text:
Can visible light or infrared radiation excite electrons?
So the limit of ionizing radiation seems to be UV light.
I'm wondering if IR light or Visible light can excite the electrons in an atom.
It has to do with the difference in energy level right?
So the closer two energy levels the lower the frequency/energy.
And that is why coordination complexes and/or transition metals are usually colored because once coordinated, the d orbitals lose their degeneracy?
I also just wanted to clarify that this excitation is different from the color we see for conjugated systems. That is because these systems when conjugated can absorb light? And so we see what is not absorbed and reflected back to us?

No, the electronic excitations for complexes and conjugated organic compounds are not different at all.  It's the same mechanism, just operating with different "input" orbitals.

Visible light can excite an electron  to produce a molecule's excited state. This happens in the rhodopsin in your eyes (causing the sensation of vision) and with very many other types of molecules with energy levels in the right range. The different energy levels involved in different types of molecules produces different colours. IR light generally does not have enough energy to promote an electron into an excited state and instead mainly excites vibrations. Photons of lower energy such as in microwaves excite molecular rotations.

i see. i thought rhodopsin in the eye occured because the molecule isomerized and that caused a conformational change in the protein.

You are correct about the retinylidene isomerisation but this is triggered by visible light, in myself and yourself roughly from about 400nm to at least 800 nm.

Different regions of the electromagnetic spectrum correspond to different atomic and molecular processes, each with one or more associated spectroscopies. Here is a general summary, with decreasing frequency/energy going from left to right:

In order to excite a valence electron, the longest wavelength or lowest energy radiation is usually in the visible region, not longer than ~700 nm. I'm not sure if there is a lower limit to this energy, or what the observed lowest energy electronic transition is, but certainly in this region, there is some overlap with the energy required to excite molecular vibrations. For some nice cases you can perform vibrational-electronic spectroscopy (usually called "vibronic" spectroscopy), where you resolve a number of vibrational transitions (peaks) as fine splitting or fine structure for a single electronic transition:
$\hspace{20ex}$
However, we do not say that "infrared radiation can excite electrons", though there is a range of frequencies where the types of excitation couple together. See also rotational-vibrational (rovibrational) spectroscopy; a nice example is hydrogen chloride in the gas phase.
The reason coordination complexes are often colored is because there are valence electronic excitations that correspond to the absorption of electromagnetic radiation in the visible region. The reason many molecules or complexes are white is because their lowest energy electronic transition is in the ultraviolet region. This is more general than losing degeneracy, and holds true for any macroscopic or microscopic system. It is not different from the color we see for conjugated systems. If the lowest energy available transition of a system is in the visible region, the system will appear colored. If it absorbs red-orange light, the opposite side of a color wheel will show you that it should appear green-blue. What you see as the system's color is the remaining color after subtracting the absorbed color. Yes, the kinds of orbitals involved are different (d-d transitions or metal-to-ligand charge transfer versus $\pi\rightarrow\pi^{*}$), but the transition energy matching is the most important thing.
For the "limit of ionizing radiation", ionization refers not to the excitation of an electron, but its removal:
$$
\ce{X + energy -> X+ + e-}
$$
The atom with the smallest first ionization energy is sodium, given as 496 kJ/mol. Converting this to a wavelength gives roughly 0.013 nm or 13 pm, which is a hard X-ray. Normally X-rays only have enough energy to excite core electrons, so this is the minimum wavelength for atomic ionization. Many atoms will require gamma rays, though I also assume that when looking at whole molecules rather than atoms, the ionization potential may be reduced.

Now to touch on the specific example of rhodopsin. It is a protein that contains retinal, which undergoes photoisomerization:

where A is the 11-cis form, and B is the all-trans form:

If we say that the ultraviolet region spans 10 to 400 nm, then retinal absorption is in the UV, not the visible. However, "the absorbance spectrum of the chromophore depends on its interactions with the opsin protein to which it is bound; different opsins produce different absorbance spectra." From that page:
\begin{array}{lcccll}
\hline
\text{Name} & \text{Abbr.} & \text{Photoreceptor} & \lambda_{\text{max}}~(\text{nm}) & \text{Color} & \text{Human variant} \\
\hline
\text{long-wave sensitive}           & \text{LWS}  & \text{cone} & 500-570  & \text{green, yellow, red}  & \text{OPN1LW "red" / OPN1MN "green"} \\
\text{short-wave sensitive 1}        & \text{SWS1} & \text{cone} & 355-445  & \text{ultraviolet, violet} & \text{OPN1SW "blue" (extinct in monotremes)} \\
\text{short-wave sensitive 2}        & \text{SWS2} & \text{cone} & 400-470  & \text{violet, blue}        & \text{(extinct in therian mammals)} \\
\text{rhodopsin-like 2}              & \text{Rh2}  & \text{cone} & 480-530  & \text{green}               & \text{(extinct in mammals)} \\
\text{rhodopsin-like 1 (vertebrate)} & \text{Rh1}  & \text{rod}  & \sim 500 & \text{blue-green}          & \text{OPN2 = Rho = human rhodopsin} \\
\hline
\end{array}
From photopsin:

The mean absorbance spectra of outer segments of the four classes of human photoreceptors. Curves labelled as follows: '498', mean of eleven rods; '420', mean of three blue-sensitive cones; '534', mean of eleven green-sensitive cones; '564', mean of nineteen red-sensitive cones.

Regarding the mechanism of opsins as triggered by the photoisomerization,

The photoisomerization induces a conformational change in the opsin protein, causing the activation of the phototransduction cascade. The opsin remains insensitive to light in the trans form. It is regenerated by the replacement of the all-trans retinal by a newly synthesized 11-cis-retinal provided from the retinal epithelial cells. Opsins are functional while bound to either chromophore, with A2-bound opsin λmax being at a longer wavelength than A1-bound opsin.

The reason for the isomerization is that upon electronic excitation, the system can approach a point on the potential energy surface where two or more geometries are degenerate; this point is called a conical intersection. That is, after exciting 11-cis-retinal, certain internal rotation barriers are lowered in energy, the nuclei begin to respond, the system passes through a conical intersection, and further relaxation of the nuclei leads to all-trans-retinal. In general, photochemical processes are very complicated.
References

Kropf, Allen; Hubbard, Ruth. The photoisomerization of retinal. Photochem. Photobiol. 1970, 12, 249-260.
Bowmaker, J. K.; Dartnall, H. J. Visual pigments of rods and cones in a human retina. J. Physiol. 1980, 298, 501-511.
Send, Robert; Sunholm, Dage. Stairway to the Conical Intersection: A Computational Study of the Retinal Isomerization. J. Phys. Chem. A 2007, 111, 8766-8773.


text:
Does rubber use one or two sulfur atoms to crosslink?
Does rubber use one or two sulfur atoms to crosslink?

It depends and can be more then two. Sulfur's usual form is $\ce{S8}$.

According to the Polymer Database,
the sulfur bridges formed in the vulcanization process of natural latex are usually comprised of 4-6 sulfur atoms, although this polyssulfide chain can be shorter if the temperature is increased.

http://polymerdatabase.com/polymer%20chemistry/Vulcanization.html

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text:
How to meausre ion interferences in solution?
Is there a way to quantitatively measure the interference of different metal ions in solution? I plan to measure concentration of Pb(II), Fe(II), and Cu(II) in solution together using a spectrophotometer.
I know that I can put multiple samples of two ions together to test for the interference trends/effects, but can I do this for all three? e.g. measure interference effects of both iron/copper on lead? Or are there too many variables to measure concentrations?
If not feasible, are there other accessible ways (besides mass-spec) to measure multiple concentrations of all three at once?
EDIT: I already plan to measure concentrations using reagents, specifically Hach TnTplus 850 and 858, and Hach CuVer 1, as they seem easy to get and use. My purpose is to accurately measure concentrations of one ion when in the presence of two other interfering ions.

Do you want to measure the concentration of each or measure how much one type of ion causes an erroneous measure of the concentration of another type ?

It doesn't satisfy your criteria perfectly, but atomic absorption spectroscopy will do it one element at a time: using flame ionization for the iron and copper, and a graphite furnace for the lead.

I want to measure the concentration of each in the presence of the other ions, i.e. measure Pb when the solution also has both Cu and Fe in it. I listed my probable reagents above in the edit. I think calculating interference curves would be helpful for this too. I was thinking that it would be easy to calculate one ion's interference on the other, but calculating interference of two ions at once would be harder. Sorry for any confusions

Also AAS is out of my budget, so simpler reagents/methods work better for me.

What you can do is to add a reagent which forms a highly coloured complex with one of the metals, this will allow you to make a measurement which is good for one of the metals. You may need to do three different experiments on three seporate aliquats from the same solution.
For example for the iron, I would suggest that you use 2,2'-bipyridine, or 2,2'-bipy as it is sometimes known. This forms a very deep red complex with ferrous iron. This would allow you to measure the iron.
You would need a series of calibration samples with different amounts of iron, please do be aware that the matrix can sometimes interfere with this measurement. I have found that oxalic acid does inhibit the formation of the iron tris bipy complex.
Maybe you should read this paper which is about 2,9-dimethyl-4,7-diphenyl phenanthroline. This is a reagent which has been reported to be good for copper determination.

Thanks. I forgot to mention that I plan on using Hach TnTplus 850 and 858, and Hach CuVer 1 already, for Pb, Fe, and Cu. They measure using PAR, 1, 10 phenanthroline, and Bathocuproin, respectively. I already know the basic process behind using reagents and spectrophotometers, but is there a way to calculate the effect of multiple ions on one ion's concentration? i.e. put a concentration of Pb into solution with Cu and Fe, and measure just the concentration of Pb using the TnT 850 PAR?

Can't you simply run standard curves for Fe in the absence and presence of either or both of the other metals? Of course, the concentration of the other metals may make a difference, and it certainly isn't feasible to run standard curves for all possible combinations and permutations.

