Notation for excited states I am reading some papers about quantum chemistry and in one I found a notation for excited states of a molecule that I don't understand at all.

In the picture below you can see the potential curves of biphenyl as a function of its torsion angle. In the description of the picture, the author speaks of states $1^1A$ and $1^1B_1$. What does he mean by these states?

As has already been mentioned in comments, those letters and numbers for each state correspond to their symmetry, specifically the symmetry of the (purely electronic) wavefunction. For more information regarding symmetry notation, see here, and see here for more discussion about how symmetry can be applied to molecular orbitals.

It might not be clear from the second question, but because the symmetry of the overall N-electron wavefunction results from the direct product of all occupied 1-electron wavefunctions (at least in Hartree-Fock), exciting electrons into molecular orbitals with different symmetries will lead to excited N-electron wavefunctions (excited states) with different symmetries.

To consider the specific case of biphenyl,

the $\tau = 0^{\circ}$ structure, corresponding to the leftmost values on the x-axis, will look like the 2D structure just above where both rings lie in the same plane. As an exercise left to the reader, you can figure out that the point group of the planar form of biphenyl is $D_{\mathrm{2h}}$. The point group $D_{\mathrm{2h}}$ has the following character table:

$$\begin{array}{c|cccccccc|cc} \hline
D_\mathrm{2h} & E & C_2(z) & C_2(y) & C_2(x) & i & \sigma(xy) & \sigma(xz) & \sigma(yz) & & \\ \hline
\mathrm{A_g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\
\mathrm{B_{1g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & xz \\
\mathrm{B_{3g}} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \\
\mathrm{A_u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\
\mathrm{B_{1u}} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\
\mathrm{B_{3u}} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline
\end{array}$$

The three-dimensional symmetry of any object can always be broken down into one or more irreducible representations (irreps) belonging to the object's point group; each irrep corresponds to a row in the point group's character table. The symmetry table is the leftmost column in each irrep row. To enumerate them,

The labels from the figure are $1^1A, 1^1B_{1}, 1^1B_{2}, 1^1B_{3}, 2^1B_{3}, 2^1B_{2}, 2^1A, 2^1B_{1}$.
The labels from the character table are $\mathrm{A_g}, \mathrm{B_{1g}}, \mathrm{B_{2g}}, \mathrm{B_{3g}}, \mathrm{A_u}, \mathrm{B_{1u}}, \mathrm{B_{2u}}, \mathrm{B_{3u}}$.

You can see that these labels are almost but not quite the same as those in the figure. A minor difference is italicized versus regular/upright text; this is a matter of typographic choice and there is no physical or chemical difference between them. There are two significant differences:

The labels in the figure are prefixed with a regular number and a superscript number. The regular number helps uniquely defines the state ordering: $1^1B_{1}$ is the first state with $^1B_{1}$ symmetry, and $7^1B_{1}$ is (hypothetically) the seventh state with $^1B_{1}$ symmetry. Without the superscript number, each symmetry label only specifies the spatial symmetry, and with the superscript defined as $2S + 1$, the spin multiplicity, each label can also specify the spin symmetry. $1^1B_{1}$ is the first singlet ($S = 0$) state with $B_{1}$ symmetry, and $1^3B_{1}$ is (hypothetically) first triplet ($S = 1$) state with $B_{1}$ symmetry. See this excellent answer for more information, in particular the figure.
The labels in the figure do not have u/g in the subscript, because for most of the biphenyl structures, where $\tau \neq 0^{\circ}$, the molecular point group is no longer $D_{\mathrm{2h}}$, but one of lower symmetry. When the rings are perfectly perpendicular to each other ($\tau = 90^{\circ}$ and $\tau = 270^{\circ}$), the molecular point group will be $D_{\mathrm{2d}}$, like hydrogen peroxide, and every other intermediate value will be even lower in symmetry: $D_{\mathrm{2}}$, similar to skewed ferrocene (rather than staggered or eclipsed). To see how the irrep and its symmetry operations from a point group of higher symmetry map to those in a point group of lower symmetry, one needs a descent table. Here, most notably, all skewed structures are missing the inversion operation that would necessitate the gerade/ungerade subscript. Because most of the structures are of lower $D_{2}$ symmetry, it is easiest to represent all structures with $D_{2}$ point group labels.

For completeness,

$$\begin{array}{c|ccccc|cc} \hline
D_\mathrm{2d} & E & 2S_4 & C_2 & 2C_2' & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & z & xy & \\
\mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$

$$\begin{array}{c|cccc|cc} \hline
D_2 & E & C_2(z) & C_2(y) & C_2(x) & & \\ \hline
\mathrm{A} & 1 & 1 & 1 & 1 & & x^2, y^2, z^2 \\
\mathrm{B_1} & 1 & 1 & -1 & -1 & z, R_z & xy \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & y, R_y & xz \\
\mathrm{B_3} & 1 & -1 & -1 & 1 & x, R_x & yz \\ \hline
\end{array}$$

See how there are fewer symmetry operations (middle set of columns) in $D_{2}$ than in $D_{\mathrm{2d}}$ than in $D_{\mathrm{2h}}$.


Why does small value of van der Waals gas constant "b" ensure easier liquefication? Larger value of b signifies that the size of the molecules is larger and hence the forces acting should become larger due to increased surface area.This should ensure easier liquefication. But my book says that smaller value of b corresponds to easier liquefication. Where am going wrong?

The book says "Chlorine is more easily liquefied than Ethane because 'a' for  chlorine is greater than 'a' for ethane but 'b' for fluorine is less than 'b' for ethane. Van der Waals constant 'a' is due to force of attraction and 'b' due to finite size of molecules. Thus, greater the value of 'a' and smaller the value of b, larger the liquefaction"

The $b$ in the Van der Waals equation of state signifies the volume that is effectively taken out per molecule by intermolecular repulsive interactions. You can imagine that a small $b$ will help the gas to liquefy, since in most ordinary liquids molecules are packed closer together than in a gas.

As written in Atkins Physical Chemistry:


Can I make my own graphene at home in my kitchen? Graphene is cool. Can I make some? What do I need? Is it at all feasible to make it in quantity?

I Googled the phrase "How to make graphene" and got this result, which contains a section on how to make it at home.

The answer is, unequivocally, yes you can. Below are excerpts from the linked page.

  (1) The first DIY method is to use a lead pencil to deposit a thick layer of graphite onto a paper. Then use ordinary sticky tape to peel off a layer of graphite from the paper. Use another piece of sticky tape to remove a layer of graphite from the first sticky tape. Then, use a third piece of unused sticky tape to remove a layer from the second piece of sticky tape, an so on. Eventually, the graphite layers will get thinner and thinner, and you will end up with graphene, which is single-layer graphite in the strict sense, or bi-layer or few-layer graphite (which acts almost like graphene in certain uses). Even though this way of making graphene is only a proof-of-concept, the sticky tape method works. It takes patience and time, but it's the DIY method which the Manchester group used in 2004. And remember, they actually won a Nobel prize for their work, so there's no messing with sticky tape!

  (2) Shear exfoliation in liquids starts by pouring powdered crystal in a liquid, and then using a shear mixer to separate (exfoliate) layers of material from the crystal. The liquid that is used in such processes is chosen so that the small graphene particles don't clump back together, and the result is a liquid suspension of graphene. The suspension can then be dried to obtain graphene nanoflakes, or it could be directly used in later technological processes, such as making of graphene coatings and others.
  So, how do you make graphene at home using this method? Well, you can replace the shear mixer with a kitchen blender (since shear mixing is a fancy word for high-tech blending), and instead of special liquids you can use water and dish detergent. The detergent is added so that the particles don't clump, and acts as a surfactant. A great source of graphite powder are graphite pencils. Simply powder a few pencil leads and add the powder to the detergent solution, then blend for a while. Unless you have access to special microscopes and other equipment, you will probably be unable to confirm the existence of graphene in your detergent solution, but there's a quick rule of thumb which you can use to estimate the particle size. Generally speaking, if the graphene particles are too big, they will sink to the bottom of the vessel, while if they are smaller, they will float to the top. Nanoparticles are so small that they will be suspended in mid-water, and this is what you're aiming for if you want to make graphene. After you make enough graphene particles, you might want to filter the suspension and leave it to dry.

  (3) If you pour a non-water-soluble liquid into water it will either float on top of the water, or the water will float on top of it, depending on whether the added liquid is heavier (more dense) or lighter (less dense) than water. A common example is oil floating on top of water. One group of scientists discovered that you can make graphene by pouring two solvents which do not mix together, such as heptane and water, into a glass and adding finely ground graphite powder and putting it in a sonication bath. You can buy heptane in most chemical supply stores, water is readily available (it's a good idea to use distilled water), and you can easily obtain graphite sticks in art shops. An ultrasound bath is needed, but these are not uncommon on ebay and the like. The resulting graphene is one to four layers thick, and is chemically stable. It can be extracted and deposited on a clean glass plate, or any other substrate. The idea is to mix the finely ground graphite with water and heptane (1:1 ratio of water and heptane) and sonicate it for a while. What happens next is that graphite flakes exfoliate at the interface between water and heptane and graphene actually starts "climbing" up the glass walls of the vial. If a hydrophilic substrate, such as a glass slide, is introduced through the interface, a thin film of graphene will climb the walls of the slide as well. The glass slide can then be extracted and after drying, you'll be left with graphene-coated glass. If you use a polyethylene vial, the resulting graphene will not climb the walls of the vial, but you can still extract it with a slide made of glass or other hydrophilic material. Graphene is deposited as a thin film on both sides of the glass plate, and its slight light-absorbing properties can be seen by looking through it (Graphene absorbs about 2.3% of visible light). The explanation of the process is that graphene self-reassembles from the small exfoliated flakes on the interface between water and heptane because the surface tension of graphene (54.8 mN/m) is nearly right between the surface tension of water (72.9 mN/m) and heptane (20.1mN/m). The thickness of the graphene film is limited by capillary forces on the interface and the excess graphene will simply fall to the bottom of the vial. The formation of thick aggregates is suppressed by the diffusion and the energy requirement needed to form a new layer. This is possibly the easiest way to make DIY graphene at home as an arbitrarily large sheet. Other methods only create graphene flakes, but these flakes could possibly be used instead of graphite flakes in this method as a refinement stage.


How to calculate the mass of phosgene in the equilibrium mixture of carbon dioxide and chlorine? Consider the reaction:
$$\ce{CO(g) + Cl2(g) <=> COCl2(g)}$$
A closed $\pu{5.69 L}$ flask containing a reaction mixture at equilibrium has $[\ce{CO}] = \pu{1.8 \times 10^-6 M}$ and $[\ce{Cl2}] = \pu{6.3 \times 10^-7 M}$. Given that the equilibrium constant $K_\mathrm{eq} = 2.9 \times 10^{10}$, calculate the mass of $\ce{COCl2}$ present in the flask.

How can I solve the question above? I thought that equilibrium constants deal with concentrations and not masses, so I'm not sure how to use it to calculate a mass.

We have an equilibrium mixture for the reaction

$$\ce{CO (g) + Cl2 (g) <=> COCl2 (g)}$$

We should recognize that

$$K_\mathrm{eq} = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$

Note that the reactants and products all have stoichiometric coefficients of 1 so we can write our $K_\mathrm{eq}$ equation out in this way.

