transform, it makes nicer looking spectra but adds no more information. Perhaps the most important point to check is  the spectral resolution of the instrument you are simulating spectra from, and adjust the width of you broadening function appropriately.


How many alternative chemical ways to transport oxygen in living creatures are known? Mammals and many other groups of animals usually transport oxygen using haemoglobin and other complex proteins, the core of which is based on an iron coordinated to a porphyrin. There are plenty of subtle differences in the protein structures but are there any notable alternatives to that central iron-porphyrin structure?
Obviously there are other chemicals that can do a similar job. Polyfluorinated hydrocarbons were once considered as a viable emergency alternative to donor-blood as they are very good at transporting oxygen. But those are not something that living creatures are likely to develop.
So the question is are there either possible or known alternative chemistries for transporting oxygen in living creatures?

Yes! A variety of metalloproteins are used for oxygen transport. These include proteins that bind to porphyrins with different central metal atoms, as well as proteins that bind the metal atoms directly, without a porphyrin ligand. Fun fact- hemocyanin, which uses two copper atoms as prosthetic groups to bind oxygen, is responsible for the blue color of horseshoe crab blood!
Here are some relevant links:
https://en.wikipedia.org/wiki/Respiratory_pigment
https://en.wikipedia.org/wiki/Hemocyanin
https://en.wikipedia.org/wiki/Hemoglobin (scroll to "other oxygen-binding proteins" near the bottom)


What is the most "important" resonance structure of SCN⁻? Numerous online references say that $\ce{SCN-}$ has two resonance structures:

I am wondering why this structure is not also possible?

I expect structure 3 to be rare because of the high formal charges, but shouldn't it be included as a possible resonance structure?

Furthermore, there is disagreement about whether structure 1 or structure 2 is more common. I would expect structure 2 to be more common because the negative charge is on the more electronegative N atom. However, this worksheet says that structure 1 is more common. On the other hand, this video says that structure 2 is more common. Which should it be?

I performed a quick calculation on the DF-BP86/def2-SVP level of theory and analysed it with Natural Resonance Theory (from the Natural Bond Orbital Theory).
This results in the following major contributors to this wave function:
$$\left[
\underset{(1)}{\overset{67.49\%}{\ce{^-S-C#N}}} \ce{<->}
\underset{(2)}{\overset{21.25\%}{\ce{S=C=N^-}}}
\right]$$

The third contributor is a weird structure with a 'long-distance-bond' between the sulfur and the nitrogen with $7.22\%$. All other contribution are neglected/discarded.

While your structure 3 is indeed a valid contributor, its actual contribution will be very small. Forcing the program to use it as a structure, it resulted in an error, as it was unable to match the orbitals to that structure. The reason for that is likely that the overlap between the sulfur and the carbon is too poor to actually be considered a good contributor. This would also explain the smaller contribution of the second structure.

Here are the localised (according to NBO) molecular orbitals:

(Colour code: blue/orange - occupied molecular orbital [Lewis];
red/yellow - virtual molecular orbital [unorccupied, non-Lewis])

The total contributions in terms of atomic orbitals to the above:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   8. (1.98209) LP ( 1) S  1            s( 79.86%)p 0.25( 20.13%)d 0.00(  0.01%)
   9. (1.77474) LP ( 2) S  1            s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
  10. (1.77474) LP ( 3) S  1            s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
  11. (1.96342) LP ( 1) N  3            s( 52.42%)p 0.91( 47.54%)d 0.00(  0.04%)
  12. (1.99743) BD ( 1) S  1- C  2
               ( 44.86%)   0.6698* S  1 s( 20.64%)p 3.81( 78.61%)d 0.04(  0.75%)
               ( 55.14%)   0.7425* C  2 s( 51.41%)p 0.94( 48.47%)d 0.00(  0.12%)
  13. (1.99846) BD ( 1) C  2- N  3
               ( 41.02%)   0.6404* C  2 s( 48.44%)p 1.06( 51.48%)d 0.00(  0.08%)
               ( 58.98%)   0.7680* N  3 s( 48.02%)p 1.08( 51.75%)d 0.00(  0.23%)
  14. (1.99735) BD ( 2) C  2- N  3
               ( 43.98%)   0.6632* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 56.02%)   0.7485* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  15. (1.99735) BD ( 3) C  2- N  3
               ( 43.98%)   0.6632* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 56.02%)   0.7485* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
 ---------------- non-Lewis ----------------------------------------------------
  16. (0.01904) BD*( 1) S  1- C  2
               ( 55.14%)   0.7425* S  1 s( 20.64%)p 3.81( 78.61%)d 0.04(  0.75%)
               ( 44.86%)  -0.6698* C  2 s( 51.41%)p 0.94( 48.47%)d 0.00(  0.12%)
  17. (0.01384) BD*( 1) C  2- N  3
               ( 58.98%)   0.7680* C  2 s( 48.44%)p 1.06( 51.48%)d 0.00(  0.08%)
               ( 41.02%)  -0.6404* N  3 s( 48.02%)p 1.08( 51.75%)d 0.00(  0.23%)
  18. (0.22011) BD*( 2) C  2- N  3
               ( 56.02%)   0.7485* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 43.98%)  -0.6632* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  19. (0.22011) BD*( 3) C  2- N  3
               ( 56.02%)   0.7485* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)

               ( 43.98%)  -0.6632* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)

On the terminology.
Alchimista already explained most of this, however, I cannot stress enough: There is no such thing as a most stable resonance structure. Therefore when you say common, you probably mean large contribution to the wave function, and when you say rare, you probably mean little contribution. None of the resonance structures can be independent from each other, as they are all hypothetical.

Please read more about that here: What is resonance, and are resonance structures real?


Calculating energy level from wavelength The transition of an electron from a higher level to a lower level results in the emission of a photon of wavelength $350.0\ \mathrm{nm}$. If the energy of the higher level is $-3.24\times10^{-19}\ \mathrm J$, calculate the energy of the lower level.

By using $E = h\nu$

I find $E = 5.679\times10^{-28}\ \mathrm J$

How do I use this value to solve the qus?

All you need are two equations and two constants (speed of light and Planck constant):
\begin{eqnarray}
\mathrm{E} & = & h\nu\\
h & = & \mathrm{6.62606957 \cdot 10^{−34}   J\cdot s}\\
\mathrm{c} & = & \lambda\nu\\
\mathrm{c} & = & 299792458\ \mathrm{m\cdot s}^{-1}\\
\end{eqnarray}

Using these data, the energy of a photon with $\lambda = 350\ \mathrm{nm}$ calculates to:

$$\mathrm{E} = \frac{6.62606957 \cdot 10^{-34} \cdot 2.99792458 \cdot 10^8}{350 \cdot 10^{-9}} \cdot \frac{\mathrm{J \cdot s \cdot m}}{\mathrm{s \cdot m}} = 0.05676 \cdot 10^{-17} \mathrm{J} = 5.676 \cdot 10^{-19} \mathrm{J}$$

This is the energy difference to the higher level.


Burning Of Coal and Black Body Radiation Coal burns with a red glow.

What is the ignition temperature of coal?

If it is T, and I take another substance which has melting point quite above T,
and heat it to T, Will it also emit red light?

Don't confuse the ignition temperature with the temperature of a burning coal. One is the temperature required to start the reaction, the other is the equilibrium temperature achieved when the substance is burning.

As for the colour of things at a given temperature, the colour will be broadly the same at the same temperature: this is black body radiation. For an interesting take on this in the context of movies, see this amusing video about why movies get it wrong on casting iron swords. A key point is that the glowing colour shown in most movies is far too cool (too orange red) for molten iron (which should be much whiter). Movies cheat and use aluminium which melts at less than 700 °C and glows orange. Iron needs ~1,600 °C and glows far whiter.


Explain why the noble gas Kr does follow the general trend for covalent radii Looking at covalent radii I know the trend states that as you move from LEFT to RIGHT the radii decreases due to increased nuclear charge. Up until you hit the noble gasses like neon. Neon does not want to form a covalent bond with itself so we have to look at the van der Waal forces to determine its radii (which is bigger than Fluorine, disproving the trend).

Up until you reach Krypton, the noble gasses tend to be slightly larger than the halogen group.

I am having a hard time understanding why the noble gas Kr does follow the general trend for covalent radii in the fourth period, even though the noble gas Ne does not.

Would it be because of the greater shielding effect?

Covalent Radii of 2nd Period and 4th Period Halogen and Noble Gase.

F
 - 0.064 nm

Ne
 - 0.070 nm

Br
 - 0.114 nm

Kr
 - 0.109 nm

Since the increased atomic radius causes a decrease in the electric interaction between electrons and nucleus, the heavier noble gas elements, such as Kr and Xe can form compounds with other elements. (KrF2, XeF4, ... &c.)

The covalent radius is quite meaningless in for elements such as Ar and Ne since its compounds seldom exist, but Kr and Xe compounds can occur with relatively high stability, so its 'covalent radius' can be meaningful.

That 0.109nm is probably a covalent radius.

Something like measuring a Kr-F bold length in KrF2 and then subtracting the covalent radius of a fluorine atom.

Trivially, the Van der Waals radius tends to be bigger than the covalent one, the noble gas atom radius seems as if it does not follow the trend, but from Krypton which the covalent radius is measurable, the data follows the general trend in periodic table.


Alcohol in the bloodstream modeled by differential equations Does anyone know any good references which use partial differential equations to model the alcohol concentration in the bloodstream? I'm a mathematician, so I'm searching for a rigorous mathematical model.

If you have any other suggestions of models in chemistry that uses partial differential equations, feel free to post them here.

One such article that I found was published by the University of South Florida. The abstract for which reads:

  Given a set of differential equations describing blood alcohol content as a function time, we integrated the
  equations to obtain a general solution. The general solution equation depends on three free parameters: the initial
  concentration of alcohol in the stomach after ingestion, the rate of alcohol absorption into the blood stream and the rate
  at which the alcohol is metabolized by the liver. We fitted our solution to experimental data to determine the unknown
  parameters for a particular subject.