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Why is the oxidation state of Co -1 in [Co(CO)4]-?
In $\ce{[Co2(CO)8]}$, most people say the charge of $\ce{Co}$ is 0. How does it become $\ce{[Co(CO)4]-}$?
Most answers would say that $\ce{Co}$'s charge becomes -1. How does this work? -1 is not listed as the oxidation state of $\ce{Co}$ or any transitional metals.

You didn't check the right list then...

Ignore list of "allowed oxidation states" for metal complexes. "allowed oxidation states" might be the most common, but definitely not the only ones the atom is allowed to be in. For example, for $\ce{Fe}$ all oxidation state from -2 to +6 are known, for $\ce{Mn}$ -3 and from -1 to +7 are known, for $\ce{Ir}$ it is -3 and from -1 to +8  with tentative +9 reported for gas-phase monocation.

In $\ce{[Co2(CO)8]}$, most people say the charge of $\ce{Co}$ is 0. How does it become $\ce{[Co(CO)4]-}$?

It is traditionally assumed that ligands do not change oxidation state of the central atom and all atoms in them have same oxidation state as in a free molecule. At the same time, the total sum of all oxidation states of the particle must be equal its charge. Assuming total charge of the particle $-1$ and neutral ligands, the charge by necessity means that oxidation state of  $\ce{Co}$ in $\ce{[Co(CO)4]-}$ is -1.
This is not the only case, there are numerous carbonyl anions with negative charges, like $\ce{[Mn(CO)5]-}$, $\ce{[Fe(CO)4]^{2-}}$. Compounds with even higher formal negative oxidation state of the central metal atom were reported, with anions like $\ce{[Mn(CO)4]^{3-}}$ and $\ce{[Cr(CO)4]^{4-}}$.
It would be a gross misunderstanding to assume that extra charge is located on the metal atom. It is siphoned onto $\ce{CO}$ ligands via $\pi$-backbonding. However, again, metal-ligand interactions are traditionally ingnored in calculation of oxidation state.
Truly negatively charged metal atoms are much rarer, but still happen. Example would be salt $\ce{[Na(15-crown-5)]Na}$ with sodium anion. Such salts are extremely powerful reducers.

Hmm, afaik rather cryptands are used for sodide as crown complexes are unstable. Nice post though.

@permeakra Would the electronic configuration of $\ce{Co^-}$ be $[\ce{Ar}] 3d^8 4s2$ or $[\ce{Ar}] 3d^{10}$?

@adosar $\ce{[Co(CO)_4]^-}$ it's $\ce{[Ad]3d^{10}}$, since the anion is properly tetrahedric. But a free anion is a different matter and I have no idea what configuration it would be in.


text:
Chlorination of Water using sodium hypochlorite
$\ce{NaClO_{(s)} + H2O <=> NaOH_{(aq)} + HClO_{(aq)}}$
This reaction is often used in swimming pools as an alternative to directly chlorinating the water.
I read that the water must be kept slightly acidic. Otherwise, $\ce{OH-}$ ions in the alkaline solution would react with the $\ce{H+}$ ions in the chloric (I) acid to form water. Consequently, $\ce{ClO-}$ ions are removed and the concentration of $\ce{H+}$ ions decreases.
So far this all seems fairly reasonable.
However, my textbook then said this:

In alkaline solution, the equilibrium moves to the left.

This contradicts what I have read online and elsewhere. Surely if the concentration of hydrogen ions decreases, the equilibrium would shift to the right to increase the concentration of the products and thus the hydrogen ions?
Thanks in advance.

Why did you introduce hydrogen ions into this equilibrium? They're not present. Therefore, if you increase the concentration of hydroxide, the equilibrium shifts left.

Reading NaOH(aq) makes me cringe. That compound disassociates completely.

@StianYttervik That's the whole point of using the aqueous phase notation...

@Zhe it is redundant for those kinds of substances. Use it for weak acids or substances that doesn't dissociate completely - then the notation carries meaning.

It is preciously with weak acids where the meaning is ambiguous... @StianYttervik

Yeah my bad. Weak acids was an erroneous example by me. It is useful when you want to be brief, to explain that something is dissolved in water for instance as $\ce{CO2(g) + Ca(OH)_2(aq)}$ which would go slightly different if it wasn't in water. But if water's presence is implied otherwise, then it is very redundant - and conveys the wrong message. In water, there is no $\ce{NaOH}$  only $\ce{Na^+}$ and $\ce{OH^-}$

This question is similar to a problem in nuclear reactor chemistry, I will get onto that soon.
If we were to combine chlorine and water then we will have an equilibrium.
$\ce{Cl2 + H2O -> HCl + HOCl}$
The HCl is a strong acid while the HOCl (hypochlorous acid) is a weak acid. By careful adjustment of the pH it is possible to determine where the equilibrium is. If we were to make the pool too acidic then the oxidant in the water would be lost and the chlorine would enter the air as Cl2. This is a waste of chlorine and also is harmful to swimmers and even sometimes the strucuture of the building.
I assume that there is some chloride present in commerial sodium hypochlorite solution as the classic way to make this chemical is to combine sodium hydroxide and chlorine. The compound is formed by the reaction below.
$\ce{2NaOH + Cl2 -> NaCl + NaOCl}$
Even if we were to have a close to chloride free solution of sodium hypochlorite then I am sure that the first equilbrium I mentioned would be important. It is important to keep in mind that hypochlorite can disproprotionate. It can be converted into chloride and chlorate. If you combine chlorine and cold dilute sodium hydroxide then you make sodium hypochlorite while if you were to use hot concentrated sodium hydroxide then you will make sodium chlorate.
$\ce{3HOCl -> 2Cl^- + ClO3^- + 3H^+}$
As hydrochloric and chloric acids are stronger than HOCl as Bronsted acids, I think that by making the system very alkaline you would favour the conversion by the disproprotionation. What I know about explosives / prolellants and pesticide chemistry is that while going along the series
$\ce{OCl- , ClO2^-, ClO3^3}$ and $\ce{ClO4^-}$
That while in terms of thermodynamics the chlorine species become better oxidants, the rate at which they can act as oxidants often becomes slower. Thus in a pool if we change from sodium hydrochloite to sodium chlorate then it would be less able to kill the germs in the pool.
Also if we made the pool too alkaline then the water would be too corrosive.
In a reactor containment with a water on the bottom such as many BWR designs the iodine chemistry is similar. When the water is acidic then the iodine will form volatile I2. But when the water is alkaline then the elemental iodine will be converted into iodide and iodate by similar reactions.

The reaction as shown goes only a little to the right. The $\ce{HClO}$ shown is poorly ionized; on the other hand, the $\ce{NaOH}$ is completely ionized. How far the reaction goes to the right will be indicated by the pH, which is probably about 10, considering that sodium carbonate is added to commercial bleach to raise the pH to 11 for greater shelf life.
When chlorinating a swimming pool, long shelf life (in the pool) is not desired. You want some more activity, even if at the expense of having some of the $\ce{NaClO}$ divert to other products.
There are two variables in the swimming pool:

the chlorination amount
the activity, which will be controlled by the pH.

So you put in the chlorine, then dial up the best pH. As far as moving the equilibrium, perhaps it is better to show the reaction as,
$$\ce{ClO- + H2O -> HClO + HO-}$$
The original sodium hypochlorite will be totally ionized; the hydrolysis reaction is the $\ce{ClO-}$ ion attracting a proton from water to form somewhat nonionized HClO and some $\ce{OH-}$ ions. Then the prospect of adding more $\ce{OH-}$ ions is easily seen to move the equilibrium back toward the left, generating more $\ce{ClO-}$ ions. Likewise, adding acid generates more $\ce{HClO}$ in the solution.


text:
Differences between Liquid and Solid Infrared Spectrometry
In general, when should I use liquid or solid IR spectrometry to analyse my compound? What should I expect to see? Would I see higher or lower frequency, broader or narrower transitions? My specific case is a tricobalt cluster, Co3(CBr)(CO)9. But I'm also curious for all compounds in general.

It depends, and your question reads like you're asking for a complete introductory lecture on IR. What's your educational background, and have you done literature research on what method people working on similar material use? If not, do so. ;-)

Some basic comments can be found in Spectrometric Identification of Organic Compounds. Although it is primarily concerned with interpretation, it has some notes on the types of instrumentation and samples that would typically be encountered in an (in)organic chemistry lab. I'll only reproduce parts that mention the quality of the obtained spectrum.

Vapor-phase spectra resemble those obtained at high dilution in a nonpolar solvent: Concentration-dependent peaks are shifted to higher frequency compared with those obtained from concentrated solutions, thin films, or the solid state (see Aldrich, 1989).
Thick samples of neat liquids usually absorb too strongly to produce a satisfactory spectrum. (...) For example, thick samples of carbon tetrachloride absorb strongly near 800 cm-1; compensation for this band (using a reference beam with pure solvent) is ineffective since strong absorption prevents any radiation from reaching the detector.
Solids are usually examined as a mull, as a pressed disk, or as a deposited glassy film. The suspended particles in a mull must be less than 2 microns (in diameter) to avoid excessive scattering of radiation. The quality of the spectrum (when using a pellet) depends on the intimacy of mixing and the reduction of the suspended particles to 2 microns or less.
In general, a dilute solution in a nonpolar solvent furnishes the best (i.e., least distorted) spectrum. Nonpolar compounds give essentially the same spectra in the condensed phase (i.e., neat liquid, a mull, a KBr disk, or a thin film) as they give in nonpolar solvents. Polar compounds, however, often show hydrogen-bonding effects in the condensed phase. Unfortunately, polar compounds are frequently insoluble in nonpolar solvents, and the spectrum must be obtained either in a condensed phase or in a polar solvent; the latter introduces the possibility of solute–solvent hydrogen bonding.