Because we are given concentrations of reactants in the equilibrium mixture, we can go ahead and plug in the values that we know and solve for $[\ce{COCl2}]$.

$$\begin{align}
2.9 \times 10^{10} &= \frac{[\ce{COCl2}]}{(\pu{1.8 \times 10^-6 M})(\pu{6.3 \times 10^-7 M})} \\
[\ce{COCl2}] &= \pu{0.033 M}
\end{align}$$

We are asked how many grams of $[\ce{COCl2}]$ is present in the reaction mixture.  Concentration and mass are related via the following equation:

$$[\ce{COCl2}] = \frac{n(\ce{COCl2})}{V} = \frac{m(\ce{COCl2})}{M(\ce{COCl2}) \cdot V}$$

or

$$m(\ce{COCl2}) = [\ce{COCl2}] \cdot M(\ce{COCl2}) \cdot V$$

where $n$ is the amount of $\ce{COCl2}$ (in g), $M$ is the molar mass (in g/mol), and $V$ is the volume (in L).

Let us plug in all of our known values and solve.

$$\begin{align}
m(\ce{COCl2}) &= (\pu{0.033 mol L-1})(\pu{98.906 g mol-1})(\pu{5.69 L}) \\
&= \pu{18.6 g}
\end{align}$$


Why do Asn-Ser and Gln-Thr have different H bonding patterns? I noticed that Asn-Ser H-bonding is different from the H-bonding between Gln-Thr.
However both have the same functional groups. I wasn't able to find any information about this. Maybe it's due to steric effects, but because these striking similarities between Asn and Gln and on the other side Ser and Thr I don't think that is an explanation (but maybe it is, I'm just guessing).

excuse me for the drawing, but I couldn't find an image showing this clear

I found these H-bonding patterns in Fundamentals of Organic Chemistry, McMurry, seventh edition.

update

Are al these H-bonds possible?

I don't have the textbook you've gotten this from, but unless there's a statement to the contrary, I suspect these are intended as examples of hydrogen bonding, rather than an exhaustive list of the only hydrogen bonding pattern which can exist.

If you start to look at real structures, you will see both "directions" of hydrogen bonding (hydroxyl donor/carbonyl acceptor and nitrogen donor/hydroxyl acceptor) in real protein structures.

Take, for example, the structure 1GKT. S297 is very clearly in a hydrogen bonding orientation with N303, with the hydrogen of the hydroxyl pointing toward the carboxylate oxygen of the asparagine. Also, Q22 is very likely in a sidechain-sidechain hydrogen bond with T29, where the hydrogen on the glutamate nitrogen is pointing toward the oxygen of the threonine.

If you spend some time looking at various experimental structures from the PDB website, you should be able to find examples of a wide range of hydrogen bonding patterns - basically, any acceptor and any donor can be paired, depending on the geometry of a particular protein.

One caveat I should mention here is that because of the way X-ray crystallography works, it's often difficult to tell which way the sidechains of the ASN and GLN residues are oriented. With electron density, the NH2 group and the =O group look very similar. Often it's a bit of guesswork on the crystallographer's part to tell which orientation they have, often relying on the layout of neighboring residues to determine the most likely hydrogen bonding network.

However, 1GKT was not determined by X-ray crystallography, but by neutron diffraction. In neutron diffraction it's very clear where the protons are located. (I picked a neutron diffraction structure as the example for this very purpose.) I also double checked that there wasn't any reason for flipping those particular ASN/GLN residues, both by manual visualization and with the MolProbity web server.


Why isn't hydration an oxidation reaction? http://chemistry2.csudh.edu/rpendarvis/OxRed.html  states,

  Addition or removal of water does not involve, by itself, an oxidation or a reduction reaction. The addition of water to an aldehyde to form a hydrate does not involve oxidation or reduction.

But in this reaction,

The oxidation states of the carbons change from

-3, -1 to -2, -2

As the oxidation states have changes, shouldn't it be considered a redox reaction?

  An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron Source

As context is always King, I will briefly mention that the original quote is about the redox chain from carboxylic acid via an aldehyde to a primary alcohol (and potentially to an alkane). In this context, the aldehyde is capable of adding water as shown in $(1)$, but this reaction does not constitute a redox process in any definition of the word (count electrons).

$$\ce{R-CH(=O) + H2O <=> R-CH(-OH)(-OH)}\tag{1}$$

In other cases, the removal or addition of water does change the oxidation states of various carbon atoms as you observed. This is not only true for the dehydration of an alcohol as in your example but e.g. also in such key reactions as the Wittig reaction or the Suzuki cross-coupling.

However, organic chemistry deals with a great number of individual synthetically useful reactions that must be classified in some way or another for humans’ benefit. In this organic chemistry classification system, only those reactions that

unilaterally oxidise a single carbon atom by means of addition of an oxygen atom or removal of hydrogen atoms; or
concurrently oxidise two carbon atoms bound to each other by the addition of oxygen or removal of hydrogen in such a manner that both oxidation states change by the same value

as oxidations. (And vice-versa for reductions. Individual exceptions most certainly will apply.) Your example is, in inorganic terms, a comproportionation which means one carbon is oxidised while the other is reduced. Not only does this not cleanly fit into ‘carbon is oxidised’ or ‘carbon is reduced’, meaning it creates a terminological mess, but it also would subsume most if not all reactions under the general redox labels which destroys the idea of classifying reactions as oxidations or reductions in organic chemistry altogether.

So: yes, in a strict (inorganic) sense it is a redox reaction, but in the interest of a more practical classification of organic reactions it is not considered one.


Predicting changes in pH of water when chlorinating using two different methods Two methods for chlorinating water are using chlorine gas $(\ce{Cl2})$ or calcium hypochlorite $(\ce{Ca(OCl)2}).$ By using chemical equations how can you predict the pH of the water brought about by the chlorinating process?

In aqueous solution, chlorine can undergo disproportionation: one atom in $\ce{Cl2}$ (oxidation state $0$) is reduced to $\ce{Cl-}$ (oxidation state $-1),$ the other is oxidized to a species in which the chlorine atom has the oxidation state $+1.$

$$\ce{Cl2 + 3H2O -> 2H3O+ + Cl- + OCl-}$$

Ask yourself how that influences the pH of the solution.

What about adding calcium hypochlorite to water? If $\ce{CaCl2}$ is the calcium salt of hydrochloric acid, what about $\ce{Ca(OCl)2}?$

 You may have guessed it: $\ce{Ca(OCl)2}$ is the calcium salt of chloric(I) acid.

You may remember that adding the soluble salt $\ce{MA}$ of a strong acid $\ce{HA}$ to water will not influence the pH of the solution.

 If $\ce{HA}$ is a strong acid, it will fully dissociate in water:
 $$\ce{HA + H2O <=>> A- + H3O+}$$
 This means in return that the corresponding base $\ce{A-}$ is very weak:
 $$\ce{A- + H2O <<=> HA +OH-}$$

Now, is $\ce{OCl-}$ a strong or a weak base? In order to decide, you might want to have a look up the $\mathrm{p}K_\mathrm{a}$ of the corresponding acid and keep in mind that for a corresponding acid-base pair in water $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14.$


How to make a sephadex slurry I want to purify crude alpha-amylase with column chromatography. I am using a spehadex 75, but for some reason I can't find any information on how to make the slurry.
I can quickly find a lot of information on how to pack and store a column, but what I really want to know is how much powder and buffer do I mix together?

Sephadex resins swell as they absorb buffer, so the amount of buffer you need to add is dependent on the amount that the specific resin you are using is expected to swell. It is also dependent on the final thickness of slurry that you want to achieve. Too thick - the resin will not be suspended well enough and will pack poorly. Too thin - the resin at the bottom will settle before enough buffer has flowed through to allow you to add more, and again you will not get good packing (and it will take a lot longer to pour).
Sigma Aldrich provides technical documents online with recommendations for preparing Sephadex slurries from dry powder, which you can find here.
Here are some the the key points. The expected swelling of the Sigma Aldrich resins is provided in units of mL per 1 g dry resin:
Sephadex G-10 Medium: 2-3 mL/g
Sephadex G-25 (all grades): 4-6 mL/g
Sephadex G-50 Fine: 9-11 mL/g
In all cases, add buffer and allow the resin to swell (1-3 hours at room temperature), then top off the buffer so that the settled swollen resin is ~75% of the total volume. That will give you a slurry of appropriate thickness when stirred.


Calculating the mass of water removed by air conditioning condensation I cannot figure out how to calculate the mass.

The question is:

  "Suppose the percent relative humidity is $\mathrm{80.0\%}$ at $\mathrm{91.4\ ^\circ F}$ ($\mathrm{33.0\ ^\circ C}$) in a house with a volume $\mathrm{245\ m^3}$. Then the air conditioner is turned on. Due to the condensation of of water vapor on the cold coils of the air conditioner, water vapor is also removed from the air as it cools. After the temperature has reached $\mathrm{77\ ^\circ F}$ ($\mathrm{25\ ^\circ C}$), the percent relative humidity is measured to be $\mathrm{15.0\ \%}$. What mass of water vapor has been removed from the house? (Reminder: take into account the difference in saturated water vapor pressure at two temperatures.)"

The answer is $6.00\ \mathrm{kg}$.

Relative humidity= (actual pressure of $\ce{H2O}$ pressure/partial pressure of $\ce{H2O}$ if saturated) * 100

So, my first step was to find the actual pressure of the initial and final conditions.

$X/355.1 = 0.80$

$X = 284.08~\mathrm{Torr}$

Then $X/23.76 = 0.15$

$X= 3.564$

$\mathrm{284.08~Torr- 3.564~Torr = 280.516~Torr}$ (water removed, I think)

I attempted to use the ideal gas formula ($pV= nRT$) but kept getting the wrong answer. How do I approach the rest of the problem?

Step 0) Realize that sometimes the answer given for a homework problem is wrong.

In this case it appears that the correct answer is $\pu{6.11 kg}$ of water, not $\pu{6.00 kg}$ of water.

The approach to the problem given below is to simply calculate the total water mass in the house before and after cooling, then subtract the latter from the former to get the mass of water removed by condensation.

Step I) Assemble parameters applicable before and after cooling:

$R =$ gas constant $= \pu{8.31m^3~bar~mol^{-1}~K^{-1}}$

$V = \pu{245m^3}$

$MW_{water}=\pu{18.0g~mol^{-1}}$

${n=\frac{PV}{RT}}$

Step II) Calculate water mass before cooling:

$T = \pu{33^oC = 306~K}$

$P_{saturation} = \mathrm{50.3~mbar^{(1)}}$

$P = 80\%$ of $P_{saturation} = \mathrm{40.2~mbar}$

${n=\mathrm{\frac{0.0402~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 306~K}=387~mol~\ce{H2O}}}$

$\mathrm{387~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=6.97~kg~\ce{H2O}}$

Step III) Calculate water mass after cooling:

$T = \mathrm{25^oC=298~K}$

$P_{saturation} = \mathrm{31.7~mbar^{(1)}}$

$P = 15\%$ of $P_{saturation} = \mathrm{4.76~mbar}$

${n=\mathrm{\frac{0.00476~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 298~K}=47.9~mol~\ce{H2O}}}$

$\mathrm{47.9~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=0.862~kg~\ce{H2O}}$

Step IV) Calculate total water removed from the house:

$$\mathrm{6.97~kg~\ce{H2O} - 0.862~kg~\ce{H2O} = 6.11~kg~\ce{H2O}}$$

Reference:
1) Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.
....


How the equation that describes evolution of alkyl nitrates has been derived? I was reading this paper, Bertman, S. B., J. M. Roberts, D. D. Parrish, M. P. Buhr, P. D. Goldan, W. C. Kuster, F. C. Fehsenfeld, S. A. Montzka, and H. Westberg (1995), Evolution of alkyl nitrates with air mass age, J. Geophys. Res., 100(D11), 22805–22813, doi:10.1029/95JD02030 (link attached), and had a hard time trying to derive equation (12) from equation (11). If anyone knows how to do that, help would be highly appreciated.