If this sounds useful to you, it can be found here.


Keeping the electronic configuration in the ground state is expensive in Ab-initio MD? In E. Sandré & A. Pasturel's An Introduction to Ab-Initio Molecular Dynamics Schemes

  If using small time steps (typically in the order of a few tenth of femto-second), the changes in the electronic configuration is small enough so that convergence of the electronic configuration is easily achieved using a few conjugate gradient steps. However, the computational cost of dynamical simulation appears to be mostly due to keeping the electronic configuration in the ground state.

I don't understand this. DFT and HF are methods that can only give electron ground state. Since it is either DFT or HF that is used in Ab-initio MD simulations, then what does this excited state is referring to?

My understanding:

First of all, the number of electron orbitals calculated by either DFT or HF approach equals the number of electrons in the system, so we get no electronic excited state.

Under Born-Oppenheimer approximation, while varying the atomic configuration, the electrons are expected to stay in their ground state.
Since the validity of BO approximation dictates no abrupt change in electronic energy.

So, the reason of 'Keeping the electronic configuration in the ground state is expensive' is that when the atomic configuration changes in MD simulation, it is hard to make the electronic energy difference small.

Is my understanding correct?

Let us first note that the calculation of the first derivative of the total DFT or HF ground state energy of a molecule with respect to its nuclear coordinates (called the gradient) depends on a converged self-consistent field (SCF) solution. The solution is obtained in an iterative fashion and is represented as a collection of molecular orbital (MO) coefficients. In some sense, they are unique for the nuclear coordinates. However, if we move the nuclei only very slightly, repeat the calculation, and compare the new set of MO coefficients with the previous ones, they will be almost identical.

Almost, that's the key here. The gradient will become more of an approximation as you move away from the nuclear coordinates for which the underlying SCF solution was originally calculated (and merely recalculate the gradient at new coordinates using old MO coefficients). One way of dealing with this problem is to take the old SCF solution as a starting point for a new SCF solution, which should converge in just a few steps. But those few SCF iterations can easily dominate the cost, given that a gradient calculation after obtaining the SCF solution typically has a computational cost on the order of 1-3 SCF iterations.$^{1}$

To summarize: I understand the sentence "However, the computational cost of dynamical simulation appears to be mostly due to keeping the electronic configuration in the ground state." in the quoted section of the question to mean: recalculation of the SCF solution is necessary at least every few steps of the molecular dynamics (MD) simulation and that recalculation is the most expensive part.

$^{1}$ A specialized quantum chemistry program, that would keep certain integrals and other intermediates across the different SCF procedures, would help somewhat with the computational cost at the expense of more on-disk storage and more uncertainty, since all those quantities implicitly depend on the nuclear coordinates.


Difference between dissociation energy and dissociation temperature for water? I was trying to find the temperature at which water dissociates into hydrogen and oxygen, and so I came across this Wikipedia page. It states that it takes $\pu{493.4 kJ/mol} + \pu{424.4 kJ/mol}$ to dissociate the bonds in a water molecule which gives a total of about $\pu{920 kJ/mol}$ or about $\pu{51 MJ/kg}$. So I tried to calculate the temperature that would give that huge amount of energy using the heat capacity of water vapor. I took $\pu{3.3 kJ/kgK}$ for water vapor as an average specific heat, and so by simply dividing $\pu{51 MJ}$ by $\pu{3.3 kJ}$, I ended up with about $\pu{15,000 K}$.
But on this other Wikipedia page, I found: "at $\pu{2200 ^\circ C}$ about three percent of all $\ce{H2O}$ molecules are dissociated into various combinations of hydrogen and oxygen atoms. At the very high temperature of $\pu{3000 ^\circ C}$ more than half of the water molecules are decomposed".
So now I am very confused. If $\pu{3000 ^\circ C}$ is enough to dissociate more than half of the water molecules, then it should be expected that almost all the molecules should dissociate at around $\pu{4000 ^\circ C}$ or so and not the $\pu{15,000 K}$ I calculated earlier.
So am I missing something here? Or is there an explanation to this?

One key is an assumption about various degrees of dissociation.

It states that it takes 493.4 kJ/mol + 424.4 kJ/mol to dissociate the bonds in a water molecule which gives a total of about 920 kJ/mol or about 51 MJ/kg.

That sounds like the energy required for $\ce{H2O <=> 2H + O}$.

But on this other Wikipedia page, I found: "at 2200 °C about three percent of all H2O molecules are dissociated into various combinations of hydrogen and oxygen atoms.

Here it sounds like they are considering less extreme dissociations than complete dispersal to individual atoms. For example, $\ce{H2O <=> H + OH}$ and perhaps even $\ce{H2O <=> H2 + O}$ or $\ce{H2O <=> 2H + \frac{1}{2}O2}$ and related equilibria may be happening.
In effect, if just one OH bond breaks in water, that only requires 493.4 kJ/mol and leads to the formation of species such as $\ce{OH}$ and $\ce{H}$ atoms.  Those species have a different heat capacity than $\ce{H2O}$ vapor.  The heat capacity page you linked to says:

The values above apply to undissociated states. At high temperatures above 1500 K dissociation becomes appreciable and pressure is a significant variable.

In the calculation you are doing you are not including a specific pressure variable either.  Heat capacity could also be a function of pressure.  Presumably the data you linked to is all for 1 atm of pressure, although it didn't really say.  Likewise, the claim that at 2200 °C, 3% of water is dissociated must also be pressure-dependent.  For which pressure is it true?
The final complexity is the entropic term.    The three dissociated atoms have 9 translational degrees of freedom but intact water only has three.
These pressure and entropy complexities matter because $\Delta G$ is what determines equilibrium, and $\Delta G = \Delta U + \Delta (PV) - T \Delta S$.  The bond dissociation energies you quote represent the $\Delta U$ term, so the calculation you are doing is implicitly assuming that $\Delta U \gg \Delta (PV) - T \Delta S$.
I think that your calculation is admirable in that it comes so close to the right answer while still neglecting all these complexities!


Simple experiment to do at home that involves acids and bases? so long story short I am in the IB Diploma and I am required to come up with an experiment to investigate pretty much anything to do with chemistry. I'm thinking about doing mine on investigating how temperature affects the pH value of a solution, but my teacher said it was too simple.

Basically, is there any reaction I can make at home with simple chemicals(vinegar, baking soda, salt, juice, etc.) that would produce a meaningful experiment and involves me measuring the pH level? Thank you guys so much.

Measure the pH+ of various foods, and make a spreadsheet listing them.

You might also measure pH+ change with time in a food, e.g. apple juice or bread dough, with a few mg of yeast sprinkled in.


How do I work out the amount of H+ ions in a sodium hydroxide and water solution? I know the $pH$, pressure, temperature and concentration of the solution. Is there a method I can use to work out the amount of $\ce{H+}$ ions?

The pH can be directly related to the concentration of $\ce{H3O+}$:

$$\text{pH}=-\log{[\ce{H3O+}]}$$
$$\Downarrow$$
$$[\ce{H3O+}]=10^{-\text{pH}}$$


Is there a way to use free software to convert SMILES strings to structures? So far I've been unsuccessful in finding an open source tool that will convert a large number (>100k) of SMILES strings to a chemical structure.
Is this really only solved by commercial applications?
I would not mind converting through a number of intermediate formats.
Ideally I'd do it with R, but that is the least of my worries.
It's also neither for personal nor academic use.

According to the website, Open Babel should do the trick: Documentation - SMILES, Sourceforge.

For example, the following code will give you a neat SVG file of the molecule benzene:

obabel -:"c1ccccc1" -O benzen.svg

If you experience problems using it, you are welcome to ask more specifically.

Alternatively, you can use a web-query from the national cancer institute. It is easily accessible by the following code

http://cactus.nci.nih.gov/chemical/structure/"structure identifier"/"representation"

For example: benzene, "structure identifier"=c1ccccc1, "representation"=image.

Another open source solution, where you can directly export the structure into a molecular editor is Avogadro. (It uses Open Babel though.)

Depending on the actual problem, however, there might already be more advanced routines.


Why does Ruthenium only have one electron in its 5s orbital in the neutral oxidation state? So recently I have undertaken some research with some organometallic chemistry involved, specifically regarding that of ruthenium-aryl compounds. Trying to understand the chemistry of the central metal ion, I noticed something weird a couple of weeks ago, from Wikipedia: Ruthenium only has one electron in that orbital ($4d^75s^1$), rather than the two one would expect a transition metal to have, going with a naive Aufbau across the periodic table.

I am aware, I think, in the 4th period period transition metals, of a tendency to promote their 4s electrons to the 3d orbitals, when possible, so that they have either "half-filled" (5 electrons) or completely filled (10 electron) shells.  These are "magic numbers" for transition metals, apparently; just another one of those heuristic things they teach you in inorganic chemistry that I wish I understood more about. But alas, ruthenium has an asymmetric 7 electrons in its 4d orbital, blasted thing.

Does anyone know anything? I tried using Google Scholar and a few databases, but when I type in "orbital" pretty much all I see is molecular orbital calculations specific to particular molecules. I'd really love to see papers pointing at some faint ideas I may be dimly aware about...

From Oxtoby et al, Principles of Modern Chemistry, 7th edi p219 and Levine, Quantum Chemistry, 5th edi, p314, the orbital energies of Sc and the elements beyond, 3d<4s. The electrostatic repulsion makes energy of 3d$^3$ higher than 3d$^1$4s$^2$.

Just my pure guess, Ru is bigger than Fe, electrostatic repulsion is smaller, so 5s$^1$. However, the 6s$^2$ electrons in Os are extra stable due to relativistic effect, so again 6s$^2$.