One of the reasons you don't want very concentrated samples (for structural determination) is that you increase the amount of intermolecular interaction present in the sample, which can change peak positions and widths. A prototypical example is water:

For the vapor (gas phase), the water molecules are almost non-interacting, leading to well-defined and narrow peaks. The structure here may still be from some small number of environments where a molecule is interacting with a handful of others, but it may also be rotational fine structure (see rovibrational spectroscopy). In the (condensed) liquid phase, these individually-identifiable peaks coalesce into one broad peak due to the many hydrogen bonding environments induced by collsion with other molecules. The ~400 cm-1 redshift of the peak center originates from the nature of a single hydrogen bond:

where the molecule on the right (the h-bond acceptor) donates electron density into the $\ce{O-H}$ bond of the molecule on the left (the h-bond donor), specifically an antibonding orbital, which weakens the bond. Weaker bonds are lower in energy, and wavenumbers (cm-1) are units of energy. Finally, for the solid phase, most common polymorphs of ice have regular structure with a finite number of hydrogen bonding environments, which leads to slightly more distinct peak structure in the solid over the liquid. These peaks are further redshifted due to the decreased $\ce{O-O}$ distance in ice over liquid water (2.76 versus 2.82 Å), leading to even greater charge transfer between water molecules. Broadening and shifting can be also seen for glycine peptides of different lengths prepared as both a solid mull and in continuous films due to the terminal $\ce{-NH3 / -NH2+}$ and backbone $\ce{-NH}$ groups.
I don't have any references for it right now, but you can also obtain peak broadening due to solute–solvent chemical exchange; I have only heard of this for hydrogen.
To summarize,

increasing the number of distinct chemical environments for a particular vibrational oscillator is going to broaden the peak associated with that oscillator, and
intermolecular interactions that directly alter the electron density of the vibrational oscillator are going to shift its peak position; whether it will redshift or blueshift roughly corresponds to a decrease or increase in bond order.

References

Silverstein, Robert M.; Webster, Francis X.; Kiemle, David J. Infrared Spectrometry. Spectrometric Identification of Organic Compounds, 7th ed.; John Wiley & Sons, Inc.: New Jersey, 2005; pp. 79-80.
http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html
Blout, Elkan R.; Linsley, Seymour G. Infrared Spectra and the Structure of Glycine and Leucine Peptides. J. Am. Chem. Soc. 1952, 74, 1946-1951.


text:
How can change in Gibbs free energy ever be positive?
I have a question regarding this formula for Gibbs free energy of a system: $$\Delta G=\Delta H-T\,\Delta S$$
According to the second law of thermodynamics: $T\,\Delta S\ge\Delta Q$ and to my understanding, $\Delta H$ is just the same as $\Delta Q$.
Now, if $\Delta Q$ will be always smaller than (or equal to) $T\,\Delta S$ and $\Delta H=\Delta Q$, then looking at the first formula again it seems that $\Delta G$ can only be smaller than (or equal to) zero.
What am I doing wrong here?

What is $\Delta Q$ in this case?

Who says $\Delta H$ is always equal to Q?

@ChesterMiller ∆H is undeniably not always equal to Q, but at an elementary level, which the OP seems to be, that approximation is used in a valid manner to explain it with ease to students, who can later learn more. It's like laying a temporary groundwork to be filled later, but at least as long as they're dealing with isobaric consitions, ∆H=Q.

Writing "∆Q" is misleading and has no meaning upon close inspection. Q does not change, it is an amount of heat transferred. The energy or entropy (H, G, U, S) may change. Heat and work do not change, as they are not properties stored in the system.

Heat doesn't change? It certainly is contained in a system. Heat is gained or lost. Work is done by or done on the system, it is a process.

You didn't do anything wrong here. This was a very good observation. This is one of the criteria of spontaneity, which I believe you haven't learnt yet.
As you correctly wrote, $\Delta G$ can only be negative or 0 for any reaction. This is actually a consequence of the second law of thermodynamics, which also goes on to say that in any spontaneous reaction the entropy change of the universe is always positive.
Essentially, the answer to your question heading is: in any spontaneous reaction, the Gibb's free energy change can never be positive. Since you took the conditions of the second law, which essentially explains spontaneity, you arrived at this result.

What am I doing wrong here?

You are missing the possibility of non-PV work.

According to the second law of thermodynamics: $T\,\Delta S\ge\Delta Q$ and to my understanding, $\Delta H$ is just the same as $\Delta Q$.

$\Delta H$ is just the same as $q$ under the following conditions: You are working at constant pressure, and the only work is expansion work. In that case, the change in entropy of the surrounding and the change in entropy of the system are described by $-\frac{q}{T}$ and $\Delta S$ (assuming the latter refers to the system). Under these conditions, the reaction or process will always go in a direction that decreases $G$, i.e. $\Delta G$ will be negative.
However, it can be positive if you do work on the system, e.g. charging a battery. So it is not a general statement that $\Delta G$ can never be positive. This is different from the absolute statement about increasing entropy, as described by the second law.

Please don't promote misconceptions by writing "∆Q". Q does not change, it is an amount of heat transferred. The energy or entropy (H, G, U, S) may change. Heat and work do not change, as they are not properties stored in the system.


text:
Why does 3-hydroxy-butan-2-one give positive Tollens' test?
Why does 3-hydroxy-butan-2-one give a positive Tollens' test despite the absence of an aldehyde group? What is the mechanism?

From the Wikipedia entry on Tollens 1

Tollens' reagent is a chemical reagent used to determine the presence of an aldehyde, aromatic aldehyde and alpha-hydroxy ketone functional groups

The substrate you have drawn is an alpha-hydroxy ketone.

answered here https://chemistry.stackexchange.com/questions/94620/in-what-way-could-benzoin-give-tollens-test

I have no access to a drawing package

@Abcd: What do you mean by "very vastly presented" and " the specific mechanism" in regard to Waylander's link? Aldehydes are oxidized by Tollens' reagent. So are some other functionalities. A positive test MAY indicate the presence of an aldehyde.


text:
Why does pH increase as a weak acid becomes more dissociated?
From Wade's Organic Chemistry [1, p. 71]:

In most cases, the $\mathrm{p}K_\mathrm{a}$ of an acid corresponds to the $\mathrm{pH}$ where the acid is about half dissociated. At a lower (more acidic) $\mathrm{pH}$, the acid is mostly undissociated; at a higher (more basic) $\mathrm{pH}$, the acid is mostly dissociated.

I'm having a hard time intuiting this concept. Consider placing $\pu{1 mol}$ of $\ce{HF}$ in $\pu{1 L}$ of water. As the $\ce{HF}$ dissociation reaches equilibrium, I'm assuming the concentration of $\ce{H3O+}$ ions in solution increases, until the maximum concentration of $\ce{H3O+}$ ions are reached at equilibrium. If the concentration of $\ce{H3O+}$ ions in water increases as the dissociation of $\ce{HF}$ proceeds, then wouldn't the $\mathrm{pH}$ of the solution decrease (become more acidic) as the acid dissociates over time?
If any one has any thoughts on how to best interpret the quoted statement, that would be very much appreciated! Any and all new perspectives are welcomed. Thanks in advance for your generous help.
References

Wade, L. G.; Simek, J. W. Organic Chemistry, 9th ed.; Pearson: Glenview, IL, 2017. ISBN 978-0-321-97137-1.

In your quote it should have mentioned that this is true only of a weak acid, although this is implied by the mention of pK. Recall that a weak acid is only partly dissociated and so the equilibrium can be pushed 'left and right' as it were by added acid or base.

A weak acid (HA) in water: $\ce{HA + H2O <=> H3O+ + A-}$, and thus,
$$K_\mathrm{a} = \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$
By Le Chatelier's principle, if you add some $\ce{H3O+}$ from an external source (see Mithoron's comment above) to increase $\mathrm{pH}$, the equilibrium favors the backward reaction in order to reduce some $\ce{A-}$. Consequently, the amount of the dissociation of acid decreases and $K_\mathrm{a}$ will remain constant.
In similar manor, if you decrease $\mathrm{pH}$ of the solution by adding some $\ce{OH-}$, that will react with equal amount of $\ce{H3O+}$. Therefore, the equilibrium favors the forward reaction in order to produce some $\ce{H3O+}$ to keep $K_\mathrm{a}$ constant. Accordingly, $\ce{[A-]}$ in the solution increases while $\ce{[HA]}$ decreases, meaning the amount of the dissociation of acid increases. At one point, $\ce{[A-]} = \ce{[HA]}$, and hence, $K_\mathrm{a} = \ce{[H3O+]}$. At that point, therefore, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$.
Applying this to your example ($\ce{HA} = \ce{HF}$), $\mathrm{p}K_\mathrm{a}$ of $\ce{HF}$ is $3.14$ ($K_\mathrm{a} = 7.2 \times 10^{-4}$).
When $\pu{1.0 mol}$ of $\ce{HF}$ is dissolve in $\pu{1.0 L}$ of deionized water (no added external acid to adjust $\mathrm{pH}$), $\ce{[HF]} = \pu{1.0 mol\cdot L^{-1}}$ and $\ce{[F-]} = \ce{[H3O+]} = \alpha \gg 1.0$. Thus,
$$\ce{[H3O+][A-]} = K_\mathrm{a} \cdot \ce{[HA]}$$
$$\alpha^2 = K_\mathrm{a} \cdot 1.0~~~~\text{or}~~~~\alpha = \sqrt{K_\mathrm{a}}= \sqrt{7.2 \times 10^{-4}}= 2.7 \times 10^{-2}$$
Therefore $\mathrm{pH}$ of the solution is as follows:
$$\mathrm{pH} = -\log \ce{[H3O+]} = -\log \alpha = -\log 2.7 \times 10^{-2}= 1.57$$
Thus, at $\mathrm{pH} = 1.57$, the ionization of the solution is: $\ce{[F-]} = \ce{[H3O+]}= \pu{2.7 \times 10^{-2} mol \cdot L^{-1}}$.
Now, if you added an external base to the solution to increase its $\mathrm{pH}$ until it becomes $\mathrm{pH} = 3.14$, then, at $\mathrm{pH} = 3.14$, $\ce{[F-]} = \ce{[HA]}= \pu{0.5 mol \cdot L^{-1}}$.

