How to determine conformation using IR In my advisor's Ph.D. research, she used IR spectroscopy to (help) determine the conformation of a compound of interest (its structure is known.)

In organic chemistry, I learned about using IR as a way to identify functional groups (and in conjunction with HNMR, CNMR, & mass spec, I was able to determine the structure of a compound.)

How can one determine conformation using IR? What are the limitations of this? I feel like it could get messy really quickly.

This is perhaps my biased view as a theorist, but I've collaborated quite a bit with a condensed-phase 2D-IR group and a little with a UV resonance Raman group, so I have some understanding of the experimental perspective as well.

For anything outside of the simplest models, like a 2- or 3-state system you can scribble on an envelope, or the simplest molecules that would display configurational dynamics, like short peptides or single amino acids, experiment strongly benefits from computation. Have fun searching the literature for tables of IR peak frequencies, looking at the assigned normal modes, (which are sometimes compound motions), and then seeing if any of those match your hopefully very similar structure. If the molecular structures are known, then for simple cases a few of the conformers can be made by hand using your favorite computer program or a partially-automated conformer search, have their geometries optimized, and harmonic vibrational frequencies calculated to at least qualitative accuracy in an afternoon. Then it's a matter of clicking around in a visualization tool to see which interesting peaks correspond to which normal modes.

If populated conformers are harder to find, such as in systems that are larger with many rotatable bonds or have substantial librational motion, an exhaustive conformer search may be too expensive in terms of computer time, and running molecular dynamics trajectories with some sort of post-trajectory clustering analysis is an alternative, since IR and Raman spectra can be generated from trajectories via velocity, dipole, and polarizability autocorrelation functions.

Example

Here is an example where a lithium cation is interacting with L-proline in two ways: 1. with the backbone-bound carboxylic acid, and 2. with the ring. Perhaps this is cheating, because it isn't a single molecule, it's two different coordination modes, but you would see something similar with an intramolecular conformational change or rearrangement such as keto-enol tautomerization.

Here are their respective vibrational spectra within the harmonic approximation.

Already without any assistance, we can see huge changes outside of the fingerprint region. This is similar to how hydrogen bonding in water changes the IR spectra when moving from molecules to the bulk. The three most notable changes going from backbone- to ring- bound are

a blueshift (increase in energy) of the carbonyl peak, moving it from 1760 to 1852 wavenumbers,
the intensity increase of the ring $\ce{C-H}$ stretches below 3000 wavenumbers, and
a few changes above 3500 wavenumbers.

1 is probably due to charge transfer into an antibonding orbital localized on the carbonyl group. However, it's also strongly coupled to the $\ce{-OH}$, so interactions there will have an indirect effect. I'm less sure about the $\ce{C-H}$ stretches, but they belong to the three hydrogens closest to the $\ce{Li+}$ (7, 11, and 14), and Mulliken charges don't support a CT idea, which isn't surprising and doesn't rule it out. For number 3, the $\ce{-OH}$ stretch moves from 3686 to 3439. This redshift is probably a result of the blueshift to the coupled mode; assuming that the total vibrational energy between conformers is similar ($E_{\text{ZPE}} = 90.100$ vs. $\pu{90.182 kcal/mol}$ here), vibrational energy tends to be redistributed evenly and in ways that make physical sense; see here for a discussion about how this simplified two-mode coupling works which might also be applicable here.

Frequencies are in wavenumbers (cm$^{-1}$) and intensities are in units of km/mol. Modes below 1200 wavenumbers are commented out.

$$
\begin{array}{rrrr}%
\hline%
\text{backbone frequencies} & \text{backbone intensities} & \text{ring frequencies} & \text{ring intensities} \\
\hline%
%% 30.91 & 2.37 & 62.75 & 4.92 \\
%% 88.36 & 6.77 & 91.13 & 2.54 \\
%% 168.98 & 24.56 & 146.67 & 15.42 \\
%% 200.22 & 35.17 & 239.15 & 53.54 \\
%% 301.33 & 5.59 & 299.49 & 10.23 \\
%% 323.29 & 67.62 & 399.53 & 47.70 \\
%% 393.10 & 18.41 & 436.61 & 77.33 \\
%% 423.78 & 49.21 & 454.79 & 10.29 \\
%% 485.16 & 69.90 & 517.44 & 4.46 \\
%% 534.48 & 56.37 & 541.24 & 57.13 \\
%% 565.71 & 45.28 & 582.68 & 14.18 \\
%% 625.82 & 27.87 & 622.71 & 9.03 \\
%% 663.98 & 12.16 & 703.37 & 26.78 \\
%% 754.48 & 22.99 & 758.60 & 12.72 \\
%% 802.57 & 16.27 & 808.16 & 50.74 \\
%% 842.01 & 5.53 & 823.92 & 48.36 \\
%% 858.14 & 14.85 & 863.84 & 7.69 \\
%% 875.87 & 27.62 & 875.21 & 13.13 \\
%% 897.13 & 13.53 & 912.31 & 2.55 \\
%% 919.58 & 0.12 & 928.34 & 12.11 \\
%% 966.10 & 12.84 & 952.79 & 7.90 \\
%% 1007.42 & 0.12 & 1017.17 & 0.53 \\
%% 1034.74 & 23.77 & 1056.15 & 1.11 \\
%% 1070.65 & 26.45 & 1116.38 & 40.56 \\
%% 1154.24 & 13.04 & 1119.11 & 9.28 \\
%% 1161.61 & 3.40 & 1148.10 & 7.38 \\
%% 1181.00 & 74.31 & 1173.28 & 7.56 \\
1213.61 & 4.92 & 1203.94 & 6.81 \\
1222.70 & 2.73 & 1228.83 & 4.07 \\
1263.58 & 152.20 & 1261.21 & 39.44 \\
1279.48 & 0.73 & 1272.66 & 12.99 \\
1314.79 & 1.01 & 1321.29 & 57.22 \\
1328.39 & 58.17 & 1342.94 & 314.05 \\
1366.92 & 47.01 & 1365.77 & 74.02 \\
1390.24 & 127.09 & 1402.33 & 19.65 \\
1424.08 & 13.57 & 1522.07 & 16.68 \\
1432.59 & 5.88 & 1561.70 & 0.40 \\
1460.15 & 3.30 & 1572.71 & 24.88 \\
1760.04 & 284.35 & 1851.86 & 270.80 \\
2976.69 & 25.42 & 2607.42 & 324.10 \\
2983.82 & 7.16 & 2806.11 & 147.60 \\
2993.71 & 4.29 & 2900.28 & 54.24 \\
3000.08 & 8.94 & 3024.05 & 5.13 \\
3039.46 & 13.36 & 3027.04 & 2.76 \\
3063.00 & 5.22 & 3061.92 & 2.04 \\
3075.69 & 4.87 & 3074.89 & 3.80 \\
3477.77 & 29.07 & 3438.98 & 235.23 \\
3685.91 & 93.01 & 3527.08 & 36.70 \\%
\hline%
\end{array}%
$$

All calculations were performed using B97-D/def2-SVP. See source for XYZ structures.


How to quickly cool a test tube without cracking it? Is there a way to safely cool a heated test tube quickly without cracking it due to sudden thermal stress in the glass? One thing that I tried was blowing on it, cooling it in slightly lukewarm water followed by cold water but this method kind of involves a "hope for the best factor". Is there a better technique?

Rather than working on your technique, check whether other test tubes are available.

Apparently, a glass-like material with a very low thermal expansion coefficent would be a good choice.

The glassblowing shop at the University of Delaware provides a nice overview of the possible materials and their properties, and there are definitely additional online resources, e.g. by manufacturers like Schott, Corning, etc.

Here's a summary (coefficients are given in $cm/cm/°C$):

soda lime glass $89 \times 10^{-7}$

Pyrex, Duran $32.5 \times 10^{-7}$

Vycor  $7.5 \times 10^{-7}$

fused quarz $5.5 \times 10^{-7}$

Please note that quarz is both pretty expensive and demands special care! Alkaline solutions and quarz are a bad conbination!


Why do ionic compounds have such high melting points and boiling points? Do they have IMAFs? Molecules have IMAFs that determine the melting and boiling points. What about with ionic substances? What keeps the different compounds together? Do they experience IMAFs? I've heard that it's the ionic bond itself, but doesn't the bond remain after melting, and even if it doesn't then how come covalent bonds don't cause the boiling and melting points in molecules?
Sorry for all the questions, main ones are in the title.

Molecules have IMAFs that determine the melting and boiling points. What about with ionic substances? What keeps the different compounds together? Do they experience IMAFs?

The problem is that you are trying to apply a paradigm that was developed for covalent compounds (the distinction between covalent bonds and intermolecular forces) to ionic compounds. They don't play by the same rules. In an ionic compound, all the ions are held together by electrostatic forces, which mean exactly the same thing as ionic bonds. There is no discrete molecular unit in an ionic compound. The word "molecule" does not apply.

Consequently, there is no such thing as an intermolecular force in an ionic compound. There are only ionic bonds.

  I've heard that it's the ionic bond itself, but doesn't the bond remain after melting

Loosely speaking, it does not remain after melting. The reason why they have such high boiling points is therefore because of the fact that you have to break the ionic bonds for it to begin melting.

  and even if it doesn't then how come covalent bonds don't cause the boiling and melting points in molecules?

Covalent bonds don't break during melting or boiling, with some exceptions (see: giant covalent compounds).


What is the effect of basis set on charge transfer? I am calculating the energy and charge transfer of a globally neutral system, composed by two opposely charged monomers (EMIM-BF4), coming closer to each other, with DFT (B3LYP) and MP2.

As a thumb rule, the larger the basis set, the better the energy result should be (given BSSE is taken in account). Is this also valid for charge transfer? What is the role of polarizability? Is there a theoretical argument to trust the results from one basis set against another one, besides the size of the basis set?

I am comparing Pople's STO (minimal bs), Pople's N-311G(p,d) (triple split bs),
Dunning-type Correlation Consistent basis (CCQ or cc-pVQZ) and APCseg-1 ( Augmented Polarization Consistent bs) and I get different results for each case when distance reaches 3 Angstrom (check the attachment).

Which one should I trust?

Thank you

Mulliken charges are not representative of charge transfer, I would argue, in any case. Mostly because of their enormous basis set dependency, as Geoff pointed out.

In any case, several points are missing here; for instance whether the geometries were optimized with each method/basis set combination. This is not trivial.

Assuming a single static geometry (or set of geometries) for all single point calculations, if you go with an MP2 approach, the larger basis set will give you the best result. This is so by definition. In fact, you should always use a large basis set with MP2 because this method converges slowly and requires a large amount of virtual orbitals. In DFT you never know, but I would try to use modern basis sets, such as Ahlrich's (RIP) "def2tzvp" (naming conventions change, but you will find it). Such basis sets are optimized for DFT, and are generally diffuse enough (as a rule of thumb, you should always include diffuse functions if you have an anion in your system).

For your goal, several methodological problems arise. First, if you are indeed optimizing the geometry, B3LYP might be very bad. Dispersion corrections or range separation might help, but you can check the B3LYP-MP2 difference to see if anything suspicious arises.

Secondly, defining charge transfer is not easy. It is not trivial in SAPT (if you want to use SAPT i'd recommend PSI4 dearly for speed, there's a paper by A. Misquitta about charge transfer in SAPT but its not trivial to implement), which is the most rigorous energy term decomposition scheme, and in principle free from intermonomer BSSE.