Quantitative calculation will be highly appreciate for justifying this hand-waving argument


pH throughout a titration Estimate the $\ce{pH}$ of the solution that results when $\ce{75.0 mL}$ of $\ce{0.275 M HCl}$ is added to $\ce{50.0 mL}$ of $\ce{0.275 M Na2HPO4 (aq)}$.

My book gives the successive $\ce{Ka}$ values for phosphoric acid as: $\ce{Ka1 = 7.6*10^{-3}, Ka2 = 6.2*10^{-8}, and Ka3 = 2.1*10^{-13}}$.

I recognized that after the stoichiometric calculation, $\ce{6.875 mmol of H3PO4}$ and $\ce{6.875 mmol of H2PO4-}$ exist in the solution. I then used the Henderson-Hasselbalch equation to determine that the resulting $\ce{pH = -log(Ka1) = 2.12}$. I think ignoring the third deprotonation is justified because the successive $\ce{Ka}$ values are greater than 3 orders of magnitude from each other.

However, then answer in the textbook is $\ce{pH = 2.80}$. What am I doing incorrectly?

The molarity of the $\ce{HCl}$ and the $\ce{Na2HPO4}$ is the same. So 50 ml of $\ce{HCl}$ converts the $\ce{Na2HPO4}$ to $\ce{NaH2PO4}$. The remaining 25 ml of $\ce{HCl}$ essentially converts half the $\ce{NaH2PO4}$ to $\ce{H3PO4}$. So the solution is as if 6.875 millimoles of $\ce{NaH2PO4}$ were mixed with 6.875 millimoles of $\ce{H3PO4}$ in a total of 100 ml of solution.

The Henderson-Hasselbalch equation is:

$$\text{pH} = \text{pKa} + \log{\dfrac{\ce{[H2PO4^-]}}{\ce{[H3PO4]}}} = 2.1192 + \log{\dfrac{0.06875}{0.06875}} = 2.1192 \ce{ ->[Rounding] } 2.12$$

Let's check....

For the phosphate species let $C_\mathrm{P}$ be the concentration of all the phosphate species. The following equations give the fraction of each species as a function of $\ce{[H^+]}$. The equations were most easily calculated using a spreadsheet.

$$[C_\mathrm{P}] = \ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$$

$$\dfrac{\ce{[H3PO4]}}{[C_\mathrm{P}]} = \dfrac{\ce{[H^+]^3}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 0.49999$$

$$\dfrac{\ce{[H2PO4^-]}}{[C_\mathrm{P}]} = \dfrac{\ce{K_\mathrm{a1}[H^+]^2}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 0.50001 $$

$$\dfrac{\ce{[HPO4^{2-}]}}{[C_\mathrm{P}]} = \dfrac{\ce{K_\mathrm{a1}K_\mathrm{a2}[H^+]}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 4.0791\times10^{-06}$$

$$\dfrac{\ce{[PO4^{3-}]}}{[C_\mathrm{P}]} = \dfrac{K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 1.1272\times10^{-16} $$

Thus the assumption that the H_H equation can be used is valid.

The textbook answer is wrong.


In metal complex electron counting, is the neutral or ionic formalism more correct? I have been practicing finding total valence electron counts by both methods, and a few points have arisen that have confused me.
1) Which method gives the true d-electron count?
From what I understand, the ionic method assumes that both electrons from a bond go to the more electronegative atom, and the bond is 100% ionic. This lines up with formal oxidation state which uses the same premise. As a result, d-electrons are usually calculated as completely lost from the metal, giving a positive formal oxidation state.
In the neutral method, the d-electron count doesn't change as all calculations assume that M-L bonds are cleaved such that both species are neutral, and metal oxidation state stays at 0.
This results in 2 different values for a d-electron count, and I was wondering if the value actually means anything. I guess it doesn't matter in the end as the same number of electrons are put into the overall MO scheme, but I was wondering if being asked the d-electron count has any point to it. This can be seen in the below example from here:

2) If the neutral formalism is used, where does the electron deficiency in positively charged complexes come from?
In the ionic method, we assume everything is charged and just take electrons out of the metal to account for the overall charge (even this seems strange, why do we take electrons from the metal when we could take them from anywhere else to give the same MO scheme?). But in the neutral formalism, if everything is assumed neutral, where are we supposed take the electrons from?
3) If the electrons don't have to be taken from the metal, what is the point of assigning metal oxidation state?
I always hear statements like "d8 metals such as Pt(II) commonly form square planar complexes", but both the assignment of d8 and Pt(II) differ when using a different formalism. So is the use of the ionic method and oxidation of the metal specifically simply a convention that allows us to compare complexes quickly by their total electron count?
It seems to me like the correct answer is somewhere in between the 2 methods (between 100% ionic and 100% covalent), and the use of one or the other is arbitrary, but any corrections would be appreciated.
Also, if oxidation state is arbitrary, is the argument that "metals with lower ionisation energies reach higher oxidation states so can form more bonds" the same as saying "metals that have lower ionisation energies generally have larger radii so can fit more covalent bonds around them"?

I think, the answer to number 1 will remove the need to answer the subsequent questions:

1) Which method gives the true d-electron count?

Neither!
You cannot, and I want to stress that, derive the oxidation state of a metal in a complex solely by looking at its ligands. You can make educated guesses and in some cases only one oxidation state is plausible but there are many cases—most notably those denoted $\ce{\{FeNO\}^n}$ where such an assingment is meaningless to impossible. Without consulting spectroscopic data, it is absolutely impossible to tell whether a $\ce{\{FeNO\}^7}$ compound is $\ce{[\overset{+II}{Fe}(NO^.)]}$ or a $\ce{[\overset{+III}{Fe}(NO^-)]}$. In fact, entire papers have been published arguing one way or the other.
This essentially renders your follow-up questions entirely moot, but I do wish to take the time to point out that the metal’s oxidation state along with the corresponding oxidation state of the ligands is theoretically important information that can predict the electrochemical behaviour of a complex. The only problem is that it cannot always be neatly predicted as shown above. It is worth noting, though, that certain complexes have been assigned metal oxidation states beyond reasonable doubts (e.g. $\ce{[Cu(NH3)4(H2O)2]}$), and in such cases the predictions of reactivity are absolutely adequate.


Is it possible to make a fuel cell that runs on triglyceride? I apologize if this is a silly question. The reason I ask is I'm wondering if it would be possible to build a fuel cell that runs on body fat. I know ones for butane exist, could they work for this? If not, is there any good reason it isn't possible?

Basically I'm imagining some kind of flexible inductive charger in your hand, or wrist for a smart watch, and connected to a fuel cell that's connected to the circulatory system and runs off body fat.

It just seems silly to lug external batteries along on long on bike rides when the whole point of the ride is to get rid of excess chemical energy anyway.

I obviously don't have the skills to pursue this, but I just want to know if the dream is silly. :)

Thanks!

It is certainly possible to oxidize fats electrochemically, but there are a few problems with the idea that I can think of:

Triglycerides do not flow freely through the blood as they are too hydrophobic and are instead carried by lipoproteins that would have to be removed. Similarly, free fatty acids are transported bound to serum albumin.
The blood concentration of triglycerides and fatty acids is relatively low most of the time in healthy individuals.
Existing fuel cells that work on fatty acids seem to be mostly microbial fuel cells, which are quite slow.

I think it would be more practical to use glucose from blood instead, using an enzymatic fuel cell, as blood glucose levels are relatively high and the oxidation is easier to accomplish (though there are still a lot of technical challenges to be resolved—biofouling, efficiency, device lifetime, etc.).


Reaction of nitrobenzene with potassium ferricyanide I came across this question in an exercise book. Can anyone give detailed answer (mechanism) of this reaction?

I am unable to proceed. I thought there maybe some redox reaction involved where $$\ce{Fe^3+ -> Fe^2+}$$ maybe taking place but then nitrobenzene would have to be oxidised. Can someone please help ?

I'd go more for an oxidation, but on the ring (hydroxylation).
My guess comes from this paper here : "The Hydroxylation of Aromatic Nitro Compounds by Alkalies" by O. C. Dermer.

This is very harsh (and somehow old) chemistry.

First, for m-dinitrobenzene and sym-trinitrobenzene, hydroxylation is reported in your conditions, that is to say alcaline conditions with the help of potassium ferricyanide as oxidizing agent (reactions A and B, see reference in the paper).

Second, the authors report the reaction C (with less than 4% yield though, according to my calculations), with a small amount of p-hydroxyazobenzene as a side-product. They did this reaction by mixing only nitrobenzene with potassium hydroxide, so I would expect the same product (o-nitrophenol) with maybe a better yield in your conditions.


Are electronic descalers a scam? It seems there are some electronic kits sold around nowadays by major retailers (e.g., Home Depot and Amazon) claimed to remove built-up scale from water pipes.

But, do they actually work? If so, then how?

Perhaps, inducing high-frequency eddy currents inside copper piping somehow lessens its ability to support limescale? Or, perhaps it gets broken down by some specific radiation (emitted by the kit) in the limescale's absorption spectrum?

These devices have been reported to work in some situations, but not in others.

The underlying theory involves altering the growth of calcium carbonate scale with an ordinary magnetic field. The magnetic field changes which form of calcium carbonate crystal (calcite, aragonite, and vaterite) is favored. Certain forms of crystal make tough scale while others can pass through a water system without scaling.

The growth and precipitation of carbonate crystals are affected by many other things such as calcium concentration, carbon dioxide concentration, temperature, pH, and the presence of other metallic ions. Exposure time to the magnetic field, magnetic field strength, and magnet orientation also have an effect.

The different forms of carbonate crystals can be tested to have specific properties under controlled laboratory conditions.  Under uncontrolled conditions, the scientific literature is full of mixed results and mixed explanations.  Because so many things affect the formation of scale, it is impossible to say if a magnet or device generating a magnetic field will work to thwart scaling in any specific situation.  It may work wonderfully in some situations but not at all in others.