Aaah, I understand now! The quote was referring to an how an external or added acid / base would influence the dissociation of the weak acid. I thought it was referring to how the pH evolves as HF dissociates (ex: how pH changes as a function of time when 1 mol of HF dissolves in 1 L of deionized water). Thanks so much for clearing this up and for going into great detail.

@Mathew Mahindaratne> if you decrease or "increase" $\pu{pH}$ of the solution by adding some $\ce{OH−}$

@Adnan AL-Amleh: When you add a strong base such as $\ce{OH-}$, it reacts completely with $\ce{H3O+}$ in the solution. As a consequence, the equilibrium is disturbed and shifted right to get back to equilibrium. It means acid dissociate more, but increased $\ce{[H3O+]}$ is much lower than  lost of $\ce{[H3O+]}$ by reacting with  $\ce{[OH-]}$ so pH increases.

Maybe need to correct:">In similar manor, if you decrease $\pu{pH}$ of the solution by adding some $\ce{OH−}$"

RE: "In most cases, the pKa of an acid corresponds to the pH where the acid is about half dissociated. ..."
In general the definition of pKa is the pH at which $\ce{[HA] = [A^-]}$. The only reasons that this would not be true is if:
(1) The pKa is less than 0 since you can't have an aqueous solution with a pH < 0.  Thus if the pKa < 0, then the acid is considered to be completely dissociated in aqueous solution. But remember that one drop of water into a liter of concentrated sulfuric acid isn't an aqueous solution.
(2) You allow for the fact that it is really activities that are at equilibrium not concentrations. So for solutions that are more than about 0.1 molar activity and concentration for a species are not the same.


text:
What does it mean for a higher glass transition temperature in terms of crystallinity?
The glass transition temperature is associated with amorphous regions in a polymer. If I have observed in a lab experiment that polymers with a higher glass transition temperature tend to have less crystalline structures, how should I interpret this?

Mostly coincidence. It's much more useful to look at and compare specific, comparable examples and explain their actual properties. E.g.  the different Tg's and crystallinities of LDPE and HDPE.

A higher glass transition indicates less energy is required for the bonds of the glass to relax their strain. This is usually due to a slower cooling rate. In a slower cooling rate regime, more time is allowed for crystallization to occur.  So while you are correct to observe a correlation, one does not cause the other but rather they have a common cause of the slower cooling rate.


text:
Determining volume of acid needed to be added to reach a specific pH
Suppose there are $\pu{0.024}$ moles of $\ce{Na2HPO4}$ in a $\pu{0.250L}$ solution. How would we calculate the amount of $\pu{0.145M}$ $\ce{HCl}$ needed to reach a $\mathrm{pH}$ of 6.60?
I have attempted to solve the problem using the following steps:

The final solution needs a proton concentration of $\pu{10^{-6.6} moles/L}$
However when we add $\ce{HCl}$, the base would react to produce phosphoric acid and $\ce{NaCl}$.
$\ce{Na2HPO4 + HCl -> H3PO4 + NaCl}$

Thus the amount of $\ce{HCl}$ needed to neutralize the weak base is equal to the moles of the base which is 0.024.
Afterwards, I attempted to find the volume of $\ce{HCl}$ needed to be added taking the $\pu{0.024 moles}$ needed into account. However, I can’t figure out how the produced phosphoric acid or amphiprotic $\ce{HPO4^2-}$  contributes to the $\mathrm{pH}$ quantitatively as no $K_\mathrm{a}/K_\mathrm{b}$ value is given in the question.

Welcome to Chemistry.SE. We’d like you to take the Tour to get familiarized with the site and our policy towards homework questions.  I'd like to point out that your chemical equation is incorrect (step 2).  Keep also in mind that the acid has three $K_\mathrm{a}$ values and you need $K_\mathrm{a2}$ to resolve this question.

For this you have to know the $pK_{a_2}$ of $\ce{H_3PO4}$.Experimental data shows, $pK_{a_2} = 7.2$.  Now suppose you have added $x$ L. of $\ce{HCl}$. Now the reaction which you have written will happen when you add $\ce{HCl}$ in excess. Initially the following reaction will happen $$\ce{Na2HPO4 + HCl -> NaH2PO4 +_ NaCl}$$ if we add acid quantitatively. So, moles of $\ce{HCl}$ added will be $x \times 0.145$, and that's why if we add the acid in such a way that the whole acid gets consumed; after the whole reaction, remaining moles of  $\ce{Na2HPO4 }$ will be $0.024 - (x \times 0.145)$, and the moles of $\ce{NaH2PO4}$ produced will be same as the acid consumed.  Now the solution contains a mixture of $\ce{HPO4^2-}$ and $\ce{H2PO4-}$, which acts as a acid buffer. Applying the Henderson-Haselbach equation we will have, $$pH = pK_a + log\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}$$ which gives us $6.6 = 7.2 + log\frac{0.024- (x \times 0.145)}{x \times 0.145}$  Solving you will get $x =0.132 L.$ So, you need to add $0.132 $ litres of acid to get that $pH$ of 6.6.

Thank you - the question did not have the acid dissociation constant and I now know that it expects us to search it.


text:
Separating two sugars according to their heavy atoms - Steered Molecular Dynamics
I am trying to separate two methyl-N,N'-diacetylchitobiose sugars using steered molecular dynamics in AMBER. I have tested out:
&rst
iat=-1,-1,
r2=2.5,
rk2=10.0,
r2a=12.5,
iresid=0,
fxyz=1,1,1,
outxyz=1,
igr1=2,5,6,8,9,11,14,16,18,20,22,24,26,27,28,32,33,35,36,38,41,43,45,47,49,51,53,54,55,59,60,
igr2=65,68,69,71,72,74,77,79,81,83,85,87,89,90,91,95,96,98,99,101,104,106,108,110,112,114,116,117,118,122,123
/

Which should apparently separate the two sugars relative to their heavy atoms from a distance (between each other) of 2.5 A to 12.5 A along the x,y and z axes and using a force constant of 10 (I think kcal/mol). The solvent I am using is water.
I ran this separation for 100 ns and the two sugars refused to separate. I am not sure if my force constant is too small, or if something else is wrong with my reference file. I have googled it and spoken to m supervisor about it, and unfortunately I have not succeeded in finding a solution.
Any help would be appreciated.

Are you sure you have specified the correct groups/residue IDs/atom numbers? If yes, then have you tried increasing the force constant to a very high value, to see whether or not the molecules separate?

The force constant was too low. I managed to fix it now.

I have realized the problem. The force constant was too low. Increasing it to 25.0 from 10.0 sorted things out.


text:
how to manufacture PTC ink
I am trying to create a PTC ink similar to Henkel ECI8120 , I researched PTC rubber and found that it was Polydimethylsiloxane (PDMS) loaded carbon nanoparticles, I was wondering if anyone here has some input on how to go about this.
I was thinking of dispersing carbon nanoparticles into the PDMS , but how do I get this to cure/solidify ?
I want to express that I am just asking for experimentation pointers/tips and not a complete solution.

I believe some formulations use a crystalline polymer such as polyethylene. I've heard pdms referenced on the internet and in wikipedia but never found it in a paper.

It would have been helpful if you have said what kind of polymer-based nanocomposites you would like to create. Yet, I found a good article for the fabrication of poly dimethylsiloxane/carbon nanofiber (CNF)-based nanocomposites. The schematic diagram of experimental procedure is illustrated in following figure (from the Ref. 1).

The experimental section in Ref.1 describes the step-wise procedure for the fabrication. It also provides a useful few more references for polymer-based nanocomposite formulations. I also included another reference (Ref.2), which may be helpful to you.

References:

S. Chowdhury, M. Olima, Y. Liu, M. Saha, J. Bergman, T. Robison, Poly dimethylsiloxane/carbon nanofiber nanocomposites: fabrication and characterization of electrical and thermal properties, International Journal of Smart and Nano Materials, 2016, 7(4), 236-247 (https://doi.org/10.1080/19475411.2016.1269027).
H. Azimi, F H. Tezel, J. Thibault, Effect of embedded activated carbon nanoparticles on the performance of polydimethylsiloxane (PDMS) membrane for pervaporation separation of butanol, J. Chem. Technol. Biotechnol., 2017, 92(12), 2901–2911 (https://onlinelibrary.wiley.com/doi/pdf/10.1002/jctb.5306).

How is the PTC behavior accomplished?  I've heard of carbon black and pdms being used to make strain sensors but dont understand the PTC behavior or how to tune that.


text:
Is a meta-sodium sulfonate group on a benzene ring electron withdrawing or donating?

I am wondering if a sodium sulfinate group would decrease or increase the electron density of the aromatic ring

I'd say it is like carboxylate group, only worse.

I agree, the S is strongly d positive and that will draw electron density out of the ring.

Related: Directing nature of the -SOR substituent in electrophilic aromatic substitution

@Gaurang not related.  Adding another oxygen atom changes a lot of things.

As $\ce{SO3H}$ is a strongly electron withdrawing group, I'd say that $\ce{SO3Na}$ is basically the same, but to a slightly less extent. The reason for this is that being a group of highly electronegative atoms both groups have strong inductive electron withdrawing effects but they are both at donating electrons via conjugation (only possible from $\ce{S\bond{2}O}$ bond). This means that the sodium sulfonate group will decrease the electron density on the aromatic ring.


text:
Linalyl acetate: polar or nonpolar?
Why is linalyl acetate nonpolar?
From the structure of the molecule I would have assumed that it was polar since it has a $\ce{C=O}$ bond and is asymmetric. What is it that makes it nonpolar?

Depends on what your definition of non-polar is. How are you defining it?

I only know the basics of nonpolar and polar molecules; distribution of charge and if the charge is distributed evenly across the molecule then the molecule is nonpolar @Waylander

There is no overall charge on the molecule though there is one region of polarity. But in terms of its properties, it will dissolve in non-polar solvents rather than polar, will run high under std tlc conditions etc which to my eyes makes it non-polar.