Defining charge-transfer is somewhat easier in some flavours of the many available EDAs, as in the Kitauma-Morokuma scheme which I think is implemented in GAMESS. Real space EDAs, as the Interacting Quantum Atoms method, might be useful here as well. I think real space charges are far more convenient than Mulliken's in any case. You see, charge-transfer is a nice term but a very diffuse quantity.

Otherwise, you might be better off by checking some more available, well defined observable. For example, the dipole moment of the dimer will capture the charge separation.

In any case, good luck with this project!


Does the structure of water change when a drop of color is added? If I add a drop of color to water, would it change its structure entirely?

No it does not change the structure of water.
Yes the H-bonds between water molecules are broken to accomodate the dye molecules. However, liquid water does not have a set shape or fix molecular arrangement. So what happens is,
The dye molecules simply diffuse out all throught and among the water molecules. The dye molecules become hydrated to some extent, that is, the charged groups and areas on the dye molecule interact with water to form H-bonds or polar bonds which allow the molecule to remain suspended among the water.

Source, General chemistry course work.


What happens to water under high pressures without possibility of escape? Knowing very little about the nature of water, wondering how it might behave at the centre of a planet or centre of an another massive gravitational body.

Could water take such pressures or might it break into separate hydrogen and oxygen to find something more accommodating for the pressure exerted, if such atmosphere would allow the volume even?

Finally, bonus points: could triple points in water play a role in keeping temperatures low under high pressures via fluid thermodynamics?

What happens to water under high pressures without the possibility of escape?
It depends on what you mean for high pressure and on the temperature, however, the water phase diagram can help you to understand what will happen. This is from Wikipedia User Cmglee:

You can see that at high-pressure water assumes a solid form (ice): you will have ice X at 100 GPa, and what is labelled as hexagonal ice XI in the diagram but is actually superionic water or ice XVIII at 1 TPa (temperature range 0 to 650 K).  These sorts of ices have different lattice and internal energies.
What would happen to water as pressures increase towards infinity without the possibility of the water escaping confinement?
Unfortunately, chemists don't deal with infinity, basically, however, there will be a point where the water molecule won't longer exist all the bonds will break.  After that probably you will have something called electron-degenerate matter have a look here, here is all about quantum mechanics, Pauli exclusion principle and black holes so on more Physic S.E. stuff.

Could triple points in water play a role in keeping temperatures low under high pressures via fluid thermodynamics?
You can also note that there are other triple points like (100 k, 62GPa) this however doesn't directly affect the properties of water but are the properties of water that determine where are the triple points. In a close system, kinetic energy is constant so I think that fluid thermodynamics doesn't matter in this case is more related to equilibria.
More info about water phase diagrams here.


What white powdery compound turn black immediately after heating? What are some white compounds that look like powder (like flour) and immediately black after being heated? It turns immediately black and creates a odorous gas.

The answer to this question partially answers yours.Let me speculate. The compound could be some crystallized organic compound, making it appear white. On heating, it degrades and becomes solid carbon, which is black, giving of some carbon dioxide and water vapour. It wouldn't have any sulfurous groups, as it would give off sulfur dioxide.


Will all thermodynamic equations have a linear dependence with moles? I'm taking chemical thermodynamics this semester, and currently seeing things like the Maxwell's Relations.

My professor never uses explicitly the variable "n" and always works with molar quantities like $\bar{V}$ and stuff like that.

I think this simplifies the algebra a great deal, and can be useful if you have the temptation to get a "molar quantity". However, I don't like it right now.

I think theres nothing that can assure me that if I were to work with the variable $n$ in my derivations, I won't somehow get a quadratic dependence of my thermodynamic equation with $n$, or logarithmic, who knows.

But it seems that there is always a linear dependence with $n$. This makes physical sense to me, however, how can I prove this to myself? Maybe this is more of a question of mathematics, but it has scientific implications as well.

Thanks.

Extensive properties actually have a linear dependence with the common magnitudes used for amount of substance.

Mathematical speaking, you work in some representation in which one magnitude is represented with a function of some variables, and the later represent other physical variables. For example $S(U,V,N)$. When a magnitude represents a extensive property, the function is a homogeneous function of of order 1 of the extensive variables, that is:
$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$

We choose $\lambda = 1/N$, so
$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$, using lower case symbols for (say molar) properties,

$ N \,S(e, v, 1) = S(E,V,N)$

So, taking in mind that your equations are formulated for finding values for your thermodynamic functions, they must be stated in the form (for the previous example):

$S(E,V,N) =$ $whatever$

If you work with molar properties you will find just:

$S(e, v, 1) =$ $whatever$ $for$ $a$ $mol$, but using the third equation:

$S(E,V,N) = N\times$ $whatever$ $for$ $a$ $mol$

So the homogeneous (order 1) property will assure you that there is linear dependence with $N$.

Note: Thermodynamic functions that also depend on intensive variables have a zero order dependence in the homogeneous sense with these variables.

EDITED:

After reading the @Curt F. answer, to avoid confusions I decided made a comment: The explaining above works in the scope of 'traditional' thermodynamics, where no surface effects are involved. There are another reasons that can leads to draw wrong conclusions about this linearity, for example, if external fields are involved. But these cases are not in the scope of any course on thermodynamics that I aware of.


Are all amino acids except glycine chiral compounds? Specifically, I was thinking of the case where the R group is a carboxylic acid functional group or an amine group. Then there would not be a chiral carbon atom. Is there a restriction on what can be an R group that I am missing?

In general, there are a variety of amino acids that are not chiral (achiral).  In physiological proteins, the amino acids are generally alpha-amino acids (one carbon between the amino and carboxylic acid groups), but there are also, beta-amino acids, gamma-amino acids, etc.

Just limiting ourselves to alpha-amino acids, other than glycine, the most interesting example of an achiral amino acid is amino malonic acid, which has been given its own three-letter abbreviation (Ama) along with the more-common alpha amino acids.  Ama has being found in the hydrolysates of E. Coli and human proteins.  See Aminomalonic acid: identification in Escherichia coli and atherosclerotic plaque PNAS vol. 81 page 722.

Also, the amino acids of protein orgin generally have an alpha H, but if this H is substituted the compound is still an alpha amino acid.  If the two side chains on the alpha carbon are the same, the compound is achiral. For example alpha-methyl alanine.  An example that occurs naturally in cranberries is 1-aminocyclopropane 1-carboxylic acid. The alpha amino acids with two side chains are sometimes referred to as alpha branched amino acids, and have been found in meteorites.

Another class of achiral alpha amino acids are those with a side chain double-bonded to the alpha carbon such as dehydroalanine.


Need help interpreting this data table (provided below) I am confused about this data table. Can someone please explain how this person got a linear alkanes of differing molar masses? (I know some basics of organic chemistry, but I'm still learning.) Are there differing compounds to each of the ones listed underneath Boiling Point? If so, where can I learn these compounds or is there a database that has a list of compounds under each one?

The first row for each number in the leftmost column represent alkanes with $n$ many carbons and aldehydes and ketones with $n-1$ carbons and one double bonded oxygen. These are all in the same row because if you replace a carbon with a double bonded oxygen, you wind up with the same mass once you remove the excess hydrogens. The next row is for alcohols, which have an $-\ce{OH}$ instead of a $=\ce{O}$, which leads to an increase in molar mass by 2.

The alkane link above links to a table, which shows the boiling points with the molecular formula for each alkane.


What is a good source for large tables on the properties of water? I'm currently doing an investigation for school on finding the Clausius-Clapeyron relation/gradient for  boiling water and comparing it to the gradients along my line of best fit. It is given by the equation,
$$
\frac{\text{d}P}{\text{d}T} = \frac{L}{T\Delta v}.
$$
I'm having trouble finding the specific volume of both water and steam (I found this for water but I'm not sure of the credibility) for $\Delta v$. I was hoping someone could reference me to a large data table for finding the specific volume, or a calculator. Thank you.

Such values for various properties of liquid water and steam can be looked up in so-called steam tables. Software versions also allow calculations (ususally interpolation) of additional values. For example, I usually use REFPROP – NIST Standard Reference Database (however, not always the latest version) or some old and simple proprietary steam table software from Siemens/KWU.

If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha.


Diastereomers or not Why is the first one not a pair of diastereomers while the second is?

The only reason I could think of was in the first $\ce{Me}$ is the same as $\ce{CH3}$ that is same groups even though differently oriented are considered same. While in second $\ce{COOH}$ and $\ce{OH}$ are obviously different. Is my reasoning correct? If it is could someone please state it in better words so that I understand the minute details.
Also I was able to reason this ONLY AFTER the correct answer was told. How to identify if a certain pair of compounds is a diastereomeric pair?

Your first set of compounds are identical.
In the Fischer projection of a chiral centre, you can rotate three of the connected groups in a clockwise/anti-clockwise direction without changing the configuration of the chiral centre.
Upon performing this operation (keeping the top fixed, rotating bottom three clockwise) on the first compound in the first set, we see that the methyl group now occupies the position of the $\ce{-CH3}$ group in the second compound and the $\ce{-CH3}$ in the first occupies the position of the methyl group of the second.
Now, $\ce{-CH3} \equiv -\text{Me}$. Therefore, the first two compounds are now identical. As yukelid stated, identical compounds are not diastereomers.
However, repeating the same for the second case does yield two different compounds that are non-mirror images yet non-superimposable. Hence they are diastereomers.
Enantiomers and diastereomers are only seen for chiral compounds since achiral compounds would have symmetry that makes them identical. So, chirality is needed for enantiomers and diastereomers.


Why do 3d orbitals have lesser energy than 4s orbitals in transition metals? This is quoted from Jim Clark's Chemguide

  For reasons which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series.

What is the reason stated above? Why does in only transition series, 4s has more energy than 3d & not in the preceeding elements?

Disclaimer: I now believe this answer to be fully incorrect.

  Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22.

  However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in Shriver & Atkins, as well as of the idea that the 3d orbitals are somehow higher in energy than the 4s orbitals. I believe it was mentioned that the energies were calculated with the outdated Thomas–Fermi–Dirac model, but cannot really remember. I will ask another question about the 3d vs 4s issue, but in the meantime I would point the reader in the direction of these articles:

  Pilar, F. L. 4s is always above 3d! Or, how to tell the orbitals from the wavefunctions. J. Chem. Educ. 1978, 55 (1), 2 DOI: 10.1021/ed055p2.
  Melrose, M. P.; Scerri, E. R. Why the 4s Orbital Is Occupied before the 3d. J. Chem. Educ. 1996, 73 (6), 498 DOI: 10.1021/ed073p498.
  Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D. Transition Metals and the Aufbau Principle. J. Chem. Educ. 1994, 71 (6), 469 DOI: 10.1021/ed071p469.
  Scerri, E. R. Transition metal configurations and limitations of the orbital approximation. J. Chem. Educ. 1989, 66 (6), 481 DOI: 10.1021/ed066p481.
  Some criticism of Atkins' books by Eric Scerri.

While Molly's answer does a good job of explaining why electrons preferentially occupy the 4s subshell over the 3d subshell (due to less inter-electron repulsion), it doesn't directly answer the question of why the order of the 3d/4s energies changes going from Ca to Sc. I stole this figure from Shriver & Atkins 5th ed:

The red line represents the energy of the 3d orbital, and the blue line the energy of the 4s orbital. You can see that up to Ca, 3d > 4s but for Sc onwards, 4s < 3d.

