Effect of magnetization on oxidation (rusting) of iron Can magenetizing a piece of iron bar slow-down (or speed-up) the oxidation (rusting) process? In other words have any influence on it ?
From what I've looked up( wikipedia, quora), it appears that the property of magnetism depends on 'spin' and also the creation of bonds does have an impact on the 'spin' (apparently has something to do with pauli's exclusion principle).
PS: Tried to ask the question on physics.stackexchange.com but did not find all the tags so decided to ask here.

OK, those who say magnetic fields have no effect can buy a relatively inexpensive Magnetizer (as I once did) and try performing a controlled experiment (as I attempted) with the rusting of iron. There is some apparent evidence that alignment of magnetic fields can promote radical activity.
This UK study on the effect of a magnetizer in reducing the amount of chlorinating pool chemicals required, is an example of a successful application. Reference, see Magnetic treatment of swimming pool water for enhanced chemical oxidation and disinfecting.
However, with respect to possible applications and significant results, apparently electrochemistry applications are at the top. For example, see Zou, P.; Leddy, J., Magnetized Nickel Electrodes for Improved Charge and Discharge Rates in Nickel Metal Hydride and Nickel Cadmium Batteries, in Electrochemical and Solid-State Letters 2006, 9, A43-A45. To quote from the article:

...magnetized electrodes yielded higher currents than the corresponding nonmagnetic electrodes by 70 and 230 %, respectively.

Source material for background includes Magnetic field induced motion behavior of gas bubbles in liquid, by Keliang Wang, Pucheng Pei, Yu Pei, Ze Ma, Huachi Xu & Dongfang Chen, in Scientific Reports 6, Article number:21068 (2016), doi:10.1038/srep21068 ), interesting article comments include:

the kinetics of oxygen bubbles is mainly proportional to intensity of the electromagnetic field. The magnetic-field induced rotational motion of oxygen bubbles in a square electrolyzer can increase liquid hydrodynamics, thus solve the problems of oxygen bubbles coalescence, and uneven distribution of electrolyte composition and temperature. These types of oxygen bubbles movement will not only improve energy saving and metal deposition for energy storage and metal refinery, but also propel object motion in application to medical and martial fields.

Note, with my experiment reported above with iron rusting, I observed significant more frothing/air bubbling activity around iron in the presence of the Magnetizer. One could look at my experiment as a crude rendition of an iron metal/air battery cell with the Magnetizer forming a magnetic iron electrode.
Finally, a link to a prior StackExchange comment on the topic here.


Elemental Silver Colour What is the colour of elemental Ag? I have doubt whether it is black or silver in colour.

In Tollens' test, we get a silvery precipitate of elemental silver - hence silver in colour.
I have read somewhere and also in the following links as a reference that it was not mere speculation

http://books.google.co.in/books?id=AFY2gKwh4AsC&pg=PA553&lpg=PA553&dq=elemental+silver+precipitate&source=bl&ots=nzfsQcBAm6&sig=jTCQk7-k0HzAByI25ESI6OSclJs&hl=en&sa=X&ei=S1d7U6aXIY-OuASHp4HIDQ&redir_esc=y#v=onepage&q=elemental%20silver%20precipitate&f=false

(towards the end)
and

http://books.google.co.in/books?id=cZL_-hyw98sC&pg=PA675&lpg=PA675&dq=elemental+silver+precipitate&source=bl&ots=DDhdhikKhU&sig=zIq-qQM7VI4i5enohZVi3oIagHw&hl=en&sa=X&ei=S1d7U6aXIY-OuASHp4HIDQ&redir_esc=y#v=onepage&q=elemental%20silver%20precipitate&f=false

Please confirm this. Thanks in advance.

Silver:

Silver sulfide precipitate is black.  You are pobably confusing a precipitate in a test for silver, with actual silver.  The precipitate in a test will be a compound of silver, such as silver sulphide which is black.

Nanophase metal particles are optically different from bulk in many cases, and can be black, but this doesn't seem to be what the question is referring to.


kinetic and thermodynamic parameters of biodiesel production from lard I want to know if someone has experimental results and calculations of the kinetic and thermodynamic parameters of Base Catalyzed Transesterification of pig fat, if not which is the method used to calculate them?

Thanks!! :)

I don't expect a significant energy difference per ester bond between glyceryl and methyl esters of a fatty acid. The way to drive such a reaction forward is by removing the desired product as the reaction proceeds. Then more product will form as the system seeks a new equilibrium.


Quick Stoich & Gasses I have a quick question in Stoich and Gasses, I however could not get any far, any answer will be appreciated. Thank You.

  A sample of $\ce{Pb(NO3)2}$ is heated, yielding $\ce{O2, NO2\ gas,\ and\ solid\ PbO}$. A
  $\ce{293 \ ml}$ gas sample, measured at $\ce{200C}$ and $\ce{1\ atm}$, is collected. What
  was the mass of the $\ce{Pb(NO3)2}$ sample?

I feel like in this situation I need to have only one gas so I can apply the Ideal Gas Law to solve and find the mass of $\ce{Pb(NO3)2}$. What should I do there are two gasses and I don't know the volumes or the masses?

$$\ce{2Pb(NO3)2 -> O2 + 4NO2 + 4PbO}$$

The equation of the reaction provides you with the composition of the gas liberated.

You assume ideal gas, which means that the gas particles are indistinguishable. This means the reaction equation could look like
$$ \ce{2Pb(NO3)2 -> 5X + 4PbO} $$
where 5 moles of gas are liberated for every 2 moles of lead nitrate.

Reforming the ideal gas equation to $n=pV/(RT)$ will lead you to the numerical value. The amount of lead nitrate will then be $$n_{\ce{Pb(NO3)2}} = \frac{2n}{5}$$

From that calculating the mass is a trivial task.


What is the reason a grain of salt alters the perceived bitterness of coffee by such a high degree? Coffee served hot can be quite bitter (presumably high pH).

Several people I have known will add a single grain of salt to the coffee - which reduces the perceived bitterness.

My question is: What is the reason a grain of salt alters the perceived bitterness of coffee by such a high degree? Can you explain the pH change that is occurring during this reaction?

Is simply a dissolution no reaction occur and as said Niculau the pH change, if occur, is absolutely negligible.

Salt in general affect taste lowering the bitterness of the food enhancing the positive sensory attributes of foods. In fact the reason of this behavior is the sodium ion $Na^+$ (even litium ion works) and his interaction with the receptors in taste buds. So sodium acetate therethically works as well, Gery Beuchamp has studied many of this salt, quoting from NCBI:

  various sodium-containing ingredients have been known to reduce the
  bitterness of certain compounds found in foods, including quinine
  hydrochloride, caffeine, magnesium sulfate, and potassium chloride

This is an article about the suppression of bitterness by sodium. From here I'm afraid is more a question of physiology.

I read this the first time here:

This, Herve. Molecular gastronomy: exploring the science of flavor. Columbia University Press, 2006 p. 94


Why is the carbon monoxide triple bond stronger than the nitrogen-nitrogen triple bond? The $\ce{CO}$ triple bond has a bond enthalpy of 1072 kJ/mol while the $\ce{N2}$ triple bond has a bond enthalpy of 945 kJ/mol, at least according to my source. The molecules are isoelectronic, so I couldn't think of any way to explain it using MO theory. I've read many explanations that did not make much sense, including

Of course because they are different elements and nitrogen nitrogen bond is balanced and oxygen is more electronegative than nitrogen so it will have a greater bond strength

So my question is why is the $\ce{CO}$ triple bond stronger than the $\ce{N2}$ triple bond?

I agree in part with Mithoron and MaxW's comments: these are different molecules, not even sharing a mutual atom (e.g. C-O vs C-C), so direct comparison is restricted.

However, Pauling's concept of electronegativity does explain why, in general, the heteronuclear A-B bond is stronger than the average of the homonuclear A-A and B-B bonds. the difference in the electronegativity of the two atoms in the bond lend a somewhat ionic character to the bond, strengthening it.


Is it possible to have an allotrope of carbon as a long unbranched chain of itself? Is it possible to have an allotrope of carbon with formula $\ce{(=C=)_n}$?

Well, let's actually leave alone the separate question like from which "monomer" would one produce such a thing (can one actually consider it being a polymer? Or a covalent crystal?), because it seems like the most suitable candidate would be dicarbon.

All in all, might it be (under some conditions) energetically favorable for carbon to form a long unbranched chain of itself?

Well, this is a rather obvious development (in retrospect, that is). We had an $sp^3$ carbon with tetrahedral bonds, and it made diamond. We had an $sp^2$ carbon with trigonal bonds, and it made graphite. What if we just had a linear $sp$ carbon?

Such thoughts have been around for more than quite a while. Theorists swarmed around that non-existent tree. No wonder that people started trying to actually obtain that linear carbon.

Not knowing anything about you, I still can claim with a good deal of confidence that the first reports of its synthesis are quite probably somewhat older than you (or me, for that matter). Then they were proved... not quite reliable. Some groups agreed with that, some disagreed, a great deal of confusion abounded, then more reports came and things became hairy. So they stayed for the past few decades and will stay for a while, I guess.

Now to the Q&A.

Does this compound exist?

· Define "exist", otherwise no meaningful answer would be possible.

Is it stable?

· Define "stable", otherwise no meaningful answer would be possible.

Is it thermodynamically stable at any P and T?

· Most researchers agree that it isn't.

Is it electrically conductive along the strands?

· Most researchers agree that it is.

Is it made of double bonds or of alternating single and triple bonds?

· The difference is less than you seem to think, to the point that it might not be a question worth pursuing.

What is its crystal system?

· Nobody knows. Chances are there are polymorphs with different crystal systems.

Is it super strong?

· Nobody knows, as with any non-existent compound.

Should it be called a polymer?

· About as much so as graphite and diamond, which means: No.


Is Henry's Law sufficient for calculating CO2 solubility? $\ce{CO2}$ reacts with water which is why it enters and exits solution slower than other gasses (if I understand correctly this is why carbonated drinks bubble gradually instead of all at once).

Henry's law is an approximation that assumes the gas solute does not react chemically with the solvent.

Is the Carbonic acid reaction significant enough that Henry's Law does not work for $\ce{CO2}$?
What about in the case of blood, where the reaction is catalyzed by the enzyme Carbonic Anhydrase?

There are multiple forms of Henry's law (see this Wikipedia entry). The net or effective Henry's Law takes into account the other pH-dependent equilibria of $\ce{CO2}$ and carbonate species in water (chemical reactions with water in your question).

In the case of blood, as you said the picture is even more complex. You would have to be careful using, for example, published values of the effective Henry's Law solubility of $\ce{CO2}$ in that case.


Purging submersible with nitrogen I do a lot of short term ocean field deployments where system electronics are self contained in a submersible enclosure. A new project I'm working on requires a long term deployment in the deep ocean. Therefore, once the enclosure is deployed, I can not just go recover the instrument if something fails. I've spoken to a lot of colleagues about how they prepare electronic enclosures and one of the standard protocols is purging the enclosure of atmospheric air. For one, any moisture in the air can condense at cold, deep depths, which can cause electrical problems. I think another reason they do it is to reduce oxygen exposure. Typically the purging is done with argon from an air cylinder. However, this is something I don't have. Now I know I can go out and get one...that's not the point of this question. What I do have is a small nitrogen generator that was needed for other projects.
The question I would like to ask here is, can I use the 99.x% dry nitrogen air from the nitrogen generator as a substitute for argon? I know argon is denser than nitrogen, so clearly is a better choice, but I'm not concerned with the better choice. I want to know if I can use what I have at my disposal to achieve the same goal: replace atmospheric air with 99.x% dry nitrogen air?