Being asymmetric has nothing to do with polarity. For instance, do you think 3-methylhexane is polar? It is a hydrocarbon but asymmetric (carbon-3 is chiral) .

While I agree that the answer in the duplicate is correct for this case as well, clearly dipole moment is a poor descriptor: I calculated it for 3-methylhexane as 0.02 D and for linalyl acetate as over 2.2 D.


text:
Transition metal or transition element?
With reference to the periodic table’s d-block, are the terms “transition metals” and “transition elements” interchangeable when describing these elements?
Are there any situations where it would be incorrect to use one phrase instead of the other?

Properly speaking, a "metal" is any substance with certain specific properties; chiefly, that it presents the charge transport behaviour of an electrical conductor. Since these properties are phase-dependent, it'd be more accurate to talk about "metallic phases" rather than "metallic substances" - as a given substance can present both metallic and non-metallic phases (to cite a famous, even if still not completely settled, example, hydrogen is thought to present a metallic phases, but not all hydrogen phases are metallic).
There are other ways to conceptualise metallic properties (such as a focus on band structure, or in terms of bonding) but these are pretty much equivalent to the definition that focuses on electrical conductor behaviour, as these are the underlying mechanisms that explain electrical conductors.
In any case, is it correct to identify certain elements as "metals", beyond talking about a specific metallic phase?
The answer is that no, it's not formally correct (IUPAC Gold Book prefers the term "transition elements"), but it is a well-established term, so when we refer to "metals" there is usually little ambiguity. Note that the term metal is also used in accepted definitions that focus on the properties and not on the elements (such as "molecular metal").

In its preference for "transition elements" IUPAC may be bracing for a paradigm shift.  Until now all d-block elements have had physical and chemical properties that qualify them as metals.  But now, with the coming of the 6d block, we could see a transition element that is gaseous at ambient conditions and thus would not have the physical properties of metals unless condensed by cooling or pressure.  We also see a mixing of late transition "metal" properties with noble gas properties here at what is normally a transition group (12), which to some extent is seen even for the 5d element mercury.  The days of just transition "metals" are passing fast.


text:
Calibration Ammonia Sensor and Gas Concentration
My name is Felipe and this is my first post here.
I'm working on a project to calibrate an electrochemical ammonia sensor. But I have a lot of doubts because I am a layman in chemistry. I'll list them and I'd be grateful if you could respond them.
The system I have here is basically a cylinder with ammonia gas of known concentration connected by a 0.5 lpm regulator and a tygon tube to a chamber where the sensor is.
Here are my doubts:

When I open the regulator the gas starts to flow into the chamber, the concentration of oxygen(or other gases) in the chamber can affect the concentration of ammonia gas? Or even if the gases mix, there is a fixed concentration of ammonia?
I've seen a lot of calibration videos. Why does nobody care about the tube length? I think as long as the tube length more is the waste of gas, right?
When I stop with the ammonia gas flow and disconnect the tube from the chamber. What are the precautions I should take with the contaminating ammonia in the air? Let's say I'm using a calibration gas of 50ppm

The calibration gas will purge the air out of your chamber, so that it contains 50 ppm NH3.  This assumes that there are no reactions that change the mixing ratio of the ammonia (e.g. reaction with water in your chamber).  The length of the tube doesn't matter unless the gas reacts with your tubing, though length will decrease the pressure of the gas coming out of the tube.

You do need to let it flow long enough so that the volume of (tube + chamber) is thoroughly purged first. If no dead volumes or recirculation paths that could be perhaps 10 volumes or more of gas flow. But, 500 sccm is a pretty good flow rate given some Tygon tubing and a small chamber. As for wasting, the bottle has lots of gas in it... As for safety, a glance at the MSDS shows you should not breathe directly from the gas line (but then, you don't say what the balance in the cylinder is so you could just asphyxiate yourself without worrying about the NH3 TLV).

Of course, I guess you could calculate a volume required to flush the tubing, but it's just easier to wait until the reading is steady.

Thank you for your answers. Does anyone know how to calculate the volume needed to flush the tubing?

@JonCuster I know the waste is really too low. But I keep wondering how it is done in companies that have a lot of sensors. You agree if I have to repeat the process thousands of times, at some point the waste begins to be relevant.

Well, I’d use a load lock and cycle many sensors in at a time...


text:
Chemistry write-ups in LaTex - is it worth drawing figures in LaTex?
Being an undergraduate chemistry student I have to write a significant amount of hand-ins e.g. lab reports. Up until now I've been using LaTex with mhchem for equations and siunitx for numbers/units, but I've not invested the time and effort in learning other, more complicated chemistry related packages. This means, that I usually draw mechanisms with ChemDraw and other figures with image editors.
So my question is: which packages is it worth spending the time on learning (and using)?
E.g. chemfig for complicated reaction schemes, modiagram for molecular orbital diagrams, bpchem for IUPAC names and common idioms, etc.

This is a great question, but it's unfortunately a poor fit for the Q&A side of the site. Please raise it in chat! I suspect there will be opinions there, on the matter.

Permalink to head of chat discussion


text:
How are alkalide ions synthesized?
How are alkalide molecules synthesized, for example, I have read in a paper that the reaction $$\ce{2Na -> Na+ + Na-}$$ is exothermic with a $\Delta H = \pu{-438 kJ/mol}$.
I have also read that in the presence of a certain type of crown ether, under specific conditions, $\ce{Na+}$ crystallizes at around $\pu{-20 ^\circ C}$.
Can someone walk me through the mechanism through which sodide, or any alkalide, in general, is actually synthesized?

I think this is the crown ether in question [2.2.2]Cryptand

And it's to do with [2.2.2]Cryptand being so really good at making complexes with (and stabilising) $\ce{Na+}$, that even the $\ce{Na-}$, can't reduce it

Also see: Alkalide (Wikipedia)

It is not $\ce{Na+}$ crystallizing,  but it is Cript-$\ce{Na+ . Na-}$. That crystalization occured at dry ice temperature, but decomposes at $\pu{83 ^\circ C}$ (https://pubs.acs.org/doi/pdf/10.1021/ja00809a060)

@Mathew Mahindaratne is it possible to separate the $\ce{Na-}$ from the crystalline solid (which I understand, separated, would be pretty unstable)?

@HKhan, you can't separate charged anything from the opposite charges under anything like typical lab conditions.  You need a plasma, and even then the only stable negative charges are electrons.

Where did you get your information regarding the exothermicity of the reaction? And I think it is also important that you state exactly the states of the reactants and products involved. As far as I know, the reaction in gas phase is definitely endothermic.

The reaction energetics
Using the method of the superposition of configurations, the electron affinity of $\ce {Na}$ was theoretically determined to be $\ce {+0.54 eV}$ $\ce {^1}$, that is, around $\ce {-52.1 kJ/mol}$. The gas-phase process
$\ce {2Na (g) -> Na^+ (g) + Na^- (g)}$
has been determined to be endothermic by $\ce {4.54 eV}$ $\ce {^2}$ while the solid-state process
$\ce {2Na (s) -> Na^+ .Na^- (s)}$
has also been estimated to be endothermic by $\ce {0.8 eV}$ $\ce {^2}$.Thus, your assertion regarding the large exothermicity of the reaction is questionable. However, it is worthy to note that the $\ce { \Delta H_f}$ and $\ce {\Delta G_f}$ for $\ce {Na^+.Cry Na^-}$, where $\ce {Cry}$ = [2.2.2] cryptand, are $\ce {-10 kJ/mol}$ and $\ce {+28 kJ/mol}$ respectively $\ce {^3}$. Your large exothermic enthalpy may likely refer to the lattice energy, i.e. for the process
$\ce {M^+.Cry (g) + M^- (g) -> M^+.Cry M^- (s)}$.
For $\ce {M = Na}$, the $\ce {\Delta H}$ and $\ce {\Delta G}$ for the above process are $\ce {-323 kJ/mol}$ and $\ce {-258 kJ/mol}$ respectively $\ce {^3}$.
Preparation of the alkalide
$\ce {Na^-}$, $\ce {K^-}$, $\ce {Rb^-}$, and $\ce {Cs^-}$ anions are stable both in suitable solvents and in crystalline solids$\ce {^3}$. The latter can be prepared either by cooling a saturated solution $\ce {^4}$ or by rapid solvent evaporation.
The principal difficulty in preparation of crystalline salts containing alkalide ions by the method of cooling a saturated solution is the low solubility of these alkali metals in the amine and ether solutions $\ce {^3}$. Without a sufficiently large concentration of the metal dissolved in solution, precipitation of the solid upon cooling would be insignificant. This difficulty was resolved by the use of crown-ether and cryptand complexes, such as those of [18]crown-6 and [2.2.2] cryptand] $\ce {^3}$. The complexating agent complexes with $\ce {M^+}$ , shifting the equilibrium (1) far to the right, significantly increasing the concentrations of the dissolved metal ions.
(1) $\ce { 2M (s) -> M^+ (sol) + M^- (sol)}$
(2) $\ce { M^+ (sol) + Cry (sol) -> M^+.Cry}$
This technique of using complexating agents was also what Dye et al. used in their synthesis in 1973 $\ce {^4}$. As reported by Dye et al., a sufficiently concentrated solution of sodium metal (in excess) dissolved in ethylamine with [2.2.2] cryptand was first prepared. The solution is then cooled to dry ice temperatures, giving a gold-coloured crystalline solid precipitate. Through thorough analysis, this precipitate was then determined to be $\ce {Na^+.Cry Na^- (s)}$ with $\ce {Cry}$ being the [2.2.2] cryptand.
References

Weiss, A. W. Theoretical Electron Affinities for Some of the Alkali and
Alkaline-Earth Elements. Phys. Rev., 1968, 166 (1), 70-74
Tehan, F. J.; Barnett, B. L.; Dye, J. L. Alkali anions. Preparation and Crystal Structure of a Compound which contains the Cryptated Sodium Cation and the Sodium Anion. J. Am. Chem. Soc., 1974, 96 (23), 7203–7208
Dye, J. L. Compounds of Alkali Metal Anions. Angew. Chem., 1979, 18 (8), 587-598
Dye, J. L.; Ceraso, J. M.; Lok, M. T.; Barnett, B. L.; Tehan, F. J. A Crystalline Salt of the Sodium Anion (Na-). J. Am. Chem. Soc., 1974, 96 (2), 608-609


text:
Sum of Bond Enthalpies in an Exothermic Reaction
The following reaction is exothermic.