As chemguide rightly points out, up to Ca, the 4s orbital is lower in energy than the 3d. The energy of an electron in an orbital is given by
$$E = -hcR\left(\frac{Z_\text{eff}}{n}\right)^2$$
where $hcR$ is a collection of constants, $Z_\text{eff}$ is the effective nuclear charge experienced by the electron, and $n$ is the principal quantum number. Since $n = 4$ for the 4s orbital and $n = 3$ for the 3d orbital, one would initially expect the 3d orbital to be lower in energy (a more negative energy). However, the 4s orbital is more penetrating than the 3d orbital; this can be seen by comparing the radial distribution functions of the two orbitals, defined as $R(r)^2 r^2$ where $R(r)$ is the radial wavefunction obtained from the Schrodinger equation:

The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron "tends to spend time" near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron penetrates the core electrons (i.e. 1s through 3p subshells) better. It is therefore shielded less than a 3d electron, which makes $Z_\text{eff}$ larger. Going from the 3d to the 4s orbital, the increase in $Z_\text{eff}$ wins ever so slightly over the increase in $n$, which makes the energy of the 4s orbital lower.

Now, going from Ca to Sc means that you are adding one more proton to the nucleus. This makes the nuclear charge larger and therefore both the 4s and 3d orbitals are stabilised (their energies decrease). The catch is that the energy of the 4s orbital decreases more slowly than that of the 3d orbital, because the 4s orbital is relatively radially diffuse (the maximum in the radial distribution function occurs at a larger value of $r$). If you have studied physics, you could think of it as the interaction between two point charges; if the distance between them is large, then increasing the magnitude of one point charge has a smaller effect on the potential energy $U = -\frac{kq_1q_2}{r}$. The faster decrease of the 3d energy also makes sense because if nuclear charge were to tend to infinity, shielding would become negligible; the orbital energies would then be entirely determined by $n$, and if this were to be the case, you would expect 3d < 4s in terms of energies, as we said at the very start.

However, in Sc, the electrons preferentially occupy the 4s subshell even though it is higher in energy, and this is also because the 4s orbital is radially diffuse - the electrons have more "personal space" and experience less repulsion. One way of putting it is that an empty 4s orbital in Sc has a higher energy than an empty 3d orbital, but a filled 4s orbital has a lower energy than a filled 3d orbital. The fact that 4s > 3d in energy also explains why, for the transition metals, the 4s electrons are removed first upon ionisation ($\ce{Sc^+}: [\ce{Ar}](3\mathrm{d})^1(4\mathrm{s})^1$.)

I just want to end off with a comment that the factors that determine the electronic configurations of d-block and f-block elements are actually very closely balanced and just a small change in one factor can lead to a completely different electronic configuration. This is why Cr and Cu have an "anomalous" configuration that maximises exchange energy, whereas we don't get carbon adopting a $(1\mathrm{s})^2(2\mathrm{s})^1(2\mathrm{p})^3$ configuration in order to have "stable half-filled shells".


Role of Alcohols(EtOH,MeOH) in nitro reduction used for Metal(Sn,Fe) I'm trying to reduce an aryl nitro group.
I understand that these reactions are radical reactions by the metal, but I don't understand
why alcohols are used as the solvent. Are they used as a source of protons, or just because they are good solvents?

LHM...Look at this mechanism that I posted as an answer to an earlier SE Chem question/answer relating to the reduction of nitrobenzene by metals.

In most reactions that occur on metal surfaces, many of the mechanistic details are not fully understood (see @Martin's comments in the earlier post).  Nonetheless, this mechanistic scheme shows several steps where a proton ($\ce{H+}$) is required.  These proton transfers are necessary to complete the reduction process.  This is why these reactions are typically run in solvents, like ethanol, that have available protons that can be donated in order to complete the reaction...ron


Why draw "up" arrows first when filling orbital diagrams? I am a 11th/12th grade student studying orbital configurations. I have gone over this lesson and it says to always draw spin up first. However, it never says why. Is drawing down spins first incorrect?

There is no real reason about that and drawing them "down" would be exactly the same as long as you draw them always in the same direction (it needs to be consistent).

The same applies to orbital lobes which can be hollow (white) or plain black depending on their relative sign. When two molecules' orbitals interact, you just consider they have the right sign for an interaction.


What is the difference in Da of ESI analytes with multiple charges? This is a follow-up question to Why are isotopes an issue in reading mass spectra?

A follow-up question: electrospray ionization (ESI) produces multiply charged species and charge can be >40 (for the purposes of m/z ratio), and assuming the difference in mass for an isotope is ~1 Da (with each additional neutron for an analyte at a charge of +1 in positive ionization mode) and assuming there is only one additional isotope, would the difference in weight for two isotopes each with a charge +10 be 0.1 Da?

More realistically if you have a molecule with Br, then the distance between isotopes:

Will be 2 $m/z$ units in case of 1 charge
And 1 $m/z$ units in case of 2 charges

You can use this information to determine if there are molecules ionized twice — their isotopes appear closer than expected.

Eventually since ESI/APCI adds $\ce{H+}$ for each charge, given original EMW was 100 Da, then:

$m/z$ in case of 1 charge: $(100+1)/1=101$ and $(102+1)/1=103$
$m/z$ in case of 2 charges: $(100+2)/2=51$ and $(102+2)/2=52$

Or in case of negative ionization you'll need to subtract 1 and 2 Da, respectively.


Pure gallium with lower melting point? I melted about 2 kg of gallium and put it in a plastic container, in order to make crystals. After letting them grow and extracting them, I let the gallium freeze at the room temperature, which is below the melting point.

However, for some reason there is a small quantity of gallium (10-20 g) that is not freezing after days. I extracted the liquid then froze it in a refrigerator. But if I put it outside again, it melts, so this is not supercooled gallium, a well documented phenomenon.

It behaves like normal gallium with slightly lower melting point, something like a Galinstan eutectic. However this is 99.99% pure gallium (that was fully solid) and I am pretty sure that I never contaminated it...

Any idea what is happening?

It's possible that the gallium is not as pure as advertised. If it has (say) some indium as an impurity, the indium-gallium alloy would have a lower melting point, potentially below room temperature.

I thought that partial freezing might have crystallized out more pure gallium, and left less pure gallium as the melt, but I'm not sure that's how indium-gallium alloys work. I made a few grams of the alloy, and I never saw any such behavior in it -- then again, I never experimented much along those lines.


Can I electrolyze saturated sodium hydroxide? Can I electrolyze saturated $\ce{NaOH(aq)}$? My guess is that I can't because in the aqueous solution, the concentration of $\ce{OH-}$ is too high and thus there isn't enough $\ce{H+}$ remaining.

At $T=25\ \mathrm{^\circ C}$, saturated sodium hydroxide solutions contain about $50\ \%$ sodium hydroxide. Hence, the concentration of $\ce{Na+}$ is about $c(\ce{Na+})=19\ \mathrm{mol\ l^{-1}}$ and the concentration of $\ce{H3O+}$ corresponds roughly to $\mathrm{pH\approx15}$, i.e. $c(\ce{H+})\approx10^{-15}\ \mathrm{mol\ l^{-1}}$ (neglecting activity coefficients).

The positive ions $\ce{Na+}$ and $\ce{H3O+}$ are attracted to the cathode.
At the cathode, the redox potentials $E$ imply that, even under the given alkaline conditions, reduction of $\ce{H+}$ to $\ce{H2}$ is still preferred over reduction of $\ce{Na+}$ to $\ce{Na}$:

$$\begin{alignat}{3}
\ce{Na+ + e- \;&<=> Na}\qquad &&E^\circ = -2.713\ \mathrm V\qquad &&E&&=E^\circ+0.05916\ \mathrm{V}\times\log_{10} [\ce{Na+}]\\
& && && &&=-2.713\ \mathrm V+0.05916\ \mathrm{V}\times\log_{10} 19\\
& && && &&=-2.64\ \mathrm V\\[6pt]
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm V\qquad &&E&&=E^\circ+0.05916\ \mathrm{V}\times\log_{10} [\ce{H+}]\\
& && && &&=0.000\ \mathrm V+0.05916\ \mathrm{V}\times\log_{10}(10^{-15})\\
& && && &&=-0.89\ \mathrm V
\end{alignat}$$

Therefore, the thermodynamically preferred reaction is the reduction of water to hydrogen according to
$$\ce{2H+ + 2e- -> H2 }$$
or in this case better (in view of the given alkaline conditions)
$$\ce{2H2O + 2e- -> H2 + 2OH- }$$

Nevertheless, under the given alkaline conditions, the concentration of $\ce{H3O+}$ is very low. The predominant positive ions are $\ce{Na+}$, which migrate to the cathode. Thus, especially when applying a high current density, the major part of the current of positive charges that reach the cathode is carried by $\ce{Na+}$. Therefore, $\ce{Na+}$ actually can be discharged at the cathode:
$$\ce{Na+ + e- -> Na}$$
The ratio of discharged $\ce{Na+}$ to $\ce{H+}$ also depends on the cathode material. For some materials, the reduction of $\ce{H+}$ to $\ce{H2}$ is kinetically inhibited, which results in a significant overpotential.

However, any generated $\ce{Na}$ immediately reacts with water (except when a mercury cathode is used):
$$\ce{2Na + 2H2O -> 2Na+ + H2 + 2OH- }$$

Therefore, the net reaction is again the above-mentioned reduction of $\ce{H2O}$ to $\ce{H2}$. The $\ce{Na+}$ ions act as electrolyte and might be considered to act as a catalyst:

$$\begin{align}
\ce{2Na+ + 2e- &-> 2Na}\\
\ce{2Na + 2H2O &-> 2Na+ + H2 + 2OH- }\\
\hline\ce{2H2O + 2e- &-> H2 + 2OH- }
\end{align}
$$


What is a 'Trace Crossing'? I'm reading about electrodeposition of conductive polymers in a chemistry textbook, and the author mentions 'trace-crossing' in the 'potentiodynamic response'. From the context I can figure out that 'potentiodynamic response' means cyclic voltammogram.  But what is the 'trace crossing'?

'Trace crossing' is in the literal sense, where 'trace' is another word for 'line'. 'Trace crossing' basically means that the line in the voltammogram crosses itself.


Why does NO have a residence time of 4 days? If $\ce{NO}$ is unstable, why does it have an atmospheric lifetime of 4 days? I know that it has a lone pair and one unpaired electron making it highly reactive. Therefore, shouldn't it react immediately in the atmosphere?

It could be possible that you're thinking of the overall lifetime of $\ce{NO_x}$ ($\ce{NO2 + NO}$), which indeed does have a longer atmospheric lifetime than $\ce{NO}$ individually and which even, at around 5 km altitude, has a lifetime of ~4 days. The two interconvert during the day due to photo-chemistry, and the ratio of the two is dependent on the photolysis rate of $\ce{NO2}$ ($j_{\ce{NO2}}$) and on the rate coefficient of the reaction between $\ce{NO}$ and $\ce{O3}$.