Using argon is, in effect, a costly physical path to remove oxygen, hence your interest in using nitrogen.
However, I suspect, a more efficient, cost-effective, and likely safe chemical path is also possible. For example, assuming a control structure (so as to mitigate potential deaths from O2 depletion, as has been reported, for example, among sewer workers working in a recently drained iron pipe), oxygen scavenging can result from a limited exposure to iron surface previously scrubbed with an acid. Here, lemon juice appears to be a friendly, pleasant, and effective choice.
Note, iron powder and ferrous salts have been used in the food industry to remove problematic oxygen (see, for example, THE USE OF OXYGEN SCAVENGERS AND ACTIVE PACKAGING TO REDUCE OXYGEN WITHIN INTERNAL PACKAGE ENVIRONMENTS).
One should be able to stoichiometrically compute, and experimentally validate, a safe amount of Fe (or its surface area) to employ. This is to limit a continuing unsafe scavenging action if, for example, subsequent maintenance work is required.
You can also employ commercially available products containing Calcium Chloride, for example, to remove water vapor.


What determines the elastic limit of a substance? from what I know, the elastic limit of a substance is the maximum amount of stress/force/pressure (a bit of confusion on my behalf here) a substance can withstand and still return to its original shape afterwards. my question is what factor(s) determine this limit?

also, I have noticed that some plastics go opaque when under stress and have found the reason for this effect in  this question however, what I do not understand is why some plastics and other synthetic materials do this before the elastic limit (and turn clear upon shrinking) and others do this after the elastic limit of the substance (and the effect is permanent). why can this process of turning opaque be reversed in materials that turn opaque before the elastic limit of the substance.

First let me clarify that there is no way to calculate the elastic limit of a substance - it has to be measured. Viscoelastic theory is pretty well-developed for only being about 150 years old, but it's not that well developed.

To answer your question, though, lots of factors affect elasticity in general and the elastic limit in particular.

It would probably be the most helpful to think about what is happening on the molecular scale. In any viscoelastic material, there is a range of deformations (strain, or the response to stress) and timescales over which molecules will go back to pretty much the same relative position that they were in before the deformation occurred. That's elasticity. What makes this happen? Each atom or molecule sits in a little potential energy well that exists because of its position relative to other atoms in the substance. Atoms like to be close to each other, but not too close. As a result, when given the opportunity they will arrange themselves so that as many atoms as possible are the optimal distance apart(resulting in the lowest energy).

When a material is deformed, we are moving atoms relative to each other. When one atom moves, its energy increases. This also changes the potential energy surface seen by its neighbors. If it moves a little bit, then the change is not usually great enough to cause the neighbors to move by much. On the other hand, for big movements, the effect can be great enough that the other nearby atoms rearrange themselves in order to find the next lowest energy level.

When the relative motion of atoms due to mechanical deformation is small enough that most of them go back to their original positions once the applied stress is removed, then we say the stress/strain was within the elastic limit.

When the atoms are permanently rearranged, the shape of the material is changed and we say that we have exceeded the elastic limit.

The ease with which atoms can move out of position (you can think of this like the depth of the wells), and their tendency to move back to the lowest energy (which can be thought of as the "steepness" of the wells) are the two main factors that affect how much stress/strain is needed to permanently deform a material. There are lots of things that can affect each of these, which is why there is no comprehensive theoretical model that can predict the elastic limit of materials based on molecular structure alone.

I can summarize a few important factors here, though:

degree of crystallinity in polymeric materials
type and number of dislocations in crystalline materials
length of time that the stress/strain is applied, in conjunction with the temperature of the material (this related to a phenomenon called time-temperature superposition)
structure of the molecules making up the material (e.g. linear vs. crosslinked or branched polymers, fcc vs. bcc crystal structure, or type of atoms in an alloy)
mesoscopic structure (e.g. crystalline grain size and shape, aggregation of nanoparticles, or long-range order in polymers)
macroscopic structure (e.g. fiber orientation in composite materials)

There are many more! And each of the ones I listed is practically an entire field of its own in materials science.

To answer the second part of your question - the cloudiness you are referring to is called "crazing" in polymer science. It is basically caused by molecules pulling apart from each other, without covalent bonds breaking. It can be reversed when it happens before the elastic limit is reached because the molecules can still go back to their original positions. If it doesn't happen until after the elastic limit, then the material is permanently deformed, and the molecules cannot return to their original position.


Why does activation energy depend on temperature? To me it makes sense that the activation energy $E_\mathrm a$, as given by the Arrhenius equation, doesn't depend on the temperature as the needed energy (potential energy) to start a reaction has nothing to do with the temperature (kinetic energy) of the molecules. And a higher temperature simply means that more molecules have sufficient energy to proceed in the reaction $E\gt E_\mathrm a$.
So I don't understand why activation energy would depend on temperature, but this equation shows that it does $E_\mathrm a=\Delta H^{\ddagger}+RT$.
Can someone explain why?

Although the question has been already answered, I will like to add a bit more of detail.
The expression you are quoting come from the thermodynamic formulation of the TST that expresses the dependence of the rate constant of temperature as
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta G^\ddagger_0}{RT}\right)$$
and can be partitioned in an entropic and enthalpic contributions
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta S^\ddagger_0}R\right)\exp \left(-\frac{-\Delta H^\ddagger_0}{RT}\right)$$
From this equation you can obtain the activation energy $E_\mathrm a$ using the relation
$$\frac{\mathrm d\ln k(t)}{\mathrm dT}=-\frac{E_\mathrm a}{RT^2}$$
Using this equation you will obtain
$$E_\mathrm a=\Delta U^\ddagger_0+RT$$
Since $H=U+pV$
$$\Delta H^\ddagger_0=\Delta U^\ddagger_0+p\,\Delta V^\ddagger_0+RT$$
For a condensed phase reaction $\Delta V^\ddagger_0\approx0$ and
$$E_\mathrm a=\Delta H^\ddagger_0+RT$$
For a gas phase reaction, $p\Delta V^\ddagger_0=\Delta n^\ddagger RT$.
For a gas phase unimolecular reaction $\Delta n^\ddagger=0$, the result is equal to that the obtained by a condensed phase reaction, while for a second-order reaction $\Delta n^\ddagger=-1$, and
$$E_\mathrm a=\Delta H^\ddagger_0+2RT$$
You can find a more detailed discussion in Keith Laidler, Chemical kinetics 3rd. ed., Harper Collins Publishers, (1987); Jeffrey I. Steinfeld, Joseph S. Francisco, William L. Hase, Chemical Kinetics and Dynamics, Prentice Hall (1989); or Paul L. Houston, Chemical Kinetics and Reaction Dynamics, McGraw-Hill (2001).


What is the mechanism of this pyrylium synthesis? In this preparation, http://www.orgsyn.org/Content/pdfs/procedures/CV5P1106.pdf, 2,4,6-trimethylpyrylium perchlorate is prepared from 4 molecules of acetic anhydride and 1 molecule of t-butanol.
However, I cannot find a mechanism of this reaction, nor can I think of one. Could someone please provide the mechanism? Thanks.

I tried to work out a mechanism from the information I could gather about the reaction. If someone notices something odd/wrong please comment and I'll fix it.

The steps are either showing the protonation or skipping directly to the protonated forms of the reactants, them being in acidic medium.

Formation of isobutylene

Not very complicated, in acidic conditions we remove the hydroxyl group from the t-butanol to form isobutylene. The latter behaves as a nucleophile in the next steps of the reaction. This step is acid catalysed, so the acid is not consumed.

Formation of Mesityl Oxide

Also relatively simple step. Nucleophile addition to the protonated C=O bond [Adn] followed by elimination of Acetic Acid.

Tautomerism

Shows the Mesityl Oxide tautomerism, important because the tautomer that will react to form the final product is the one in the right (with the double bond in the terminal carbon).

Enol Formation

Here a Enol is formed by a nucleophilic attack of the Mesityl Oxide over another Acetic Anhydride molecule (protonated by the acidic medium). In the last step I added the deprotonation of the more acidic oxygen to respect Pauling's rule of charges by keeping an overall +1 charge in the molecule. The left oxygen is more acidic since the other oxygen can better stabilize the charge with delocalization. I wrote the molecule with wavy bonds because I'm unsure about the stereospecificity of the step, but maybe the Z conformation is favored due to the charges on the oxygens repelling themselves?

Cyclization

Final step is cyclization of the intermediate, by nucleophilic attack of the carbon by the deprotonated oxygen. The intermediate is dehydrated and rearranged to form the final product (releasing a proton to reach aromatic status). This step had an intermediate with +2 charge, which I'm a little skeptical about due to the Pauling's rule of charges, so I also draw the mechanism as a concerted step:

If the rule of charges isn't that strict, maybe the non-concerted path is a closer description of what's happening, but I'm keeping both here for now. If anyone thinks one of the two is the right one, comment and I'll update the answer.


Why is the ionization enthalpy of francium greater than that of cesium? Why is the ionization enthalpy of francium greater than that of cesium, even though it has a larger size? I found no Google result regarding this.

The reason is the same as in the case of barium and radium, a contraction of core electron orbitals with increasing effective nuclear charge. This also changes the shielding (from nuclear charge) of electrons in outer orbitals, and in the end, all $s$ orbitals are contracted and lowered in energy. $s$ orbitals are particularly affected because they have no nodal plane at the nucleus.

This means that the single valence electron in the $7s$ orbital of $\ce{Fr}$ is closer to the nucleus and thus more tightly bound than the $6s$ electron of $\ce{Cs}$. More energy is therefore required to remove the $7s$ electron to form $\ce{Fr+}$ than in the case of $\ce{Cs}$, whose $6s$ electron is more loosely bound.


Why does bond length decrease with increased basis set quality for SCF methods? From Reviews in Computational Chemistry Vol. 5, for $\ce{H2O}$ using HF, MP2, CCSD, and CCSD(T), all lengths in angstroms:

$$
\small
\begin{array}{rcccc}
\text{Basis}  & \text{HF} & \text{MP2} & \text{CCSD} & \text{CCSD(T)} \\
\hline
\text{STO-3G} & 0.989 & 1.013 & 1.028 & 1.028   \\
\text{3-21G}  & 0.967 & 0.989 & 0.993 & 0.994   \\
\text{DZ}     & 0.951 & 0.979 & 0.979 & 0.980   \\
\text{6-31G*} & 0.947 & 0.969 & 0.969 & 0.971   \\
\text{DZP}    & 0.944 & 0.963 & 0.961 & 0.962   \\
\text{TZ2P}   & 0.941 & 0.958 & 0.956 & 0.959   \\
\text{Experimental} & 0.957
\end{array}
$$

In case one is skeptical due to effects from the lone pairs and the changing bond angle, here are results for $\ce{BH}$ using Hartree-Fock:

$$
\small
\begin{array}{rc}
\text{Basis}  & r_{e} \\
\hline
\text{cc-pVDZ} & 1.235950 \\
\text{cc-pVTZ} & 1.222062 \\
\text{cc-pVQZ} & 1.220306 \\
\text{cc-pV5Z} & 1.220010 \\
\text{cc-pV6Z} & 1.219877
\end{array}
$$

and for $\ce{LiH}$ using Hartree-Fock:

$$
\small
\begin{array}{rc}
\text{Basis}  & r_{e} \\
\hline
\text{cc-pVDZ} & 1.619017 \\
\text{cc-pVTZ} & 1.607865 \\
\text{cc-pVQZ} & 1.605530 \\
\text{cc-pV5Z} & 1.605519
\end{array}
$$

Going on the NIST CCCBDB and looking at calculated geometries, I cannot find a counterexample, at least when considering only HF, as adding dynamic correlation introduces an effect to counterbalance this bond shortening, since electrons can then properly repel each other to levels beyond the mean-field approximation, and DFT offers no insight beyond HF here.