In an exothermic reaction, heat is given out to the surroundings so the enthalpy of the reaction mixture should decrease.
Hence $\sum$ Reactant Bond Enthalpies $> \sum$ Product Bond Enthalpies
Yet I have just read that:
$\sum$ Reactant Bond Enthalpies $< \sum$ Product Bond Enthalpies
which does not make sense to me.
Am I right or wrong, please explain?

EDIT
I see where I made a mistake.
The products have a lower enthalpy so they are in a more stable state. Therefore more energy is required to break the bonds in the products so the sum of the bond enthalpies of the products is greater than that of the reactants.
I think this is right.

The statement

$\sum$ Reactant Bond Enthalpies $< \sum$ Product Bond Enthalpies

is correct, since bond energies or enthalpies are usually defined as energy or enthalpy required to break a bond. See for example the IUPAC Gold Book entries on bond energy and bond-dissociation energy.
Values taken from “Bond Dissociation Energies”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL. for the example that is given in the question are as follows.
$$\begin{align}
D(\ce{I-I})+D(\ce{H-H})&=(152.25\pm0.57)\ \mathrm{kJ\ mol^{-1}}+(435.7799\pm0.0001)\ \mathrm{kJ\ mol^{-1}}\\
&=(588.03\pm0.57)\ \mathrm{kJ\ mol^{-1}}
\end{align}$$
$$\begin{align}
2\times D(\ce{H-I})&=2\times(298.26\pm0.10)\ \mathrm{kJ\ mol^{-1}}\\
&=(596.52\pm0.20)\ \mathrm{kJ\ mol^{-1}}
\end{align}$$


text:
Are ortho, meta and para used in IUPAC nomenclature?
I was presented with the following question in an online test:

To avoid any confusion, here are the arranged options:

A. 2-Chlorotoluene
B. 1-Chloro-2-methylbenzene
C. B and D
D. O-Chlorotoluene
Correct answer: (C) B and D.
Explanation: Chloro is given priority so numbered as 1

My question is: Are the ortho, meta and para prefixes part of the official IUPAC nomenclature or not?
I chose B, what should be the correct answer to this question?

.. if taking typography into account, all names in the image are wrong.

@mykhal: No, options b and c (in the image) are correct.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book),

The locants o, m, p are no longer recommended; the numerical locants '1,2-', '1,3-', and '1,4-' must be used in substitutive names. However, as an exception, the three isomers of xylene are still recognized as o-, m-, and p-xylene in general IUPAC nomenclature (see P-22.1.3).

and

The letters o, m, and p have been used in place of ortho, meta, and para, respectively, to designate the 1,2-, 1,3-, and 1,4- isomers of disubstituted benzene. This usage is strongly discouraged and is not used in preferred IUPAC names.

The preferred IUPAC name for the compound that is given in the question is 1-chloro-2-methylbenzene.

The name 2-chlorotoluene may be used in general nomenclature.


text:
Where can I find the total enthalpy and heat of dissolution in water of a given components?
I don't really have a background in chemistry but I'm actually studying in aerospace engineering, sorry if this question may seem trivial.
I am interested in the decomposition process of the hydrogen peroxide ($\ce{H2O2}$) and I need to find the total enthalpies of $\ce{H2O2}$, $\ce{H2O (l)}$, $\ce{H2O (g)}$ and $\ce{O2 (g)}$ and the heat of dissolution ($\Delta H_{\text{dis}}$)of $\ce{H2O2}$ in water, as a function of the solution concentration of $\ce{H2O2}$.
Firstly, I tried to find the values of the total enthalpies, but I couldn't find them. However, I found a lot of values about the $\Delta H^0$ and $h^o_f$. I can't understand if I can retrieve the value of $H$ with the $\Delta H^0$ and/or $h^o_f$.
Secondly, I can't find a good answer to another question. Does the $H$ is a function of the pressure or not? If so, how can I find the value of $H$ at a given pressure and a given temperature?

See Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics or Moran et al, Fundamentals of Engineering Thermodynamics to get the heat of reaction from heats of formation, and to get the effect of pressure on enthalpy.

It's very unlikely you actually need the enthalpy.  Like energy (of which it is a variation), absolute enthalpy is almost unmeasureable, normally one deals only in enthalpy changes.  In your case I would hazard a guess that you're  interested in things like dH_rxn, the heat of reaction for the decomposition of H2O2.  You can readily calculate this from the dH0 of the components.  The latter are called the "standard heats of formation" and they are combined with "Hess's Law."  Googling either of those terms should lead you to explanations of how the data you can find can be combined into the data you want. How exactly you do it will depend on what quantity you want, and the thermodynamic conditions (e.g. T and p) under which you want it, and the accuracy you need.
Yes, H is generally a function of pressure and temperature.  The starting search term there would probably be "derivatives of the enthalpy" because you would generally calculate H at one (T,p) from H0 at some other (T0,po). You'll find the derivatives can be related to material properties like the heat capacity, compressibility, and coefficient of thermal expansion.  For changes over significant temperature ranges these can be integrated, provided they are known as functions of temperature, or you are willing to use approximate models (like the ideal gas model) that have formulae.

Thank you for your answer. I actually find the heat of reaction for the decomposition of H2O2, but I do not understand how to relate this value to the temperature of the products. That's why I tried to use another technique to find this value. This technique requires the total enthlapies.

If you mean, relate dH at one temperature to dH at another, then start with derivatives of the enthalpy.  If you mean how a release of a certain amount of heat will change the temperature of something, including the reactants, then what you need is the heat capacity.


text:
'Alcohol Free' toiletries
Sometimes I see 'alcohol free' written in big letters on toiletry bottles, such as on deodorants and shaving foams.
You would think that the toiletries bearing these words contain 100% no alcohol, but looking at the ingredients suggests otherwise - I still see alcohols in these products.
So does 'alcohol free' only refer to one type of alcohol not being present in the toiletry, such as a type of alcohol which causes irritation etc?

Might answer the question: https://www.fda.gov/Cosmetics/Labeling/Claims/ucm2005201.htm

Consumer product labeling is not chemistry. Don’t get me started on ‘organic’...

@Tyberius If you turn that into an answer, you'll get my upvote :3

@JonCuster Yeah, chemists and non-chemists perceive 'organic' products slightly differently...

Summarizing from the FDA website (some of this info could vary by country, but I think the general ideas should hold anywhere) commercial use of the term "alcohol" doesn't coincide with its use in chemistry. Cosmetic products use "alcohol" to mean ethanol, so you may find other alcohol compounds in a product that is ostensibly "$100\%$ alcohol free".


text:
Generalized reaction rate and derivation
Most textbooks give the reaction rate
$$r = k C_\ce{A}^a C_\ce{B}^b  \tag{1}$$
for the elementary process $\ce{a A + b B -> P}$. However, expression (1) is not valid in the general case: e.g. it is lacking fluctuations, small-size corrections, memory effects, relativistic corrections,...
I have seen some derivations of the rate expression (1) using kinetic theory, Langevin approach, or chemical master equation, but the derivations introduced further approximations: e.g. the kinetic theory approach considered only binary collisions in diluted gases under local thermodynamic equilibrium.
Consider the process $\ce{A -> P}$, according to (1) the rate expression would be
$$r = k C_\ce{A}  \tag{2}$$
However this expression not only predicts an incorrect stationary state, but it does not represent the real concentration in the system, because $C_\ce{A}$ is only an averaged concentration and not the instantaneous concentration of species $\ce{A}$ in the system. In a first approach the above rate is replaced by
$$r = k \tilde{C}_\ce{A} + \tilde{f} \tag{3}$$
where $\tilde{C}_\ce{A}$ represents the real concentration in the system and $\tilde{f}$ is a stochastic chemical 'force'. Now (3) represents the rate of the real concentration and it gives the correct stationary limit. (3) satisfies a fluctuations-dissipation theorem that relates $\langle\tilde{f}(t)\tilde{f}(t')\rangle$ to the rate constant $k$. This relation is used in experimental methods that measure chemical fluctuations to obtain the rate constant.
Equation (3) continues being an approximation, a generalization could be the introduction of memory effects, which could give something as
$$r = \int_0^t \kappa(s) \tilde{C}_\ce{A}(t-s) ds + \tilde{F} \tag{4}$$
where the kernel $\kappa$ and the generalized stochastic 'force' $\tilde{F}$ have complex expressions, but (4) continues being an approximation. Consider now the process $\ce{2A -> P}$, according to (1) the rate expression would be
$$r = k C_\ce{A}^2  \tag{5}$$
but this is not correct in general. Introduce the volume factors into $k$ to obtain
$$r = K N_\ce{A}^2  \tag{6}$$
and you can check this predicts a rate $r=K$ when $N_\ce{A}=1$, which is obviously impossible because at least two molecules are needed for the step. The rate (6) is invalid for mesoscopic systems (e.g. reactions involving small amounts of matter in living cells) and it is also incorrect for large systems in the long time limit, when $N_\ce{A} \rightarrow 0$. A corrected rate is given by
$$r = K N_\ce{A} (N_\ce{A} -1)  \tag{7}$$
which cannot be derived from (1). However, (7) can be derived in combinatorial kinetics, which postulates the next rate expression for an elementary process $\ce{a A + b B -> P}$
$$r = K \frac{N_\ce{A}!}{(N_\ce{A} - a)!} \frac{N_\ce{B}!}{(N_\ce{B} - b)!}  \tag{8}$$
But again (8) is an approximation.
Each rate expression I find in the literature is valid only for some specific cases, but fails for others.
What is the general expression for the reaction rate for an elementary process and how can the expression be derived from first-principles?