You have to quote a bit more of the paper for context, namely the reaction scheme. In the paper, it is given as

$$\ce{RH ->[\beta k_\mathrm{A}] RONO2 ->[k_\mathrm{B}] P}$$

The first thing to notice is that the concentration of $\ce{RH}$ as a function of time is given by the first-order integrated rate law

$$[\ce{RH}]_t = [\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)$$

and therefore equation (11) becomes

$$\frac{\mathrm{d}[\ce{RONO2}]}{\mathrm{d}t} = \beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)- k_\mathrm{B}[\ce{RONO2}]$$

This is a ordinary differential equation and there are many ways of solving them (this is hardly the place to discuss ODEs; consult a maths book if you are interested). In this case, the way I did it is to multiply throughout by $\exp(k_\mathrm{B}t)$. I am going to use $c$ to represent $[\ce{RONO2}]$ to simplify the equation a little bit:

$$\begin{align}
\frac{\mathrm{d}c}{\mathrm{d}t} &= \beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)- k_\mathrm{B}c \\
\exp(k_\mathrm{B}t)\frac{\mathrm{d}c}{\mathrm{d}t} + \exp(k_\mathrm{B}t)k_\mathrm{B}c &= \exp(k_\mathrm{B}t)\beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t) \\
&= \beta k_\mathrm{A}[\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t]
\end{align}$$

Now if you consider the derivative of $\exp(k_\mathrm{B}t)c$ with respect to $t$, we have (by the product rule, and bearing in mind that $c$ itself is a function of $t$)

$$\frac{\mathrm{d}}{\mathrm{d}t}[c\exp(k_\mathrm{B}t)] = \frac{\mathrm{d}c}{\mathrm{d}t}\exp(k_\mathrm{B}t) + ck_\mathrm{B}\exp(k_\mathrm{B}t)$$

which is the same as the left-hand side of the equation above. The thing we multiplied by throughout, $\exp(k_\mathrm{B}t)$, is called an integrating factor. So, we can now integrate both sides:

$$\begin{align}
\frac{\mathrm{d}[c\exp(k_\mathrm{B}t)]}{\mathrm{d}t} &= \beta k_\mathrm{A}[\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] \\
\int_0^t \frac{\mathrm{d}[c\exp(k_\mathrm{B}t)]}{\mathrm{d}t}\mathrm{d}t &= \beta k_\mathrm{A}[\ce{RH}]_0 \int_0^t \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] \mathrm{d}t \\
\left[c\exp(k_\mathrm{B}t)\right]_{t=0}^{t} &= \beta k_\mathrm{A}[\ce{RH}]_0 \left(\frac{\exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - 1}{k_\mathrm{B}-\beta k_\mathrm{A}}\right) \\
c_t\exp(k_\mathrm{B}t) - c_0 &= \beta k_\mathrm{A}[\ce{RH}]_0 \left(\frac{\exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - 1}{k_\mathrm{B}-\beta k_\mathrm{A}}\right) \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - [\ce{RH}]_0) \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_t \exp(k_\mathrm{B}t) - [\ce{RH}]_t\exp(\beta k_\mathrm{A}t)) \\
c_t\exp(k_\mathrm{B}t) &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_t \exp(k_\mathrm{B}t) - [\ce{RH}]_t\exp(\beta k_\mathrm{A}t)) + c_0 \\
\frac{c_t}{[\ce{RH}]_t}\exp(k_\mathrm{B}t) &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (\exp(k_\mathrm{B}t) - \exp(\beta k_\mathrm{A}t)) + \frac{c_0}{[\ce{RH}]_t} \\
\frac{c_t}{[\ce{RH}]_t} &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{\exp(-k_\mathrm{B}t)c_0}{[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)} \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{c_0}{[\ce{RH}]_0}\exp[(\beta k_\mathrm{A} - k_\mathrm{B})t] \\
\end{align}$$

I have shown quite a lot of the intermediate steps, if anything seems puzzling just work through it, it is really just algebraic manipulation.

Replacing $c$ with $[\ce{RONO2}]$ gets you to nearly your desired equation

$$\frac{[\ce{RONO2}]_t}{[\ce{RH}]_t} = \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{[\ce{RONO2}]_0}{[\ce{RH}]_0}\exp[(\beta k_\mathrm{A} - k_\mathrm{B})t]$$

The only difference in the paper is that $\beta$ was sometimes dropped (i.e. some of the $\beta k_\mathrm{A}$'s simply became $k_\mathrm{A}$). I glanced through the paper and can't find a rationale for it, but I will admit I did not look at it in great detail.


Is there proof that the exact exchange correlation functional is an analytic function? Mathematics show that some problems cannot be solved analytically. Could this be true for the exact exchange correlation functional?

Most textbooks state that the form of the exact exchange correlation functional is unknown. It is said to be an unknown mathematical object. Does this mean that said functional is an analytical function? And if so, what is the proof that shows that the functional is analytical?

See my colleagues work, it seems to be the answer you want.

http://journals.aps.org/pra/abstract/10.1103/PhysRevA.93.042511


Why is Thorium (At. no. 90) an f-block element at all? Being an exception of the Aufbau principle, Thorium has an electronic configuration of $\ce{[Rn]}\mathrm{7s^25f^06d^2}$ instead of the expected $\ce{[Rn]}\mathrm{7s^2 5f^2 6d^0}$.

Two other elements, Lanthanum and Actinium, also showed such an anomaly as $\mathrm{6s^2 4f^0 5d^1}$ and $\mathrm{7s^2 5f^0 6d^1}$, respectively. But we placed them both in the d-block of the periodic table.

Since, even in Thorium, the last electrons occupied d-orbitals only (disobeying Aufbau), then why is it placed in the f-block of the periodic table. I mean, it should be placed in the d-block, shouldn't it?

Edit: Note that here, I don't intend to ask or know how or why the 2 electrons disobeyed the Aufbau, and occupied the higher energy 5d instead of lower energy 4f.

Rather, I want to know only that why it is kept in the f-block when the last orbitals which the electrons occupied were of d-subshell.

The bulk metal actually does involve the $f$ orbitals, distinguishing it from what would be its congeners in Group 4 of the $d$ block.  From Wikipedia:

Despite the anomalous electron configuration for gaseous thorium atoms, metallic thorium shows significant 5f involvement. A hypothetical metallic state of thorium that had the [Rn]6d27s2 configuration with the 5f orbitals above the Fermi level should be hexagonal close packed like the group 4 elements titanium, zirconium, and hafnium, and not face-centred cubic as it actually is. The actual crystal structure can only be explained when the 5f states are invoked, proving that thorium, and not protactinium, acts as the first actinide metallurgically.[1]

Cited reference:
1.
Johansson, B.; Abuja, R.; Eriksson, O.; et al. (1995). "Anomalous fcc crystal structure of thorium metal". Physical Review Letters. 75 (2): 280–283.


Alcoholic proof and temperature The US defines the "proof" of an alcoholic beverage to be twice the alcohol per volume percentage at 60° F (about 15.5° C)

As someone who does not have a chemistry background, I am wondering what the 60°F requirement means.

Given a mixture of alcohol and water that is 80 proof (40% alcohol by volume at 15.5°C), what is the alcohol by volume at room temperature 20°C?

Briefly, since ethyl alcohol expands more rapidly with increasing temperature than does water, then measuring the mix by volume at a warmer temperature would give you less alcohol and more water, because the alcohol would be less by weight for the same volume.

Thermal Coefficient of Expansion (approximate, measured at room temperature)

Alcohol, ethyl (ethanol):   0.00109 / K

Water:                            0.00021 / K

Using these constants, you can figure out the answer to your question. But note the wording: given a mixture. If the EtOH/H2O is already mixed, changing the temperature does not change the concentration or proof, and you would have to separate the ingredients to measure their respective volumes. BTW the volume of the mix is slightly less than the sum of the volumes of its ingredients.

Moral: if your measuring your drinks by shot-glass rather than with a scale, you'll get a weaker drink in a warm bar ;-)


Radial Probability Distribution Curve versus ψ² versus r curve for 1s orbitals My question is basic, but I have already referred to a couple of books.

This is an excerpt from Linus Pauling's book:

  the most probable distance of the electron from the nucleus, which is the value of $r$ at which $D(r)$ has its maximum value, is seen from Figure 21-1 to be $a=\pu{0.529 \mathring{A}}$, which is just the radius of the normal Bohr orbit for hydrogen.
  [Figure 21-1 is the Radial Probability Distribution Curve (RPDC) for the 1s orbital, see below.]

And on the next page he states that:

  The function ${\psi^2_{100}}$ has its maximum value at $r=0$, showing that the most probable position for the electron is in the immediate neighborhood of the nucleus; that is, the chance that the electron lie in a small volume element very near the nucleus is larger than the chance that it lie in a volume element of the same size at a greater distance from the nucleus.

Don't these two statements contradict each other? $\psi^2$ versus $r$ says that the probability density is highest at $r=0$. RPDC says that the probability is highest at $r=\pu{0.529 \mathring{A}}$. Why is there such a stark difference between probability density and probability? When $r$ tends to $0+$ we observe high electron probability density (wrt the $\psi^2$ curve) but minimized electron probability (wrt the RPDF curve)

I do understand, however, that Radial Probability Distribution Curve represents probability and $\psi^2$ versus $r$ represents probability density. But why is there a difference in these two?

For reference:

It is very important to note here that $R(r)$ is just the radial part of wavefunction. Wavefunction does not describe any observable. The Born interpretation says that the probability density of finding electron between any two points $x_1$ and $x_2$ is given by
$$\rho=\psi\psi^*.$$
While the probability is
$$P =\int_{x_1}^{x_2}{\psi\psi^* \mathrm{d}x}.$$
(This is only in one dimension. In spherical coordinates you must integrate over the volume).
Why is there a difference between these two quantities? I think it will be easier to understand using an analogy. Consider a rod of unit length in which the charge varies as a function of $x$. Say the function is
$$q(x)=x\mathrm{e}^x-1$$
The average charge of the rod is $0$. While the average charge density of this rod is $\mathrm{e}-2$. Where average is given by
$$\langle f \rangle=\frac{\int^{b}_{a}{f \mathrm{d}x}}{b-a}.$$
I hope you have understood the difference between function density and the function itself. The radial probability and probability density are something similar. By seeing the graphs you can to a certain extent predict the average of the probability density and probability. In the second graph it is $0$ while in the second graph it is close to the peak which is $a_0$.


Density of Ni vs Cu using unit cell How might one determine whether Ni is more dense than Cu using only the atomic masses of Ni and Cu, and the fact that Ni has a simple cubic unit cell, and Cu has a face-centered unit cell. You are not given the radii of each element. I feel like this question cannot be answered with the given information only, considering as Ni and Cu have very close densities (0.06 g/cm^3 difference).

I think you need to make at least one more assumption along the way to determining the relative densities. But let’s take it stepwise:

We are given a primitive cubic cell for nickel and a face-centred cube for copper.
We therefore conclude that per unit cell one atom of nickel and three atoms of copper are present.
We know from the periodic table that copper is a bit heavier than nickel ($63.5$ versus $58.7$)
Therefore, it is clear that copper’s unit cell is more than four times as heavy as nickel’s. But we don’t know their sizes.
From mathematical considerations we can also calculate how much of each unit cell is filled with atoms ($74~\%$ for fcc; somewhat less which I am too lazy to calculate now for primitive cubic) — this again means that there should be more weight per volume in the copper unit cell.
What seems to be lacking in my eyes is a way to compare the unit cell sizes, whether by atom size or unit cell dimensions. However, since nickel and copper are right next to each other in the periodic table, it may be safe to assume their sizes to be almost identical.
If that assumption is safe, we can use either route to clearly show how copper’s density is higher. But note that if copper’s atoms were significantly larger than nickel’s, this reasoning would break down extremely quickly.