The best way to tell if such a device is worthwhile would be to see if several people using your same water supply report it as having worked lately.  If water management practices change (e. g. factors affecting the formation of scale are controlled) then such a device may cease to work.  If no one using your same supply can attest to its operation then this would be a risky purchase.


Why doesn't iron(III) iodide exist? $$\ce{Fe2O3 + 6HI -> 2FeI2 + I2 + 3H2O}$$

Why don't we get $\ce{FeI3}$? After all, iron's oxidation state is $+3$ in the reagent.

Should one just memorize that up to bromine, it's $\ce{FeX3}$, and below it's $\ce{FeX2}$?

The standard reduction potentials for the following half reactions can be found here.

$$
\begin{align}
\ce{Fe^3+(aq) + e- &-> Fe^2+(aq)} &\quad E^\circ &= \pu{+0.77 V} \\
\ce{I2(s) + 2 e- &-> 2 I-(aq)} &\quad E^\circ &= \pu{+0.54V} \\
\ce{Br2(l) + 2 e- &-> 2 Br-(aq)} &\quad E^\circ &= \pu{+1.07V} \\
\ce{Cl2(g) + 2 e- &-> 2 Cl-(aq)} &\quad E^\circ &= \pu{+1.36V}
\end{align}
$$

You can see from this that only iodide is a strong enough reducing agent to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$ at standard conditions. Even with non-standard concentrations it will be very difficult to get bromide to do the reduction because the difference in electrode potential is large.

The trend in electrode potentials for the halogens can be explained in terms of the increasing electronegativity going from iodine to chlorine which increases the first electron affinity. It just so happens that the crossover point with the iron reduction is between iodine and bromine.


Why is only one lone pair in imidazole delocalised? I was wondering why only one of the nitrogen lone pairs in imidazole is involved in the delocalized π-system. Specifically, I was wondering why it is that lone pair B in particular is delocalized, instead of lone pair A:

Here is my conjecture about why B's lone pair is delocalized while A's lone pair cannot be.

If we delocalize A, we'd get extreme ring strain; we'd make the geometry around that nitrogen $\ce{sp}$. This implies a $120^\circ$ bond angle. That's problematic.
Effective $\unicode[Times]{x3C0}$-orbital overlap (conjugation) only occurs when the p-lobes are planar — i.e. parallel with each other. If the nitrogen with the B lone pair kept its B lone pair, then that nitrogen would be tetrahedral and that would distort the planarity of the molecule; everything else in the molecule appears to be $\ce{sp^2}$ hybridized.
We also run into formal charge issues; we get a positive formal charge on the "A" nitrogen and a negative formal charge on the B nitrogen. While this isn't a dealbreaker, it's not preferable, as there is separation of like formal charges.
Anything I miss? Any points I mangled?

The correct answer to this question is more direct, and not listed in your items. The non-bonding electron pair in nitrogen 3 is in an orbital perpendicular to the π-bonding p orbitals of all other atoms in the imidazole ring. Thus, it does not have the appropriate geometry to overlap with other orbitals forming π-bonds, and does not participate in any resonance effects.

This effect of orbital geometry is often invoked to explain the difference in basicity between pyridine and pyrrole. In the former, the nitrogen lone pair is of type 3 and does not participate in resonance, and so it has a basicity comparable to regular amines and imines. However, in the latter, the lone pair is of type 1 and does participate in the aromaticity of the ring. Protonation of the nitrogen atom would require localization of the electron pair on the nitrogen, breaking most of the resonance effect. This is energetically unfavourable, so pyrrole displays a much lower basicity than expected for amines or imines.

This section on pyridine has a little more information on why the type A lone pair does not delocalize, including a helpful picture. Notice that the lone pair is in an $\mathrm{sp^2}$ orbital parallel to the plane of the ring, while the π-bonded system is perpendicular.

I can make the following observations on your conjectures:

1) I don't think this applies, actually. The fact that orbital overlap is disallowed in the first place precludes the possibility of even thinking what would happen to the geometry of the molecule if the electron pair in nitrogen 3 formed a π-bond; it just doesn't.

However, if the molecule had a boron atom instead of carbon in position 2 (with an empty $\mathrm{sp^2}$ orbital pointing in the same direction as the full $\mathrm{sp^2}$ orbital in nitrogen 3), then I'm not sure what would happen. A more simple example of this type of structure would be 1,2-azaborine (of which only the 1,2-dihydro derivative seems to have been isolated as of yet).

I could imagine canonical structures in which the electron pair on nitrogen links with the boron atom, forming an additional bond (analogous to benzyne, which is isoelectronic with 1,2-azaborine). However, to form an extra N–B π-bond, then either the bond will be a $\pi(\mathrm{sp^2}\text{–}\mathrm{sp^2})$ bond (which is unusual as far as I know), or both the nitrogen and boron atoms would have to partially rehybridize into a mix of $\mathrm{sp^2}$ and $\mathrm{sp}$ configurations to allow a $\pi(\text{p–p})$ bond, which would likely imply an increase in bond angle and a huge increase in ring strain. Neither option seems particularly good, so I suspect such canonical structures would contribute little to the actual bonding picture in the boron-substituted ring.

2) The nature of the nitrogen atom which participates in resonance is actually something between $\mathrm{sp^2}$ and $\mathrm{sp^3}$. The concept of hybridization isn't very adept at explaining resonance naturally.

I'm also not happy with how much my arguments rely on hybridization, but I don't know other effective means to explain what's going on.


Catalyst and Inhibitors home experiments My teacher assigned me a science experience involving catalysts and inhibitors at home.
I think I understand the principle of it, but I fail to find possible experiments that I can do at home with a thermostat and some home substances.
The experience:
3 reactions...
1 with the catalyst,
1 with the inhibitor,
1 without anything.
I will compare them by measuring the reaction speed with a factor (preferably the heat variation).
EDIT: I can also measure the volume of the product.
Thank you.

You might look up the decomposition of hydrogen peroxide, $\ce{H2O2}$. The 3% solution sold as antiseptic should be safe to use at home, as would be some readily available catalysts and inhibitors.
Horseradish might be used for a catalyst... but don't eat it in the lab.
To measure speed of reaction, measure gas evolution with an inverted, liquid-filled graduated cylinder or small measuring cup over time.
You can have fun burning things (carefully) in the evolved $\ce{O2}$, too.


Iron(II) oxide on stainless steel? I noticed after some use that a stainless steel spoon that I have started forming $\ce{FeO}$. Stainless steel is not supposed to rust like this. I don't know of any black iron hydroxides, chromium oxides, or chromium hydroxides so that lead me to deduce that it is Iron(II) Oxide.

But why would stainless steel oxidize? I mean it does not make any sense since I was told that stainless steel does not oxidize.

And how can I get rid of the oxide? Putting the spoon in phosphoric acid perhaps since phosphoric acid is a really good rust remover?

It was probably a poor grade of stainless steel. The chromium oxidizes before the iron therefore you obtain a thin layer of protection. In poor grades of stainless steel, I have seen rust form over a period of time or during exposure to non ideal conditions such as high humidity. This is probably because the manufacturer added the minimum chromium required to meet standards, and so you get a smaller ensemble of chromium oxide and more iron oxide.

You can treat the metal with phosphoric acid to remove the rust. You can also treat the stainless steel with concentrated sulfuric acid to produce a phosphate layer for additional protection with minimal loss of material (they do this in industry)


Why doesn't the boiling point of an aqueous solution depend on intermolecular attraction? I understand that boiling point is a colligative property for aqueous solutions, and that it empirically depends on the mole fraction of the solute, rather than the identity of the solute. I understand the van't Hoff factor / ionic strength and why it matters. I also get that an increase in entropy is the primary driving force for boiling point elevation.

My question is, why does the influence of intermolecular attraction go out the window for solutions, as opposed to pure substances? Shouldn't the identity of the specific dissolved species - that is, their charge and their size, affect the intermolecular attractions in the solution? Shouldn't greater intermolecular attraction increase the boiling point by reducing vapor pressure, just as it does for pure substances? For example, neutral molecular solutes as opposed to strong electrolytes, or large ions like iodide vs smaller ones like fluoride.

Is the effect simply too small to matter compared to the entropy factor, or am I missing something theoretically?

GOOD QUESTION! That is the nastiest p-chem question I've seen.  Atkins, Physical Chemistry 2nd edition (the readable one), pp 228-232pp.  Boiling point elevation is presented as purely an entropy effect, given two assumptions:

1) The solute is non-volatile.
2) The non-volatile solute does not dissolve in the solid solvent.
Both assumptions are explored in Chapter 10.

A solute increases the entropy of the solution but does not change the entropy of the vapor at a given P and T.  A higher T is then necessary to give the same entropy difference.  I am not satisfied that this wholly addresses a strongly interactive solute versus a non-interactive solute versus a strongly negatively interactive solute,

http://www1.lsbu.ac.uk/water/kosmos.html
Kosmotropes and chaotropes

http://www.amazon.com/Physical-Chemistry-Peter-Atkins/dp/1429218126
Atkins 9th edition is current.
"There is almost no chemical concept that Atkins can't make unnecessarily confusing"
http://www.amazon.com/Physical-Chemistry-A-Molecular-Approach/dp/0935702997/ref=zg_bs_16052591_2
Library - it's pricey.
http://www.amazon.com/Thermodynamics-Engineering-Approach-Yunus-Cengel/dp/0073398179/ref=dp_ob_title_bk
Library - it's very pricey.