One potential reason that doesn't receive much attention is intramolecular BSSE, where one part of a molecule borrows the basis functions from another. However, in the case of my calculations, I would argue that only cc-pVDZ could be considered a "deficient" basis, and the trend is still seen when leaving it out. I also believe that this is more plausible for molecules with distinct "domains" that can interact with each other, like one amino acid in a peptide forming a hydrogen bond with another AA, etc.

The current rationale I have is that as the basis set becomes more complete, and functions with higher angular momentum (polarization functions) are included, the overlap between the atomic wave functions is increased. This increased overlap results in higher electron density in the bonding region, which strengthens it, causing it to become shorter, like going from ethane to ethene to ethylene.

Specifically, I'm looking for a literature reference where an explanation is given for the phenomenon, not just examples with or without comparison to experiment.

I think that your rationale is right. I would think it in terms of energy instead to make a little more precise. Thinking about a diatomic molecule, the nuclei-nuclei repulsion function of distance does not change with the basis set used. So the difference is strictly in terms of electronic part. At large distance the atoms behave more similar to an isolated atom, so few spherical harmonics are enough (for a relative description). As atoms get closer it is needed more flexibility to represent the very different probability density that there is in the space region located between atoms. With a smaller basis set this wave function is approximated poorer. If so, it is expected that the energy rises in such region if one consider the variational principle.

Here I calculated for the oxygen molecule in the region close to equilibrium distance (about 1.2 A) the difference in electronic energy between HF/STO-3G and HF/6-31* basis set:

The behavior close to 1.4A should be just an artifact in convergence (I did not take care about that).

Sadly I never found a discussion about that in the literature. Up to my understanding there is not to much that can be said about that.


Determining whether something is an acid, base, or amphoteric I have to find whether these compounds are acids, bases, or amphoteric: $\ce{CO3^{-2}}$, $\ce{HPO4^{2-}}$, $\ce{H3O^{+}}$ and $\ce{NH4^{+}}$.

I know the first can obviously act as a base, but others can gain or lose $\ce{H^{+}}$, so I think they are amphoteric. However, according to my book, I'm wrong. Can you help me clear my doubt? Thank you very much!

An amphoteric substance can act both as an acid and a base. That is, an amphoteric substance can donate protons and also accept protons.

$\ce{CO3^{-2}}$ does not have any protons so it cannot act as an acid. It is the conjugate base of $\ce{HCO3^{-}}$.

$\ce{HPO4^-{2}}$ is the conjugate base of $\ce{H2PO4^{-}}$. It can accept a proton to go back to $\ce{H2PO4^{-}}$ or it can donate a proton to become $\ce{PO4^{-3}}$. Thus, it is amphoteric.

$\ce{H3O^{+}}$ is the conjugate acid of $\ce{H2O}$. It can donate a proton to go back to $\ce{H2O}$ but it cannot accept any other protons. Thus, it not amphoteric.

$\ce{NH4^{+}}$ is similar to $\ce{H2O}$. In this case it is the conjugate acid of $\ce{NH3}$. As such it can donate a proton to go back to $\ce{NH3}$ but it cannot accept any protons. It is not amphoteric.

So, out of your list only $\ce{HPO4^-{2}}$ is apmhoteric.


Entropy Change for mixing water at two different temperatures Assume that the heat capacity of water is independent of the temperature.  Calculate the net entropy change when 1 mole of water at 0 degrees Celsius is mixed with 1 mole of water at 100 degrees Celsius. Assume that the heat capacity of water is (4.184 J/(K-g))(18 g/mol) = 75.3 J/(K-mol) and that the heat capacity of the calorimeter is negligible.

I was wondering if the following general methodology of my attempt was correct.

In attempting to solve this, I first found that the equilibrium temperature was (273+373)/2 = 323 K.

Then, I calculated the entropy by using the equation S = (heat capacity)*ln(T_eq/T_initial) for both 0 degrees and 100 degrees.

Afterwards, I added the two values for entropy and obtained 1.827 J/(mol-K) as the answer.

.

Well, your procedure is entirely correct. There's only one problem: the units. Since the number of moles of water is specified, your final answer should not include $\text{mol}^{-1}$. And since there's exactly $1 \text{ mol}$ of both the cold and hot water, the numerical answer should be the same.

I don't know where your error arises from but I suspect it may be because you used the units of $C_{p,m}$ (molar heat capacity at constant pressure) instead of $C_p$ (heat capacity at constant pressure). The former is an intrinsic property, whereas the latter is an extrinsic one, equal to the former multiplied by the amount of substance: $C_p = n\cdot C_{p,m}$. And their relation to the $\Delta S$ of heating at constant pressure:

$$\begin{align}
\Delta S &= C_p\ln{\left(\frac{T_2}{T_1} \right)} & &\text{Units: J K}^{-1} \\
\Delta S_m &= C_{p,m}\ln{\left(\frac{T_2}{T_1} \right)} & &\text{Units: J K}^{-1}\text{ mol}^{-1}
\end{align}$$

I'm entering nitpicking mode again so I shall also pedantically point out that as far as this equation is concerned, entropy comes with a triangle in front of it. Meaning you write $\Delta S$ instead of $S$.


Which isomers (cis or trans) are more water soluble? As a general sense I know that cis isomers are less symmetrical and more polar in nature whereas trans isomers are less polar. Now given those information, would it be appropriate to assume that cis isomers are more water soluble owing to their polarity?

Also there was a discussion in our class where my teacher told us that between the cis and trans Butenedioic acids, the cis acid is not water soluble on account of the fact that it undergoes dehydration and the carbonyl groups are bonded by an oxygen atom in between eliminating hydrogen bonds (??).

I would like to know if there is any constant character when it comes to solubility of cis and trans isomers.

That's not quite correct and can be quite easily to be shown as wrong by looking up data tables and/or MSDS. Fumaric acid (the common name of the trans-isomer) is less soluble, with a solubility in water of 4.9 g/L whereas maleic acid (the cis-isomer) has a solubility of 478.8 g/L.

Maleic anhydride would also react in water to form maleic acid.


Infrared Spectroscopy of N2O Would you be able to identify the stretch and bend in the spectrum? I can not find much information about the $\ce{N2O}$.It's solved, does anyone know the symmetry of these vibrations?

I don't know what flex and link voltage are; however, here's the rest of it as per this site at UCLA and this homework solution key from Illinois State:

The $\ce{N-N}$ stretch is at $\pu{2200 cm-1}$, the $\ce{N-O}$ stretch is at $\pu{1290 cm-1}$, and the doubly-degenerate $\ce{N-N-O}$ bend is at $\pu{590 cm-1}$.


Gas expansion through adiabatic turbine Gas expands through an adiabatic turbine.  The inlet stream flows at 100 bar and 600 K, while the outlet stream is at 20 bar and 445 K. The constant-pressure molar heat capacity of the gas is $30.0 + 0.02T$, where the molar heat capacity is in J/(mol K) and the temperature is in K.  The process is at steady state.

Assume the gas is non-ideal and follows the following equation:

$$p=\frac{RT}{V_\mathrm m-b}+\frac{ap^2}{T\left(V_\mathrm m-b\right)}\tag{1}$$

with $\displaystyle a=0.001\ \frac{\mathrm{m^3\ K}}{\mathrm{bar\ mol}}$

and $\displaystyle b=0.00008\ \frac{\mathrm{m^3}}{\mathrm{mol}}$

For this gas the differential molar enthalpy change is:

$$\mathrm dH_\mathrm m=C_{\mathrm m,p}\mathrm dT+\left[V_\mathrm m-T\left(\frac{\partial V_\mathrm m}{\partial T}\right)_p\right]\mathrm dp\tag{2}$$

Assume that the ideal expression for the molar heat capacity given is still valid, but that it is only true as $p$ goes to 0.

And the question is; what is the shaft work of the turbine?

This question is from a thermodynamics course and to solve it students were sort of guided through a solution.
First substitute equation (1) into equation (2) and find a general expression for the differential molar enthalpy change of the non-ideal gas.
Then draw a $pT$ diagram of an appropriate path to evaluate the enthalpy change of the process, and just integrate along the path to find the answer.

I still can't figure out how this was supposed to be done.

Thanks to anyone who bothered to read/think about this problem as it is kind of long.

You are aware that, from the open system version of the first law of thermodynamics, the shaft work per unit mass must be equal to the change in enthalpy per unit mass, correct?  So you need to determine the change in enthalpy per unit mass between the gas entering the turbine and the gas exiting the turbine.  Do we agree so far?

We know that enthalpy is a function of state.  So we need to figure out a convenient path between state 1 at 100 bar, 600 K, and state 2 at 20 bar ,445 K such that the equation provided can easily be applied to determine the change in enthalpy.  Consider the following 3 part path:

Hold the temperature at 600 K and lower the pressure to say 1 bar, where the gas behaves as an ideal gas.  Use the equation provided with dT set to 0 to evaluate $\Delta h$ for this step
Hold the pressure constant at 1 atm, and use the ideal gas heat capacity to calculate $\Delta h$ for cooling the gas down from 600K to 445K.
Hold the temperature constant at 445K and raise the pressure from 1 bar to 20 bars.  Use the equation provided with dT set to 0 to evaluate $\Delta h$  for this step.

The sum of the enthalpy changes for these three steps adds up to the enthalpy change for the overall change.


Please explain the following graph for a quantum mechanical harmonic oscillator Graphs such as the above keep coming up when talking about harmonic oscillators in a quantum mechanical sense. However, I simply cannot make sense of them. What does each line represent why are they waves and what is the parabola? Also, I see sometimes graphs such as:

These seem to follow on from the initial graph. It's something to do with the recovery of classical behavior but I don't understand the first graph so stand no chance of getting the second ones at high n - perhaps you could explain the relation between the first and second graph please? Thanks

Background

Let's start by considering the case of 2 hydrogen atoms separated by a large distance.  If they happen to be on a trajectory that brings them closer together they can interact with one another.  If they get close enough together they can form a bond and create a hydrogen molecule.  This process is pictured in the following graph.

image source

This "potential well" is often referred to as a "Morse potential".  Out on the right edge of the curve (say 5 angstroms), we have our 2 separated hydrogen atoms.  As the distance between them lessens, stabilization occurs, they can fall into the well and form a hydrogen molecule.

The red lines in the Morse potential represent different vibrational states of the hydrogen molecule.  Initially, the hydrogen atom is formed in one of the higher vibrational states.  It can either absorb more energy through collisions and go back to 2 dissociated atoms or it can give away some excess energy through collisions and fall to a lower vibrational level.

The classical view of the hydrogen atom is two weights connected by a spring.  The more energy in the system, the wider the oscillation of the spring.  With lower energy the oscillations are smaller.  The energy of the spring system (y-axis) as a function of separation (x-axis) is given by Hooke's Law as

$$E(x)={1\over2}kx^2$$

This is really an equation that describes a parabola, so the lower part of a Morse potential can be accurately described by a parabola.  Your top drawing represents the lower part of the Morse potential diagram.

While we have used hydrogen as our model in the above explanation, the same reasoning applies to any bond formation (and subsequently bond stretch) process.  For example, we could describe a $\ce{C-H}$ bond stretch in a completely analogous manner.

Your Questions

  What does each line represent

Different vibrational levels, as explained above

  why are they waves

Each energy state (vibrational level) is described by its own wavefunction ($\Psi_n$).  The square of the wavefunction (${\Psi_{n}^2}$) describes the probability of finding the electron at a certain point on the x-coordinate for that specific vibrational energy level. From the solution to the wavefunction we find that as we go to higher and higher energy levels, the number of nodes (a point where the wavefunction = 0) in our wave continually increases.  The square of the wavefunction produces positive values (the "upside-down" parabolas). We always have the same number of nodes in the wavefunction and the probability distribution.  For example, in the lowest vibrational energy level (n=0) there is no node in either the wavefunction or the probability distribution.  In the n=1 vibrational level there is one node in the wavefunction and probability distribution.

  what is the parabola?