As $j_{\ce{NO2}}$ is constant with altitude, the ratio between $\ce{NO}$ and $\ce{NO2}$ is dependent upon the reaction rate coefficient between $\ce{NO}$ and $\ce{O3}$. This, in turn, is governed by the temperature, and under lower temperatures, the reaction rate is slower. Combined with a decrease in the number concentration of ozone (due to the decrease in the number concentration of air with increasing altitude, if we assume a constant mixing ratio of ozone throughout the troposphere, of course), the ratio favours $\ce{NO}$ over $\ce{NO2}$.

At higher altitudes, then, you do see a situation where there is more $\ce{NO}$ than $\ce{NO2}$, and where the overall lifetime of $\ce{NO_x}$ can be higher than the 1 to 2 days found at the surface (reaching, sometimes, almost 12 days in the upper troposphere).


Does the ammonium ion hydrogen bond with water? What about oxonium ions, such as hydronium ion? Do these hydrogen bond with water?

If we see hydrogen bonding as a purely electrostatic phenomenon, then why not? Ammonium ion still has nitrogen bearing a partial negative charge. Of course it can interact with positively polarized hydrogens in an electrostatic way.

However, we also know that hydrogen bonding is partially covalent. So, it seems unlikely that the ammonium ion would be able to form any additional covalent bonds.

There is a detailed consideration of this issue in  Hydrogen Bonding of the Ammonium Ion Journal of Molecular Structure 345 (1995) 77-81.

First, the try to be clear about what they mean by "hydrogen bond":

  For the purposes of the discussion, a hydrogen
  bond is taken to be a specific short-range interaction
  in which the N-H bonds within the
  ammonium ion are significantly affected by the
  surrounding ions, and not simply electrostatic
  attraction.

They conclude that ammonium ions rotate too freely in water for significant hydrogen bonding to exist.

In gas phase calculations, they find that one water molecule hydrogen bonds to one ammonium ion (in other words, the dimer is hydrogen bonded), based upon the lengthening of the N-H covalent bond, but as you add 2, 3, and 4 water molecules, the N-H bond length returns to the free ammonium N-H bond length and the O --- H length increases.


How to calculate molar concentration from equilibrium equations with no initial concentration given? I have the equations:
\begin{align}
\ce{CO2(g) + H2O(l) &<=> H2CO3(aq)} &\quad K_1 &= 10^{-1.47} \tag{1}\\
\ce{H2CO3(aq) &<=> H+(aq) + HCO3-(aq)} &\quad K_2 &= 10^{-6.35} \tag{2}\\
\ce{HCO3-(aq) &<=> H+(aq) + CO3^{2-}(aq)} &\quad K_3 &= 10^{-10.33} \tag{3}\\
\end{align}
Atmospheric $\ce{CO2}$ is at $\pu{10^{-3.5} atm}$ and the solution is pure water.
So far I have been able to find the equilibrium equations and have got the concentration of $\ce{H2CO3}$ to be $10^{-4.97}.$ However, from this I do not know where to go in order to find the concentration of bicarbonate ($\ce{HCO3-}$).
I have the equilibrium equations as:
\begin{align}
\tag{1}
  \frac{[\ce{H2CO3}]}{[\ce{CO2}][\ce{H2O}]} &=
  K_{\ce{CO2}} = 10^{-1.47}\\
\tag{2}
  \frac{[\ce{H^+}][\ce{HCO3^-}]}{[\ce{H2CO3}]} &=
  K_1 = 10^{-6.35}\\
\tag{3}
  \frac{[\ce{H^+}][\ce{CO3^2-}]}{[\ce{HCO3^-}]} &=
  K_2 = 10^{-10.33}
\end{align}
By multiplying $10^{-1.47}$ by $10^{-3.5}$ I got $10^{-4.97}$ as previously stated.
From there I get $\ce{HCO3-}$ to be $10^{-6.35}+\frac{10^{-4.97}}{\ce{H+}}$.
However I do not know how to find $\ce{H+}$.

I will continue with the data of $[\ce{H2CO3}] = \pu{10^{-4.97} M}.$
Now, as the $K_2$ of $\ce{H2CO3}$ is very small as compared to its $K_1,$ we can assume that all the $\ce{H+}$ will come from the first dissociation of $\ce{H2CO3}.$
$$
\begin{array}{lccc}
& \ce{&H2CO3 &<=> &H+(aq) &+ &HCO3-(aq)} \\
&\text{Initial} & 10^{-4.97} && 0 && 0 \\
&\text{Final} & 10^{-4.97}(1-\alpha) && 10^{-4.97}\alpha && 10^{-4.97}\alpha
\end{array}
$$
$$K_1 = 10^{-6.35}$$
$$10^{-6.35} = \frac{(10^{-4.97}\alpha)^2}{10^{-4.97}(1-\alpha)}$$
As $K_1 \ll 1,$ $(1 - \alpha)\approx 1:$
$$\alpha = \sqrt{\frac{10^{-6.35}}{10^{-4.97}}}$$
$$
\begin{align}
[\ce{HCO3-}] &= \ce{[H+}] = 10^{-4.97}\alpha \\
  &= \sqrt{10^{-6.35}\times 10^{-4.97}} \\
  &= 10^{-5.66} \approx \pu{2.19E-6 M}
\end{align}
$$
Note, that I have not considered self ionization of water because the $\mathrm{pH}$ of this solution will be slightly less than $6$ $(\mathrm{pH} = 5.66).$ Which means even after accounting for self ionization of water, the results won't change much. This is a good approximate of what  we needed. And if you want to calculate $[\ce{CO3^2-}]$ than just calculate it in a similar way to which you calculated $[\ce{H2CO3}],$ because now you have $K_2,$ $[\ce{HCO3-}]$ and $[\ce{H+}].$


Why is a very small peak with larger m/z not considered to be the molecular ion? With regards to the following spectrum:

When asked for the M/Z for the molecular ion, the peak at 58 was taken, not 59.

What causes the peak at 59, and why isn't it taken as the molecular ion?

Without knowing more details, it is hard to guess, but at this $m/z$, it seems likely that the peak is the result of one $^{12}\ce{C}$ being substituted by one $^{13}\ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum.

Note that with more carbons in a larger molecule, it becomes more likely that at least one of them is a $^{13}\ce{C}$. Thus, the spectra become more complicated that way.

This whole analysis goes out the window if this is not an organic molecule, though similar patterns may arise for other molecules - it depends on the isotopes and their distribution.


Do magnesium and tin have the same volume? Having the same atomic radii of 145 pm, should one mole of tin and one mole of magnesium take up the same volume? Since tin is much heavier, that means tin is much denser than magnesium. Is this true, or is there an error in my thought process.

No, that is not how it works.

Atomic radius means the radius of an atom. A single atom. It does not in and by itself mean that a solid chunk of a material has a particular density. A solid chunk of material has many atoms, and they are arranged in certain ways and sometimes more than one way of this arrangement is possible.

Tin and Magnesium are very different. Magnesium is arranged in a hexagonal close packed structure, basically just repeating units of skewed (rombohedric) prisms. This can be hard to imagine, but imagine just stacking perfect spheres so they rest on each other and likely you will end up at a close packing arrangement. (there are only two*, try to find them. Hint: one has three layers before it repeats the bottom, the other has only two layers then it repeats)

*: There are actually 4, but you'll need a lot of layers and a lot of spheres to find those.

Tin has two different ones, so already here your question has to be no'ed without further investigation, but lets anyway. The first one, grey tin (${\alpha}$) has a diamond like structure. I am finding it hard to describe them in a meaningful way, but they have examples over at the excellent Wikipedia. The other one, tin (${\beta}$) which is white has a body centered tetragonal - which is basically stack 4 - put 1 in the middle and then stack 4 on top of that again (yes, the spheres will roll off, but there are so many of them per layer that hardly any of them do.) The white one is amusing, if you bend it it screams like you are hurting it. Sounds absolutely disgusting, like nails on a chalkboard. This is related to it's structure.

Summary:
When you put a whole mole (ha... ha...) of atoms together, they organize in structures, and it is the number of atoms per volume that you have to multiply with the atomic weight to get the density of the chunk. Fortunately for most kinds of structures, you don't have to do this, someone already has done the work, by actually measuring the densities of crystals and made tables. Just look it up.


Imaginary Bonding Interactions Usually in chemistry, we deal with bonding interactions. That is, if I have the diatomic A-A molecule or A-B molecule, there's a favorable interaction (i.e., a bond) and a prototypical MO diagram like this:

As I'd say in a lecture, one orbital goes down in energy, one orbital goes up in energy. I can use Hückel theory to define the energies of the bonding and anti-bonding interactions:

$$e_{ab} = e_a + \frac{H_{ab}^2}{e_a - e_b}$$
$$e^*_{ab} = e_b + \frac{H_{ab}^2}{e_b - e_a}$$

Now, consider if $H_{ab}$ is imaginary.. we'd use the same equations to get a very unusual MO diagram:

How would you describe such an interaction? It's clearly not the traditional favorable bonding. It's not a non-bonding interaction. And it's not the traditional anti-bonding case like $\ce{He2}$, because if I put 2 electrons into this system, the anti-bonding orbital isn't filled.

I ask the question because in a survey we've performed using DFT, we find many such combinations.

I'm leaning towards calling it an imaginary bond, but if there are known discussions, I'd be curious to read references.

The short answer is that I need to come up with better wording to describe the interaction and that Hückel is surprisingly bad for these systems.

I'll give a slightly longer answer for now, and add a reference as we submit the paper.

First, an example system:

So "A" is a strongly electron-deficient dinitrothiophene ring (an electron acceptor) and "B" is a strongly electron-rich diaminothiophene ring. The figure in the question is a subset of the entire MO diagram, focusing on the HOMOs of the monomers and the HOMO and HOMO-1 of the dimer.

Why do I think this is a strange system? Well, the particle-in-a-box and Hückel models are usually pretty good at explaining conjugated organic molecules. If you make a longer conjugated system, the HOMO should go up in energy, and the HOMO-LUMO gap goes down (delocalization).

The alternative, localization, would have the orbitals remain at the same energy level:

Both of those "limits" can be explained in the Hückel model by changing the $H_{ab}$ or $\beta$ parameter (depending on how you write the equation, the symbol is different). In the localized case, $H_{ab} = 0$.

What's strange about these strong donor-acceptor cases is that as I said, it's not really delocalized, or localized.

Instead, as one comment correctly deduced, the DFT calculation (B3LYP/6-31G*) shows a large degree of charge transfer, but not necessarily localization. The dipole moment computed to be >7 Debye.

If one attempts to use Hückel to analyze these results, you must infer that $H_{ab}$ is imaginary. This indicates an "open system" because electrons are indeed transferred - from monomer "B" to monomer "A" creating the large dipole moment.

My overly-simplified question was probably a bit confusing, but:

It's interesting that no simple model captures charge transfer unless you allow Hückel to have imaginary (non-physical) $H_{ab}$ parameters. (NB, my colleagues aren't too happy with this concept, suggesting it simply indicates Hückel is a poor choice.)
It's surprising that in prototypical conjugated systems, I can't (yet) find a discussion about this alternative to delocalization and localization. We find it commonly as the difference in energy level increases.


The time-dependent pH value in a fermentation I want to know if someone has supporting tips or solutions to my problem:
In biotechnology, fermentations are used to grow microorganisms. These need a specific media/broth in which temperature, pH, substrate and many other factors conclude to an optimum. Only if the surrounding broth is in optimal condition for the individual microorganism, the microorganism will grow as best as possible.