Which one of the compounds listed would we use in order to get the highest concentration of sulphate ion in a saturated solution? Solutions are prepared by dissolving a sulphate salt in water. Which one of the compounds listed would we use in order to get the highest concentration of sulphate ion, SO42-, in a saturated solution? What would be the concentration of sulphate ion in this case?
A. $\ce{CaSO4}$    $K_{sp}$ = 7.1E-5
B. $\ce{SrSO4}$    $K_{sp}$ = 3.4E-7
C. $\ce{BaSO4}$    $K_{sp}$ = 1.1E-10
D. $\ce{RaSO4}$    $K_{sp}$ = 4.0E-11

My first instinct was to simply look at the $K_{sp}$ values.. Since $\ce{CaSO4}$ has the largest $K_{sp}$ value, it should be the most soluble, right?
But then I looked at a data page referring to the solubility of common compounds in water (Please see image. According to the table, $\ce{SO4^2-}$ has low solubility with $Ca$, $Sr$, and $Ba$.
My question is- does the solubility depend on the $K_{sp}$ value (was my first instinct right), or is the table correct?
Any ideas would  be appreciated.

The $K_{sp}$ values are the way to go .. the other information you mentioned are just some convenient rules of thumb for predicting relative solubilities of compounds when you don't know the $K_{sp}$ values.  Also, the 4th choice in your list .. radium .. probably also belongs on that list, since it's the next alkali earth element after barium, but it's included because it's a radioactive element, and people don't commonly use or encounter it.


net charge nucleobases at alkaline conditions While doing anion exchange chromatography of a short oligonucleotide, I have noticed that at $\rm pH=12$, the oligo retains longer on the column than the same oligo- at $\rm pH=8$. I thought the reason for this could be the change in the net charge of nucleobases.

Could anyone suggest what is the net charge of RNA nucleobases (adenine, guanine, cytosine, uracil) at $\rm pH=12$ (denaturing conditions)?

The conjugate acids of the adenine and guanine have pKa's between 9-10. Guanine also has a nitrogen with a pKa of 12.3, which would be mostly protonated at pH 12. The pKa for the phosphoric acid proton of a nucleotide is significantly lower than 7.

Therefore, in a pH 8 buffer, the basic nitrogens of each adenine and guanine are protonated, while the phosphate is deprotonated. This gives an overall net charge closer to zero than you would see at pH 12.

In a pH 12 buffer, where the pH is above the pKa of both the phosphoric acid protons and the nitrogen conjugate acid protons (except the last one on each guanine), all of these positions would be deprotonated. Therefore, the formal charge on the nitrogens would be zero (again, except the most basic nitrogen on guanine), but the phosphates would still have a charge of -1 each. This would lead to a net charge of significantly higher charge magnitude because there would be very few positive charges to balance out the negative charges.

At pH 12, the net charge should be approximately: $$(n_N +n_u) - (n_c + n_g).$$
Approximate net charge at pH 8: $$n_N - (n_a + n_c + 2n_g).$$

nN= number of nucleotides

nu= number of uracils/thymines

nc= number of cytosines

ng= number of guanosines

na= number of adenosines


deltaS of a decomposition reaction Suppose we had the following reaction:

$$\ce{Polymer ($n$~units)_{(aq)} -> Polymer($a$~units)_{(aq)} + Polymer($n-a$~ units)_{(aq)}}$$

Without knowing the specific monomer unit of the polymer, and assuming the polymer is straight chain, is there anyway to determine the ΔS of the reaction? I mean, I know we can determine the sign (which will almost certainly be positive), but is there a way to make even a rough approximation of magnitude? We can assume the conformational entropy is negligible, i.e. the polymers are fairly rigid.

The largest contribution to the entropy change will be the translational component, which is easy to evaluate using the Sackur-Tetrode equation
$$S^\circ_{\rm{trans}}=R\ln(bM^{3/2}T^{3/2})$$
where $b$ is a collection of constants with a value of 3.75 (mol/gK)$^{3/2}$ $M$ is the mass in g/mol, $T$ is temperature in Kelvin, and the standard state is 1 M.

Due to the slowly varying logarithmic dependence on $M$ and $T$ this contribution is usually on the order of 100's of J/molK.

While the translational contribution is the largest, rotational and vibrational contributions can be sizable so this will give you an order of magnitude estimation at best.

Derivation starting from the wikipedia equation
$$S^\circ=kN \left( \ln \left( \frac{V^\circ}{N} \left( \frac{4\pi mU}{3h^2N}\right)^{3/2} \right)+\frac{5}{2}\right) $$
$V^\circ$ = 1 L = 0.001 $m^3$, $N=N_A$, $U=\tfrac{3}{2}N_AkT$, $M=mN_A$
$$S^\circ=R  \ln \left( \left[e^{5/2}\frac{V^\circ}{N_A} \left( \frac{4\pi \tfrac{3}{2}k}{3h^2N_A}\right)^{3/2}\right]M^{3/2}T^{3/2} \right) $$
The term in square brackets corresponds to $b$ and using SI units $b=1.20\times10^5$ mol(kg/K)$^{3/2}$ which corresponds to 3.75 mol(g/K)$^{3/2}$


What does the commercial chemical composition notation "Percent dry, wet with water"mean? I am not a chemist, I develop software. I'm hoping for a general explanation I can understand in order to assist me with a project I'm working on.

Customers viewing a product online such as peroxide see 97% (dry wt.), wet with 25% water.

What does this mean? What would you expect the product to be in the form of? A powder, liquid/gel, solid in water? Any specific type of container?

I believe there is no general rule.
In your case (assuming you mean this) it is a solid. materials can have 25% and even more water and still be solids.


Potential Energy Curve from Hartree-Fock Calculations I am currently working on some quantum chemical calculations, in particular looking at dissociation energy curves generated from Hartree-Fock, CI, CCSD and MP2 methods for diatomic molecules. Now, I was always under the impression that in order to get a potential energy surface  (in the Born-Oppenheimer approximation)  one needs the energies for the electronic problem and the nuclear-nuclear interaction:
$$E(R) + V_{\mathrm{NN}}(R) = U(R)$$
Where $E(R)$ is the energy obtained from the solution of the electronic Schrödinger equation, $V_{\mathrm{NN}}(R)$ the internuclear coulomb potential and $U(R)$ the potential energy surface. I have now seen multiple publications and books that draw potential energy curves using only $E(R)$.
So my question: Are these publications/books wrong, or do I have a misunderstanding of how the PES is computed under the BOA?
Example publication: Dutta, A.; Sherrill, C. D. Full configuration interaction potential energy curves for breaking bonds to hydrogen: An assessment of single-reference correlation methods. J. Chem. Phys. 2003, 118 (4), 1610–1619. DOI: 10.1063/1.1531658.

The energy calculated by a quantum chemistry program is essential always the total potential energy $U(R)$.  For example Fig 1 and Fig 4 of Dutta and Sherrill referenced in the question both plot the total sum of electronic and nuclear energies $U(R)=E(R)+V_{NN}(R)$.
For an example, have a look at a PES I have generated for $\ce{H2}$, using Restricted Hartree-Fock and the STO-3G basis set using PySCF.  The characteristic PES, with a single minimum is actually the total sum of energies $U(R)$, labelled the "Potential Energy Surface"

N.B. The zero bond-length electronic energy is calculated using a helium nucleus as the program doesn't allow superposition of atoms.


Why does dry ice "dissolve" so quickly in water? I just stumbled across an ancient video of some people putting a 60 pound block of dry ice in a swimming pool. Pretty cool stuff.

It's pretty interesting to me that in the video, the dry ice sublimates much quicker in the water than in the air.

I am wondering why this might be? Is it possible that water is acting as some sort of catalyst? If so, how can I picture this on a microscopic level? (That might not be possible to answer but the catalyst portion is.)

I was thinking that the $\ce{CO2}$ might be motivated to form carbonic acid, and by Le Chatelier's principle, the dry ice would sublimate even faster.

I'm not so sure that makes sense though...

I don't know. Any thoughts?

It's just about heat capacity + heat conductance, nothing more. $\ce{CO2}$ sublimates much quicker in the water because water is much better at heat exchange than air. Anybody who ever experienced swimming in water at 15°C and walking in air at 15°C can attest to that.

Catalysts for heat exchange are usually made of copper and other metals. (Also, usually they are not called catalysts.) You have one in your computer, attached to the CPU.

As to formation of carbonic acid, it hardly has any influence on the process (though it does happen to a certain extent, of course). Most $\ce{CO2}$ escapes in gaseous form anyway.


Work done by gas while expanding against constant external pressure? This is the graph for the same between pressure (y) and volume (x):-

Here, why do we assume the gas pressure to be same as the external pressure everytime in during the process of expansion?

In a reversible process, the pressure of the gas is uniform throughout, and the force per unit area that the gas is exerting on the piston face is equal to its pressure, as determined from its equation of state.  So the work is equal to the force integrated over the displacement, or, equivalently, the uniform gas pressure integrated over change in volume.

In an irreversible process, the pressure of the gas is not uniform throughout, but, at the piston face, the force per unit area that the gas exerts on the piston face must match the force per unit area that the piston face exerts on the gas, by Newton's third law.  But, in this case, because the gas pressure is not uniform throughout (and also because viscous stresses contribute to the force per unit area of the piston), the force per unit area at the piston face cannot be determined from the equation of state.  However, if the force per unit area at the piston face is imposed by specifying external conditions on the system, the work that the gas does can still be calculated as the integral of the external pressure integrated over the volume change.

So, to summarize, in both reversible and irreversible processes, the force per unit area exerted by the gas on the piston face is equal to the externally applied pressure and the work is equal to the external pressure integrated over the volume change.  However, in the case of a reversible process, the uniform gas pressure can be determined from the equation of state of the gas, but, for an irreversible process, it cannot (and needs to be determined by controlling the external pressure in some way).


Calculating phosphorescence quantum yield using the given datas' A molecular excited state has a fluorescence lifetime of 10 nanoseconds and a phosphorescence life time of 10 microseconds. There are no other processes, what's the phosphorescence quantum yield?