Explained in the background section


Wavelength-dependence of refractive index in internal reflection spectroscopy? In many literature sources (web example), only a single value for the refractive index is assumed for the infrared element, and another is typically assumed for the sample substance. However, should not the refractive index be wavelength-dependent? So the wavelength-dependence of the penetration depth,

\begin{equation}
d_p = \frac{\lambda}{2\pi n_1 \sqrt{\sin^2\theta - (n_2/n_1)^2}}
\end{equation}

would not only be in the numerator, but also in the denominator: $n_1 = n_1(\lambda)$ and $n_2 = n_2(\lambda)$? Here $d_p$ is the penetration depth, $\lambda$ is wavelength, $\theta$ is the incident angle, and $n_1$ and $n_2$ are the refractive indices of the infrared element and sample, respectively.

Yes it should, but it doesn't matter much.

In physics, roughly speaking, everything depends on everything. We wouldn't be able to do a thing if we couldn't tell what is important and what isn't. Now let's see how much the refractive index actually changes as $\lambda$ goes, say, from 400 to 800nm (thus covering the entire visible range).

(source)

Looks like $n$ of a heavy flint glass goes from 1.85 to 1.75 (that's a change of 5%, in most other materials even less), while $\lambda$ changes by... how many percent?


Do antimicrobial abilities of copper boilers decrease over time? As copper boilers age and develop oxide buildup, does this lessen the copper's antimicrobial abilities in killing bacteria? I was thinking since that coating would be on the copper, there would be less contact between it and the water, thus killing less bacteria.

It is possible that the ability to release soluble copper ions could be reduced with age.
I am assuming, for example, that a very small amount of the cuprous oxide coating could react with, say, carbonic or hypochlorous acid (from chlorinated water) to create a poorly soluble basic salt.
However, with time, to quote a source, namely Handbook of Industrial Water Treatment, Boiler Water Systems, Chapter 12 Boiler Deposits: Occurence And Control Contact:

DEPOSITS:
Common feedwater contaminants that can form boiler deposits include calcium, magnesium, iron, copper, aluminum, silica, and (to a lesser extent) silt and oil. Most deposits can be classified as one of two types (Figure 12-1)...scale that crystallized directly onto tube surfaces
sludge deposits that precipitated elsewhere and were transported to the metal surface by the flowing water
Scale is formed by salts that have limited solubility but are not totally insoluble in boiler water. These salts reach the deposit site in a soluble form and precipitate when concentrated by evaporation. The precipitates formed usually have a fairly homogeneous composition and crystal structure.

So, this growing deposit could likely reduce the concentration of copper ions that in even micro quantities has been reported as having an oligodynamic effect.
For example, per a reference a 2011 work published in Applied Environmental Microbiology, Metallic Copper as an Antimicrobial Surface, to quote:

The antimicrobial activity of copper and copper alloys is now well established, and copper has recently been registered at the U.S. Environmental Protection Agency as the first solid antimicrobial material...In connection to these new applications of copper, it is important to understand the mechanism of contact killing since it may bear on central issues...

As to possible mechanics, the article further notes:

In these enzymes, copper serves as an electron donor/acceptor by alternating between the redox states Cu(I) and Cu(II) (15)... the redox properties of copper can also cause cellular damage. A number or mechanisms have been suggested. Reactive hydroxyl radicals can be generated in a Fenton-type reaction:

$\ce{ Cu+ + H2O2 -> Cu++ + OH- + .OH}$ (1)

The extremely reactive hydroxyl radical can participate in a number of reactions detrimental to cellular molecules, such as the oxidation of proteins and lipids (45).

So at least partially, suspected Fenton-like properties of copper could contribute to its reputed antimicrobial properties. However, this requires a surface presence of either metal copper or cuprous salt, but a growing scale coating may be inhibiting to this action.


Why does a smaller amount of propanol burn slower than a much larger amount? Why does $30~\mathrm{mL}$ of $\ce{C3H7OH}$ burn in 1 second (inside a water jug) compared to a sixth of the amount: $5~\mathrm{mL}$ of $\ce{C3H7OH}$ (on a plate) burn in 180 seconds? Why does the smaller amount take longer to burn the propanol?

For "burning" you need oxygen because actually when a compound is burning it's what is called an oxidation following this reaction

$$\ce{C3H7OH + 5O2 -> 3CO2 + 4H2O}$$

with propanol called the combustible and oxygen is called the comburant. Both are required as well as a power source to start the combustion process.

So in a plate you have the air surrounding the compound then you have practically an endless amount of oxygen available.

So the reaction will stop when all the Propanol will be consumed.

Hence, in a jug, this time the limiting reactive is the oxygen because you have less oxygen supply. Then the combustion will stop when all the oxygen will be consumed in the jug.

So, to sum up, in the jug the oxygen will be consumed faster than the propanol in the plate, that's why the combustion last longer in the plate even if you have more combustible.


Establish the formula of an oxide of chromium An oxide of chromium used in chrome plating has a formula mass of
  100.0 u and contains four atoms per formula unit. Establish the formula of this compound, with a minimum of calculation.

  Express your answer as a chemical formula.

So the question tells me that the mass of the Chromium Oxide is 100.0 amu and has 4 atoms for very unit. Thus, the question wants to know a formula that has 4 atoms? Right?

Hence, if $\ce{Cr}$'s (Chromium) molecular mass is 51.9961 and since the question is stating that we are dealing with an oxide, which would mean that it is a ionic compound (Chromium gives two electrons to Oxygen [15.999 amu]), wouldn't the answer be $\ce{Cr2O2}$; but that would equal 135.9902 grams.

On the other hand, if I am supposed to view a formula unit as an empirical formula, wouldn't that require me to write the simplest formula for a compound?

Yet, $\ce{CrO}$ would give me (51.9961 + 15.999) 67.9951 g/mol and that is not correct, or is it?

Rounding judiciously, the mass of chromium is 52 amu and oxygen is 16 amu.

Our compound is CrxOy and has 4 atoms.  The sum of those 4 atoms' mass is 100 amu.

x + y = 4

52x + 16y = 100

Solve the system of equations to find the answer.  Alternatively, you can just guess and check.

Any species with 2 chromium atoms must have a mass greater than 100 amu, since Cr2 = 104 amu and we haven't included any oxygens.  Thus we know for certain that there is only 1 chromium atom in the compound.  The remaining three must be oxygen.  I leave it to you to check that the answer is correct.

I think your confusion stems from assuming that something called "chromium oxide" must have 1 chromium atom for every 1 oxygen atom, and that isn't the case.


How does ammonium chloride increase the solubility of magnesium hydroxide in water? In my textbook it is written that $\ce{Mg(OH)2}$ is slightly soluble in water, and it forms a milky solution which is called "milk of magnesia".

Its solubility can be increased by the addition of $\ce{NH4Cl}.$ But how? There seems to be no reason.

There is a strong reason. But it is the chemical reaction, not just a better dissolution.

$\ce{Mg(OH)2}$ is a base with the limited solubility, defined by $K_\mathrm{sp}=[\ce{Mg^2+}][\ce{OH-}]^2$

$\ce{NH4+}$ ion, created by $\ce{NH4Cl}$ dissolution, acts as a  weak acid:

$$\ce{NH4+ + H2O <<=> NH3 + H3O+}$$

with $\mathrm{p}K_\mathrm{a}=9.25$

$\ce{OH-}$ ions formed by dissolution of $\ce{Mg(OH)2}$ are eliminated by recombination

$$\ce{OH- + H3O+ <=>> 2 H2O}$$

what supports the dissolution by keeping the product of ion concentrations below the $K_\mathrm{sp}$.

Effectively, there is ongoing equilibrium:

$$\ce{Mg(OH)2 v + 2 NH4+ <=> Mg^2+ + 2 NH3 + 2 H2O}$$


Molecular kinetic interpretation of pressure Regarding the molecular kinetic interpretation of pressure, we arrive at a point in the $F=\frac{\Delta P}{\Delta t}=\frac{2mvx}{\Delta t}$, that is to say, the force exerted in the shock on the wall is calculated by dividing the change in the amount of movement of the wall by that time, which I understand should be the time of duration of the shock, but nevertheless the time used is the time between two shocks, the time in which the molecule travels the distance of the edge of the vessel for one side and returns...this does not contradict the definition of impulse that refers to the duration of the shock??

The derivation can be obtained without having to explicitly consider the time of impact. Suppose the molecules have kinetic energy $E=mc^2/2$ where velocity is $c$ and mass $m$. The molecules are in a fixed box of volume $v=xyz$. The molecule can move in any direction and so has components $u, \;v,\; w$ along each axis meaning that $c^2=u^2+v^2+w^2$ but as no direction is special (the gas fills the volume uniformly) then we can let $c^2=3u^2$.

The pressure on the vessel's walls is due to the change in momentum and for a wall collision after moving along the $\pm x$ direction is $2mu$. A molecule will on average hit opposite walls $u/2x$ times per second when moving in the $\pm x$ direction. By Newtons second law the rate of change in momentum equals force, thus a molecule exerts a force $2mu \cdot u/2x=mu^2/x$ on the face of the container perpendicular to $x$ which has area $yz$. As pressure is force /area and as the collision area is $yz$ then $p=mu^2/(xyz)\equiv mu^2/V$. Finally, substituting for $u$ and allowing for $N$ molecules the pressure is $p=Nmc^2/3V$ i.e pressure is twice the kinetic energy divided by three times the volume, $\displaystyle p=\frac{2E}{3V}$


How to separate HCl and NaOH from an aqueous solution of NaCl? At home I make experiments on salt (as it is easily available at home).
One day I came to know that on mixing salt and water I will get Hydrochloric Acid and Sodium Hydroxide. So I became really excited, but my excitement broke into small pieces because I don't know how to separate $\ce{HCl}$ and $\ce{NaOH}$ from the solution of $\ce{NaCl}$ and water.

The solution can be electrolyzed to produce chlorine and sodium hydroxide solution.  The chlorine can be added to water to produce HCl and HOCl.


Why is trans-cyclooctene chiral? How does trans-cyclooctene exhibit chirality if there are no stereocenters?

Related follow-on questions:

Are all higher cycloalkenes chiral?
Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?

Very interesting question! The key word you are looking for is planar chirality. In trans-cyclooctene, the polymethylene bridge can either go "in front of" or go "behind" the plane of the double bond, assuming you fix the double bond and the two hydrogens in place.

As pointed out by @jerepierre, they are considered different molecules due to a high-energy barrier which prevents the interconversion. Cyclooctene is the first cycloalkene to have both stable cis- and trans- isomers. The chain in trans-cyclooctene is not long enough to swing over the double bond. As the chain gets longer, the energy barrier to rotation decreases.

Here are the two mirror images of trans-cyclooctene (image source: own work).

These two molecules are mirror images of each other but are not superimposable. Therefore trans-cyclooctene is chiral despite not having a chiral center.

Source

Edit: this is the first time I learned that a chiral molecule does not have to have a chiral center. After making some searches online, I feel a need to expand the answer to clarify the concept of chirality.

A chiral molecule is one that has a non-superimposable mirror image. Mathematically a molecule is chiral if it is not symmetric under an improper rotation. Chirality arises due to:

point chirality: typically a carbon center with four different substituents;
axial chirality: such as allenes with different substituents on each carbon (see this question);
planar chirality: such as the case of trans-cyclooctene;
inherent chirality: due to the presence of a curvature in a structure that would be devoid of symmetry axes in any bidimensional representation, such as fullerenes.