Free API to view 2D representation of Molecules I'm looking for a free API that can accept a SMILE and produce the 2D structure of a molecule. I would like to include this in a web page.

I have collection of SMILES and would like to have the 2D structure available as part of a database. I tried generating images using RDKit but I can't add that these images to any data frame or tabular data structure. Any suggestions on how to display the 2D representation of and any SMILE queried on the web page will be appreciated.

Open Babel can convert all kinds of input into pictures with 2D structures.


In equilibrium, if bonds are constantly being created and broken, where does the energy that does so come from? I'm a student who recently learned about equilibrium and how it is not a state where nothing is made, but rather, a state where the product is decomposing at the same rate as the reactants are combining. If breaking and creating bonds takes energy, where does this energy come from? Does it have to be added to the system? Also, if a closed system, say, a glass jar, reaches equilibrium, and heat slowly leeches out of the jar, does the balance of equilibrium shift, or does the reaction slow down but stay in equilibrium?

The energy comes from the molecules themselves. See, they have certain energy of thermal motion. You might have heard of the ${3\over2}kT$ thing, but that's only an average value. Molecules have different energies, described by the so-called Maxwell distribution. Those with higher energy might take part in some reactions which are not possible for the entire bulk.

As for your second question, it's option 1: the equilibrium typically will shift. (Though the reaction might slow down a bit, too.)


XPS analysis intrepretation A lab presented me with an elemental composition (mainly metals in char) derived from an XPS analyzer and mentioned that binding energies suggested a list of compounds (various oxides, sulfides and silicates etc). Is there a way to show the composition in terms of these compounds instead of just the elements? Only familiar in bucket chemistry (aka chem eng), these analyzers are a bit of a black box to me even after reading up on them.

Also, is analyzing the surface of a batch of tiny particles <<1 micron somewhat tricky to get reliable results?

This is a bit late, but I'll add an answer anyway. XPS is already a method that can be used for the speciation, so you should be able to differentiate between the different the different oxides of the same metal or metalloid. This is because the work function to eject an electron from a shell of the metal centre is dependent on its chemical environment.


Thermochemistry questions regarding enthalpy I have a few questions about heat of combustion. I have two different scenarios that I was to present:
Scenario 1: Given heat of combustion in the equation
$\ce{CH4 + 2O2 -> CO2 + H2O}$ and $\Delta H =\pu{-889kJ mol-1}$
Q1: Would $1$ mol methane produce $\pu{889 kJ}$; whereas, $1$ mol of oxygen would produce $\pu{444.5 kJ}$?
Scenario 2: Given the energy produced per mol of a molecular compound
$\ce{C2H6O + 3O2 -> 2CO2 + 3H2O}$ and $\Delta H=?$
However, we are given that the oxygen produces $\pu{400 kJ}$ of energy.
Q2: Would that mean that the $\Delta H=3 \times400=\pu{1200 kJ mol-1}$?
It would be really great if someone could explain the theories behind thermochemistry.

You should not separate the species on the left and assign enthalpy variations to each of them. On Scenario 1, you say that 1 mol of methane reacts with 2 mols of oxygen, and the variation of enthalpy for this is −889 kJ (since it is 1 mol of methane). If you had 2 mols of methane, and 4 mols of oxygen, $\Delta$H would be 2*(-889) kJ, and so on. As you mentioned, if only 1 mol of oxygen were available, only 0.5 mol of methane would react and $\Delta$H would be 0.5*(-889) kJ
So, for Scenario 2, you said it was given that

the oxygen produces 400kJ of energy

This probably means that, for each mol of oxygen that reacts with ethanol, 400 kJ of energy is released. So yes, if you wanted the $\Delta$H for the reaction, you should multiply by 3 and get that $\Delta$H = -1200 kJ mol-1.
This is in agreement with the measured molar enthalpy of combustion of ethanol (-1211 kJ mol-1)


Introduction to modeling chemical reactions I am looking for introduction to modeling of chemical reactions. I think there is the base approach, where concentrations of chemical species are given, plus ratios of each possible reaction / outcome. Is there a paper lightly explaining internal mechanics of such reactions, and their modeling via system of differential equations?

The topic you are asking about is called reaction dynamics, and is closely related to chemical kinetics (the former is more focussed on atomistic events, reactive and nonreactive collisions, while the later usually describes the mathematical treatment of the kinetic laws governing species concentrations). It was the topic of the 1986 Nobel Prize in Chemistry.

It is covered by most physical chemistry textbooks, and more in depth by specialized textbooks on that matter. It is sometimes taught together with basics of statistical mechanics or statistical thermodynamics. Check your favorite (or local) university for a list of reading material on this topic!


Why are carbonyl compunds more reactive towards catalytic hydrogenation than alkynes and alkenes? Alkynes are more reactive towards catalytic hydrogenation than that of alkenes, because they are more unsaturated. This is the correct reasoning, right?
But why are carbonyl compounds more reactive towards catalytic hydrogenation than alkynes? The $\ce{C=O}$ in carbonyl group is polar, but how does that affect reactivity?

Example:

Answer: Option (D), $c>a>b$

That is not true in general: the stronger nature of the C=O bonds (larger bond energy) means that they are not as easily hydrogenated as C=C bonds. However, your compound (c) is not a regular carbonyl group, but an acyl chloride, which is a quite reactive species and thus very easily hydrogenated.

For a bit of reading, see here the section on “Catalytic Hydrogenation”:

  As a rule, the carbonyl group does not add hydrogen as readily as do the carbon-carbon double and triple bonds. Thus, it is fairly easy to reduce an alkene or alkyne function without affecting any carbonyl functions in the same molecule. By using a platinum catalyst and increased temperature and pressure, it is possible to reduce aldehydes and ketones to alcohols, but carboxylic acids, esters and amides are comparatively unreactive. The exceptional reactivity of acyl halides, on the other hand, facilitates their reduction under mild conditions, by using a poisoned palladium catalyst similar to that used for the partial reduction of alkynes to alkenes. This reduction stops at the aldehyde stage, providing us with a useful two-step procedure for converting carboxylic acids to aldehydes, as reaction #1 below demonstrates. Equivalent reductions of anhydrides have not been reported, but we might speculate that they would be reduced more easily than esters. The only other reduction of a carboxylic acid derivative that is widely used is that of nitriles to 1º-amines.


Using bleach and water to shine pebbles I soaked a large jar full of pebbles (from a river) in bleach water for a couple of years and the rocks came out super smooth and shiny when dried. I have tried to repeat this process again but am not having any luck with it this time. Have you ever heard of this and do you know if there is a certain ratio of water to bleach that works? Perhaps I put something else in the water the first time and caused a chemical reaction?

Here is a picture of the rocks that became smooth and shiny after a couple of years in water with bleach added for keeping the water clean and clear. And a picture of the rocks I have done the same to for a year. An interesting note is that fossils such as Petoskey Stones do not change at all.

I've collected rocks and tumble polished rocks. It is not possible that just sitting in bleach water tumble polished the stones. You can leave the stones in bleach water till you die and they still won't get polished. The only explanation is that you bought polished stones and put them into the bleach water. Over a period of time you just forgot what type of stones that you started with.

The other point is that "river rocks" (rocks smoothed in a river...) will look much smoother when wet because the water wets the surface and "hides" some of the surface roughness. So dry Petoskey Stones just look sort of mottled gray stones, but when wet the coral pattern shows nicely.


What is the charge of Silicon dioxide? What is the charge of Silicon dioxide ions? I can't find it anywhere.

It seems to be 0?

Also is there anywhere I can find these on my own?

Like what @GeoffHutchison said,

$\ce{SiO2}$ is not an ion. It is a network solid. Its net charge is zero.

From charge balance, since oxygen holds a -2 formal charge, then silicon must hold a +4 formal charge in order to balance out. $(+4) + (+2 \times -2) = 0$


Comparison of enol content How can we find the order of percentage of enol form
of the compounds in the following equilibria?

Percentage of enol form in x is highest as due to aromaticity. But how can we compare that for y and z. The compound z is non-aromatic but compound y has resonance.

You may also think of hydrogen bonding here. See that in case of y, intramolecular hydrogen bonding is possible and hence greater percentage of enol content is of y than z. You may note that this is a general factor of esters (in this case a cyclic ester) in determining enol content percentage and that this factor may not always be the 'winning' one (as you can compare between x and y).


What would happen if I combined this polymer and this metal (assuming I can) So let me explain, I recently found out about a material called sorbothane (which is a polymer). Now, sorbothane is special because not only does it highly efficiently absorb shock, but also highly absorbs vibration.

However, when it does this, it temporarily loses its shape, so I was wondering, what would happen if I combined it with a strong yet light metal like titanium? (Note: I am 17 and don’t know a whole lot about chemistry.)

You can combine materials in two ways: chemically and mechanically.
On average, if you want to the desired physical properties to add up, you want to go mechanical way, not the chemical one.

Once the reaction takes place, the there is absolutely no reason to assume the product preserves properties of both precursors. It might sometimes, but more often than not it's a completely different beast.
Speaking of retaining physical properties in terms of logic and addition operation, in chemistry both $\text{true} + \text{true} = \text{false}$ and $2 + 3 = -42$ are possible.
Mixing water and sodium won't result in liquid metal.

What you probably want to look into instead is mechanical combination, or, more precisely, composite materials such as fiberglass or carbon plastics alternative where titanium would act as a 2D or 3D supporting mesh for sorbothane, though not reacting with it chemically.
Such poly-phase materials are already used for monolithic composite armor plates.


What substance(s) will prevent superglue from hardening in its tube? Superglue often sets in the tube, so I was wondering if ethanol or even water could stop it from hardening and going to waste. Ideas, anyone?

Based on what I gathered from this Wikipedia article on super glue :

Super glue is typically composed of a class of compounds called
 cyanoacrylates.

Cyanoacrylates are capable of taking part in this process we call "Polymerization", where chains of cyanoacrylate molecules, called "polymers" ( Poly- Many, mere- parts) are formed. Each constituent cyanoacrylate molecule in the polymer is called a "monomer" ( Mono- One, mere - part).

The chemical structure of methyl-2-cyanoacrylate:

Polymerization is known to occur for a number of reasons, and can follow a number of pathways. (So to keep this brief, I'll just skim over how methyl-2-cyanoacrylate polymerizes). Wikipedia provides the following mechanism:

Over there, the Nu bit is a common abbreviation we use for a nucleophile in chemistry. It is the presence of a nucleophile, that catalyzes the polymerization of methyl-2-cyanoacrylate.

So when you squeeze out a little superglue, how does it harden? Air around you, as well as the surface you're applying the superglue on will, inevitably, have trace quantities of water. To a small extent, water splits to provide positively charged hydrogen ions (essentially, it's just a proton) and negatively charged hydroxide ions (OH). It's the hydroxide ions that play the role of the nucleophile in the above-depicted mechanism.

So ironically, when super glue "dries out", it just transiently combined with (a part of) a water molecule. That is, as @MaxW points out in the comments, water catalyzes the "drying" of superglue.

Since it's water that causes superglue to harden/set, manufacturers go all out to make sure the packaging is water/air-tight. Ever noticed, that when you buy a tube of superglue, it has a seal at the mouth which you've gotta pierce before you can use it? That seal is what prevents the glue from drying out while it's still in the (unopened) tube. Once you pierce the seal, it's only a matter of days before a sufficient amount of water (courtesy: Atmospheric moisture) manages to seep in and render render the remaining superglue useless.