I divided the above two lifetimes naively to get $0.1\%$.  I believe we can use the equation $\phi_f = k_r \times \tau_f$, and similarly for $\phi_p$, then divide the two. But I am not sure at all. Can anyone guide me how should I solve this question and is my answer correct?

The triplet can only be formed by intersystem crossing from the singlet. The fluorescence yield is $\phi_f=k_f^0/(k_f^0+k_i)$ where $k_i$ is the intersystem crossing rate constant to form the triplet and $k_f^0$ the radiative rate constant for fluorescence. The fluorescence lifetime is $\tau=1/(k_f^0+k_i)=10\cdot 10^{-9}$. The triplet yield is $1-\phi_f$. The triplet lifetime is not important and as the triplet only phosphoresces (you mention no other processes, so no, non-radiative, pathway from triplet) all the  molecules formed must phosphoresce (and it does not matter how long this takes as its the only pathway out of the triplet.)  However, you still have one piece of evidence missing, namely $k_f^0$.


Which Oxygen atom in HCOOH (formic/methanoic acid) does Carbon donate its electrons to, to obtain a partial positive charge? I was wondering if the Carbon atom in HCOOH (methanoic/formic acid) forms a positive partial charge by donating its electrons to both the Oxygen atoms, since they both possess a higher electronegativity value, or just one of them.

I was also curious about how the single bond and double bond between Carbon and the Oxygen atoms would affect the electron donation.

Electronic effects are cumulative. For example, the central carbon atom in an orthoester like HC(OEt)3 is surrounded by 3 oxygen atoms which add up.

In the case of formic acid (which is what you described in your question), yes the carbon atom is donating its electron density to BOTH oxygen atoms. However, the oxygen atom involved in the C=O bond will attract more electrons, because the double-bond is shorter.


Why is the molecular peak not observed in the mass spectrum of 2,2-dimethylbutane? Why isn't the molecular weight of 2,2-dimethylbutane on the mass spectrum? I can tell that a $\ce{-CH3}$ group is removed just by the math, but I do not know the reason why. The molecular weight is 86 and I know it has to do with fragmentation.

The answer probably has to do with how the mass spectrum was obtained.

Most mass spectra that are provided online, including the ones on the NIST WebBook, are electron ionisation mass spectra. That means that you put the sample in a chamber, accelerate electrons to a really high speed (giving them approximately $70~\mathrm{eV}$ of energy), and then bombard the sample with those high-energy electrons in order to ionise the sample.

The ionisation of these organic molecules doesn't take that much energy - $15~\mathrm{eV}$ is a good approximation. So you have an extra $55~\mathrm{eV}$ of energy in each molecule. How can the molecule get rid of this excess energy? The simplest way is to break chemical bonds; a typical bond strength lies between $3$ and $10~\mathrm{eV}$.

Branched compounds in general do tend to fragment more easily than straight-chain compounds. The reason is simple. If you think about n-hexane, no matter which bond you cleave, you're going to end up with a primary cation (or radical). However, if you cleave at one of the branching points of 2,2-dimethylbutane, you'll get a tertiary cation (or radical). A more stable product means that fragmentation occurs more readily.

However, the fact that the molecular ion is not observed is not really unique to 2,2-dimethylbutane. Lots of EI mass spectra don't have the $\ce{M+}$ peak. If you want to observe the $\ce{M+}$ peak, it's a better idea to use a gentler ionisation technique, such as chemical ionisation or electrospray ionisation.


Fish Fertilizer acid / base reactions Hi Everyone new to chemistry and learning lots but I had a question that I am having trouble working out.
I work for a vegetable grower and we make our own fertilizer.
My question is around the fish fertilizer. The brief description of the process is we use fish meal and mix it with phosphoric acid and enough water to make a paste. Let it sit a week then add, depending on volume a good amount of KOH. We also add a some skim milk and a small amount of molasses. This process among other things converts the protein in the fish to plant available nitrogen and gives us phosphorus and potassium.
I would love to understand all the chemical processes that are happening here and would like to understand how we are converting the protein in the fish meal to nitrogen. Is it phosphoric acid or KOH that is doing the work and how does the amount of these acids effect the process? Why is such a small amount of acid but a large amount of base used?
I have done my own research but coming up short in understanding what is going on and what the acid and base are doing.

Cells of fish contain proteins, and proteins are made of long chain of amino-acids, linked together by peptide bonds. In the presence of an acid (like phosphoric acid in your experiment or hydrochloric acid in our stomach), the peptide bonds are destroyed and the free aminoacids are released, so that they can be immediately assimilated by living organisms, like roots of plants if these acids are added to the ground as fertilizers.
But the phosphoric acid is not destroyed in the process. It still remains in the aminoacids mixture. As most acids, it is a corrosive substance that must be neutralized before being disposed in the ground. This neutralization is made with KOH, by one of the reactions $$\ce{H3PO4 + 2 KOH -> K2HPO4 + 2 H2O}$$ $$\ce{H3PO4 + 3 KOH -> K3PO4 + 3 H2O}$$ Whatever the reaction chosen, phosphoric acid is now transformed into a potassium phosphate, which can be assimilated. The advantage of this process is that now the fertilizer contains amino acids (bringing Nitrogen N to the soil), plus a salt bringing two important elements for fertilizing a soil : phosphorus P and potassium K. The three fundamental elements (N, P, K) for the agriculture are present in this fertilizer
To go back to the use of KOH, the equation shows that the amount of KOH to be added is twice or three times the amount of the initial phosphoric acid. That is why you have to add a huge amount of KOH to your mixture.


Lone pairs in MO theory? I have learned that non-bonding MOs are lone pairs. However when considering the MO diagram for the $O_2$ molecule, I do not see any non bonding orbitals yet there are two  lone pairs on each oxygen atom. Where do these come from?

Short answer:

Non-bonding MOs are usually lone pairs, but a lone pair does not necessarily mean there is a non-bonding MO.

Long answer:

You are trying to match what you learned from Lewis structures to the orbital picture. In general, that does not work (there may be exceptions).

Let's look at the MO diagram:

It is tempting to assign the bonding $\sigma_{2p}$ to the first (sigma) bond in $\rm O_2$. The two $\pi_{2p}$ bonding orbitals would then results in 2 $\pi$ bonds, hence we get a O-O triple bond. The further two electrons in the anti-bonding $\pi_{2p}^*$ orbitals could be assigned to one oxygen each, thus we have a biradical. The remaining two electron lone pairs might be assigned to the two $\sigma_{2s}$ orbitals. This is actually one of the two valid Lewis structures for $\rm O_2$.
(The other possible Lewis structure has only a double bond and two lone pairs at each oxygen atom.)

The problem is that the lone pairs are ascribed to only one atom, but the MOs are delocalized over the molecule. Thus there is no direct relationship between them. A non-bonding orbital is usually just an atomic orbital, which you may thus ascribe to a lone pair. But that does not mean each lone pair leads to a non-bonding orbital.

It is however possible, to localize the molecular orbitals (by making appropriate linear combinations of MOs) in such a way you get atomic orbitals, which you might match again with the lone pairs. The thing here is, that there are many different sets of orbitals which are all valid descriptions of the bonding situations.

Also note, that MO diagrams are just a simplification as well.

Source of the MO Diagram


Hydrogen bonding in alcohols vs amines Consider an alcohol and an amine compound with roughly the same molar mass. If I understand correctly, the boiling point for the alcohol is greater for two reasons:

The $\ce{O-H}$ bond is more polar than $\ce{N-H}$ because oxygen is more electronegative than nitrogen.
The oxygen atom has two lone pairs of electrons to act as hydrogen acceptors, whereas the nitrogen has only one.

The question is, which one of these reasons (if they are valid) contributes more to the increase in boiling point?

Both reasons given are valid and contribut to alcohols having higher boiling points than amines, with one caveat. It is not stated whether the amine is primary, secondary or tertiary, and a tertiary amine has no $\ce{N-H}$ bond.

Although the $\ce{O-H}$ bond is more polar than the $\ce{N-H}$ bond and can therefore form somewhat stronger hydrogen bonds, the fact that an alcohol has two lone pairs is the main reason for the higher boiling point.

Even though a primary amine has two hydrogens that can act as hydrogen bond donors, their already weaker $\delta +$ is spread among two hydrogens, making those hydrogen bonds very weak, effectively limiting them to act as a single hydrogen bond donor.

So, in general, an amine can only participate as a single hydrogen bond acceptor and donor (except for tertiary amines), while an alcohol's two lone pairs and single hydrogen allow for a much stronger hydrogen bonding network.


Similarity between Ecell and equilibrium constant Kc $$\ce{Cu^{2+} + Zn<=>Zn^{2+} + Cu}$$

$K_c=1.9*10^{37}$

$E_{cell}=1.1V$

I know that they both suggest that the forward reaction is going to completion but can someone please tell me a bit more about what both these values are suggesting and how these two values concord with each other.

Also, many books quote such reactions in only one direction and gives a value for $E_{cell}$. However, isn't $E_{cell}$ just another way of illustrating the condition of an equilibrium (i.e. in which direction the reaction would occur if left alone)... so how can a one way reation be shown

$E^\circ_{cell}$ and $K_c$ are related. Both indicate which direction of the reaction is spontaneous. Both tell you something about the position of equilibrium.