Salt + Sodium Bicarbonate releasing water in carpet I had some fleas issues so I read about putting salt and sodium bicarbonate in the carpet and in the edges with the purpose to reduce the humidity and dehydrate eggs and larvae. After few days, the weather changed, to humid, rainy and foggy. UK weather... Suddenly, out of nowhere I found all the floors and carpets humid and specially in areas where the mixture was more abundant, there were quite a lot of drops of water.
I wonder why. Probably the hygroscopic property of salt and sodium bicarbonate, in a change in humidity or atmospheric pressure released the water accumulated during several days. I have no idea.

I've never heard of sodium chloride/sodium bicarbonate being an effective dessicant. Pure sodium chloride is not hygroscopic at all.

However, in air sodium bicarbonate slowly decomposes to sodium carbonate which in fact is hygroscopic. If sodium carbonate, and hygroscopic impurities in table salt, such as magnesium chloride or calcium chloride, deliquesce in air of high humidity to give small droplets then these droplets start to grow due to a high concentration of dissolved salts (including sodium chloride and bicarbonate). The effect is reversible at low humidity or higher temperature.


Can there be active THC be trapped in the evaporator? In relation to the vaporization of THC, when smoked, is it possible that there is any active THC residue that becomes trapped inside the evaporator after the flame goes out?

Wouldn't some of the vapor get stuck inside because the instrument is not a hollow tube?

When burning a blunt, joint, doobie or whatever, THC evaporates (bp of tetrahydrocannabinoles is around 155 °C). Depending on

the length of the "object" , related to the temperature gradient between the front and the mouth piece, and
the type of the "filling" and its adsorptive properties

a part of the vaporized material might condense and/or get absorbed in a colder region of the filling. Similar effects, such as the enrichment of nicotine in the stump of cigarettes have examined.

It is however unlikely that any of the transported and reabsorbed components can be inhaled from a cold "reactor". The relevant components are no longer in the vapour phase and the vapour pressure of the tetrahydrocannabinoles will be too small.


Proof for second-order perturbation term of energy In perturbation theory, if
\begin{align}{\hat {H}} &= {\hat {H}^{(0)}} +  λ{\hat {H}^{(1)}} \tag{1}\\
{ψ} &= {ψ^{(0)}} +  λ{ψ^{(1)}} +λ^2{ψ^{(2)}} + ... \tag{2}\\
{E} &= {E^{(0)}} +  λ{E^{(1)}} +λ^2{E^{(2)}} + ... \tag{3}\\
\end{align}
then, by applying these three equations on ground state $ψ_0$ and using the general time-independent ${\hat {H}{ψ} = Eψ}$ , it can be shown that:[1]
$$E_0^{(2)} = \sum_{n \neq 0}\frac{|H_{n0}|^2}{E_{0}^{{(0)}}-E_n^{(0)}} \tag{4}$$
The proof can be found in quoted reference; it is too long and not shown here. There is one step reads:

\begin{align}\int{ψ^{(0)*}_0}&\hat {H}^{(0)}{ψ^{(2)}_0} dτ + \int{ψ^{(0)*}_0}\hat {H}^{(1)}{ψ^{(1)}_0} dτ = \\ &E^{(2)}_0\int{ψ^{(0)*}_0}{ψ^{(0)}_0} dτ + E^{(1)}_0\int{ψ^{(0)*}_0}{ψ^{(1)}_0} dτ + E^{(0)}_0\int{ψ^{(0)*}_0}{ψ^{(2)}_0} dτ\tag{5}\end{align}
the first and last terms cancel.

I don't see why the first term (on LHS) is equal to the last term (on RHS). Based on the first three equations above, we can only conclude
$$\hat {H}^{(0)}{ψ^{(0)}_0} = {E}^{(0)}_0{ψ^{(0)}_0} \tag{6}$$
Why is $\hat {H}^{(0)}{ψ^{(2)}_0} = {E}^{(0)}_0{ψ^{(2)}_0} $? Also, why is ${ψ^{(2)}_0}$ also an eigenfunction of $\hat {H}^{(0)}$, with same eigenvalue as that of ${ψ^{(0)}_0}$? I believe it has something to do with linearity of quantum mechanics, but not able to prove it.

Reference
[1]: Peter Atkins, Julio de Paula. Physical Chemistry (8th Edition). OUP. 2006. Page 314-315.

As the unperturbed Hamiltonian is Hermitian it follows that
$$\int{ψ^{(0)*}_0}\hat {H}^{(0)}{ψ^{(2)}_0} dτ  = \left[\int{ψ^{(2)*}_0}\hat {H}^{(0)}{ψ^{(0)}_0} dτ\right]^*=\left[\int{ψ^{(2)*}_0}E^{(0)}_0{ψ^{(0)}_0} dτ\right]^*=E^{(0)}_0\int{ψ^{(0)*}_0}{ψ^{(2)}_0} dτ$$
by the definition of Hermiticity and the fact that all eigenvalues of a Hermitian operator are real.


Concentration Inquiry I just found a random example so I could highlight my inquiry:

  A chemist wants to make .5L of a 20M solutionusing $\ce{NaOH}$. The solution will be made from solid $\ce{NaOH}$ and water. How many grams of
  $\ce{NaOH}$ are needed to make this solution?"

In this case, I am finding the moles of $\ce{NaOH}$ that would equate to the number of moles in 0.5L of a 20M solution. But, is it saying that (x grams of $\ce{NaOH}$ + x volume of $\ce{H2O}$) make up the total 0.5L ? or am I adding x grams of $\ce{NaOH}$ to .5L of $\ce{H2O}$ and not considering the changes in volume?

If the final volume of the solution is .5L then what is important here is how many moles/grams of the substance would be used to make the solution. In general when making a solution, the solution is raised to the final volume after the mixture. So whether .5L of $\ce{H2O}$ is used, or .4998L of $\ce{H2O}$ is used doesn't matter.


how protons(H+) react with matter and cause the solution to be acid? For example if my hands were put in a HCl solution how does the protons in the solution which in high amounts will interfere/react with my hand(skin->proteins..) and cause it to burn...or how protons react with a plastic bag and cause it to melt.

Well HCl won't desolve all plastics. In case of your hand there are complex proteins involved containing a variety of elements e.g, carbon, nitrogen, oxygen..., when your hand comes in contact of a proton, the proton is a electrophile and seeks negative charge centers, where ever on a protein it finds the negative charge center e.g, on oxygen atom in a hydroxyl(-OH) group in some amino acid in some protein in your hand, it will attach itself to the oxygen and eventually causing it(oxygen) to loose off from the protein as water, because the ion H20^+ isn't stable and the newly formed Oxygen Hydrogen bond isn't easy to break, so the part breaks off as water. millions of such reactions or even trillions happen eventually corroding off the hand or al large part of hand exposed to protons in a mobile medium.
An acid burn isn't actually a burn but corrosion of proteins, and it doesn't either melt things away rather disintegrates them.


Why does Ethylene Glycol have higher boiling point than Propylene Glycol? Looking at the two alcohols, both have hydrogen bonding and are singly bonded throughout the molecule. However, propylene glycol has a larger electron cloud which to my knowledge would increase its London dispersion forces and thus its boiling point. I already know both molecules exhibit hydrogen bonding since they're both alcohols.
However, it seems that ethylene glycol has a higher boiling point despite propylene glycol seemingly having stronger IMFs. I believe this has something to do with molecular shape but I can't seem to find the definitive answer explaining why ethylene glycol has a higher boiling point.
I've used https://pubchem.ncbi.nlm.nih.gov/compound/Propylene-glycol#section=RTECS-Number and https://pubchem.ncbi.nlm.nih.gov/compound/1_2-Ethanediol#section=Odor for comparison of the two alcohols.

This is all about linearity of the molecule and steric hindrance. Consider the linear structure of ethylene glycol versus the branched structure of propylene glycol. EG can "stack" more easily and therefore engage in more intermolecular bonding than PG. More attraction means more energy required to vaporize. On the other hand, 1,3 propane diol, which is the linear isomer of propylene glycol, has a higher boiling boiling point than EG at around 213C. To further illustrate - 1,4 butanediol (linear) has a higher boiling point than 1,3 butanediol (branched).


Ab initio molecular properties which can be experimentally measured and don't require derivatives/gradients I'm attempting to teach myself HF and DFT by implementing various algorithms, but I've run into a snag: many molecular properties which are experimentally measurable appear to require taking gradients, which is considerable extra work (from a coding perspective). Are there any properties where I can start with a known optimized geometry, run my HF or DFT implementation, and then compare the results to experimental values? Pedagogically, I'd like to be able to motivate DFT by showing it clearly outperforms HF SCF in some experimental regime.

Many thermochemistry benchmarks, like the GMTKN55 (homepage listing most results, DOI), use fixed geometries and require only ground state total energies. Some of the subsets probably have back-corrected experimental references, but you would have to check each paper, I'm afraid. Note that for some properties, a dispersion correction is vital.


Vapor Pressure, Equillibrium, and Increasing Pressure I'm trying to make sure my understanding of vapor pressure and transients is correct.

Let's assume I have a thermally-controlled sealed environment which started out as the regular air (i.e. atmospheric gasses and their respective ratios) with no gaseous water or ethanol at 101.325 kPa, 30 C. This container will maintain a temperature of 30 C after the start of this experiment.

Inside this environment there are two containers which can be remotely opened.

When I open the first container, exposing the pure water inside with a saturation vapor pressure (First question: is that the right term?) of 4.2455 kPa (from Wikipedia) and wait until it reaches equilibrium (100% humidity), does this mean this is the new state of the container?

Pressure: 101.325 + 4.2455 = 105.5705 kPa
Partial-Pressure of H2O: 4.2455 kPa

Then I open the second container, exposing pure isopropyl alcohol inside with a saturation vapor pressure of 10.555 kPa (from WolframAlpha) and wait until it reaches equilibrium again, does this mean this is the new state of the container?

Pressure: 101.325 + 4.2455 + 10.555 = 116.1255 kPa
Partial-Pressure of H2O: 4.2455 kPa
Partial-Pressure of Ethanol: 10.555 kPa

So, in the end, the pressure of the vessel increased each time a new liquid was introduced.

In this hypothetically-ideal environment, would the pressures simply rise until it reached that perfectly equilibrium? If so, I expect in a real environment the pressure would rise but there would always be a cold spot somewhere which caused the vapor pressure (in that specific area) to drop, leading to water condensing, leading to the partial pressure being lower as the air mixes, leading to more evaporation; is this correct?

Some notes:

Assume that there is enough of each liquid to reach equilibrium without running out of liquid.
The sealed environment effectively adds energy to counteract the heat of vaporization of each liquid, and also prevents the gasses from releasing/absorbing energy into/from the external environment or from changing due to pressure changes. It is essentially a magic make-everything-this-one-temperature machine.

There are a couple of gotchas' with the problem.

The volumes of the liquids and gases are specified. So 1 drop of water in a million gallon tank of dry air isn't going to be enough water to reach saturation.

You stated that "This container will maintain a temperature of 30 C after the start of this experiment." However the problem statement doesn't explicitly state that the contents of the container will be in thermal equilibrium with the container. That does seem to be a reasonable interpretation however. (If the evaporation had taken place adiabatically, then the liquid and gas would have to cool to account for the evaporation.)

There is another significant factor here. The water would absorb ethanol, and the ethanol would absorb water. You can't really figure this out without knowing all the volumes, or at least the relative volumes.