Note that in case of superglue, the manufacturers don't (typically) add anything to keep the superglue from drying...they just put on a seal and cross their fingers. Whereas your typical liquid/PVA glue is a water based emulsion. When PVA glue is squeezed out of the tube and onto a surface it, literally, "dries out" by losing water.

Once superglue is set, it is set... there's no turning back O:)

You can't revert it to usable superglue through any (simple) means.

However, since you now know that it's water that's responsible for superglue "hardening", if you want to make the superglue left in the (opened) tube last longer, store it an an air tight jar that's been packed with a desiccant, like silica gel ;)


Chemical equilibria and changes in concentration I learned that if concentration of reactants or products is changed, the position of equilibrium changes but value of $K_c$ remains the same.
This just does not make intuitive sense to me... I mean whatever concentration of reactants I start with does not matter, $K_c$ remains same....?

Also, can someone please elaborate the term position  of equilibrium

Assuming the following reaction:

$$\ce{A + B <=> C + D}$$

Where:

$$K_\text{c}=1.23\times10^{-2}$$

We say that:

$$K_\text{c}=\frac{\ce{[C]*[D]}}{\ce{[A]*[B]}}=1.23\times10^{-2}$$

And the reaction will be at equilibrium for any concentrations (actually, it should be activities, which are in essence effective concentrations) of A, B, C and D that satisfy the above.

I can understand your confusion. What they mean is... If you have a reaction at equilibrium, and remove half the reactants, then the reaction will proceed towards reactants, until equilibrium is once again reached.

  Their way of saying it:

You remove reactants, and the position of equilibrium is shifted towards the left, because the reaction will now proceed towards the left hand side to restore equilibrium.

It's confusing, because equilibrium occurs at the same conditions as before, so you haven't really shifted the equilibrium position in the way it sounds like.


Acids / Bases - Need some clarity for final exam I'm studying for my final exam, and these are the questions I got wrong on the mid term, could anyone help correct me? I know people are against giving answers for homework, so I must add that is now exam period for Universities, so this is only for my further understanding!
Thanks so much!

My attempts will be in this format

1)The pH of a 0.2M unknown base X OH is 8.15

a)Calculate [OH-]

10^(-8.15) = 7E-9M

b)What is the % disassociation

(7E-9/0.2M) x 100 = 3.5E-6%

c)What is the Kb of the base?

This one I don't even know where to begin..

2)Calculate the pH of the following solutions

a) 0.10M KCN, Ka (HCN) = 6.2E-10 (Hint: KCN is the salt of a weak acid and a strong base)

x = sqrt(Ka*c)
x = sqrt(6.2E-10 * 0.1) = 7.8E-6
-log(7.8E-6) = pH = 5.1

1a) First use $pOH = 14 - pH$

Then use: $pOH = \log([OH^-])$

b) Simply use: $\dfrac{\text{dissociated}\ [OH^-]}{\text{original}\ [OH^-]}$

c) Use $K_b = \dfrac{[X^+][OH^-]}{[XOH]}$

2) Dissolving $KCN$ in water gives:
$\ce{KCN \rightarrow K^+ + CN^-}$

The following equilibrium reaction is thus formed

$\ce{CN^- + H2O \rightleftharpoons HCN + OH^-}$

Which is:

$K_b = \dfrac{[HCN][OH^-]}{[CN^-]}$

From the $K_a$, you can get the $K_b$ through $K_w\ =\ K_aK_b$ so that:

$\dfrac{K_w}{K_a} = \dfrac{[HCN][OH^-]}{[CN^-]}$

Let the amount of $CN^-$ that reacts to be $x\ M$, thus $[CN^-]$ becomes $0.1-x$ and you get $x\ M$ of both $[HCN]$ and $[OH^-]$. Solve for $x$ to get $[OH^-]$ and from there, $pH$ should be easy to calculate.

3) Use $Ksp = [Ba^{2+}][F^-]^2 = 1.7\times 10^{-6}$ and solve for $[Ba^{2+}]$.


Mass transfert in process chemistry engineering, two film theory I'm having troubles to understand a part of my lesson. I have no idea where to ask my question, but it is consider as chemistry so here we go.

Let me introduce some notations :
$x$ and $y$ are the mole fractions
of the same compound but respectively in liquid phase and in gas
phase.
$k_{c_l}$ and $k_{c_g}$ are the coefficient of exchange of matter in
$\ce{m.s^{-1}}$
$N_{1n}$ is the molar flux density of the compound $1$ in the normal position

Let now me introduce (translating from french to english) you the part of the lessons in which I have troubles :
No phenomenas of production or consommation of the compound $1$ occur in the interface $I$, and we are in stationnary state then the molar flux density is the same in both parts and we have $$N_{1n}=c_lk_{c_l}\left(x_1^{I}-x_1\right)=c_gk_{c_g}\left(y_1-y_1^{I}\right)$$
The fluids at the interface are supposed to be in thermodynamical equilibrum then we have $$y_1^I=K_1^{eq}x_1^I$$

We can for example define a fictive gas phase which will be in equilibrium with the liquid phase (or the contrary). The mole fraction of the compound $1$ is by definition $$y_1^*=K_1^{eq}x_1$$
The total driving force is then $$\begin{align} (y_1-y_1^*)&=(y_1-y_1^I)+(y_1^I-y_1^*)\\ &=(y_1-y_1^I)+ K_1^{eq}\left(x_1^I-x_1\right)\end{align}$$ if we suppose to have a unique average equilibrum constant in the liquid film.
Now I'm having troubles their :
We can then define a global transfer constant in the ralation to the driving force in gas phase : $$\frac{N_{1n}}{K_{c_g}c_g}=\frac{N_{1n}}{k_{c_g}c_g}+\frac{N_{1n}K_1^{eq}}{k_{c_l}c_l}$$
I really don't understand this last relation, if someone can help me, I would be glad. I apologize in advance for the troubles you can have too, to understand what I tried to translate here.
Thank you
PS : I have really no idea for the tags, feel free to edit them

You already derived this relation (eq 1),
\begin{align} (y_1-y_1^*)&=(y_1-y_1^I)+(y_1^I-y_1^*)\\ &=(y_1-y_1^I)+ K_1^{eq}\left(x_1^I-x_1\right)\end{align}

Now, to keep the molar flux density between the gas phase concentration $y_1$ and fictitious conc. on the other side $y_1^*$ we assumed a fictitious overall transfer coefficient $K_{cg}$. So, now we have another relation for molar flux density
$$N_{1n}=K_{cg}c_g\times(y_1-y_1^*)$$
that follows,
$$y_1-y_1^*=\frac{N_{1n}}{K_{cg}c_g}$$
Similarly,
$$y_1-y_1^I=\frac{N_{1n}}{k_{cI}c_I}$$
and
$$x_1^I-x_1=\frac{N_{1n}}{k_{cg}c_g}$$
Plug these above three relations into eq 1 and you will get your last equation.


Is 1,2- Hydride shift possible in ketonic cations? The question in my worksheet asks me to predict the result of the reaction between benezene and $\ce{(CH3CH2CH2CO)2O}$ in presence of anhydrous $\ce{AlCl3}$.

My attempt:

After complex formation with $\ce{AlCl3}$ we get this carbocation: $\ce{CH3CH2CH2CO+}$ that has two resonance structures.

But, if we consider hydride shift we get $\ce{CH3CH2C^+HC=O}$ which is not only stabilised by resonance but also by the $\text{+I}$ effect of ethyl group.

However, the answer is $\ce{PhCOCH2CH2CH3}$ which isn't formed by the second carbocation with hydride shift .

So I'd like to know if hydride shift is possible in ketonic carbocations or not? And if it is, is the answer wrong?

The carbocation doesn't rearrange since oxygen donates it's lone pair and the resulting carbocation is quite stable since the octet of all the atoms is complete.


Why is the pOH of 1M HCl 14? Why is the pOH of 1 molar HCl 14 in an aqueous solution?

I was learning about strong bases and strong acids in chemistry within this context:

We have 1L of pure water at 25°C, and the pH of water is 7 and the pOH of water is 7, so it is a neutral solution.

Then we add 1 mole of HCl into the water. HCl, because it is such a strong acid, disassociates completely in the water. After some basic calculations, we found that the pH of HCl was 0 and the pOH was 14, but the way he showed this was using the water's pH and pOH - does this make sense?

So we found that the pH of water is 0, and then he did something funny and got a pOH of 14. Why is the pOH of this solution 14?

Here is the video, video mark, 10:02: Khan Academy - pH & pOH of Strong Acids and Bases

In the moment from the video that you referred to, Sal was using the equation $\rm pK_w=pH+pOH$ to find the $\rm pOH$ of the solution. $\rm pK_w$ is equal to 14, since $K_w$ is equal to $1\times 10^{-14}$, giving us the new equation $14=\rm pH+pOH$. We know that the $\rm pH$ of a 1M solution of $\ce{HCl}$ is zero from earlier on in the video. This gives us that $14=0+\rm pOH$, or in words, that the $\rm pOH$ must be equal to 14.


Is it safe to mix calcium carbonate and sodium chloride? I know zero about chemistry, and Google has not offered up a plain answer, so I beeseech thee; Is it safe to mix calcium carbonate and sodium chloride to make a simple mouthwash?  Calcium carbonate for the acid neutralizing, and sodium chloride for disinfectant.

Safe in context of chemistry, physiology or long term health care prevention ?

With zero knowledge, it is safer to stay with commercial products, until you know, what you are doing.

For mouthwash, calcium carbonate is quite ineffective, being insoluble solid. Better would be solution of sodium bicarbonate ( baking soda ), present in some toothpastes as well.

Salt is not great desinfectant and must be used in really high concentration, that can be very disturbing if swallowed.


Would the concentration of oxygen around fire increase as we get close to the fire (from the outside) Since fire require oxygen, wouldn't a fire, like those in the American forest along west coast, constantly draws in oxygen from around? And if that is so, would the concentration of oxygen around the fire higher than farther away?

A fire has no way to directly "draw in" oxygen for it to burn. It consumes what oxygen is in the immediate vicinity of the flame, depleting the air in oxygen compared to the concentration of oxygen in air further away. This sets up a concentration gradient such that oxygen will diffuse into the depleted region near the fire.

An even more important process than diffusion of oxygen into depleted air is convection driven by the heat of the fire. Oxygen-poor air that has been in the fire gets hot and rises, drawing fresh air into the base of the fire. Note that this still isn't oxygen being drawn to the fire, but a mass of fresh air flowing into the lower pressure area at the fire resulting from convection.

The bottom line is that a fire consumes oxygen, so the region around the fire will be slightly depleted in oxygen, not enriched as you suggest.


How to calculate the volume of oxygen in a reaction between hydrogen peroxide and bleach using this apparatus I know that the reaction between bleach and hydrogen peroxide produces oxygen but using this apparatus and information  I need to state how to calculate the volume of oxygen released. From my initial thoughts, I thought that the volume of oxygen will simply be the volume of the gas collected, however, the answer is said to be the total volume of gas collected subtract $20\ \mathrm{cm^3}$. I fail to understand this as I believe that the $20\ \mathrm{cm^3}$ of the hydrogen peroxide reacts with the bleach completely and even if not, will still nly produce oxygen and not any other gas, making the use of subtraction unnecessary and wrong for our answer.

This is how it works in my head:

If instead of peroxide in the syringe, you just had $\pu{20cm^{3}}$ of air, then injecting the air will cause a reading of $\pu{20cm^{3}}$ to happen.  The water gets pushed out of the way.