However, $E^\circ_{cell}$ is a directly measurable value. If you construct an electrochemical cell as using a zinc electrode immersed in a 1.0 M solution of say zinc chloride and a copper electrode immersed in a 1.0 M solution of copper (II) chloride with a salt bride of sodium chloride connecting the solutions, a voltmeter connected between the two electrodes will read the value of $E^\circ_{cell}$, at least initially. The voltage will change as the reaction progresses, but the initial voltageis $E^\circ_{cell}$ which is defined for standard conditions of 1.0 M concentration, 1 bar of pressure, and 25 $^\circ$C.

$$\ce{Zn|ZnCl2}(1.0 \text{ M})\parallel\ce{CuCl2}(1.0 \text{ M})|\ce{Cu} \ \ \ \ E^\circ_{cell}=1.1\text{ V}$$

The relationship between $E^\circ_{cell}$ and $K_c$ is the Nernst Equation. Note the use of both $E^\circ_{cell}$, the standard cell potential and $E_{cell}$, the observed cell potential under other conditions. The remainder of the alphabet soup is defined below.
$$E_{cell}=E^\circ_{cell}-\frac{RT}{zF}\ln{Q}$$

$E^\circ_{cell}$ - Standard cell potential
$E_{cell}$ - nonstandard cell potential at other conditions
$R$ - the ideal gas constant (which appears here because it is equal to Avogadro's number times Boltzmann's constant
$T$ - the temperature in kelvins
$z$ - the number of moles of electrons being transferred in this redox process. If we wanted to use the number of individual electrons, then we would need to replace $R$ with $k_B$
$F$ - Faraday's constant
$Q$ - the reaction quotient

We need to use $Q$ in the Nernst equation, because this equation can relate the cell potential to any range of conditions, not just equilibrium.

  At equilibrium $Q = K_c$ and $E_{cell} = 0 \text{ V}$ (there is no net change in the reaction composition, so there is no potential difference across the half-cells.

Thus, the Nernst Equation becomes:

$$0=E^\circ_{cell}-\frac{RT}{zF}\ln{K_c}$$
$$E^\circ_{cell} = \frac{RT}{zF}\ln{K_c}$$

  To determine the values of $E^\circ_{cell}$ and $K_c$ for the reverse reaction...

For $E^\circ_{cell}$:

$$E^\circ_{cell} = -E^\circ_{rev}$$

For $K_c$

$$K_{rev} = \frac{1}{K_c}$$


Definition of the B3LYP functional in common QC programs I wanted to use custom functionals with Gaussian and came up with some interesting definitions of the B3LYP functional within Gaussian, Orca and Turbomole, ...

$$\small\begin{array}{lcccccc}
\hline
  &  \text{Gaussian}  &  \text{ORCA}  & \text{Turbomole}  &  \text{NWChem} & \text{Molpro} & \text{GAMESS(US)}\\
\hline
\text{HF}  &  0.2  &  0.2  &  0.2  &  0.2  &  0.2  &  0.2\\
\text{Slater}  &  0.8  &  0.8  &  0.8  &  0.8  &  \color{red}{0.08}  &  0.8\\
\text{Becke}  &  0.72  &  0.72  &    0.72  &  0.72  &  0.72  &  0.72\\
\text{VWN-x}  &  {\color{red}1}  &  {\color{red}{1}}  &  0.19  &  0.19  &  0.19  &  0.19\\
\text{LYP}  &  0.81  &  0.81  &   0.81  &  0.81  &  0.81  &  0.81\\
\hline
\end{array}$$

It seems to be clear, that all programs define it somehow similar, but there are also some points that made me curious about the definitions. Let's start with the overall structure of this functional:

$$\text{XC-Functional}=a E_x^\text{local}+(1-a)E_x^\text{HF}+b E_x^\text{non-local}+c E_c^\text{non-local}+(1-c) E_c^\text{local}$$

The Gaussian documentation for DFT inputs specifies how one could change the amounts of each "sub-functional" by using the IOp-Statements through:

IOp(3/76=mmmmmnnnnn) IOp(3/77=mmmmmnnnnn) IOp(3/78=mmmmmnnnnn)

Where mmmmm and nnnnn will get divided by 10000 and 3/76=$P_1P_2$, 3/77=$P_3P_4$ and 3/88=$P_5P_6$. Those $P_{1,..,6}$ values are part of the following equation, which is a modified version of the equation above.

$$\text{XC-Functional}=P_2 E_X^\text{HF} + P_1 \left(P_4 E_X^\text{Slater} + P_3\Delta E_x^\text{non-local}\right) + P_6 E_C^\text{local} + P_5 \Delta E_C^\text{non-local}$$

Combining both, this yields: $P_1 = \{1,0\}$, $P_2 = 1-a$, $P_3 = b$, $P_4 = a$, $P_5 = c$ and $P_6 = 1-c$.

Now looking at Gaussian and ORCA, they both use 100% VWN and 81% of LYP, which should actually be combined to give 100% (instead of 181%) as is done by all other mentioned programs through using 19% VWN and 81% LYP.

It seems, that I am lacking some important information to understand that definitions in both, Gaussian and ORCA, and I hope seriously, that someone can enlighten me.

PS My discussion with Martin came up with the possibility, namely that there might be some missing brackets in the second equation, giving the following:

$$\text{XC-Functional}=P_2 E_X^\text{HF} + P_1 \left(P_4 E_X^\text{Slater} + P_3\Delta E_x^\text{non-local}\right) + P_6 \mathbf{{\color{red}(}}(1-P_5)E_C^\text{local} + P_5 \Delta E_C^\text{non-local}\mathbf{{\color{red})}}$$

Appendix: Information from the program output or the documentary

Gaussian 09 Rev. A.02 output

IExCor=  402 DFT=T Ex=B+HF Corr=LYP ExCW=0 ScaHFX=  0.200000
ScaDFX=  0.800000  0.720000  1.000000  0.810000

ORCA 3.0.3 output

Fraction HF Exchange ScalHFX         ....  0.200000
Scaling of DF-GGA-X  ScalDFX         ....  0.720000
Scaling of DF-GGA-C  ScalDFC         ....  0.810000
Scaling of DF-LDA-C  ScalLDAC        ....  1.000000
Perturbative correction              ....  0.000000
NL short-range parameter             ....  4.800000

Turbomole 7 define

 b3-lyp          | HYB  | 0.8S+0.72B88   | 0.19VWN(V)     | 1-3,5,6,10
                 |      | +0.2HF         | +0.81LYP       |
 b3-lyp_Gaussian | HYB  | 0.8S+0.72B88   | 0.19VWN(III)   | 1-3,5,6,10
                 |      | +0.2HF         | +0.81LYP       |

NWChem
Molpro

B3LYP   EXACT:B88:DIRAC:LYP:VWN5    0.2:0.72:0.08:0.81:0.19

the 0.08 is probably a typo
GAMESS(US)

As a former contributor to Orca who had access to Turbomole sources before that, I can say with confidence that the internal definitions of B3LYP and B3LYP_gaussian are consistent among them and Gaussian (where available). So no calculations should go wrong.

While I cannot speak for Gaussian, for Orca I am certain that the ScalLDAC parameter is badly printed here (and not relevant for many of the newer functionals).

As to what you identified as a typo in Molpro -- It may be that B88 was shoe-horned into the more mainstream definition of GGAs, which can be loosely worded as "a multiplicative correction to the LDA energy" (the energy density, actually). Becke, on the other hand, described the B88 functional originally as an additive correction ($\Delta E$). NWChem makes the distinction explicit in the keyword nonlocal, see the manual section on exchange-correlation potentials. So for Molpro, the exchange accounting should read: $0.2+0.72+0.08=1$, whereas the others use: $0.2+0.8=1$ and keep the $0.72$ implicit.


Uses of Fourier / Laplace transforms in chemistry (apart from spectroscopy) I'm just now learning about the Fourier transform, which seems like a pretty useful tool, and I know it has uses in spectroscopy (e.g. FTIR) since that is what Google shows when searching for applications of the Fourier transform in chemistry. Interestingly, I found out that the Laplace transform can be used in
chemical kinetics.
I was wondering if there were any other cool uses for the Fourier/Laplace transform in chemistry?
And if so, any links to resources to learn more about that particular use would be appreciated.

For the principle «what is small, gets large; what is large, gets small» when applying Fourier transformation and inverse Fourier transformation, it is one important tool in crystallography to relate between the regular arrangement of molecules in crystals at Angstrom scale, and the intensity recordings on film (old school) and electronic detector (very often area CCD) where signals are much wider apart from each other.  (This is often is referred to the relationship between direct and reciprocal space, too.)  Thus, scrutinizing the diffraction pattern's symmetry from special directions (figure b) below) allows the trained eye to relate to possible symmetries of the crystalline sample (a) , and the ratio of vector lengths describing the smallest unit of a crystal structure, the unit cell.  Today, assessing position and intensities of these data is typically assisted by a computer.

(credit, open access)
As shown by Tsutaoka et al. (Eur. J. Phys. 35 (2014) 055021, doi 10.1088/0143-0807/35/5/055021), it is possible to simulate such patterns with a laser pointer passing two film gratings slightly rotated to each other; the 2D crystal is then build by the points where the two gratings S1 and S2 intersect each other (note the vector ratio $\vec{a} / \vec{b}$ vs. $\vec{a^*} / \vec{b^*}$, and angles enclosed), e.g.

ImageMagick is one program allowing you to experiment with FT on 2D images but recommend to consult it altogether with the above mentioned open access publication by Aubert and Lecomte (J. Appl. Cryst. (2007) 40, 1153-1165, doi 10.1107/S0021889807043622) for additional examples and background.


How to draw resonance structures for CH3COCN What are the resonance structures of $\ce{CH3COCN}$ molecule?

Here are the significant resonance structures for pyruvonitrile.


Why is the mass of 1 mole of an atom/molecule equal to its gram atomic/molecular mass?(proof) Why do we take the mass of 1 mole of an atom to be equal to its gram atomic mass and the mass of 1 molecule to be equal to its gram molecular mass? I want a mathematical rigorous proof. There are other questions on this site dealing with this question but none of them have the "proof".