Now if you assume that there is enough liquid to saturate the gas phase, that the system (gas phase and liquid phase) absorbs heat from outside the system, then the vapor pressure of the liquid at $30\ ^\circ\mathrm{C}$  is added to the internal pressure of the gas phase before the evaporation takes place.


How to calculate molar mass of (6)UO2+2D and (6)UO2OH+D? For a hydrogeochemistry project, I want to calculate the molar masses of some compounds. I came across a problem when encountering species like (6)UO2+2D (aq) and (6)UO2OH+D(aq). I don't know what the (6) and the +2D mean.

I started off with IPC-MS results for several trace elements and fed them to VMinteq for the speciation modelling. I tried to google the species, but that query produced no useful results. I am no chemist and I don't know "how" or "where" to search efficiently for an answer.

EDIT: another (6) appeared in FA1-UO2(6)(aq)

These notations signify someone performed calculations in Visual MINTEQ accounting for dissolved organic matter (DOM) using a NICA-Donnan modeling method.
From the manual nicadonnan.pdf bundled with the installer:

In Visual MINTEQ, a “D” suffix is used to identify a diffuse-layer species such as a Donnan species.
[…]
The numbers 6, 7, 8 and 9 have no conceptual significance; they are only used by Visual MINTEQ to distinguish different humic components, and they reflect the order in which the components appear on the NICA-Donnan menu (with 6 as the starting number).

Considering the default numbering layout

and that FA1 and FA2 refer to fulvic acids containing carboxylic (FA1) and phenolic groups (FA2), OP's species can be interpreted as such:
(6)UO2+2D(aq): Weakly (electrostatically) bound uranyl $\ce{UO2^2+}$ to fulvic acid in the aqueous phase.
(6)UO2OH+D(aq): Weakly (electrostatically) bound uranyl monohydroxide $\ce{UO2(OH)+}$ to fulvic acid in the aqueous phase.
FA1-UO2(6)(aq): Organically complexed uranyl $\ce{UO2^2+}$ to carboxylic fulvate in the aqueous phase.
Several examples from the manual:

FA1-Zn(aq), FA2-Zn(aq): Organically complexed Zn to dissolved fulvic acid. Sites 1 and 2  refer to carboxylic and phenolic functional groups, respectively.
FA1-Zn(s), FA2-Zn(s): Organically complexed Zn to fulvic acid in the solid phase.
HA1-Zn(s), HA2-Zn(s): Organically complexed Zn to humic acid in the solid phase.
(8)Zn+2D: Weakly (electrostatically) bound Zn to dissolved fulvic acid.
Zn+2D(s)(6): Weakly (electrostatically) bound Zn to fulvic acid in the solid phase.
Zn+2D(s)(7): Weakly (electrostatically) bound Zn to humic acid in the solid phase.


When to use -ate and -ite for naming oxyanions? I'm now learning about nomenclature. In particular, oxyanions.

Basically, when you have an anion that is a combination of a non-metal with oxygen.

According to my book:

  It ends with -ate for the most common oxyanions of the element. It
  ends with -ite for the oxyanions that have the same charge but with
  one less atom of oxygen.

  Example:

  $\ce{NO_3^- \implies}$ Nitrate

  $\ce{NO_2^- \implies}$ Nitrite

Ok, if you give me an oxyanion that ends with -ate, I know that I can simply remove one oxygen atom and I will get the -ite.

But, what I don't understand is how does the book know that $\ce{NO_3^-}$ is Nitrate: how does the book know that $\ce{NO_3^-}$ is "the most common oxyanion for the element". How does it know that a charge of $-1$ and $3$ oxygen atoms create "the most common" Nitrogen oxyanion?

In other words, what does it even mean by "most common"?

"Most common" means the forms most frequently encountered in the entire scope of chemistry, which I'm pretty sure corresponds loosely to the most thermodynamically stable oxidation state. It is not always the case that the '-ate' ion is the most stable, however—as an example, see the links for chlorine below).

A detailed answer necessitates noting that the nomenclature extends beyond just '-ate' and '-ite'.  Most p-block elements that form oxo-anions form a whole series of them, each with the central atom in an oxidation state two away from each neighbor.  For all elements except the lightest ($\ce{C}$ and $\ce{N}$) and heaviest ($\ce{As}$, $\ce{Se}$, $\ce{Te}$, etc.) , usually the series is considered to contain four members, though it is not guaranteed that all are stable or able to be characterized (e.g., "bromite" and "hyposulfite").

As noted in the comments to another answer, in addition to the '-ite' and '-ate' suffixes, there are also the 'hypo-' and 'per-' prefixes, where 'hypo-' only is used with '-ite', and 'per-' is only used with '-ate'.  These four combinations are used to span the series of oxo-anions for each element:

Phosphorus: perphosphate ($\ce{PO5^{3-}}$), phosphate ($\ce{PO4^{3-}}$), phosphite ($\ce{PO3^{3-}}$), hypophosphite ($\ce{H2PO2^{-}}$)
Sulfur: persulfate ($\ce{SO5^{2-}}$), sulfate ($\ce{SO4^{2-}}$), sulfite ($\ce{SO3^{2-}}$), hyposulfite ("$\ce{SO2^{2-}}$")
Chlorine: perchlorate ($\ce{ClO4^{-}}$), chlorate ($\ce{ClO3^{-}}$), chlorite ($\ce{ClO2^{-}}$), hypochlorite ($\ce{ClO^{-}}$)
Bromine: perbromate ($\ce{BrO4^{-}}$), bromate ($\ce{BrO3^{-}}$), bromite ("$\ce{BrO2^{-}}$"), hypobromite ($\ce{BrO^{-}}$)
Iodine: periodate ($\ce{IO4^{-}}$), iodate ($\ce{IO3^{-}}$), iodite ($\ce{IO2^{-}}$), hypoiodite ($\ce{IO^{-}}$)

For carbon, 'carbonate' was given preference, per the 'most common' rubric, as the only known oxo-anion. For nitrogen, the use of the prefixes was avoided, for what I assume was the sake of simplicity.  (I would argue that nitrate and nitrite are both common enough in the natural world that without such a deciding factor there would have been stiff competition for the '-ate' suffix.)

Possible irregularities in structure within a series include peroxo-anions at high numbers of bound oxygens (e.g., I believe persulfate and "perphosphate" are both peroxo species) and variable numbers of bound oxygens for a given oxidation state in anions of heavier elements (e.g., metaperiodate, $\ce{IO4-}$, versus orthoperiodate, $\ce{IO6^{5-}}$, both of which contain heptavalent iodine). As well, there is the possibility for other, non-oxo-anion compounds of the central atoms with oxygen such as chlorine dioxide, $\ce{ClO2}$; nitrogen dioxide, $\ce{NO2}$; sulfur dioxide, $\ce{SO2}$, and trioxide, $\ce{SO3}$; and (of course) carbon dioxide, $\ce{CO2}$.

You can find references to oxo-anions of arsenic, selenium, antimony, and tellurium also (links are to the '-ates'), but to my mind, these tend to edge more toward the behavior of oxo-anion-forming metals, which either only have a single appreciably stable oxo-anion (chromate, molybdate, tungstate, etc.), or break this (hypo-)-ite/(per-)-ate paradigm pretty badly (see, e.g., permanganate, $\ce{MnO4-}$, versus manganate, $\ce{MnO4^{2-}}$).

If you really want to blow your mind, take a look at the polymeric oxo-anions like polyphosphate; or at the at-present theoretical-only orthocarbonate; or at iron, for which apparently all three of the known oxo-anions are referred to as 'ferrate'.


How are there two C3 rotation axes in ammonia? If we look at the character table for the $C_\mathrm{3v}$ point group (which $\ce{NH3}$ is an example of), we see that these are the symmetry elements ${E, 2 C_3, 3 \sigma_v}$ present in the molecule. My question is — why are there $2 C_3$ rotation axes? Shouldn't there be only one, which is perpendicular to the plane containing the three hydrogens?
The $C_\mathrm{3v}$ character table:
$$\begin{array}{c|ccc|cc} \hline
C_\mathrm{3v} & E & 2C_3 & 3\sigma_\mathrm{v} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & z & x^2+y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & -1 & R_z &  \\
\mathrm{E} & 2 & -1 & 0 & (x,y), (R_x,R_y) & (x^2-y^2,xy), (xz,yz) \\ \hline
\end{array}$$

In a character table for $C_{3v}$, $ 2C_3$, means that the element is actually 2 $C_3$ elements in the same conjugacy class.  That means the two $C_3$ operations are related in some way.
In fact, it means they share the same single axis.  One $C_3$ operation is a $120^\circ$ turn around the axis, and the other $C_3$ operation is a $240^\circ$ turn around the same axis in the same direction - or equivalently a $120^\circ$ turn around the same axis in the opposite direction.
In the image below the two elements $C_3$ and $C_3^{-1}$ are the two symmetry elements of the $2C_3$ class, which are rotations about the same axis out of the plain of the image.

Symmetry elements of the $C_{3v}$ point group
Image taken from Fig S5 in supplementary information of T. Karin, et al, Phys. Rev. B 94, 041201


How to find the volume of oxygen release by the decomposition if potassium chlorate? In the reaction:
     $$\ce {2KClO3 -> 2KCl + 3O2}$$
What is the volume of oxygen released under NTP conditions when $36.75~\mathrm{g}$ of $\ce {KClO3}$ is heated?

I tried to find the answer this way:

$2~\mathrm{mol}$ $\ce{KClO3}$ form $3~\mathrm{mol}$ $\ce{O2}$
$0.3~\mathrm{mol}$ $\ce{KClO3}$ form $0.45~\mathrm{mol}$ $\ce {O2}$, i.e. $36.75~\mathrm{g}~\ce{KClO3} = 0.3~\mathrm{mol}~\ce{KClO3}$

I don't know how to convert moles into litre. Can we convert mass directly to volume in $\mathrm{L}$. I read somewhere that $1~\mathrm{L} = 1~\mathrm{kg}$.

1 kg for 1 liter is, in fact, liquid water at room temperature.

You need to use the ideal gas law $\ce {PV=nRT}$.


pKa differences and reactivity I know a lot of chemists like to quote $\mathrm{p}K_\mathrm{a}$ values and subtract them to reinforce some notion about some acid-base reaction being highly favorable or highly unfavorable or whatnot.

For example, the $\mathrm{p}K_\mathrm{a}$ of acetic acid is about 5 and the $\mathrm{p}K_\mathrm{a}$ of LDA's conjugate acid is about 36. 36 minus 5 gives us 31 "zeros" of favorability.

However, $\mathrm{p}K_\mathrm{a}$ values themselves don't tell you what solvent you are using … can using this method blindly lead to problems when working across solvents? And as far as I know, a lot of $\mathrm{p}K_\mathrm{a}$ values are measured in water … but in organic chemistry we don't just use water as a solvent. So does this method even have general validity?

An important case of acidity constants changing with solvent, with consequences in organic chemistry, can be found for compounds with hydroxyls, especially water. It is possible to deduce that hydroxyl compounds are more acidic in protic solvents compared to aprotic solvents, which is attributed to an abnormally high stabilization of the resulting conjugate base via strong $\ce{O^{-}\ ^...\  H-O}$ hydrogen bonds.

The difference is most clear for water. Water molecules solvated by water itself have a $\mathrm{p}K_\mathrm{a}$ of 15.7 in ambient conditions, but if the unusual stabilization from hydrogen bonding were removed, they would have a $\mathrm{p}K_\mathrm{a}$ of around 28. In other words, $\ce{H2O}$ is a weaker acid by some 10 orders of magnitude in non-protic solvents compared to water, and correspondingly, its conjugate base $\ce{OH^{-}}$ is a stronger base by the same factor. This means hydroxide ions can effect quantitative deprotonation of even rather weak acids, including the hydrocarbons cyclopentadiene, indene and fluorene, in solvents such as 1,2-dimethoxyethane.