Injecting peroxide would do the same thing (approximately).  After the reaction, there is still saltwater and leftover reactants taking up space.  The oxygen produced will take up additional space (and push the water away) because it is a gas and it wants to expand.


How does sugammadex (a modified cyclodextrin) draw rocuronium into its lipophilic cavity? Sugammadex is a modified cyclodextrin: 8 member sugar ring with carboxyl thioether extensions. Rocuronium is an aminosteroid, 4 steroid rings and one protonated amine group.

Sugammadex binds rocuronium really tightly. The reason for this is straightforward - lipophilic interactions with the inner cavity of the cyclodextrin, which is the perfect size because of the thioether extensions. The negatively charged carboxylate groups lock into place with the protonated amine.

But the question asks how the rocuronium is initially drawn into the cavity, which makes less sense. I've only found one reference to this in the literature, in a figure caption that says the carboxylate groups interact with the four steroid rings.

But how? The rings are not aromatic, so there is no anion-pi interaction. One would think the positively charged amine would interact with the carboxylate groups far more often, leading to the wrong orientation. If anything, the hydrophobic rings should be repulsed by the negative charge?

Figure 6: Encapsulation process: Sugammadex carboxyl groups interact with steroidal rings A, B, C, and D, drawing the aminosteroidal NMBA molecule into the cavity where additional non-covalent attractions hold the molecule securely in place.

Reference:NCBI

After consulting with a professor, the answer seems obvious.

The two negative charges repel each other when the free cyclodextrin molecule is in solution, resulting in an "open" conformation that has the carboxylate groups facing "out". The thioether chains interact with the rocuronium as it is drawn into the large open face of the molecule. As the positively charged nitrogen moves into place, the carboxylate groups "close", and lock the drug into place.


Is it a "Bary center" or a "Bari center"? (Studying Coordination Chemistry, "Crystal Field Theory" at school)

Apparently, different sources spell this differently.

The Wikipedia article on Crystal Field Stabilization Energy, calls it the "Barycenter" (as does my school textbook)

However, the actual Wikipedia article on the "Barycenter" refers to, (as far as I can tell) the center of mass of two mutually orbiting bodies. Whereas the "Bary/Baricenter" I'm talking about is the zero-energy reference level in octahedral complexes.

Another book I have at hand (Concise Inorganic Chemistry Lee, J.D), prominently uses the term "Bari centre".

The center/centre distinction aside, what I've been able to glean from my books (and the internet), is that this term is spelled in four different ways (Wah! Such gross inconsistency in literature!):

Barycenter
Bary center
Baricenter
Bari center

But which one's correct (or at least, a more preferred term in Chemistry-literature)?

It is safe to say that Barycenter is the correct spelling for the sought word. The other versions are artifact of the original word as some authors preferred to name (especially in the internet).
Barycenter (or Barycentre in British English) is a commonly used term in astronomy and in chemistry (in Crystal Field Theory) as well. In both fields, it is used to explain center of mass balance and most of the literature has used the aforementioned spelling. For example, NASA explains the question "What Is a Barycenter?" as:

We say that planets orbit stars, but that’s not the whole truth. Planets and stars actually orbit around their common center of mass. This common center of mass is called the barycenter. Barycenters also help astronomers search for planets beyond our solar system!

In Crystal Field Theory (CFT), electrons in the $\mathrm{d}$-orbitals are repelled to different degrees by the ligands. The degeneracy of the $\mathrm{d}$-orbitals is lifted. The Barycenter is the average energy of the $\mathrm{d}$-orbitals (Coordination Chemistry II: Theories of Electronic Structure and Crystal Field Theory):


Is there a more direct way to show this identity for the projection operator? The following question I jotted down from an online quantum chemistry pdf.

Show that
$$i\hbar \frac{\partial P_j}{\partial t}= [H,P_j]$$

in which $P_j$ is the projection operator and $H$ is the Hamiltonian. We are given the usual time-dependent form of Schrodinger's equation.
What I did: I supposed that this was an application of Ehrenfest's theorem because it involved a commutator and derivatives. $P_j$ does not explicitly involve time and so Ehrenfest's theorem reduces to$^*$
$$\frac{d}{dt}\langle\psi|P_j|\psi\rangle=\frac{1}{i\hbar}\langle\psi|[H,P]|\psi\rangle $$
But assuming I think w.l.o.g. that $\psi$ is normalized we can pre-multiplyby $|\psi\rangle$ to get (using $|\psi\rangle\langle\psi|=1$)
$$i\hbar\frac{d}{dt}P_j|\psi\rangle~= [H,P_j]|\psi\rangle$$
If this works (I don't assume that), my question is whether there is a direct way to do this, using the definition of $P_j$, and if so whether it would not also be an appeal to this theorem?
$^*$ Using eq. (9) from this note.

Before I go into how this result can be derived, I want to point out a mistake in your attempt. $|\psi\rangle\langle\psi|\neq1$, but rather $\sum_i|i\rangle\langle i|=1$, where the sum of is over a complete set of states. Basically, this result only holds when you sum over all the basis vectors that span the space you are studying, not just for any single vector in the space. Its easiest to understand this in analogy to more conventional vectors. Assuming a column vector $v$ is normalized, than $v^\dagger v=1$, which is the inner/dot product of the vector with itself. This is analogous to $\langle\psi|\psi\rangle=1$ in bra-ket notation. If we reverse the order of the product, $vv^\dagger=\mathbf A$ where $\mathbf A$ is some matrix. This operation is called an outer product and is akin to $|\psi\rangle\langle\psi|$. As it turns out, for a complete orthonormal basis, the sum of these outer products has to equal the identity matrix $\mathbf1$. We can make sense of this by recognizing that the sum is a sum of projectors onto all the basis vectors, which in QM might be all of the possible eigenstates of the system. Your wavefunction has to be some linear combination of these basis vectors, so if your project your wavefunction onto all the eigenstates and then sum those projections, you should get your original wavefunction back, hence the sum of the all the projectors is the identity matrix.

For the actual derivation, the result proceeds directly from the time dependent Schrodinger equation (TDSE) $$i\hbar\frac{\partial\psi}{\partial t}=H\psi\to i\hbar\frac{\partial|{j}\rangle}{\partial t}=H|j\rangle$$

where in the second half of the equation, I have written the equation for the state $j$ that the projection operator references. Writing the projection operator explicitly as $P_j=|j\rangle\langle j|$, we can rewrite your original
expression as $$i\hbar\frac{\partial(|{j}\rangle\langle j|)}{\partial t}=i\hbar\bigg[\frac{\partial|{j}\rangle}{\partial t}\langle j|+|j\rangle\frac{\partial\langle{j}|}{\partial t}\bigg]$$ where the equality is due to the product rule. The first term contains $i\hbar\frac{\partial|{j}\rangle}{\partial t}$ which we know from the TDSE equals $H|j\rangle$, so the first term becomes $H|j\rangle\langle j|=HP_j$. To find the other term, we take the complex conjugate transpose of the TDSE to obtain $-i\hbar\frac{\partial\langle j|}{\partial t}=\langle j|H$, which leads to the second term being $-P_jH$ and thus $$i\hbar\frac{\partial P_j}{dt}=[H,P_j]$$


What is the IUPAC preferred name of this tetracyclic natural product? On page 819 of Vollhardt, Organic Chemistry 7th Ed., the following "subunit of the antibacterial platensimycin" appears. I have attempted to apply von Baeyer and skeletal replacement ('a') nomenclature but am stuck deciding between numberings. Here are some vector drawings:

Is von Baeyer the preferred nomenclature for this molecule? and, if so, what is its correct numbering and preferred IUPAC name?

The pubchem website has given Platensimycin IUPAC name, $3$-$[3$-$[(1S,5S,6R,7S,9S,10S)$-$5,9$-dimethyl-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-en-$5$-yl]propanoylamino]-$2,4$-dihydroxybenzoic acid (see attached picture):

Accordingly, since https://pubchem.ncbi.nlm.nih.gov is usually updating data according to the changes, I believe the name of the cyclic system (without 5,9-dimethyl groups) OP is asking is same as one in the comment by @KarstenTheis (including the correct stereochemistry): $(1S,5S,6R,7S,9S,10S)$-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-ene (see the blue numbering in the picture).

Late correction: After exchanging few comments with Loong, I realized that the given system should be treated as a chemical compound. Accordingly, the name should have been: $(1S,5S,6R,7S,9S,10S)$-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-en-$4$-one (as suggested by Loong).


Why do hairs (rubbed with plastic comb) fallen on water seems to be sticking? I was observing from many days, how hairs seems to be sticking to water, when made to fall on it. I made an attempt to know what really was going on. I kept a jug of stationary water under the light, and kept my dry hairs combing, to make hairs fell on it. I observed once again that hairs appeared to be sticking to water.

I started thinking about the cause for this observation. At first, what went to my mind was about hairs being rubbed with comb. Usually hairs get positively charged (I don't know exactly. I thought it because plastic becomes negatively charged when rubbed with most of  other suitable surface) when rubbed with plastic comb. We also know that water has more negative charge because of oxygen atom (I came to know about this from the diagram of wiki encyclopedia-electronegativity). So, I thought positively charged hairs might be attracted by negative charge nature of water molecules, which might be making hairs to stick with water.

Is it really because of positively charged hairs and negatively charged water molecules, which is making hairs to be sticking to water?

It doesn't have to do as much with the combing as it does with surface tension. Water on the hair clumps them together, because smaller clumps have more surface area (net surface area) and thus more surface energy. Breaking these large clumps, therefore, requires energy input, which gives us the impression that they "stick" together.

As for sticking to the water individually, that's surface tension again. The surface is able to withstand a bit of weight before collapsing and letting the object drop through. Hairs are oily, and thus hydrophobic. The repulsion potential energy that would be formed / needed when it sinks outweighs the energy required to bend the surface (and get enough tension force to balance gravity). This is how water skeeters work.


What is the solvent and what is the solute? This question is and I am learning about mixtures, solutions, solutes, solvents, and other terms related to these words.

I know that a solute is the substance being mixed with or dissolved into the solvent.

Substances and mixtures: are these solvents and solutes correct?

coffee with milk - solvent:coffee , solute:milk
salt water - solvent:water , solute:salt
air freshener - solvent:air , solute:scent
brass - solvent:zinc , solute:copper

In a binary homogeneous mixture, the component present in the larger amount is generally called the "solvent". The component present in the smaller amount is generally the "solute".

In your first example, however, it is more complicated as coffee is already a complex solution (but mainly water). Milk is not a homogeneous mixture. It is a fat/protein emulsion (also mainly water).

The other three are solutions: salt water is a solution of a solid dissolved in a liquid (which is the most familiar type), air freshener (one dispersed) is a solution of one gas in others (but air itself is a mixture), and brass is a solid solution.


What is responsible for making the bonds identical? Here is an MO diagram of methane found on the internet:

There are 4 bonds here, one involving the 2s of carbon, three involving the 2p of carbon, and they have different energies.

However, I am told that the four bonds are actually equal.

Then what made them equal? Also, I am told that hybridisation does not exist.

The hydrogens are all equivalent. When thinking about MOT, you have to throw almost everything you thought you knew about bonding out the window. The electrons are not localized between any two pair of atoms—they are truly molecular orbitals. This graphic nicely shows the phasing of the molecular orbitals of methane:

$\hspace{4cm}$

[source]

The lowest energy orbital is a linear addition of the carbon $\mathrm{2s}$ orbital and all four of the hydrogen $\mathrm{1s}$ orbitals. The pair of electrons occupying this orbital is not specific to any one hydrogen, and bonds all the atoms in methane together (even the hydrogens to the other hydrogens). The other bonding orbitals are of degenerate energy and result from linear combinations of the carbon $\mathrm{2p}$ orbitals and the hydrogen $\mathrm{1s}$ orbitals.