What you've written is a macroscopic observable, so it is the general expression, and it is generally valid for the class of reactions for which it is designed (which is a bit circular, I realize, but that's the nature of empirical statements).  All the complications of which you speak vanish in the thermodynamic limit.  It sounds like what you're after is some kind of kinetic theory expression, and you're unsatisfied with dilute gas approximations, so maybe the place to start is with general formulations in the liquid state like http://rspa.royalsocietypublishing.org/content/188/1012/10

Read my answer to this question https://chemistry.stackexchange.com/questions/81019/why-is-the-rate-of-a-reaction-proportional-to-the-concentrations-of-reactants-ra as it largely covers the points you raise

@ChristopherGrayce I didn't write a macroscopic observable ($r$ could be a mesoscopic rate) and the thermodynamic limit is an approximation used to simplify some formulas, but no real system has infinite volume and infinite number of particles. I am realy looking for a dynamical rate expression, not a kinetic one, but will look the reference. Thanks

@porphyrin The first part of your answer uses collision theory with the usual simplifications (including the one that I mentioned explicitly as "local thermodynamic equilibrium"). The second part mentions the approach taken by Gillespie, which is based in chemical master equation (I also mentioned this equation) which built in many simplifications including the ignoring of non-Markovian effects in the system.

Ok, I assume also that you know the book by Erdi & Toth (Math models of chem reactions) then in that case you should probably start by calculating reaction dynamics trajectories where you can control all parameters then average over quantum states at your choice of energy distribution.

@ChristopherGrayce The paper you gave is the original BBGKY approach. Besides this approach focusing on the distribution function (I am asking about the observables), the approach was shown to break in certain dense situations in the sixties, because the approach assumes an analyticity condition for the transport coefficients that is not generally valid.

@porphyrin I cannot access that book, but I checked the TOC and read some open chapters in GoogleBooks. I didn't find a general rate expression without the limitations mentioned in my question. It has been some time since my last check of reaction dynamics literature, but I recall authors then just assuming some simplified rate equation e.g. (1) and then using approx quantum evolution methods such S-matrix to obtain k for specific systems.

You can calculate anything from the distribution functions.  If you don't know how, perhaps you need to take a step back and study some more basic statistical mechanics.  Try Hansen & McDonald's classic book on the subject, extremely accessible and clear.

@ChristopherGrayce If you know the answer to the question you could just put it here instead make the partridge dizzy.

If I knew the answer to the question I would write a paper and send it to PNAS or Physical Review and become famous.


text:
Why doesn't this question violate the conservation of mass?

Combustion of of $0.255 \space g$ of isopropyl alcohol produces $0.561
 \space g$ of $\ce{CO_2}$ and $0.306 \space g$ of $\ce{H_2O}$.  Determine the
  emperical formula of isopropyl alcohol.

I'm not asking how to solve this problem, but why this doesn't violate the conservation of mass?  Shouldn't the initial combusted mass equal the sum of the resultant masses?

Because the problem statement indirectly includes an additional reactant.
At an introductory chemistry level, the term combustion implies reaction with $\ce{O2}$. Thus, your missing mass should show up when you introduce a stoichiometric amount of oxygen to your reactants.


text:
Why do electron donating groups stabilize free radicals?
Why do electron donating groups stabilize free radicals? I have read this fact in various websites and it is not mentioned why. Isn't a free radical like a highly reactive individual molecule? Why does it require electrons? And won't the requirement of electrons depend on the octet of the central atom of the free radical?

Throughout this answer, I'm assuming that you mean carbon free radicals, as you tagged this with organic chemistry.
A free radical is actually more than just a molecule. It is somewhere between an ion and a stable molecule, but has no formal charge. However, it is highly reactive, as you mentioned. You've also identified the right reason for the need of electrons, it is the octet rule, at least at an elementary level. They behave as electrophiles, in order to complete their octet.
A general free radical of carbon has 7 valence electrons, (like the methyl, ethyl free radical etc.) and while other free radicals of carbon do exist (like carbene, benzyne, and a few I don't know about), most of these are hypo octet, i.e. they have fewer electrons than necessary to complete their octet.
Hence, they get stabilised by electron releasing groups, which "quench" the requirement of electrons for the free radicals. This can happen via induction, or by resonance,or even by hyperconjugation (which is basically the same as resonance).
I hope this answers your question.


text:
How to determine the order of acidity of the following dimethyl nitrophenols

This is how I went about it, although I got a wrong answer:

3,5-Dimethyl-4-nitrophenol (2) shows the nitro group to be sterically hindered, due to which it should go out of plane and this will be an example of the steric inhibition of resonance. Hence, there will only be a weak withdrawing inductive effect, and no resonance. Thus, this should be the least acidic.
4,5-Dimethyl-2-nitrophenol (3) shows a hydrogen bonding between the nitro and the hydroxyl group, due to which there will be a lower tendency to release the proton. Hence, even this should be less acidic.
Between 2,6-dimethyl-3-nitrophenol (1) and 2,4-dimethyl-5-nitrophenol (4), (1) should be weakly acidic due to the hyperconjugation destabilisation caused by the two methyl groups, and hence the answer should be (4), with lower hyperconjugation, lower induction from the methyl groups, and higher induction from the nitro group (no resonance at the meta position).

However, the answer is given to be (2), which I ruled out first.
Can someone please explain what is happening, and where I've gone wrong?

partially answered here https://chemistry.stackexchange.com/questions/54008/acidity-order-of-nitrophenols?rq=1

The right answer should be 3 as the $\ce{NO2}$ group has a negative mesomeric effect, which is only possible with 3. In 2, due to steric inhibition of resonance, it exhibits only -I effect, and although 3 contains H-bonding, it slightly affects acidity.

So hydrogen bonding is less effective than the -M effect, you say? Can we say that although the hydrogen bonding decreased the acidity slightly, it is still greater than the -I effect groups due to the -M effect? However, this answer doesn't match with the given answer either...

Turns out that 2 is slightly more acidic than 3.
$\ce{pKa(2)  = 8.25 @ 25 ºC}$  source:10.1021/ja01187a054
$\ce{pKa(3) = 8.55 @ 28 ºC}$  source:10.1021/ja01161a043
I could not find data for compounds 1 and 4.

$\ce{-NO2}$ is an electron-withdrawing group and $\ce{-CH3}$ is an electron-donating group and the effect of these groups can be observed at ortho- and para- position of the ring with respect to the functional group attached to the ring.

in the above resonating structure of toluene you can see -ve charge in o-, p- position, similarly in nitrobenzene you can see positive charge in o-, p- positions, and $\ce{-OH}$ group attached to electron deficient carbon or positively charged carbon release $\ce{H+}$ more easily. Considering all this condition second option is the only option where carbon to which $\ce{-OH}$ is attached is getting positive charge and not getting a negative charge, so $\ce{-OH}$ is ready to loose $\ce{H+}$.

Are you familiar with Steric Inhibition of Resonance ?


text:
How is phosphorus-32 produced?
I'm doing an assignment where I have to research a radioisotope of my choice. I decided to do phosphorus-32 and my assignment requires me to explain how the isotope is made. There is very minimal information on the production of phosphorus-32 so I'm in a little trouble.
A brief explanation would be much appreciated!

Start here: https://en.m.wikipedia.org/wiki/Cosmogenic_nuclide.

The Evaluated Nuclear Structure Data File site at the National Nuclear Data Center (nndc.bnl.gov) has data on about a dozen different nuclear reactions that result in 32P. Or the Nuclear Data Sheet at https://www.sciencedirect.com/science/article/pii/S0090375211000767

The most important nuclear reactions for the generation of anthropogenic P-32 are
Cl-35 (n, α) P-32,
S-32 (n, p) P-32, and
P-31 (n, γ) P-32.
Since especially chloride, sulfate, and phosphate can occur as impurities in water, typical P-32 activity concentrations in the reactor coolant of light water reactors are in the range of several kBq per litre. (Sources for impurities in the reactor water of BWR’s can be leaking condensers; in the primary coolant of PWR’s, phosphorous compounds in the added boric acid. Sulfate might also be released from cation exchange resins.)
In the laboratory, P-32 can be produced (even as an experiment with students) using the S-32 (n, p) P-32 reaction with fast neutrons. A sample of pressed elemental sulfur is placed close to a neutron source for several days. After that, the sulfur is washed with boiling water. The conversion of the α-sulfur allotrope to β-sulfur at 95.3 °C helps to release the phosphorous compounds from the solid sulfur matrix to the water. Thus, the resulting P-32 solution can be made carrier-free.


text:
What's the difference between demixing temperature and cloud point?
I am learning something about polymer solution which has LCST phenomenon. But I met researchers used demixing temperature and cloud point to describe the transformation of this kind of solution from solutable into insolutable state. But for me, they seem the same. So I ask it here to learn what is the difference between these two concepts?

Cloud point typically refers to the LCST (or UCST) in solution. Demixing temperature usually refers to the LCST (or UCST) of a polymer mixture or copolymer in bulk.


text:
Why does hydride acidity increase across period and down group in periodic table?
Why exactly does hydride acidity increase across period and down group in periodic table? What is the explanation with respect to electrons? I can't figure this out because for 1st period etc H is an anion while for the right and middle of the periodic table H is a cation in compounds formed. I feel confused. Does this have something to do with electronegativity of element that forms hydride?