Unfortunately, compounds containing hydroxide ions have somewhat limited use as superbases in non-protic solvents because the solubility of common hydroxide bases such as $\ce{NaOH}$ or $\ce{KOH}$ is rather small. Other hydroxides such as quaternary ammonium hydroxides have appreciable solubility in non-protic solvents, but they are always found in the form of hydrates. Attempts to prepare anhydrous $\ce{R_4N^+OH^{-}}$ inevitably result in decomposition, because the crystallized water solvates the hydroxide ion and tames its reactivity; once the water is removed, the bare hydroxide ion immediately attacks the quaternary ammonium cation creating an alcohol and a tertiary amine.

As mentioned previously, other compounds containing hydroxyls also show changes in $\mathrm{p}K_\mathrm{a}$ comparing water and non-protic solvents, though the difference tends to be smaller. For ethanol, the difference is at most about 4 units of $\mathrm{p}K$.


Do electric charges increase activation barriers? The activation energy governs the rate of reaction, and is thought to be equal to the energy needed to break the bonds needed to initiate the reaction. But if we have a reaction like:

$$\ce{A +B -> C}$$

And another like:

$$\ce{A+ + B+ -> C^2+}$$

Would the former have a lower activation energy since there is no repulsion between the molecules? Assume the same bonds are being broken in each case.

Here is a specific example from biochemistry. In the hydrolysis of ATP, the first step is the attack of a hydroxide anion at one of the phosphates. The triphosphate portion of ATP has multiple negative charges at neutral pH, so this reaction is very slow.

When catalyzed by an enzyme, the enzyme often binds to a magnesium cation in such location that it shields some of the negative charge. This makes the reaction proceed faster.

  Do electric charges increase activation barriers?

Yes, they do. And there are various strategies to lower the activation barrier in these cases. One is providing a counter ion as described above. Another is to change the charge via acid or base catalysis, i.e. protonating or deprotonating one or the other reactant to decrease electrostatic repulsion.

  Would the former have a lower activation energy since there is no repulsion between the molecules?

Yes, and it would be sufficient to remove one of the charges in the latter to lower the activation barrier because electrostatic repulsion requires both particles to be charged.


Writing equation of reaction in stoichiometry of redox titrations While solving some questions on the stoichiometry on redox titrations, I come across this question:

  Calculate the volume of $0.05\ \mathrm{M} \ \ce{KMnO4}$ solution required to oxidize completely $2.70\ \mathrm{g}$ of oxalic acid in acidic medium.

I know how to solve this question. The only problem I am facing is that the reaction equation is not given. How do I know what the products of this reaction would be? Can someone help me in how to write the equation of the reaction by the information given in the question.

See the following steps:

We know what $\ce{KMnO4}$ will do by reading the question:

  Calculate the volume of […] $\ce{KMnO4}$ solution required to oxidize completely […] oxalic acid.

We need the structure of oxalic acid. (This is knowledge required to answer the question, no way around.)

Structure of oxalic acid. Image taken from Wikipedia, where a full list of authors is available.
Determine oxidation numbers of all the atoms involved. Shortcut: hydrogen has $\mathrm{+I}$, oxygen has $\mathrm{-II}$ and carbon has $\mathrm{+III}$.
Determine what can be oxidised. Only carbon really can.
Determine what it can be oxidised to. The final product must have $\ce{C^{IV}}$, which is most common as carbon dioxide or carbonate.
Solve the redox equation using either $\ce{CO2}$ or $\ce{CO3^2-}$. Both will work.


What causes rapid nitrogen inversion? When I asked my teacher whether a nitrogen bonded to 3 unique groups counted as a chiral center, they said that it did not because nitrogen undergoes rapid inversion at room temperature. What causes this? And what other atoms are capable of doing this?

In the second row of the periodic table, elements have relatively small differences between the size their $\mathrm s$- and $\mathrm p$-orbitals. Therefore, the orbitals of $\ce{NR3}$ can go from $\mathrm {sp}^3$ to $\mathrm {sp}^2$ with relatively little energy increase, so an amine can become planar and then reorient with the inverted stereochemistry. The same occurs with carbanions. I'm not certain, but I'm guessing oxonium cations do this, as well, and I would even guess that it would happen more quickly due to oxygen's smaller size.
The reason we don't see this with $\ce{NF3}$ is because orbitals with more $\mathrm p$ character are more affected by induction, so the highly electronegative fluorine atoms would lose electron density during inversion, which would be energetically unfavorable (to the point that it doesn't occur readily).
Chiral phosphines do not invert because the size difference between $\mathrm s$- and $\mathrm p$-orbitals for larger atoms is greater, so converting $\ce{PR3}$ $\mathrm{sp}^3$ orbitals to $\mathrm {sp}^2$ would require much contraction and, therefore, more energy than for the analogous amine.


How to Determine the Rate Law knowing time and concentrations Consider the hypothetical reaction described by the following equation:

$$\ce{3 A + 2 B -> C}$$

Reagent B was added in excess and the following time concentration data was obtained.

\begin{array}{cr}
A\ [\mathrm{10^{-3}\ mol\cdot L^{-1}}] & \mathrm{time [s]} \\ \hline
5.0 & 10\\
2.9 & 50\\
1.9 & 80\\
1.1 & 120\\
0.7 & 150\\
\end{array}

(i) Derive the relevant integrated rate law and hence find the rate constant.

I am thinking that to solve this question you just sub the values into the integrate rate laws and see if it produces a linear relationship.

Is this correct? Also could you please suggest a faster method of solving this type of question.

Thank you.

You could plot the data to determine the order of the reaction.  I would use Microsoft Excel to do this.  Reproduce the data and then add columns that calculate the $\ln [A]$ and $\frac{1}{[A]}$.  Then plot $[A]$, $\ln [A]$ ,and $\frac{1}{[A]}$ vs. time (I would use separate plots).  The plot that gives the "best" linear fit (which you can assess using the $R^2$ value), corresponds to the appropriate integrated rate law.

If $[A]$ vs. time is linear, then the reaction is zero order.
If $\ln [A]$ vs. time is linear, then the reaction is first order.
If $\frac{1}{[A]}$ vs. time is linear, then the reaction is second
order.

Hope this helps.


How is it possible, that the hydrolysis of ethers proceed via a SN2 mechanism? Ethers on hydrolysis with dilute $\ce{H2SO4}$ under pressure produce alcohols.
This is basically a type of acidic cleavage of ethers by the SN2 mechanism.

(source)

There is inversion of configuration.
Now if $\ce{Nu}$ is $\ce{OH-}$ its called hydrolysis of ether.

Is it right, that for SN2 the leaving group must be more stable than the approaching group? In case of ether hydrolysis the leaving alkoxy group is less stable than the entering $\ce{OH-}$ group.

So how does the reaction succeed? Can someone please explain?

While the mechanism may look simple in the way you described it there, it doesn’t work exactly like that.

First, you are in the presence of (diluted) $\ce{H2SO4}$ so aside from extremely short-lived accidental self-deprotonation of water, $\ce{OH-}$ ions will not be around. Therefore, $\ce{OH-}$ cannot act as an attacking group, whatever the mechanism.

Second, an alcoholate is one of the worst leaving groups in $\mathrm{S_N}$ reactions one can imagine. The only case where you will frequently see leaving alcoholates is for epoxide openings. But that’s not a problem here, we have $\ce{H2SO4}$. We can protonate the ether oxygen so that the leaving group will now be an alcohol.

Third, since $\ce{OH-}$ cannot be the attacking species, we need to choose from the nucleophiles present. Either $\ce{HSO4-/SO4^2-}$ or $\ce{H2O}$ can attack the $\alpha$ carbon (in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ manner). In the former case, a later attack of $\ce{H2O}$ to liberate the sulphate needs to be assumed. Both attacks work, but neither nucleophile is strong, so the driving force must be the liberated alcohol.

Fourth, the attacking $\ce{H2O}$ will create a protonated alcohol. This needs to deprotonate to give the free alcohol. Thereby the proton we used in second (step 1 of the mechanism, if you wish) is reliberated and therefore the reaction is acid-catalysed.

Finally, one needs to consider the possibility of equilibrium. The reaction can proceed in both directions. Usually by the choice of solvent one can direct it into one or the other. Hydrolysis is usually achieved with a large surplus of water, driving the reaction towards the alcohols by Le Chatelier’s principle. The reverse reaction would be triggered by removal of water or high concentrations of either alcohol.


Drying a distillate liquid So I'm carrying out the distillation of some essential oil and I have to remove the water content in it. The usual method is using diethyl ether but I'm looking for a greener method.

What can I do?

I would suggest adding anhydrous $\ce{MgSO4}$ to your distillate, like Nicolau Saker Neto already mentioned in his comments. Mix the suspension thoroughly (e.g. by stirring) and look how long the salt remains dispersed. When it clumps up and settles quickly due to the formation of hydrates, there is still some water in your mixture, and you need to repeat the procedure until the added $\ce{MgSO4}$ remains dispersed as fine particles for a longer time (it should look similar to a snow storm). After the salt has settled, you can recover the dried oil with a pipette. Because magnesium sulfate is not listed as toxic or environmentally hazardous, it can be considered a "greener" alternative to, for example, copper(II) sulfate or diethyl ether (the latter is additionally highly flammable).


Solution of a first- and second-order consecutive reaction scheme I have a compartmental model system X, a simplified version of which, looks like this:

$$
\text{System }x: \enspace \text{STO} \xrightleftharpoons[k_2]{k_1}  \text{blood} \xrightarrow[]{k_3} \text{urine}
\equiv
x_1 \xrightleftharpoons[k_2]{k_1} x_2 \xrightarrow[]{k_3} x_3
$$

The system of differential equations is then:
$$
\frac{dx_1}{dt} = -k_1 x_1 + k_2 x_2 \\
\frac{dx_2}{dt} = k_1 x_1 - k_2 x_2 - k_3 x_2\\
\frac{dx_3}{dt} = k_3 x_2 \\
$$

I can solve this system of equations using the following matrix approach.

First, I write the rate matrix [R]. From [R] one can obtain a new matrix [A] by first replacing each diagonal element of [R] by the negative of the sum of each of row elements, and then transposing it:
$$
[R] =
\begin{bmatrix}
    0            & k_2    & 0 \\
    k_1          & 0      & k_3 \\
    0            & 0      & 0
\end{bmatrix} \\
[A'] =
\begin{bmatrix}
    -k_2            & k_2    & 0 \\
    k_1          & -(k_1 + k_3)      & k_3 \\
    0            & 0      & 0
\end{bmatrix} \\
[A] =
\begin{bmatrix}
    -k_2            & k_1    & 0 \\
    k_1          & -(k_1 + k_3)      & 0 \\
    0            & k_3      & 0
\end{bmatrix} \\
$$

I can calculate the amount in each compartment by doing the following:

$$
x(t) = e^{[A]t}x(0) \enspace \text{where} \enspace x(0) = \begin{bmatrix} 0 \\ 1 \\0
 \end{bmatrix}
$$

In python:

RMatrix = model_matrix.as_matrix()
row, col = np.diag_indices_from(RMatrix)
RMatrix[row, col] = -(RMatrix.sum(axis=1)-RMatrix[row,col])
AMatrix = RMatrix.T

def content(t):
    cont = np.dot(linalg.expm(t*AMatrix), x0))

This method is working well for me.

Now, I have a little more complicated model where reactants in compartments 1 and 2 of Systems X and Y combine to get product in System Z.