$\def\p{\mathrm{p^+}}\def\n{\mathrm{n}}$
Mass of one proton = $m_\p$ and mass of one neutron = $m_\n$. (Masses are in unified atomic mass unit-$\mathrm{u}$)

C-12 isotope consists of 6 $\p$ and 6 $\n$. Therefore, mass of C-12 isotope is equal to 6 $m_\p$ + 6 $m_\n$. Since $m_\p$≈ $m_\n$, the mass of C-12 can be replaced by 12 $m_\p$. (electrons are not taken into account since they have negligible mass)

1 $\mathrm{u}$ is defined as 1/12 of the atomic mass of C-12 isotope.

⟹1 $\mathrm{u}$ = 1/12 × (6 $m_\p$ + 6 $m_\n$)

⟹ 1 $\mathrm{u}$ ≈  1/12 ×12 $m_\p$

⟹ 1 $\mathrm{u}$ ≈  1 $m_\p$   ...equation(1)

Now,

12 g of C-12 has $\pu{N_A}$ particles and constitutes 1 mole of it. (according to the definition of 1 mole)

⟹ 1 mole of (6 $\p$ + 6 $\n$) have a mass of 12 g.

⟹ 1 mole of (12 $\p$) have a mass of 12 g.

⟹ 1 mole of $\p$ have a mass of 1 g

⟹ 1 mole of $\p$ × 1 $m_\p$ = 1 g

⟹ 1 mole of $\p$ × 1 $\mathrm{u}$ = 1 g

⟹ 1 mole of $n$ $\p$ × 1 $\mathrm{u}$ = $n$ g

Therefore, mass of 1 mole of $n$ $\p$ is equal to $n$ g

Now, the final part,

Let an atom have $x$ $\p$  and $y$ $\n$ . The atom can be considered to be $(x+y)$ $\p$  for our purposes.

1 mole of $(x+y)$ $\p$ have a mass of $(x+y)$ g.

Thus the mass of 1 mole of atoms of an element is equal to its gram atomic mass.

Molecular mass of an element or a compound is equal to the sum of the atomic masses of each type of it's constituent atoms.
Thus the mass of 1 mole of molecules of an element or a compound is equal to the gram molecular mass.


What exactly is a reducing equivalent and what does it do? I've encountered this term in the context of cellular metabolism, but I can't seem to find an explanation of what a reducing equivalent is or why it is named this way. Wikipedia was not really helpful, and most other online sources seem to use the same phrasing (which I am struggling to parse).

Is a reducing equivalent something that is itself reduced, or is it something that causes reduction (thereby being itself oxidized in the process)? The Wikipedia article uses the expression "which transfer the equivalent of one electron" which has left me all the more confused.

I would be most grateful for an explanation or links to material I could read to clarify this for myself.

The number of "equivalents" of a reagent refers to how much of the reagent there is relative to another reactant of interest. To use a concrete example, if we are interested in the $\mathrm{S_N2}$ transformation $$\ce{CH3Cl -> CH3Br},$$ a possible way of bringing this about would be to treat the starting material (methyl chloride) with a certain amount of bromide ion. If we carry out the reaction on 0.5 mol of methyl chloride, in theory we would only need 0.5 mol of bromide ion, because of the stoichiometry of the reaction:

$$\ce{CH3Cl + Br- -> CH3Br + Cl-}$$

In this case, 0.5 mol of bromide ion would be referred to as being "1 equivalent" of bromide (relative to methyl chloride, which is the reactant of interest).

We could also say that for the complete combustion of hexane, we would (theoretically) need "9.5 equivalents" of oxygen gas:

$$\ce{2C6H14 + 19O2 ->12CO2 + 14H2O}$$

In the case of a redox half-reaction, you usually have electrons in either the reactants or the products:

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

(this highly simplified reaction is relevant in, e.g., cytochrome c's role as an electron carrier in oxidative phosphorylation). Under biological conditions, this is rather unrealistic, as electrons don't actually exist in aqueous solution, nor is there any electrolysis going on. So, you need something else to act as a supplier of electrons, i.e. a reducing agent, such as NADH.

$$\ce{NADH + H+ -> NAD+ + 2H+ + 2e-}$$

In this case, because 1 mol of NADH provides 2 mol of "electrons", you could say that NADH acts as two reducing equivalents. Here, NADH is acting as a reducing agent, and is itself oxidised to NAD. [I don't agree with the Wikipedia page about hydride ion being one reducing equivalent; it should be two.]


What is the exact reason that compounds such as carbon dioxide cannot interact with other compounds through dipole-dipole interactions? Whilst I understand that $\ce{CO2}$ has a linear structure, resulting in a vector sum of the dipoles of 0, I do not understand why the dipoles formed at the two oxygen atoms as a result of their increased electronegativity cannot interact with other molecules via dipole-dipole forces. Do the vectors equaling 0 somehow negate the present dipoles?

As the other answers have indicated, CO2 has no NET dipole moment. However it does have two dipoles pointed in opposite directions (as OP keeps mentioning). This means that CO2 can possibly interact through higher moments, such as the quadrupole moment. The mathematical procedure behind this is known as the multipole expansion. It is important to note however, the higher order multipoles get weaker and weaker by a factor of 1/R approximately, so quadrupole interactions are usually ignored for example.


Obtain precipitated iron? Is it possible to obtain pure, precipitated iron with no oxidation by some chemical process?

Alternatively is it possible to electrically purify iron using an anode and cathode?

Is it possible to obtain pure, precipitated iron with no oxidation by some chemical process?

Very pure iron can be obtained by pyrolysis of iron pentacarbonyl.  Iron metal obtained in this manner is in fact called carbonyl iron and is commercially available.

$$\ce{Fe(CO)5 -> Fe(s) + 5 CO}$$

Doing this yourself is very dangerous, however, because iron pentacarbonyl is volatile and toxic.  The decomposition reaction to produce iron requires high temperatures, and the byproduct is carbon monoxide gas, which is also notoriously toxic.

  Alternatively is it possible to electrically purify iron using an anode and cathode?

This is also possible.  I don't know about other salts of iron, but to form iron from iron oxides, very high temperatures are required.  The other answer mentioned Dr. Sadoway's work.  Let me highlight a recent article of his unveiling new anode materials for electrolytic decomposition of iron oxide.  At temperatures as "low" as 1538°C, molten iron oxide can by electrolyzed to pure oxygen gas and to molten iron metal using electrochemistry.

Water is not an appropriate solvent for electrochemical formation of iron metal.  At potentials low enough to favor $\ce{Fe^0}$ formation from $\ce{Fe^{+2}}$, water is unstable and would be reduced to hydrogen gas.


Techniques to extract lipids I am planning an experiment on how to isolate casein and part of the process involves removing any lipids that may still be in it after filtering the precipitated casein out of the milk. I have read that one can use 95% ethanol to dissolve the lipids into it, since like-dissolves-like. This would not remove all lipids though and thus a 25 mL 1:1 solution of diethyl ether would be necessary to remove the remaining lipids for a certain amount of the collected precipitate.

What exactly occurs in the reaction when lipids are dissolved into the alcohol? What other alcohols could be used? How does one figure out how much alcohol they should use and at what concentration?

These questions may seem silly to some, yet I have conducted a considerable amount of research online trying to figure it out myself — to no avail though. I appreciate all well-intentioned answers.

Casein is a mixture of phosphoproteins found in milk that accounts for approximately 80% of the total protein content of cow milk (cow milk contains 3.3% total protein). Hence, it is easy to prepare casein as a pure protein from cow milk.

Casein exists in milk as the water soluble calcium salt of a phosphoprotein (UC, Davis). Thus, it is usually prepared by acid precipitation. Acid treatment removes the calcium cation, leaving a water insoluble phosphoprotein. Note that casein is amphoteric, i.e., it forms salts with both acids and alkalines.

Generally, acetic acid is used in the casein precipitation, because its mildness and less harsh property compared to hydrochloric acid. Although, most people extract casein precipitation in the cold acid solutions, such practice would lead to the formation of protein with poor physical properties. Softness and rather incomplete separation of the casein are problems encountered.

If the precipitation is performed in warm solutions, the casein separates as a single large colloid, with complete separation, leaving a yellow solution of whey behind (which primarily consists of milk-sugar, lactose).

Keep in mind that usual casein precipitation is comparatively easy but delicate. If too much acid is added or if the acid is added too quickly, or if the acid is too strong, part of casein mass will redissolve in whey. Also, casein may be extracted from whole milk. Disadvantages are that whole milk contains a relatively large concentration of fat. This fat will be physically included in the casein curd as it separates. Thus, I'd recommend to use skim milk (if possible), which would be one solution for your co-precipitation of fat problem.

A procedure for casein precipitation in UC, Davis is as follows:

First, prepare a solution of $\pu{7.0 mL}$ glacial acetic acid in $\pu{50.0 mL}$ deionized water. This is your acid solution to be used for the precipitation.

Place $\pu{600.0 mL}$ of skim milk in a $1$-$\mathrm{L}$ beaker. Warm the milk to $\pu{110 ^{\circ}F}$ (approximately $\pu{43 ^{\circ}C}$; Note that temperature should not be $\gt \pu{55 ^{\circ}C}$). At this temperature, add the acetic acid solution (prepared earlier) drop-wise and with stirring. At a certain point, casein will separate as an amorphous mass leaving clear supernatant (light yellowish). Gather the solid from the beaker with a glass stirring rod or separate it using suction filtration.

Caution: Do not add any more acid than necessary to obtain a good curd. A few minutes must be given between acid additions to allow for stabilization.

Press the removed mass of casein between paper tissues to remove as much water as possible. Place the mass of casein in a food blender with $\pu{500 mL}$ of water. Suspend the casein in the water by blending as much as possible.

Separate the casein from the water with a filter and wash the cake with 2-propanol. If 2-propanol is not available use ethanol or some low molecular weight alcohol.

Let the casein powder air dry. You may wash the dried cake with two portion of ether if necessary. You may suspend the dry powder in ether and let it soak well before filtering.

Yield of casein in $\pu{600.0 mL}$ of skim milk is reported as $\pu{13.8 g}$.