Yes, but that is somewhat circular reasoning, as electronegativity is largely just putting a numerical value to the trends in properties you are observing.  A better answer is to compare the effective nuclear charge holding on to the valence electrons as you go across or down the Table.  Going across the table, each electron gets added to the same shell, which means it shields the other electrons from the added proton poorly.  Consequently, Z* (the positive charge effectively felt by the outermost electrons) increases, and these electrons are more tightly held.  At some point the atom holds onto not only its own valence electrons, but those of the H atom when the H atom leaves, and so the H atom leaves as H+.  Conversely, at the far left of the table, the Group 1A atoms hold on so loosely to their lone electron that the H atom is able to leave with the metal's electrons, as H-.
Going down the table, you add a whole electron shell between the nucleus and outermost electrons with each period.  That is counterbalanced a bit by the fact that the shells become more closely spaced, but nevertheless for main group elements the extra shielding of the extra shells usually matters more, and so effective nuclear charge falls going down a group.
Finally, bear in mind the "acidity" of a hydride is the tendency of the compound to break up with the H atom leaving as H+.  So the greater the Z* of the partner, the more likely the H atom leaves without its electron.

you said that down a group effective nuclear charge falls due to the extra shielding of the extra shells- so then shouldn't acidic nature decrease and not increase down the group? Because acidic nature depends on the effective nuclear charge of the atom attached to Hydrogen.

How strong is the bonding between iodine and hydrogen, versus fluorine and hydrogen?  Does covalent bonding overlap have anything to do with the trend going down most groups?


text:
Why is dp/dv zero for a real gas in critical conditions?
I have read this fact in various websites but cannot understand the reason. What is an intuitive explanation of this?

This is merely a shard of a fact which does not make much sense in and by itself. After all, in systems with gas/liquid equilibrium there is nothing really special about $\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. On the contrary, this is pretty typical. See all those points where the blue lines (isotherms) are horizontal? They make up a whole huge area on the diagram.

(source)
Indeed, if you have a liquid in equilibrium with its saturated vapor, then the pressure does not depend on volume (in an isothermal processes, that is). As you compress it a little more, some gas turns to liquid; as you release it, some liquid evaporates. The pressure still remains the same - namely, the saturated vapor pressure at given temperature. Things start changing only when you pull the piston all the way out or push it all the way in.
Now, the two-phase area, huge as it is, does not comprise all diagram. If you heat the gas past the critical temperature, you'll get an isotherm of a different sort (it is not shown on the picture, so just imagine one, positioned above the rest of them). It will be all smooth and monotonic, without those horizontal parts.
Well, then there must be a temperature where this behavior changes, and that's what we call the critical temperature. Sure enough, it produces a very special isotherm: there is only one point on it where $\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. (That's what we call the critical point, BTW.) Moreover, since it is an inflection point, it also has $\left(\dfrac{\partial^2\mathfrak p}{\partial V^2}\right)_T=0$. Now that's the fact one should remember about the critical point.
The peculiar behavior of pretty much all physical properties around the critical point gives rise to various weird phenomena like critical opalescence, but that's another story.
So it goes.


text:
irreversible adiabatic process against constant external pressure?
My book has these formulae for irreversible adiabatic process against a constant external pressure $p_ext$ : $$ T_f = [\frac{C_v+(p_{ext}/p_1)R}{C_v+(p_{ext}/p_2)R}]T_i$$. It is also mentioned that when external pressure = p2 $$ T_f = [\frac{C_v+(p_{ext}/p_1)R}{C_p}]T_i$$. How do I get these formulae?
I am getting confused because they are saying external pressure. In general when irreversible work don't we take into account only the external pressure and not the internal pressure? And why exactly is this so? I have assumed so far that this is because the internal pressure differs only slightly from the external pressure. Is this so? Would someone please explain this?

In this irreversible adiabetic process, only the work done on the system will be =$-P_{ext}dV$. Because in a reversible process we actually consider $P_{ext} = P_{int} + dP$, and hence we neglect the $dPdV $ term as it is negligible.  Considering this, if we go by the definition of adiabetic process and first law of thermodynamics, we will have $$0 = nC_vdT  + P_{ext}dV$$ as the heat changed in an adiabetic process is $0$. Now simply taking one term to other side and integrating we have $$nC_v \int_{T_1}^{T_2} dT = -P_{ext} \int_{V_1}^{V_2} dV$$  where $P_1,V_1,T_1$ are initial conditions, and $P_2, V_2,T_2$ are final conditions of the system.Now we put $V_1 = \frac{nRT_1}{P_1}$ and $V_2 = \frac{nRT_2}{P_2}$, so we will have $$C_v(T_2 -T_1) = -P_{ext}R(\frac{T_2}{P_2}-\frac{T_1}{P_1})$$ now taking $T_2$ associated terms to one side and $T_1$ related terms to other side we have $$T_2(C_v + \frac{P_{ext}R}{P_2}) = T_1( C_v + \frac{P_{ext}R}{P_1})$$ and thus we have our desired result as, $$T_2=[\frac{C_v + (\frac{P_{ext}}{P_1})R}{C_v + (\frac{P_{ext}}{P_2})R}]T_1$$

The ideal gas law describes the behavior of an "ideal gas" only under thermodynamic equilibrium conditions.  For a rapid irreversible expansion or compression, the gas is not close to thermodynamic equilibrium, and its behavior is much more complicated.  It depends not only on the volume of the gas but also on the rate of change of volume.  To a very crude approximation, the force F that the gas exerts on its surroundings (which, by Newton's 3rd law is also equal to the force the surroundings exert on the gas $P_{ext}A$) can be described by:  $$\frac{F}{A}=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}=P_{ext}$$where k is proportional to the viscosity of the gas.  This viscosity term in the equation is responsible for the irreversibility of the behavior.  Note that at very slow rates of volumetric change, the equation reduces to the usual form of the ideal gas law.
If the external force of the surroundings on the gas is constant, the complexities associated with the above equation can be circumvented, and the work done by the gas on its surroundings can be directly calculated from $W=P_{ext}(V_2-V_1)$.  If the expansion is stopped before the gas has had a chance to equilibrate, the gas can ultimately equilibrate to a final pressure $p_2$ different from $P_{ext}$.  But that will not change the amount of work that has been done, since all displacement hss ceased once the final volume $V_2$ has been attained.
Between the initial and final thermodynamic equilibrium states of the system, we have from the 1st law of thermodynamics, $$\Delta U=nC_v(T_2-T_1)=-P_{ext}(V_2-V_1)$$
The remainder of the development is the same as that presented by @Soumik Das.


text:
What is the proper description/name of this amide/ester structure?

What is the proper description/name of this amide/ester structure?

According to the current (corrected) version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the names of esters are generally formed by placing the alcoholic component in front of the name of the acid component as a separate word (e.g. ethyl acetate).

P-65.6.1 General methodology
  Neutral salts and esters are both named using the name of the anion derived from the name of the acid. Anion names are formed by changing an ‘-ic acid’ ending of an acid name to ‘-ate’ and an ‘-ous acid’ ending of an acid name to ‘-ite’. Then, salts are named using the names of cations, and esters the names of organyl groups, cited as separate words in front of the name of the anion.

In this case, the corresponding acid is a carbamic acid.

P-65.2.1.1 The contracted name ‘carbamic acid’ (from carbonamidic acid), for $\ce{H2NCO-OH}$, (…), are retained and are the preferred IUPAC names.

Therefore, the methyl ester is named as methyl carbamate.
The complete name of the compound that is given in the question is
methyl [6-(propylsulfanyl)-1H-benzimidazol-2-yl]carbamate.

A similar example given in the Blue Book is 2-hydroxypropyl (2-aminoethyl)carbamate.

In IUPAC speak it is a carbamic acid methyl ester.
The IUPAC name for the compound drawn according to Chemspider structure search is Carbamic acid, N-[5-(propylthio)-1H-benzimidazol-2-yl]-, methyl ester  1
Most working chemists would call it a methyl carbamate

A common alternative name is urethane. I wouldn't list all the possible alternatives, but this one is used commonly (particularly in polymer/adhesive industry), so I thought it would be important to mention. This isn't at all common in most fields of chemistry, as far as I know, but if trying to understand chemistry in the world around us, I believe it's useful information without being the information overload that listing all possible names would be.

In IUPAC speak it is still methyl carbamate, actually.

Furthermore, in preferred IUPAC speak, it is propylsulfanyl instead of propylthio and the locant N isn’t necessary when the name is written out in full. Also note that, as drawn in the question, the compound is the 6-(propylsulfanyl) and not the 5-(propylsulfanyl) isomer.


text:
Application of BET to micropore
'application of the BET equation should be restricted to the range where the term n(1 – p/p0) continuously increases with p/p0'see 5.2.2 in the reference
This is one of the criterias IUPAC proposes for BET to be applied to micropores.
What is the physical meanning of n(1 – p/p0) and why does it need to increase?

OK, I was tracking a little the meaning of it. Rouquerol's adsorption book traces it to the paper DOI: 10.1016/S0167-2991(07)80008-5
From this paper, it goes to the Keii article in Anal. Chem. (https://pubs.acs.org/doi/10.1021/ac50154a063) but in here I could not find it clearly. The other possible reference, is [6] J.B. Parra, J.C. de Sousa, R.C. Bansal, J.J. Pis and J.A. Pajares, Adsorption Sci. Tech., 11 (1994) 51. Which I don't have the book now to check... if I remember, I will look at it tomorrow back at the office.

I checked the paper by Parra et al. and it also does not explain much. Just mentions using the alternative linear form by Keii which is more sensitive to deviations from linearity.

I checked Keii's paper. I think the increasing of n(1 – p/p0) means the forming of monolayer. When it stops increasing, the second layer starts to form.


text:
Capillary action only works with glass tubes?
I tried to reproduce the capillary action and measure how high the water rises. So far, I've tried unsuccessfully with a regular drinking straw and a very narrow plastic tube with an internal diameter of about 0.5mm. Does the tube have to be glass for it to work?

Few things in chemistry are truly unique. The tube does not have to be glass; it only has to be wettable, which glass typically is and plastic typically isn't.
In the tubes made of non-wettable materials, capillary action works the other way: the liquid will go down rather than up, akin to Hg on the picture provided.
 (source)
So it goes.

Thanks, I'll try it with the glass tube inside a weather thermometer.