$X + Y \rightarrow Z$, with a reaction constant of $k_R$.

$$
\begin{align}
&x_1 & \xrightleftharpoons[k_2]{k_1} \enspace & x_2 &\xrightarrow[]{k_3} & x_3 \\
&+  & \ &  + \\
&y_1 & \xrightleftharpoons[k_5]{k_4} \enspace & y_2 &\xrightarrow[]{k_6} & y_3 \\
&\downarrow^{k_R} & \  & \downarrow^{k_R} \\
&z_1 & \xrightleftharpoons[k_8]{k_7} \enspace  & z_2 &\xrightarrow[]{k_9} & z_3 \\
\end{align}
$$

, and the corresponding system of differential equations would be:

$$
\begin{align}
\frac{dx_1}{dt} &= -k_1 x_1 + k_2 x_2 - k_R x_1 y_1 \\
\frac{dx_2}{dt} &= k_1 x_1 - k_2 x_2 - k_3 x_2 -  k_R x_2 y_2\\
\frac{dx_3}{dt} &= k_3 x_2 \\
\frac{dy_1}{dt} &= -k_4 y_1 + k_5 y_2 - k_R x_1 y_1 \\
\vdots& \\
\frac{dz_z}{dt} &= -k_7 z_1 + k_8 z_2 + k_R x_1 y_1 \\
\end{align}
$$

I am struggling with a method to solve this system of differential equations (1st and 2nd order) to calculate the amount in each compartment at a certain time t, given the initial conditions, $k_R$, and the transfer rates $k_1$, $k_2$, $k_3$, etc...

Can I solve it using the matrix method like the one above for a system of first order differential equations? What other options in Python do I have?

Thanks in advance!

Look for equation solvers in Numpy/ Scipy and load at the top of your script scipy.integrate import odeint and import numpy as np . There are examples of how to numerically solve differential equations in the examples and on line. The Master equation approach does not work for second order steps. There is also the Gillespie method which is a sort of Monte-Carlo method see this answer which has an example code Probability of the single molecule reaction to happen


Understanding entropy / spontaneity I have issues understanding the spontaneity for these processes:

Which of the following processes are spontaneous:

(c) alignment of iron filings in a magnetic field;

(d) the reaction of sodium metal with chlorine gas to form sodium chloride;

The answer given for both (c) and (d) is that the entropy of the system decreases.

This has me puzzled because I don't think that even answers the question.
Clearly I would think both (c) and (d) are spontaneous processes.  For (c), the iron will naturally align itself in the magnetic field. For (d), I know the reverse reaction is non-spontaneous (table salt does not normally disassociate), so the forward reaction must be spontaneous.

Assuming I am correct, how do I reconcile this with the book answer?

The entropy of the universe increases for any spontaneous process.  If we assume the entropy of the system is negative (like the book does) then the entropy of the surroundings must be positive and greater.  I don't know what that would mean in this context.

I agree that it makes sense that the entropy of the system is negative.  The iron filings are moving to a more ordered state in (c).  For (d) there are more micro-states originally since there were more moles of reactants.

Based on your provided context, your book's "answers" are indeed misleading and non-answers.

Clearly I would think both (c) and (d) are spontaneous processes. For (c), the iron will naturally align itself in the magnetic field. For (d), I know the reverse reaction is non-spontaneous (table salt does not normally disassociate), so the forward reaction must be spontaneous.
The entropy of the universe increases for any spontaneous process. If we assume the entropy of the system is negative (like the book does) then the entropy of the surroundings must be positive and greater.
I agree that it makes sense that the entropy of the system is negative. The iron filings are moving to a more ordered state in (c). For (d) there are more micro-states originally since there were more moles of reactants.

These statements are correct. As you've argued, the change in entropy of the surroundings must be positive for these reactions to be spontaneous. The surroundings constitute a thermal reservoir, so the change in entropy is simply proportional to the heat evolved during the process, which implies that heat must be transferred from the system to the surroundings.
For the formation of sodium chloride, this is quite intuitive: sodium chloride is very stable and has a large, negative enthalpy of formation. The case of magnetization is less obvious, but it can be shown (see Ex 17.4 in Blundell and Blundell's Concepts in Thermal Physics) that an isothermal increase in the magnetic field leads to evolution of heat into the surroundings, and the same reasoning as above applies.


Decomposition of CaCO3 in a sealed vessel For the equilibrium

$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$

$\Delta H^\ominus = 178\ \mathrm{kJ\ mol^{-1}}$ and $\Delta S^\ominus = 160\ \mathrm{J\ mol^{-1}}$.

I'm asked to suppose that $1\ \mathrm{mol}$ of $\ce{CaCO3}$ and $1\ \mathrm{mol}$ of $\ce{CO2}$ are heated reversibly in a closed vessel under a pressure of one bar.

How much $\ce{CaCO3}$ is present in the container when the temperature reaches $1100\ \mathrm K$?

I know that at this temperature $\Delta G^\ominus = 2\ \mathrm{kJ\ mol^{-1}}$ and $K_p = 0.80$. I'm thinking that since the equilibrium constant $K = p(\ce{CO2})/p^\ominus$ is less than $1$, then the $\ce{CO2}$ would combine with any $\ce{CaO}$ around to make $\ce{CaCO3}$, but since there isn't any the composition must remain unchanged at $1\ \mathrm{mol}$ $\ce{CaCO3}$ and $1\ \mathrm{mol}$ $\ce{CO2}$. Is this right?

Now suppose the temperature reaches $1120\ \mathrm K$, at which $K_p = 1.14$. Some of the $\ce{CaCO3}$ will start to decompose, right? Suppose $n$ mol do, so that I have $1-n$ mol of CaCO3, $n$ mol of $\ce{CaO}$ and $1+n$ mol of $\ce{CO2}$.

How do I find $n$?

You are right that at $1100~\mathrm{K}$ there will not be any decomposition of $\ce{CaCO3}$ in the presence of $\ce{CO2}$.

But, as the temperature increases further to $1120~\mathrm{K}$, $\ce{CaCO3}$ starts decomposing because at this temperature $Q_\mathrm p < K_\mathrm p$.

As the reaction is taking place in a closed container, by viewing the reaction mixture as an ideal gas mixture, we can calculate the amount of $\ce{CaCO3}$ decomposed at $1120~\mathrm{K}$:

$$\mathrm P(\ce{CO2}) \times V = 1 \times R \times 1100 \tag1$$
$$\mathrm P(\ce{CO2}) = 0.80$$
$$\mathrm P(\ce{CO2}) \times V = (1+n) \times R \times 1120 \tag2$$
$$\mathrm P(\ce{CO2}) = 1.14$$

On solving, $n= 0.399$.


Can I perform a stable isotope analysis at home? I'm interested in measuring the stable isotope ratios of some bird feathers. I'm aware that a mass spectrometer is required to measure stable isotope ratios. Is it possible to measure stable isotope ratios at home, using your own 'DIY' spectrometer?

I am assuming that you wish to measure isotopes of carbon.  It is unknown to me how you will extract sample from a bird feather that is suitable for measurement in a spectrometer, and does not taint the isotope ratios. The spectrometers that can distinguish between isotopes are generally quite sensitive and complicated. So the literal answer to your question is "NO, you can not do it at home with a DIY spectrometer."

But, you might be able to purchase a spectrometer. You might be able to differentiate carbon isotopes in carbonyl compounds with infrared spectroscopy. (But don't count on it, unless you are somehow making a gaseous product such as $\ce{CO}$ or $\ce{CO2}$.)

The usual way of measuring carbon isotopes is via mass spectrometry. You can probably get a crude measurement with a machine that costs $20,000 (on eBay) but then you must get it calibrated and keep it supplied with carrier gas, etc.  In short, it will bankrupt you.

User permeakra has the best idea: send a sample to a lab.  But, I am not sure of any lab will allow you to send samples through their spectrometer, not knowing anything about your reputation or the volume of business that they might expect to get from you.


Intramolecular lactonisation in an unsaturated carboxylic acid I am having some trouble figuring out how this reaction proceeds:

Attempt 1: The most obvious thing to do is deprotonate the acid group. I tried using the carboxylate group to attack the alkene, leaving the molecule negatively charged and not really knowing what to do next! If this happened, water would certainly protonate it and then it wouldn't be unsaturated, so it can't be that simple.

Attempt 2: Using the alkene to deprotonate the acid, there will be a positive charge at (one of) the alkene carbons. Then I tried using hydroxide to add in an alcohol group but to no avail.

Is there something special going on because of the cyclic structure?

The reaction as you are given turns up no results in SciFinder, meaning it has not been reported in any scientific journal. It is thus the imagination of whoever set the test.

However, closely related reactions are reported and these may come as no surprise to the advanced chemist. They mainly differ in that iodine $\ce{I2}$ is added to the reaction mixture. The mechanism is called iodolactonisation.

Basically, iodine is attacked by the double bond to form an iodonium ion (whether or not this is actually a three-membered ring shall be deemed irrelevant in the context of this answer). This is then rear-side attacked in an intramolecular $\mathrm{S_N2}$ reaction by the deprotonated carboxyl group to form the five-membered lactone kinetically. In a subsequent step, the iodo substituent is removed in an $\mathrm{E2}$ type reaction with a base to regenerate the double bond; due to steric constraints relating to the required anti-periplanar transition state, this newly-formed double bond is shifted along the six-membered ring.

This is described for your compound multiple times in the literature using different bases, for example by Khanjin et al.[1] They use DBU as a milder alternative.

Reference

[1]: N. A. Khanjin, J.P. Snyder, F. M. Menger, J. Am. Chem. Soc. 1999, 121, 11831–11846. DOI: 10.1021/ja992453d.


Hybridization of carbon in CO2 I know that if there exists a carbon atom with single bonds only (with other atoms),then it is $\mathrm{sp^3}$ hybridized. And if at least one of the bond is double bond then its $\mathrm{sp^2}$ hybridized (not talking about two double bond on same carbon atom). And if there is at least one triple bond then its $\mathrm{sp}$ hybridized. But then carbon in carbon dioxide should be $\mathrm{sp^2}$ hybridized. But if I am right our teacher told it was $\mathrm{sp}$ hybridized. How is that possible if the carbon atom bears no triple bond?

Hybridization is determined by molecular geometry. Carbons participating in triple bonds like those in acetylene have two regions of electron density. Carbon dioxide is a linear molecule, and also has two regions of electron density. The fact that both have two regions of electron density are what make these carbons $\mathrm{sp}$ hybridized.

To reenforce hybridization is determined by molecular geometry, consider the benzyne intermediate. It has a triple bond, but is $\mathrm{sp^2}$ hybridized because it is too strained to become linear and only have two regions of electron density.


Activities of Solutes - especially acids and bases What is the concentration of hydrochloric acid needed to have a solution with the activity of the hydrogen proton = 1? I read 1.18 M somewhere but I am unable to find a corroborating source. Also, how can the activity of the hydrogen proton = 1 when it doesn't exist by itself, suggesting that its activity should = 0 (since it can't act as an independent particle)?

Also how would one get to this result? Does one use a high-grade pH meter? Don't analytical-level pH meters measure activities rather than concentrations of hydronium ion?

For almost the last 100 years, the definition of pH has been -log of hydrogen ion activity.  Prior to 1920, the definition was -log of hydrogen ion concentration.

Hydrogen ion activity or concentration does not refer to bare protons, but solvated protons, as $\ce{H_3O^+, H_9O_4+}$ etc.

See Determination of Hydrogen Single Ion Activity Coefficients in Aqueous HCl Solutions at 25°C for authoritative data.


Actual meaning of equilibrium What is actually meant by that concentration of reactants and products remain constant in equilibrium?

For an equation
$\ce{Reactant <=> Product}$, at equilibrium, the rate of reactants forming products is the same as the rate of products converting back to reactants. Therefore, the concentration of products and reactants remain constant.