In my media the broth is buffered with $[\ce{HA}] = \ce{H2PO4^-}$ and $[\ce{A-}] = \ce{HPO4^2-}$.

While the fermentation is running the pH is corrected by $2\ M\ \ce{NaOH}$ which is pumped into the broth. Via balances, the weight is recorded and the input can be calculated.

I know the initial concentrations of my buffer solutions and the concentration of the base that is dropped into at every minute of record.

How can I calcute the pH for an individual timepoint in my aqueous fermentation? My tries via the Henderson-Hasselbach-formula did not work so far...
(Neglecting other surroundings such as stirring, the other media components, the biomass' metabolites, temperature influence on the pKs, etc.),

Thanks in advance!

The pH is not corrected mathematically, it is stabilized chemically by addition of NaOH when the pH sensors get out of range. Your pH is controlled to be constant over the fermentation period (+/- a little bit).

The acid produced by the fermentation is neutralized by addition of NaOH to keep the pH constant. What you can calculate from the data you get is the rate of acid production. The moles of acid produced will equal the number of moles of NaOH added, to a good approximation. The pKa of H2PO4- is 7.2 and the pKa of acetic acid (a typical carboxylic acid) is 4.76, so the acid produced will be "strong" - i.e., fairly acidic in your medium, which appears to be aimed for about pH = 7 - and the NaOH is strongly basic, so they will balance out (approximately).

Interestingly, sodium acetate has a pH range of 8.5 - 9.9 in real life, according to SDS. Calculation is so much more pleasant, where you can calculate pH to three decimals and feel good about it (until you test it in real life and have to calibrate your meter, etc, etc.)


Depicting a mechanism of peptide bond formation with protonated and deprotonated amino group I've noticed that some textbooks and video lectures use an amino acid wherein the amino group has two hydrogens, and others use an amino acid with three hydrogens in the amino group (in the formation of a peptide bond).
How come there are two variations? And also, if the starting amino acid has three hydrogens in its amino group, what is the mechanism for the part that the nitrogen bonded to the carbonyl ends up having one hydrogen attached to it?
I worked up the mechanism on the former condition, but starting with three hydrogens, I don't quite get it. I hope you someone can enlighten me.

In neutral aqueous solution, the amino group is protonated (3 bound hydrogen atoms ) and the carboxilic acid group deprotonated (no bound hydrogen atom). In other solvents, you might have both groups in the neutral form the name amino acid suggests.
The mechanism of peptide bond formation is complex and involves intermediates with tRNA in the cell and starting materials distinct from amino acids for synthesis in the lab.


What products will the following reactions yield? Assume these 2 reactions. I must write the Lewis structures of the products.

$$\ce{C6H5COOH + CH3CH2OH ->[\ce{H+}] ?}$$
(comment: A benzene ring + $\ce{CO_2H}$ on one end, I don't know its name)
$$\ce{CH3CH2OH + (CH3CO)2O ->[\ce{H+}]?}$$ (comment: acetic anhydride)

My approach:

For the first reaction I will get acetic anhydride (source)

For the second reaction I will get aspirin and acetic acid.

Questions:

Is the above correct?
Is there a connection between the two reactions? Is it an intermediate step of some sort? What is the importance between the two reactions?
Why is $\ce{H+}$ added?
What is the benzene ring and the $\ce{CO2H}$ called?

No, I hesitate to give the correct products since this is homework, but maybe some of what's below will help
Yes, the two reactions are both esterifications; converting an acid, an acid anhydride, an acid chloride, etc., to the corresponding ester
The proton is a catalyst, it is not consumed in the reaction.  It protonates the carbonyl oxygen, making the carbonyl carbon more susceptible to nucleophilic attack
benzoic acid


Is paper or water the stationary phase in paper chromatography? I've been looking at about 10-ish sources, and half of them say that paper is the stationary phase, and the other half say that water is the stationary phase. I understand that water is on the paper through humidity, but is there a clear stationary phase, or is it both cellulose and water (if that's possible)? Does the stationary phase depend on the humidity of the room?

The stationary phase in chromatography is the one that doesn’t move according to the eyes of a macroscopic (i.e. human researcher) observer. (That complicated way to put it was to prevent anybody raising any relativism arguments.)

Obviously, the paper does not move through the water but the water does through the paper. You should discard the five-ish sources that state water be the stationary phase as they obviously have no clue about chromatography.


If Oxygen has a lower electron affinity than Sulfur, then why doesn't sulfur has a lower electron affinity than selenium? I've read that oxygen has a lower electron affinity (as shown the picture below), because it has a smaller atomic radius than sulfur and thus the electrons experience significant electron-electron repulsion.
However, I was wondering... Why is this not the case for sulfur and selenium or chlorine and bromine? Because of sulfur's and chlorine's smaller atomic radius compared to selenium and bromine, the incoming electron should also experience a slight electron-electron repulsion (just like it does with oxygen), so $\ce{Se}$ and $\ce{Br}$ should have a higher electron affinity.

This is likely due to second period elements' being quite small, so electron-electron repulsion is much more significant than in a third period element. The general trend is that EA is more positive as you move down the periodic table since effective nuclear charge is the same, but the electrons are farther away from the nucleus. Maybe it's easier to consider the second period the anomaly...


How does a mixture not retain its chemical identity? My book says that matter can be classified as an element, compound, or mixture. It further says that a pure substance (or substance) is matter that has distinct properties and a composition that doesn't very from sample to sample. So far, so good. Then it says that

  all substances are either elements or compounds

An element is one type of atom (something that "cannot be decomposed into simpler substances") and a compound is composed of two or more elements; these both clearly are substances. Then, it says that

  Mixtures are combinations of two or more substances in which each substance retains its chemical identity.

This is where I'm confused. How are mixtures not substances? They have distinct properties, and their composition doesn't vary from sample to sample, if you set a standard for the sample. You can create any arbitrary mixture, and you can create any arbitrary compound, and if you follow the "recipe" for either you will always get the same thing, as far as I know.

Where is my logic flawed?

A good example of the difference might be to look at water, $\ce{H2}$, and $\ce{O2}$.

If you have gaseous $\ce{H2}$, and $\ce{O2}$ in a container you have a mixture; you can change it to have $99\%$ oxygen and $1\%$ hydrogen or vice versa or any ratio in between, but it is still a mixture of the two gases. These mixtures will have different properties depending on the ratio of the two gases. Each gas still has its own properties and the mixture exhibits some combination of them depending on the ratio.

If they are allowed to react to form water, now we have a substance. Regardless of how we make water, it will always have the same composition (two hydrogens for each oxygen) and the same properties. Importantly, these properties are completely different from hydrogen, oxygen, or any mixture of the two.


Comparing rates of hydrolysis in cyclohexane ring (involving neighbouring group participation) I need to compare the rates of hydrolysis of these three compounds.

MY ATTEMPT:

By definition hydrolysis is either $\ce{S_N1}$ or elimination.But water being a weak base (and weak nucleophile) $\ce{S_N1}$ is preferable.
$\ce{PhS^-}$ is a better leaving group than $\ce{Cl-}$ due to mesomeric effects.
Comparing with the middle compound the leftmost and rightmost compound will certainly have higher rate of hydrolysis due to presence of good leaving group $\ce{PhS^-}$
But how to compare between the leftmost and rightmost compound's rates?

The third reaction infact occurs via neighbouring group
  participation.

(https://en.wikipedia.org/wiki/Neighbouring_group_participation)

In the third compound $\ce{S}$ atom attacks carbon containing $\ce{Cl}$ by $\ce{SN_{NGP}}$ forming sulphonium ion. An attack by $\ce{H2O}$ on the least hindered carbon of the three member ring occurs. In this case both the carbopns in three member ring will be similar though. So the third compound has highest rate of reaction.

(I'd like to add "This can only happen in the trans configuration, however it requires ring flip into the higher-energy trans-diaxial conformation." as pointed out by - @orthocresol in the comments to the orginal question.)

  Adding as a proof that trans reacts faster than cis form during NGP in cyclohexane ring.
  http://meskvmcollege.org/upload/study_material/30275.pdf (See page 10
  of the pdf)

The first compound has second highest rate. Anchimeric assistance cannot occur in cis form. Negative inductive effect by $\ce{Cl}$ enhances the rate. Reaction mechanism is $\ce{S_N1}$ and water itself acts as nucleophile.
The central compound has the lowest rate among the three. $\ce{S_N1}$ mechanism occurs and $\ce{Cl}$ is replaced by $\ce{OH}$ where water itself acts as nucleophile.

  So summing up: $$\ce{rate_3>rate_1>rate_2}$$


Temperature change and gas evolution for vinegar and baking soda experiment This is sort of 2 questions in 1, but I think they are related.

Could someone please explain why this endothermic reaction doesn't further decrease the temperature when the amount of baking soda is increased?

The amount of vinegar is constant, varying amounts of sodium bicarbonate is independent.

Attempt 1: 100ml vinegar and 1 teaspoon baking soda, temperature went from 16 degrees Celsius to 12 degrees to Celsius in <10 seconds.

Attempt 2: 100ml vinegar and 2 teaspoon baking soda, temperature went from 16 degrees Celsius to 12 degrees to Celsius in <10 seconds.

Attempt 3: 100ml vinegar and 3 teaspoon baking soda, temperature went from 16 degrees Celsius to 12 degrees to Celsius in <10 seconds.

Also, the volume of the effervescing gas collected into a balloon was the same for 1-3 teaspoons of baking soda. once again vinegar is 100ml for all 3 attempts.

I expected a further decrease in temperature as more baking soda is used.
I expected more gas when more baking soda is used.

What happened?

Although your observations make it pretty clear that you are using enough sodium bicarbonate to consume the acetic acid, even with $1$ teaspoon of sodium bicarbonate, we can also do a rough calculation to validate this hypothesis. It's a "rough" calculation because we don't know the concentration of acetic acid in your vinegar.

Vinegar is typically about $\pu{5\%}$ acetic acid by volume, but can be as high as $\pu{20\%}$. For our calculation, we will estimate that there is $\pu{5 g}$ of acetic acid in $\pu{100 ml}$ of vinegar, and use $\pu{6 g}$ as the mass for $1$ teaspoon of sodium bicarbonate, per your comment. Then, using the molecular weights of acetic acid ($\pu{60 g/mol}$) and sodium bicarbonate ($\pu{84 g/mol}$), we can calculate the following molar quantities of the reagents:

acetic acid: $\pu{0.083 mol}$ per $\pu{100 ml}$ vinegar
sodium bicarbonate: $\pu{0.071 mol}$ per teaspoon of baking soda

If our estimate of the acetic acid concentration is correct, it is close enough to a $1:1$ molar ratio with the sodium bicarbonate that you may not have noticed the difference in your measurements. In this case there is $\pu{17\%}$ more acetic acid than sodium bicarbonate. This means that the addition of excess sodium bicarbonate (i.e. $2$ teaspoons) should result in a $\pu{17\%}$ greater temperature difference, which would be $\pu{4.7^oC}$ rather than $\pu{4^oC}$. Since you are only reading your thermometer to $\pu{+/- 1^oC}$, these could be said to be the same and it is perfectly reasonable that you wouldn't have noticed the difference. It is not stated how the volume of evolved carbon dioxide was measured, but it would likely be even more difficult to notice this small change by the change in observed carbon dioxide volume.