Naturally a molecular orbital that results from a combination of a $\mathrm{s}$ orbital will be of lower energy than one than one that results from a $\mathrm{p}$ orbital.


Heterolytic vs homolytic cleavage of the C−H bond in cyclopentadiene Consider the cleavage of the indicated hydrogen in cyclopentadiene:

The hydrogen indicated will be easily removed as $\ce{H^.}$ or $\ce{H^+}$. Assume a nonpolar medium.

My question was that would aromaticity give enough stabilization for hydrogen to be more easily removed as $\ce{H+}$?

The cleavage of the C-H bond could be envisioned to take place in two steps:

Homolytic cleavage to hydrogen atom plus cyclopentadienyl radical
Ionization of the hydrogen atom to $\ce{H+}$ and addition of the electron to the cyclopentadienyl radical

Ionization of the hydrogen atom requires 313.8 kcal/mol. Aromatic stabilization, calculated from unexpected differences between calculated and experimental heats of hydrogenation, yield an "extraordinary 36 kcal/mol" greater stability than expected. Well, 36 kcal/mol could be extraordinary under certain circumstances, but compared to an ionization energy, it's not very large.

The molecular orbital situation for the cyclopentadienyl radical isn't exactly non-aromatic, it's just not as fully aromatic as it could be. The five pi electrons of the radical are all bonding, but there's room for one more bonding electron. The question is whether the extra stabilization of that electron is sufficient to cause the ionization of the hydrogen atom. A typical carbon-carbon single bond has a dissociation energy of about 100 kcal/mol and a pi bond would be a bit less. A bond would involve two electrons, so one electron would be responsible for about half that, or less than 50 kcal/mol.

So between the "extraordinary" stabilization of benzene (36 kcal/mol) and the extra bond energy of the sixth electron (50 kcal/mol), I don't think there is enough energy available to ionize the hydrogen atom. Homolytic cleavage would be easier than heterolytic. However, if you provide an ionic solvent that stabilizes the proton and cyclopentadienyl anion, circumstances could change completely.


Are CuCl and Cu2Cl2 equivalent? I have seen $\ce{CuCl}$ and $\ce{Cu2Cl2}$ used interchangeably; they seem to be equivalent. If that is the case, then why does the latter have the '$2$' subscripts?

$\ce{Cu2Cl2}$ is actually the dimer of cuprous(I) chloride $\ce{CuCl}$. Cuprous chloride dimerises and also trimerises and tetramerises in gas phase. The molecular structures of the species are measured with Gas-Phase Electron Diffraction and Quantum Chemical Calculations. Some information about the various species of cuprous chloride is given here:[1]

  The molecular geometry of gaseous cuprous chloride oligomers was determined by gas-phase electron diffraction at two different temperatures. Quantum chemical calculations were also performed for $\ce{Cu_nCl_n (n=1–4)}$ molecules. A complex vapor composition was found in both experiments. Molecules of $\ce{Cu3Cl3}$ and $\ce{Cu4Cl4}$ were present at the lower temperature (689 K), while dimeric molecules ($\ce{Cu2Cl2}$) were found in addition to the trimers and tetramers at the higher temperature (1333 K). All $\ce{Cu_nCl_n}$ species were found to have planar rings by both experiment and computation. The bond lengths from electron diffraction ($r_\mathrm g$) at 689 K are 2.166±0.008 Å and 2.141±0.008 Å and the $\ce{Cu-Cl-Cu}$ bond angles are 73.9±0.6° and 88.0±0.6° for the trimer and the tetramer, respectively. At 1333 K the bond lengths are 2.254±0.011 Å, 2.180±0.011 Å, and 2.155±0.011 Å, and the $\ce{Cu-Cl-Cu}$ bond angles 67.3±1.1°, 74.4±1.1°, and 83.6±1.1° for the dimer, trimer, and tetramer, respectively.

Reference

(1) Hargittai, M.; Schwerdtfeger, P.; Réffy, B.; Brown, R. The Molecular Structure of Different Species of Cuprous Chloride from Gas-Phase Electron Diffraction and Quantum Chemical Calculations. Chem. Eur. J. 2003, 9 (1), 327–333. DOI: 10.1002/chem.200390027


Algebraic treatment of equilibrium This is from Laitinen and Harris, "Chemical Analysis, An Advanced Text & Reference" which describes a simple acid-base equilibrium. They derive a relation in the end directly. I haven't seen this expression in many books which teach solution equilibria. The authors cite an article from the Journal of Chemical Education for the expressions (4 & 5). I didn't find article very useful. How should we derive the equations 4 and 5 in a better way than the author's explanation in italics?

The authors start from very simple equations:

  $\ce{HA + H2O <=> H3O^+ + A-}\tag{1}$

  Conventionally,

  $\ce{HA <=> H+ + A^-}\tag{2}$

  and the ionization of water,
  $\ce{H2O <=> H+ + OH-}\tag{3}$

  If $C_{HA}$ and $C_{A^-}$ are the analytical concentrations of $\ce{HA}$ and $\ce{A^-}$, and if $\ce{[HA]}$ and $\ce{[A^-]}$ are the equilibrium concentrations, the relations...

  $\ce{[HA]} = C_{HA} - \ce{([H^+] - [OH^-])}\tag{4}$

  $\ce{[A^-]} = C_{A^-} + \ce{([H^+] - [OH^-])}\tag{5}$

The authors explain ...

"result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration."

There is only one source of hydroxyl ions, ionization of water (Eq. 3 in the OP). Ionized water contributes equal amounts of hydroxyl and hydronium ion. Similarly, dissociation of acid $\ce{HA}$ results in equal amounts of hydronium and of conjugate base $\ce{A-}$. Unless a salt of the conjugate base is explicitly added then the acid is the only source of $\ce{A-}$. In absence of salt the total hydronium concentration in solution is therefore

$$\ce{[H+]}=\ce{[OH-]}+\ce{[A-]}$$

Note that this equation reflects the charge balance condition.

Solving for $\ce{[A-]}$

$$\ce{[A-]} = \ce{[H+]}-\ce{[OH-]}\tag{1}$$

We also know that the original amount of acid equals the sum of undissociated acid and conjugate base:

$$C_\ce{HA}=\ce{[HA]}+\ce{[A-]}$$

Note that this equation reflects the mass balance condition.

Solving for $\ce{[HA]}$

$$\ce{[HA]}=C_\ce{HA}-\ce{[A-]} \tag{2}$$

Substituting Eq. (1) into Eq. (2) we obtain

$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]}) \tag{3}$$

The derivation that lead to this expression is not general. It was assumed that no conjugate salt was added to the solution ($C_\ce{A-}=0$). If a completely dissociating salt $\ce{MA}$ is added at concentration $C_\ce{A-}$ then it is necessary to modify the mass balance expression:

$$C_\ce{HA}+C_\ce{A-}=\ce{[HA]}+\ce{[A-]}$$

and the charge balance expression:

$$\ce{[H+]}+\ce{[M+]}=\ce{[OH-]}+\ce{[A-]}$$

or since $\ce{[M+]}=C_\ce{A-}$

$$\ce{[H+]}+C_\ce{A-} =\ce{[OH-]}+\ce{[A-]}$$

Solving for $\ce{[A-]}$

$$\ce{[A-]} = C_\ce{A-} + \ce{[H+]}-\ce{[OH-]}$$

Combining the two balance expressions leads once again to Eq. (3):

$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]})$$

By the way, the quote you cite should read:

"result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration."

It might be that "typo" ("3" written instead of "2") that led to the confusion.


Thiosulphate acidic decomposition I just came across this reaction

$\ce{Na2S2O3 + 2HCl -> 2NaCl + SO2 + S + H2O}$

I was wondering if it was possible to predict the products of this reaction if I didn't know it (or any inorganic reaction for that matter).

Can we use arrow-pushing (as we do in organic chemistry) to get the products of an inorganic reaction? Can we devise a mechanism to find the products? It seems unrealistic to me how anyone could learn the many reaction of inorganic chemistry.

The general answer to this question is no, one cannot simply predict the products of any reaction.  How do we know that one of the products is not $\ce{S2O6^{2-}}$ or perhaps even $\ce{S2O5Cl2^{2-}}$?

The best that can be done without any guidance from experiment is to compute the thermodynamically favored products.  One does this (at standard temperature and pressure) by writing down all plausible products and then computing the Gibbs free energy ($\Delta G$) for the reactions.  The one with the smallest $\Delta G$ is the thermodynamically preferred.

But be warned:  there are mechanistic and kinetic factors to take into account.  The thermodynamically favored reaction may, as a practical matter, not take place.  The standard example is diamond --> graphite, which is thermodynamically allowed, but don't hold your breath waiting for it to happen.


Clarification on vapor pressure If I have a glass of water, and the vapor pressure is the point of equilibrium where the liquid molecules are becoming gases and the gas molecules are colliding and becoming liquid at the same rate, does that mean none of the liquid is lost? I feel as if I am misunderstanding this concept because when I leave my cup of water on the table and come back to it later on I usually notice that there's less water in the cup because it evaporated.

Yes, the vapor pressure is the point of equilibrium between the liquid and the vapor. But air moves about so the liquid never saturates the air and an equilibrium is never reached. So more and more water evaporates until it is all gone.


Synthesis of diethyloxalate from ethene Outline the steps involved in the following transformation: $\ce{C2H4}$ to $\ce{(COOC2H5)2}$ (diethyl oxalate).

I am familiar with the transformation of $\ce{C2H4}$ to ethyl acetate $\ce{CH3COOC2H5}$ by first converting to ethanol using water and acid, then using an oxidising agent such as $\ce{KMnO4}$ to convert ethanol to ethanoic acid and subsequently reacting ethanol and ethanoic acid together.
I'm not sure if the process needed to get to the diethyl oxalate is similar to this.

Conversion  1

Convert ethene to ethanol.
Ethene is mixed with steam and passed over a catalyst consisting of solid silicon dioxide coated with phosphoric(V) acid. The temperature used is 300 °C and the pressure is about 60 to 70 atmospheres (chemguide —
The mechanism for the acid catalysed hydration of ethene).
$$\ce{CH2=CH2 + H2O <=>[H3PO4] CH3CH2OH}$$

React the ethanol with oxalic acid to get diethyl oxalate.
$$\ce{(COOH)2 + 2C2H5OH ->[CCl4,\Delta] (COOC2H5)2 + 2 H2O}$$
The details of this reaction is given on Organic Syntheses — Ethyl Oxalate and in the paper by Jewel and Butts [1].

Conversion 2
Convert ethene to oxalic acid by first converting to ethyne and then making oxalic acid from it. Use the oxalic acid to make diethyl oxalate by reacting it with absolute alcohol.
$$\ce{CH2=CH2 ->[Br2][CCl4] Br-CH2-CH2-Br ->[alc. KOH][\Delta] CH#CH ->[alc. KMnO4] (COOH)2}$$
Reference

Jewel, P. W.; Butts, J. S. A Method for the Preparation of Diethyl Oxalate. J. Am. Chem. Soc. 1931, 53 (9), 3560–3561. DOI: 10.1021/ja01360a052.


Cyclopentadiene/etc. hydrophobicity after nitrogen atom substitution at a double bond? If I take a compound like cyclopentadiene ($\ce{C5H6}$) and I substitute a nitrogen for a carbon at one of the $\ce{C=C}$ bonds (making a $\ce{C=N}$ bond), how does this typically change the hydrophobicity of the compound?