Summary, TL;DR:
Your observations indicate that a single teaspoon of baking soda is consuming all of the acetic acid present in the vinegar, so when additional baking soda is added, there is no more acetic acid to react with and thus there is no indication of further reaction. Making some simple assumptions about the concentration of vinegar and mass of sodium bicarbonate in one teaspoon allows us to confirm that this is the case, within the detection limits of your method (measurement of temperature change and volume of carbon dioxide evolution).


NaCl electrolysis with copper electrodes Considering the aqueous $\ce{NaCl}$ is already concentrated. If we were to use copper electrodes instead of inert graphite electrodes:
Would copper react with the evolved chlorine gas to give $\ce{CuCl2}$ in solution, whilst still giving hydrogen gas at the cathode? Since in my experimentation, I obtained a teal blue green colored salt that was soluble in water. I do not have appropriate lab equipment to test for the identity of the salt, so I am hoping someone here could shed some light on the reactions taking place and the identity of the obtained substance. Thanks.

Electrolysis of sodium chloride solution produces results at each electrode. At the cathode, water is reduced to $\ce{OH-}$ and makes the solution more alkaline. Copper hydroxide is a blue, bulky precipitate, feebly acidic, and soluble in concentrated alkali hydroxide solutions. So if you got copper ions into solution at the anode, you could form some cupric hydroxide which is reported to form blue polymeric species in solution [1].
Crystals from concentration of this solution will likely be $\ce{NaCl}$ contaminated with some $\ce{CuCl2}$. Examination with a microscope could show whether $\ce{CuCl2}$ has formed, as the blue-green rhombic dihydrate (because, after all, it is in water), or, if you actually did enough electrolysis to do more than color the water, whether you precipitated the yellow-green hexagonal basic salt $\ce{CuCl2.Cu(OH)2}$ or the hydroxide or some mixture [2].
Although cuprous chloride is highly insoluble in water, it does form complexes in concentrated chloride solutions: the main species in $1\rm{M}$ $\ce{KCl}$ is
$\ce{CuCl2^-}$ [1]. So you really have a complex situation with anode/cathode, $\ce{OH-}$/$\ce{Cl2}$, $\ce{Cu^0}$/$\ce{Cu+}$/$\ce{Cu^{+2}}$, $\ce{CuCl2}\cdot2\ce{H2O}$/$\ce{Cu(OH)2}$, and polymeric complexes with $\ce{OH-}$ and $\ce{Cl-}$, and various combinations of products depending on how long you do the electrolysis and, of course, whether you emit $\ce{O2}$ or $\ce{Cl2}$.
Of course, that's what makes chemistry so much fun. There are many variables, some of which you can control.
[1]: Advanced Inorganic Chemistry, Cotton and Wilkinson, 1966
[2]: CRC Handbook, 62$^{\rm nd}$ ed.


How does the atomic structure of crystals determines its shapes? for example, table salt has an a cubic structure in the atomic level, and they are often cube like in the visible level. I don't see any reason that the atomic crystal structure should reflect the crystal's shape. Can someone explain this to me.

It’s all about minimizing the energy of the system. Imagine a 2D example:

You can imagine the green circles as the cations ($\ce{Na+}$) and the blue ones as anions ($\ce{Cl-}$). Where is the next chloride you add going to go?

Not a great choice – it doesn’t really balance. The nearest blue ion is already balancing its charge with two other green ions and so can’t help the new ion balance its charge as well.
This is on the grid but still has all the same problems where the new ion doesn’t get its charge nicely balanced because the nearest blue ion is already donating part of its charge to two other ions.
This is the best option because it can share partial charges with two other blue ions. It gets more shared charge and the whole thing ends up being lower energy.

This pattern will keep happening with every addition until you get a macro-scale cube. It is always lower energy to fit new ions into gaps then to branch out and make new structures.

Ions that aren’t part of a complete unit cell will develop surface charges. These surface charges are better at attracting ions that are precipitating onto the surface from solution.


Specific heat for tetraatomic gas molecules I came across a question which involved non linear tetra atomic gasses.
In it, the value of specific heat capacity at constant volume was given as
$9R,$ added up as

$$\frac 3 2 R + \frac 3 2 R + 6R = 9R$$

Now I know $6R$ is $C_V$ for triatomic gasses, but why do we add the two $\frac 3 2 R$ terms?

Also, how does being linear or non-linear affect the specific heat capacity of a gas?
Explanations with diagrams showing it in 3-D coordinate system would be helpful.

The heat capacity comprises contributions from translational motion (3 terms for each of $x, y, z$) plus (whole body) rotational motion ( 3 axis directions). Each motion accounts for $(1/2)R$ by equipartition theorem making the $(3/2)R + (3/2)R$ above.

Then there is a contribution from the fact that the molecule is vibrating. Non-linear molecules have $3N-6=6$ vibrational modes. Each vibration has $(1/2)R$ contribution from its potential energy and the same from its kinetic energy which in total is the $6R$ in your question.


Why is hydrobromic acid a stronger acid than hypobromous acid? If my understanding is correct, bromous acid ($\ce{HBrO2}$) is a stronger acid than hypobromous acid ($\ce{HBrO}$) because the additional electronegative oxygen atom draws the electron away from the hydrogen atom, making dissociation for $\ce{H+}$ easier.

Then why is hydrobromic acid stronger than hypobromous acid, despite the latter having one more electronegative atom to pull the electron? Are other factors involved here? And how do we decide which factor weighs more?

$\mathrm{p}K\mathrm{a}$ data:

$\ce{HBr}=-9.00$  (source)
$\ce{HBrO}=8.65$ (source)

The structure of $\ce{HBr}$ and $\ce{HBrO}$ is different, and also the place from where the proton leaves the molecule is also diferent.So, we have to take account of this and also the stability of the conjugate base formed after deprotonation.In $\ce{HBrO, }$ the proton is attached to the $\ce{O}$ atom of the molecule.The molecule can lose proton as, $$\ce{H -O -Br <=> H^+ + (O -Br)^-}$$ The negative charge here mainly resides on the Oxygen, which makes $3$ lone pairs on $\ce{O}$ atom, adjacent to that, there is also electron rich $\ce{Br}$ atom, which also has $3$ lone pairs on it, so there are inter electronic repulsions. For the conjugate base to get stabilisation, the weak $\ce{-I}$ effect of $\ce{Br}$ atom is not at all sufficient. Instead, if the lone pairs can somehow delocalise that will contribute better to stability of the conjugate base. But the lone pair resides on the $\ce{2p}$ orbital of Oxygen, which can't at all effectively overlap with filled$\ce{4p}$ orbital of $\ce{Br}$, and  $\ce{4d}$ orbitals are hugely high in energy. So, due to high electronic repulsion, no chances of delocalisation, and very weak $-I$ effect of $\ce{Br}$ can't make the conjugate base stable at all. So, $\ce{HOBr}$ has high $\ce{pK_a}$.But if you come to $\ce{HBr}$, the bond between $\ce{H}$ and $\ce{Br}$ is very weak. More over if you dissolve it in a sloution, the weak bond is easily broken, and the individual cations and anions can get extensively solvated which makes the  whole process of dissociation thermodynamically very much favourable.That's why, the $\ce{pK_a}$ of $\ce{HBr}$ is very low which means it can dissociate  very easily in a solvent.


difference between absorption spectroscopy and extinction spectroscopy can anyone tell me the difference between absorption spectroscopy and extinction spectroscopy in terms of experiment? and how to get extinction spectroscopy, how to get absorption spectroscopy? Thank you so much.

There should not be any difference in experiment, because the extinction is something you have to calculate by measuring the transmittance/absorbance of a given frequency.

The transmittance $T$ of a material is the ratio between the received intensity of a certain frequency to the transmitted intensity of the same frequency. $$T=I/I_0$$
The absorbance $A$ is defined as the decadic logarithm of the inverse transmittance. $$A=\log_{10}\left(I_0/I\right)=\log_{10}\left(1/T\right)$$

By using Beer-Lambert’s law $$I=I_0\operatorname{e}^{-\varepsilon~c~d}$$ one can calculate the extinction $\varepsilon$ from the measured transmittance/absorbance by $$\varepsilon=\frac{A}{c~d}$$ where $c$ is the concentration of the absorbing material and $d$ is the distance that the light has to travel in that given material.


Is the vinegar and baking soda experiment safe for children? To be more specific, could the carbon dioxide fumes produced in the reaction be harmful to a small child? I don't have a strong background in chemistry, and I was prompted to ask this question when it was suggested by a friend that a mixture of 2 cups of vinegar and baking soda is commonly used to euthanize rodents. I'm aware rodents are more sensitive than humans, but I was concerned that a small child could inadvertantly breathe enough fumes to cause harm. I think it would be a great experiment to introduce children to chemistry if not unsafe. Thanks for any advice or guidance.

A basic safety rule in chemistry is "don't breathe fumes". Maximum waft a sample of evolved gas to the nose with your hand. Given a controlled environment, adequate size vessels and small quantities and adequate ventilation, it could be a good introduction to chemistry. However - start in the way you should continue - safety glasses, safe operating procedures etc.. BTW, red cabbage makes a good indicator for this.


What are the decay products of HTO? Tritium, symbol $\ce{T}$, is hydrogen with a mass of $3\ \mathrm{amu}$. It is radioactive and undergoes $\beta$ decay. Which of the following could be the only products after a quantity of $\ce{HTO}$ undergoes decay?

$\ce{HeOH}$
$\ce{H2O}$, $\ce{O2}$ and $\ce{He}$
$\ce{H2O}$, $\ce{H2}$ and $\ce{He}$

The answer is 2 only, however I am finding it difficult to understand how I could have worked this out. Tritium undergoes beta decay, so: $$\ce{^3_1H+ -> ^3_2He^{2+} + e- + \bar {\nu}_e}$$
I had assumed it wouldn't be possible for the electron emitted in beta decay to become an orbital electron. I have seen the following question: What does HTO decay into?, but it doesn't answer my specific query. I could state that the helium ions produced gain 2 electrons from the oxygen ions to become helium, with these oxygen atoms then reacting to form $\ce{O2}$ and some of the hydrogen combining with oxygen to form water, leading to answer 2, but this feels somewhat arbitrary. What approach could be taken to solve this problem?

I don’t consider your working to be arbitrary at all. Let’s first consider the β decay including the associated $\ce{OH}$ part of $\ce{HTO}$:

$$\ce{_1^3THO -> _2^3He^2+ + OH- + e-^ + \bar{\nu_e}}\tag{1}$$

As you suggested, the electron created in equation (1) won’t stick around, it is emitted into the environment, leaving us with a very reactive $\ce{He^2+}$ ion. Obviously, that is a strong oxidant as it wants to get reduced itself. The closest electron sources are $\ce{OH-}$ and $\ce{HTO}$, so we can consider a number of other reactions. However, it makes sense to organise it in a simple redox reaction. The reduced component is obviously going to be $\ce{He^2+}$ which only leaves $\ce{HTO}$ being the oxidised component — and since both hydrogen and tritium are in the $\mathrm{+I}$ oxidation state in $\ce{HTO}$, it must be oxygen which is oxidised to either peroxide or elemental oxygen. Indeed, peroxides are not stable and typically disproportionate into oxygen and water anyway, so we might as well just consider the formation of oxygen. Here goes:

$$\textbf{Reduction:}\\
\begin{gather}\ce{He^2+ + 2 e- -> He^{\pm 0}}\tag{2}\end{gather}$$
$$\textbf{Oxidation:}\\
\begin{align}\ce{2HTO^{-II} &-> O2^{\pm0} + 4 e-}\tag{3.1}\\
\ce{2 HTO & -> O2 + 4 e- + 2 H+ + 2 T+}\tag{3.2}\end{align}$$
$$\textbf{Redox:}\\
\begin{gather}\ce{2 He^2+ + 2 HTO -> 2 He + O2 + 2 H+ + 2 T+}\tag{4}\end{gather}$$

Equation (3) and (4) would be similar if I had chosen to oxidise only to $\ce{H2O2}$ and then would have been followed by an equation (5) to show the disproportionation of hydrogen peroxide:

$$\ce{2 H2O2 -> O2 + 2 H2O}\tag{5}$$

In the end, you will arrive more or less at the same substances.