The problem is that I'm not really a chemist.  My thought here is that nitrogen is more electronegative than carbon, so this will polarize the bond, induce a dipole, and make the molecule more hydrophilic.  In general, it seems to me that mixing together atoms of differing electronegativity in alkane or alkene/etc. compounds should create dipoles that can interact with polar solvent molecules, improving hydrophilicity.

As an additional thought, how would the nitrogen substitution effect stacking interactions between the substituted ring and a regular molecule of cyclopentadiene?  My guess here would be that hydrophobic effects would cause a more favorable pi-stacking interaction between two unsubstituted copies of cyclopentadiene relative to cyclopentadiene and the aforementioned and described nitrogen substituted version of the molecule.

To clarify, by "stacking interactions" I mean the sort of $\ce{C=C}$ stacking interactions that lead to cycloaddition reactions, not that the two rings will lie on top of one-another (as in the case of something like a pair of benzene molecules).

The cyclic unsaturated imine (Schiff base) will be more hydrophilic for the nitrogen lone pair hydrogen bonding and being a Lewis base toward water.  You do know that CpH [4 + 2} Diels-Alder reaction reversibly thermally dimerizes, right?  The endo-dimer is an exo-flip then hydrogenation away from JP-10 cruise missile fuel.

The cyclic imine, ${2H}$-pyrrole is an isomer of pyrrole.  One expects it will be labile toward proton migration in base.

http://www.chemspider.com/Chemical-Structure.7330.html
http://www.chemspider.com/Chemical-Structure.5256848.html
Click the Properties tab, etc.


Energy balance in battery thermodynamics I am trying to understand the derivation of a general energy balance in battery thermodynamics. The following relation is frequently found to determine the heat generation of a battery:

$\dot{Q} = \dot{Q}_\text{rev} + \dot{Q}_\text{irrev} = IT\,\frac{\mathrm dE_0}{\mathrm dT} + I(E-E_0)$

where $\dot{Q}$ is the heat generation, $I$ current, $T$ temperature and $E$ cell voltage. The index $0$ denotes the open circuit voltage. Furthermore, $\Delta S = zF\,\frac{\mathrm dE_0}{\mathrm dT}$, with $F$ as Faraday constant and $z$ as number of exchanged electrons.

The derivation is as follows:

First law of thermodynamics: $\mathrm dU = \mathrm dQ - \mathrm dW \tag{1}$ with $\mathrm dW = p\,\mathrm dV + \mathrm dW_\mathrm{el}$

$\mathrm dH = \mathrm dU + p\,\mathrm dV + V\,\mathrm dp \tag{2}$

Substituting (2) in (1):

$\mathrm dH - p\,\mathrm dV -V\,\mathrm dp = \mathrm dQ - p\,\mathrm dV - \mathrm dW_\mathrm{el} \tag{3}$

with $\mathrm dp=0$ follows:

$\mathrm dH = \mathrm dQ -\mathrm dW_\mathrm{el}$, differentiation with respect to time:

$$\frac{\mathrm dH}{\mathrm dt} = \frac{\mathrm dQ}{\mathrm dt} - \frac{\mathrm dW_\mathrm{el}}{\mathrm dt} = \dot{Q} - EI$$

So far, everything is fine. But now:

$$H = G + TS \rightarrow \frac{\mathrm dH}{\mathrm dt} = \frac{\mathrm dG}{\mathrm dt} + T\,\frac{\mathrm dS}{\mathrm dt} + \frac{\mathrm dT}{\mathrm dt}S$$

Here my first problem in understanding arises: In general $\frac{\mathrm dG}{\mathrm dt}$ and $\frac{\mathrm dT}{\mathrm dt}S$ cancels each other, since, $S=-\frac{\mathrm dG}{\mathrm dT}$ and $\frac{\mathrm dG}{\mathrm dt}$ can be expanded with $\mathrm dT$: $\frac{\mathrm dT}{\mathrm dt}S = -\frac{\mathrm dT}{\mathrm dt}\frac{\mathrm dG}{\mathrm dT}$. This means, that the influence of the open circuit voltage in the given energy balance would vanish.

I assume, that this is prevented by simply stating isothermal conditions: $\frac{\mathrm dT}{\mathrm dt}S = 0$. But this makes no sense to me, since the whole purpose of this calculation is the temperature increase with time during cycling a cell. Can somebody explain this to me?

Now if $\mathrm dT=0$ the derivation would proceed like this:

$$\frac{\mathrm dG}{\mathrm dt} + T\,\frac{\mathrm dS}{\mathrm dt} = \dot{Q} - EI$$

$$\frac{\mathrm d(-znFE_0)}{\mathrm dt} + T\,\frac{\mathrm d\left(znF\,\frac{\mathrm dE_0}{\mathrm dT}\right)}{\mathrm dt} = \dot{Q} - EI$$
where $znF=C$ is the charge and $\frac{\mathrm dC}{\mathrm dt} = I$

And here my second problem arises: to get to the equation at the very beginning of this post, it is necessary to set $E_0$ and $\frac{\mathrm dE_0}{\mathrm dT}$ constant, so that the chain rule does not apply. And I do not understand why this should be valid?

Can anybody help me with this?

The answer to this problem explains how to arrive at

$$ dQ_p = - dw_{ele} +  dG + TdS $$

This expression is generally true for electrical work at constant T and p. From it the differential equation in the OP can be derived by taking the time derivative on both sides and making appropriate substitutions.

  $\frac{\mathrm dG}{\mathrm dt}$ can be expanded with dT:  $\frac{\mathrm dT}{\mathrm dt}S = -\frac{\mathrm dT}{\mathrm dt}\frac{\mathrm dG}{\mathrm dT}$

But remember that you defined

$G=H-TS$

and you should use that when computing $\frac{\mathrm dG}{\mathrm dt}$. But introducing that just results in a circular argument (a lot of terms cancelling to no use).

  to get to the equation at the very beginning of this post, it is necessary to set $E_0$ and $dE_0/dT$ constant

That's ok. $E_0$ is the voltage when no current is running (that is, battery is not being depleted, as there is no reaction, and therefore composition is constant over time), and is thus constant with respect to time.

  But this makes no sense to me, since the whole purpose of this calculation is the temperature increase with time during cycling a cell.

That's tricky and I do not have a complete answer. The equation derived is for a specific T. To solve for the time-dependence of the temperature evidently requires additional steps, presumably incorporating expressions for the heat capacity.


What does the value of the pH meter mean when there aren't any hydronium ions in the substance being tested? What does the value of the pH meter mean when there aren't any hydronium ions in the substance being tested?
How does the pH meter show a value which fluctuates as non-aqueous solvents are being tested? I thought that the pH meter only measured hydronium ions. Is there any way that this value actually means something in terms of acidity and basicity?

Nowadays you have to differentiate between pH measurements with non-glass sensors and with glass sensor. In the typical pH meter with a glass sensor, there could be many things that could go wrong resulting in biased results. In general, some electrodes may be sensitive to interfering ions such as $\ce{Ag+}$. So some potential could be generated also by other ions not only hydronium ions. However, the most common thing is that you are actually reading biased measurements because you are perturbing the status of your instrument.

Interaction of the solvent with the materials of your pHmeter

First thing you have to consider is the interaction of the solvent with the materials of your pH meter. You can easily damage some part of it or dehydrate the glass membrane of your electrode.

Solubility of the electrolyte in the junction

Then you have to bear in mind that in your glass pHmeter there is a junction that leaks an electrolyte for forming the saline bridge if this electrolyte is not soluble in your solvent the "circuit" will not close and you will have a wrong measurement.


Are there other Density Functional Theories other than Kohn Sham DFT? For some reason I thought that the term DFT and Kohn Sham DFT were interchangeable. Kohn Sham DFT is certainly one of the most popular. But are there other types of DFT that are in use today? I have been able to find orbital free DFT is gaining momentum (http://www.sciencedirect.com/science/article/pii/S0022509607000221)
But are there more DFT types?

KS-DFT is the basis for all orbital-based DFT Formulations. In it you have different specialities (constricted DFT, TD-DFT, rs-DFT to name a few).

KS-DFT is used for molecular problems, because of atom bonds etc.

TD-DFT is used for time-dependend problems like excitation energies, polarizabilities etc.

RS-DFT is using range-separated functionals (long-range, short-range) to handle effects better in the long-range domain.

Addition by @pentavalentcarbon:
Current cDFT implementations are based on KS-DFT and perform a real-space partitioning of the electron density.

Plane Wave DFT (no orbitals, electron density calculated with plane waves, was the first DFT) is used for periodic problems like metals etc.

DFTB (Density Functional Tight Binding) is good for Molecular Dynamics since its an approximate DFT and allows handling of larger systems and greater timescales.
Addition by @pentavalentcarbon:
A good analogy for DFTB would be that it is similar to semiempirical methods (PM7, other NDDO parametrizations) where certain integral matrix elements are replaced with precomputed, fitted parameters, but there is a minimal basis set present.

KS-DFT is dominant because as computational chemists we are mostly working with molecules. As that we need a simulation of atom bonds. There is a good Book about DFT which I can only recommend (A chemists guide to Density Functional Theory): http://www.goodreads.com/book/show/70706.A_Chemist_s_Guide_to_Density_Functional_Theory


Solubility of compounds in water Why is $\ce{NaCl}$ more soluble in water than $\ce{NaOH}$? Well, this is a question from Chemical Equilibrium and the question hints for explanation in terms of Le Chatelier's principle.

It is not. $\ce{NaOH}$ is more soluble ($>100~\mathrm{g}/100~\mathrm{ml}$) then $\ce{NaCl}$ ($<30 ~\mathrm{g}/100~\mathrm{ml}$). $\ce{NaOH}$ when dissolved produces $\ce{OH^-}$ anion, that has relatively small size and forms specific hydrogen bonds (where it is bonded to relatively small positively charged hydrogen) very well. In solid $\ce{NaOH}$ it has to form hydrogen bonds with quite big $\ce{Na^+}$ cation.


Pseudo 1st order Kinetic reaction The pseudo 1st order reaction happens when one of the reactant is in large excess and its concentration doesnt change during the reaction .

Is there any other such situation when a certain higher order reaction turns out to become a pseudo first order kinetic reaction (except the situation where an N$^{th}$ order reaction has N-1 species in large excess) ?

Example from biochemistry

The Michaelis-Menten equation for enzyme kinetics is an example.

$v=\frac{k_{cat}E_0S}{K_S+S}$ where $S$ is substrate concentration and $E$ in enzyme concentration.

At low $S$ ($S\ll K_S$), the reaction is 2nd order with $v=\frac{k_{cat}}{K_S}E_0S$, i.e. first order in both enzyme and in substrate.  However at high $S$ ($S\gg K_S$), the reaction is first order only in enzyme with $v=k_{cat}E_0$, and is zero-order in substrate.  The substrate concentration doesn't have to be in molar excess to other reactants.

Example from process control

In chemical engineering, nonlinear systems such as higher-order reactions occuring in a continuous process at steady-state are often modeled using linear (which you can think of as first-order reactions) systems as an approximation for developing process control techniques.  A MATLAB documentation page has some good information.

Another example, somewhat contrived

As a last example, consider an unstable molecule $A$.  It can decompose to $B$ by two entirely separate elementary mechanisms.  The first is $\ce{A + A -> B + B}$.  The second is $\ce{A -> B}$.  At high concentrations of $A$, the second-order kinetics will dominate and the decomposition of $A$ will be second order.  But lower the concentration enough, and the reaction will become first order in $A$ because the rate the bimolecular reaction will be dwarfed by the unimolecular route at low $A$.