Note that to end up with hydrogen gas, you would have needed a reductive environment capable of reducing $\ce{H^{+I}}$ to $\ce{H2^{\pm0}}$ — not the case if you have a severly electron-desiring ion like $\ce{He^2+}$ around.


Why do double bonds cause kinks in fatty acid chains? Fatty acids can either be saturated or unsaturated. If they are saturated, they stack easily and form solids at room temperature. If they are unsaturated, however, double bonds are introduced, and these double bonds introduce kinks into the structure of the fatty acid molecule. In a different question on this site, it was suggested that the reason why double bonds cause kinks to form is that they interrupt the zig zag shape of the molecule.

However, what was left unexplained is how double bonds interrupt the zig zag shape. If we were dealing with a cis fatty acid, and both hydrogens were located on top, why would this lead to a kink?

Instead of zig-zag, zig-zag, zig-zag etc. it now goes zig-zag, zig, zig-zag, zig- etc (the bold zig is due to the cis double bond). So the direction of the alkyl chain on one side of the cis double bond is different from that of the alkyl chain on the other. You could partially correct for that "wrong turn" by rotating a single bond by $180^\circ$, but that changes a staggered conformation into an eclipsed one.

Here is a link to a 3D model: http://www.biotopics.co.uk/jsmol/fattyacids.html and a snapshot in case the link goes away:


Why does potential energy relate to exothermic reactions in this way? I am not understanding how potential energy (which is the energy that results from position or configuration) relate to exothermic reactions. We say that exothermic reactions give off heat and so lose energy but I don't see how I can use this definition of potential energy to say that potential energy is lost from the system of an exothermic reaction. It might be my definition for potential energy that is wrong, I don't know.

Energy is a useful concept for rationalising the capability of doing work or heating; other than that, it is purely a theoretical concept.
As Feynman brilliantly pointed out,

  It is important to realise that in physics today, we have no knowledge of what energy is.
  We do not have a picture that energy comes in little blobs of a definite amount.
  It is not that way.
  However, there are formulas for calculating some numerical quantity, and when we add it all together it gives “28”—always the same number.
  It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.

But let's get back to something answerable.

Energy is (a) conserved and (b) convertible into different forms.
If you got any system in a state $q_1$ with potential energy $U(q_1)$ and it changed into state $q_2$ with potential energy $U(q_2)$, then if $\Delta U = U(q_2) - U(q_1) < 0 $, energy has been lost to the surroundings.
This loss could be lots of things, most of the time it is just heat (kinetic motion).
This loss, in the big picture, is just a transfer or conversion.

Let's get to a very simple example, hydrogen combustion:

$$\ce{2 H2 + O2 -> 2 H2O}$$

Since energy can't be produced, we say this reaction releases large amounts of heat.
But, where does it come from?

All that the equation above tells us is the atomic rearrangement that has been taken place: in the left hand side we have two $\ce{H-H}$ bonds and a single $\ce{O=O}$ bond, in the right hand side we got four $\ce{O-H}$ bonds.
So quantifying the bond energies in each side gives us:

     |  Average bond energy (kJ/mol)
-----|-------------------------------
     |        |   Left    |   Right
Bond | Single | hand side | hand side
-----|--------|-----------|----------
H-H  |  -432  | 2 x -432  |    0
O=O  |  -495  |  -495     |    0
O-H  |  -467  |    0      | 4 x -467
--------------|-----------|----------
   Total      |   -1359   |  -1868

This means we get $-1868 - (-1359) = -509$ kJ for two moles of $\ce{H2}$, i.e., $-254.5$ kJ/mol of heat is released during the combustion of hydrogen.

Wikipedia says the correct value is $-242$ kJ/mol, so our calculation has an error of around 5%.

So heat energy has been released and it came from bond breaking and forming.
The energy stored in a bond could be called latent or potential, if you will.

It all boils down to conservation and convertibility.


Meaning of the 1s orbital diagram I'm a little confused about the interpretation of the 1s orbital diagram.
I know that the outer boundary is where the wave function of the 1s orbital has a constant value, which means means there's a constant probability of finding an electron on that outer boundary.

The reason I'm confused is because, isn't there an infinite number of distances from the nucleus where there's a constant probability of finding an electron around a cross-sectional slice of the sphere?

What I mean is, if we choose r, meaning distance from the nucleus as 50 pm, you could draw a circle representing that boundary as r = 50pm. You could do the same with an infinite amount of distances from the nucleus and have the same result.

Basically I'm confused as to how/why the boundary is chosen -- does the probability density need to go beneath a certain value for the boundary to be defined? Or is it arbitrary? If r < infinity, there's a finite probability of finding an electron at that distance. So doesn't this mean that the boundary is pretty meaningless?

Graphically what i mean is:

Why not:

?

Your help is greatly appreciated :D

You are right: the outer boundary is quite arbitrary. There is no any intrinsic threshold; the probability just gradually decreases lower and lower, but never reaches 0. You may draw a sphere so as to have the electron inside with 90% probability, or 95%, or 99%, or any other value as you see fit.

This, in particular, is the reason why atomic radii are somewhat arbitrary too.


Improving efficiency of absorption refrigerator (ammonia+hydrogen) Apologies if this is not the right forum for this question, I checked the Physics side and there's no questions on this, whereas there's at least one other question here about absorption refrigeration.

I have a very standard camping fridge which runs on the absorption cycle, I'm making a leap of informed guesswork and assuming it's ammonia + hydrogen based.

I'm investigating ways the efficiency or performance of the system could be improved without major surgery but I'm neither a chemist nor physicist (nor surgeon).

Here's the system as explained by the manufacturer:

And here's the actual unit, minus some lagging round the boiler:

So you can see it's pretty basic, they haven't changed the design in decades.

My main train of thought, based on the bleeding obvious, is that either more heat could be added at the boiler end, or more cooling (basically, a fan) could be added to the condenser.

However, I don't know enough about how the coolant behaves to know if this would be of any benefit - adding heat could just ruin everything, cooling beyond a certain point may make no difference, etc. etc.

Additionally, I don't know if things could be done at other points in the cycle (cooling or heating other areas such as the absorber being the obvious easy wins) to improve matters.

I would also like to add that this is mostly driven by my curiosity for tinkering with things, I realise that the real-world answer to a better fridge is to buy a compressor type one - but they're expensive (seriously, compressor camping fridges are more than a full-size household fridge+freezer) and I won't learn anything that way ;)

EDIT: Thanks for the responses, ultimately the thing was never going to be viable running on 12v so we bought a compressor-type which is about 10x more efficient (in electricity) by comparison.

I've come here for basically the same reason, however I have some suggestions that I've done and I believe they have helped.
And my fridge is basically the same as your, just a bigger model, same manufacturer.
Firstly I should suggest messing with the process (heat etc) would not be a good idea as I would assume the manufacturer would have optimized the design years ago, and I have heard over heating the boiler can cause catastrophic and irrepairable damage .
My fridge has a freezer box (shoebox size, quite useless) within the fridge and a heat sink next to it for the fridge part. I took the door off the freezer and cut the base out of the box, that made a huge difference, infact with it set on high on a cool day of 20c or less items in the bottom of the fridge will freeze.
But as the camping car is used mostly in summer and it's hot that doesn't happen then.
If you need a freezer use the engel.
The fridge is mounted permanently into the camping car with a door on the outside to access the back of the fridge, it has vents top and bottom, on the inside of the bottom door vents I've installed 2 mini computer 12v power fans, they are very week but do set up a positive upward circulation of air over the absorber/condenser. As these use so little electricity they can be connected into your auxiliary battery without fear of draining it.
I think the added circulation helps on hot days.
And finally if you can't avoid parking it with the rear of the fridge on the sun exposed wall try a shade cloth over the area.
Hope that helps.


Why is the Haber process carried out at such high temperatures? On a large scale, ammonia is prepared via the Haber process:

$$\ce{N2(g) + 3H2(g)->2NH3(g)} \qquad \Delta _\mathrm{f}H^\circ = -46.1~\mathrm{kJ \cdot mol^{-1}}$$

The optimum conditions for the productions of ammonia are a pressure of $200~\mathrm{atm}$ and a temperature of about $700~\mathrm{K}$.

The process obviously is exothermic and $700~\mathrm{K}$ is, by no means, a low temperature. Shouldn't the temperature be much lower for optimum production of ammonia?

As others have pointed out, it is purely kinetics, but you may still wonder, why.

For a reaction to actually occur (in both directions) and thus for an equilibrium to be reached, you need to overcome the activation energy. In the case of the Haber-Bosch process, this involves breaking the highly stable $\ce{N#N}$ triple bond. Even with the catalysts used, the energy required to break apart $\ce{N2}$ is still enormous. Therefore, a lower temperature may give a better yield of ammonia theoretically (i.e. based on equilibrium and Le Châtelier considerations) but the reaction speed would be a lot slower.

Even if you consider a batch-wise process of generating ammonia (which, as orthocresol points out, isn’t the case), it is more efficient to run two batches in half the time for two sets of $15~\%$ than to run a single batch for twice the time to get an overall yield of maybe (note: This yield is a ballpark estimate) $25~\%$ — with a batch size where $1~\%$ yield is equivalent to $100~\mathrm{kg}~\ce{NH3}$, two runs at higher temperature give $3~\mathrm{t}\ \ce{NH3}$ and one run at lower temperature $2.5~\mathrm{t}\ \ce{NH3}$ in the same time frame.


Can I use litmus paper to test unknown liquid that contains a concentrated strong acid? Can I use litmus paper to test unknown liquid that contains a concentrated strong acid (example $\ce{H2SO4}$)? Can concentrated strong acid destroy litmus paper?

Yes, concentrated strong acid can destroy litmus paper.

Paper is mostly cellulose, which is a carbohydrate polymer. Concentrated sulfuric acid will pull the hydrogen and oxygen out of it as water, leaving carbon behind. The indicator dye that's supposed to change color probably won't fare too well, either.

You can dilute the unknown by adding a drop or two of it to a small amount of distilled water, then use the litmus paper to check that.

A search for "concentrated sulfuric acid on paper" will yield many images and videos showing what the acid does to paper.


How does the Mosher ester method allow determination of absolute configuration of secondary alcohols One of the challenges in synthesis is the correct assignment of absolute stereochemistry. This has been asked previously on chem.SE (here and here), but this question specifically asks about the determination of hydroxyl stereocentres using the Mosher ester analysis.

Depending on the type of reaction carried out, this can often be done by inspection (for example, if a reaction forms a pair of diasteromers, they are often easily distinguished by NMR, assuming you know the stereochemistry of the initial stereocentre).

Where no chiral centres were previously present in the molecule (or where pre-existing chiral centres are too remote from the new centre to be relevant), assignment of absolute configuration can be challenging. Whilst x-ray crystallography can be used, many small molecules do not crystallise (or at least not without a significant amount of effort).

One method for assignment of absolute stereochemistry of newly formed hydroxyl centres is the Mosher ester analysis, which works by forming a pair of diastereomeric esters, the NMR shifts of which can be used to determine absolute stereochemistry.

How does this method work? There are many diasteromeric esters that could be formed, yet only a specific few are used for this purpose, and equally, how come there is a need to form both diasteromers of the Mosher esters, when the change in shifts could be observed by just forming a single diasteromer.

The NMR spectra of two enantiomeric molecules are necessarily identical (just as all properties of enantiomers are identical in the absence of an external chiral environment). In order to tell the enantiomers apart, some additional chirality must be introduced. Two methods exist for this in NMR spectroscopy:

Chiral solvating agents, such as lanthanide shift reagents
Chiral derivatising agents, such as Mosher esters

In modern organic synthesis, the use of chiral derivatising agents is far more common, and the process can be summarised as follows:

Outline of the Mosher ester analysis. Taken from Organic Spectroscopy and Structure Elucidation lecture notes (not available online)

a. The alcohol of unknown configuration is derivatised by forming both the (R) and (S) Mosher esters. This can either be done under Steiglic conditions (DCC,DMAP) or via initial formation of the acid chloride.

b. Proton NMR spectra are ran of the two diasteromeric esters. The analysis is often ran on the crude esters, therefore the acquisition of COSY/HSQC is also useful to aid in assignment.

c. The difference between the chemical shifts of the (R) and (S) esters are tabulated (by convention we consider R-S, it makes no fundamental difference to the outcome however).  Note that the method does not consider the shift of the underivatised alcohol at all, however the peaks for the (R) and (S) ester must be unambiguously assigned. Note also that we do not take into account the oxymethine proton (the shifts of this are erratic and don't contribute to the analysis).

d. The configuration is assigned by considering which side of the molecule is most effected by the presence of the Mosher ester.

Step d is often the hardest to grasp conceptually. When we make the Mosher esters, what we're counting on is that they adopt a specific conformation (which they do). In this conformation, the phenyl group of the Mosher ester will shield part of the molecule (which is different depending on which enantiomer of the Mosher ester we're looking at). Consider the following example, in which the side shielded by the phenyl ring has a lower delta overall:

Shielding by the Mosher ester. Taken from Harvard CHEM2016 lecture notes, Eugene Kwan

The following conformational diagram is also (possibly more) instructive:

Shielding by the Mosher ester. Source unknown

The only way to actually understand the Mosher ester method is to carry it out (ideally with something of unknown stereochemistry so you can't convince yourself of the answer simply because you know its the case.

The next best thing is a worked example:

Outline of the Mosher ester analysis. Taken from Organic Spectroscopy and Structure Elucidation lecture notes (not available online)

Looking at the two spectra, it becomes apparent that (R)-(S) is positive on the LHS methyl group, and negative on the protons on the RHS propyl group. Fitting this to the model, we can assign the sterocente of the secondary alcohol as (R).

The reason both Mosher esters should be formed is due to the fact that change in chemical shift between one Mosher ester and the free alcohol is often not huge. Other derivitising agents (such as AMA) can be carried out with formation of a single ester.


Phenol vs 4-hydroxypyridine - Enol Content Which has a higher enol content; phenol or 4-hydroxypyridine?

The solution to this question in my book says that phenol has a higher enol content due to aromaticity. But 4-hydroxypyridine is also stabilized by resonance. So how does that affect enol content?

Your solution manual correctly predict that phenol has a higher enol content due to aromaticity. For example, you won't see strong carbonyl peak around $\pu{1700 cm-1}$ in FT-IR, which proves keto-form does not exists. It is 100% enol-form as phenol.
On the other hand, 4-pyridone is a different story. The dominance form, whether it is 4-pyridone or 4-hydroxypyridine, would be depend on which phase the compound is in. For example, the Ref.1 states that:

In the gas phase, there is a preference of the hydroxy form in both 2- and 4-hydroxypyridine, while in a nonpolar solvents such as cyclohexane and chloroform, the two tautomers exist in comparable amounts. However, the two equilibria are shifted entirely to the right in favor of the oxo-form, 2- and 4-pyridone, respectively in polar solvents as well as in the crystalline state.

Thus, it is safe to say, it is only in the gas phase, 4-hydroxypyridine dominate. However, in liquid phases, both form exist where predominance is depending on the solvent. In solid phase, 4-pyridone dominate. Since, both phenol and 4-pyridone are solid at ambient conditions, phenol has more enol form than 4-pyridone, which exist in keto-form as a solid.
Note that a public database set like Tautobase, implemented in DataWarrior, may be used to train yourself, too.
Reference:

Jiali Gao, Lei Shao, "Polarization Effects on the Tautomeric Equilibria of 2- and 4-Hydroxypyridine in Aqueous and Organic Solution," J. Phys. Chem. 1994, 98(51), 13772–13779 (https://doi.org/10.1021/j100102a051).
Oya Wahl, Thomas Sander, "Tautobase: An Open Tautomer Database", Chem. Inf. Model. 2020, 60, 1085–1089 (doi 10.1021/acs.jcim.0c00035).


Why does ZnO turn yellow when it is hot? Why does ZnO's color become yellow when it is hot? I have read that it is because of physical properties, but I need a more elaborate explanation.

The thermochromism of $\ce{ZnO}$ results from a minor loss of oxygen upon heating to temperatures around 800 °C, i.e. a non-stoichiometric $\ce{Zn$_{1+x}$O}$ with $x = 7 \times 10^{-5}$ is formed.

Under air, this effect is reversible. Heating (and cooling) of the material while hooked up to a vacuum pump might result in a more persistent colour change.


Is ionic bond just formed by "electrostatic" interaction between two oppositely charged ions? According to the definition of ionic bond:

  An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.

If we consider the above figure showing formation of ionic bond, due to the transfer of electron from sodium atom to fluorine.It demonstrates clearly, how electrostatic interaction takes place.
In general, electrostatic interaction takes place between two stationary charges.Moreover, electrons would not be static as demonstrated by the above figure.Electrons will be moving at high speed, producing magnetic field around them, due to their motion; and even will also have electric field around them.Thus, we can expect electromagnetic interaction between oppositely charged ions along with electrostatic interaction due to stationary protons.So, does it mean that ionic bond is due to electromagnetic interaction and not just by electrostatic interaction?
[All the statements made are up to my view on the concept.Any correction advisory is welcome]

REFERENCES

http://en.wikipedia.org/wiki/Ionic_bond   (definition and diagram has been extracted from this link)
http://en.wikipedia.org/wiki/Electromagnetic_radiation (provides notion of charge motion effects)
http://en.wikipedia.org/wiki/Electrostatics (gives notion of electrostatic force)

You are absolutely correct that the electron has an angular momentum, which can be thought of like a magnetic field and which interacts with magnetic fields.
In "light" elements, however, the magnetic interaction is almost completely negligible.

The Hyperphysics page on the Zeeman Effect shows that in a hydrogen atom, a magnetic field of one tesla (which is fairly strong) splits energy levels by $5.78 \times 10^{-5} eV$, or roughly one-millionth of the ionization energy of a hydrogen atom.

From just having done calcuations on the anomalous Zeeman effect for homework (you can use the formula in the link I provided and check it yourself), I can tell you that in a potassium atom, the energy levels of a p-electron split by 0.006 Hartree/Tesla, or 1/10,000 of the ionization energy of a hydrogen atom.

In other words, while there is a magnetic field (or what can be treated as one) generated by electron motion, it's fairly safe to ignore it in lighter elements.

As far as heavy elements go...well, you undergo what's called j-j coupling, caused by the fact that the electrons are going so fast that their motion becomes relativistic. To describe that motion requires the Dirac equation, which I have not studied in any level of detail. Instinct says that there are other considerations which are much more significant than magnetic interactions, but I have no proof whatsoever to back it up.


I'm a layman, but I'd like to know to to use GAMESS for some specific tasks I have Avogadro and GAMESS. I modeled the sulfuric acid molecule using Avogadro, then I asked it to generate an input for GAMMES to calculate the transitional state (I'm hoping this will get me somehow closer to my computational modeling of real chemistry). But GAMESS says the job finished with an error, and I don't know why.

It's worth noticing Avogadro seems to spell one of the commands wrong, and I have to manually correct it. Avogadro issues the command \$STATTP but GAMESS does not recognize this command, so I change it to \$STSTEP, and it ends with an error, and I try the other keyword possible \$STPT and I also get an error. Here's my 2 inputs and outputs:

http://pastebin.com/TabWC9wr

This is what happens when I do what you told me, @Mitradip Das:

INPUT

!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=STO NGAUSS=3 $END
 $CONTRL SCFTYP=RHF RUNTYP=SADPOINT $END
 $STATPT OPTTOL=0,0001 NSTEP=20 $END

 $DATA
Title
C1
S    16.0    -3.61564     3.32108     0.04978
O     8.0    -4.51377     4.03518    -0.82228
O     8.0    -2.75016     3.96121     1.00661
O     8.0    -2.67033     2.39549    -0.85711
O     8.0    -4.49699     2.21900     0.81326
H     1.0    -5.20320     1.93219     0.19859
H     1.0    -1.81232     2.32072    -0.38885
 $END

OUTPUT (showing only the error part, full output can be viewed here (http://pastebin.com/HRh1fS4d)

 CONSTANT LIST TOO LONG
 **** ERROR READING INPUT GROUP $STATPT   *****
 THE PROBLEM IS WITH THIS INPUT LINE, NEAR THE X MARKER
 $STATPT OPTTOL=0,0001 NSTEP=20 $END
                       X
 THE ONLY KEYWORDS ACCEPTED IN THIS GROUP ARE:
  METHOD    UPHESS    NSTEP     NNEG      IFOLOW    DXMAX
  RMIN      RMAX      RLIM      RESTAR    HESS      OPTTOL
  IHREP     HSSEND    IFREEZ    NPRT      NPUN      PURIFY
  ITBMAT    MOVIE     TRUPD     TRMAX     TRMIN     STPT
  STSTEP    PROJCT    IFCART    NPRTHS    KDIAGH    IACTAT
  IHMCON    FHMCON    SHMCON
 TROUBLE INTERPRETING $STATPT
 OBTAINING INITIAL HESSIAN, HESS=READ
 ***** ERROR, YOU DID NOT SUPPLY A $HESS GROUP.
 EXECUTION OF GAMESS TERMINATED -ABNORMALLY- AT Wed Jul 27 12:23:49 2016
                   580000  WORDS OF DYNAMIC MEMORY USED
 STEP CPU TIME =     0.00 TOTAL CPU TIME =        0.0 (    0.0 MIN)
 TOTAL WALL CLOCK TIME=        0.0 SECONDS, CPU UTILIZATION IS 100.01%

job aborted:

For GAMESS, the command is not $\$STATTP$, but it is $\$STATPT$. Also, please change the commas to dot, otherwise it considers it to be a part of the number.

Please refer to the GAMESS manual for better understanding. The link is: www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf