Why does ozone disproportionate on reaction with potassium iodide? In this redox reaction, iodide ion is oxidized to iodine and ozone reduces to hydroxide ion.

$$\ce{2KI + O3 + H2O -> I2 + O2 + 2KOH}$$

This seems like a redox reaction since iodide ion is getting oxidized and ozone is getting reduced. Then why is $O_{2}$ formed in this reaction?

The reaction equation does not really give information about the full mechanism, but one possibility is that the ozone is first reduced to ozonide ion, which is known in compounds with heavier alkali metals like potassium,; then the ozonide ion disprportionates to give diatomic oxygen:

$\ce{O3}+\ce{e-}\rightarrow \ce{O3-}$

$\ce{O3-}+\ce{e-}+\ce{H2O}\rightarrow 2 \ce{OH-}+\ce{O2}$

Note that the reducing agent is not specified as iodide ion.  Any mild reducing agent can act similarly.


Why does copper gain electrons in redox reactions, when it instead has a tendency to lose them? Cu has tendency to lose electrons. But in some redox reactions, it gains 2 electrons and become neutral. Why is it so? How can it gain them?

As you already said it, it only gains electrons when it's oxidized, not like a Cu(-II) would form or so. Therefore it's just regular redox-chemistry. Copper is somewhat noble, or at least it has a mediocre redox-potential. When dissolved in acids it only reacts with oxidizing ones like nitric acid. On the other hand if you have Cu(II) and add an iron nail, the redox-potential of iron is much lower making it a good reducing agent. But that's basically the same for every element. Once you have an oxidizing agent it will be oxidized, if you add a reducing agent it will be reduced. Gold would do the same, sodium as well if you used lithium to reduce although gold doesn't readily give off it's electrons and sodium will do it quite easily. It really depends on the other redox-system you add.

You may also take a look at the redox-series:

Here the stronger reducing ones are on top (but depending on the artist it may also be the other way). Here it means the upper one will go from left to right if the lower one goes from right to left. To take an example:

Zn will form Zn(II) and Cu(II) will form Cu(0) in a system where both are mixed. But Cu will form Cu(II) and Pt(II) will form Pt(0) if those are mixed.


Identify the structure of compound with C₉H₁₀ formula using IR and NMR data Use IR and NMR data to define the structure of the molecule, the formula of the given structure $\ce{C9H10}.$

On what basis do we conclude that its structure is that (the answer) only? On what points can we identify the structure of this molecule?
I have found out that since its frequency is just less than $\pu{3000 cm-1}$ it is an alkane. Also, the approximate value of the chemical shift is near $7.2,$ hence it is aromatic.
But what I did not get was the meaning of two peaks at $7.2$ and the shifts at $2.9$ and $2.1.$

The molecular formula of the compound is $\ce{C9H10}$. The number of unsaturations in the compound of $\ce{C_nH_m}$ is $\frac{2n+1-n}{2}$. Thus, the number of unsaturations in $\ce{C9H10}$ is $\frac{2\times 9+1-10}{2}=5$. You already found out that compound is aromatic by FT-IR, and hence $\ce{C6}$ of the formula counted out with 4 unsaturaations (3 double bond and a ring for aromatic nucleus).The remaining number of unsaturarion is $5-4=1$, and remaining number of $\ce{C}$ in the formula is 3. The remaining unsaturation could be a double bond or a ring.
When look at $\ce{^1H}$-$\mathrm{NMR}$ spectrum, chemical shifts of all but aromatic peaks are less than $\pu{3.00 ppm}$. Therefore, it is safe to conclude that the compound is a aromatic bicyclic rather than aromatic alkene. Therefore, the compound is indane:

The protons on 1-$\ce{C}$ and 3-$\ce{C}$ are chemically and magnetically equivalent, and benzylic, hence the their relevant signal would be down-field to that of protons on 2-$\ce{C}$, which is usual acyclic aliphatic carbon. The benzylic protons have two vicinal protons, thus the corresponding peak is a triplet. Meanwhile, protons on 2-$\ce{C}$ have four vicinal protons (benzylic protons), thus the corresponding peak is a quintet (5 peaks).
Note: The integration given is a ratio, not correct number of protons.


Where does the thermal energy come from in a chemical change when two substances react with one another? We're learning about chemical reactions in class and we recently did an experiment where we combined copper and nitric acid. Both were at relatively room temperature and when they were combined the temperature rose very quickly. Isn't this contradicting the first law of thermodynamics? If the thermal energy is neither created nor destroyed but only transferred than where did all that extra thermal energy come from? please help!

The energy was in the chemicals, in the form of bonds. Energy is liberated when there is bond formation. As the reaction between $\ce{Cu}$ and $\ce{HNO3}$ is feasible at room temperature so it happened.

And another thing, room temperature is a nice temperature for that reaction because first we need to break bonds of $\ce{HNO3}$ to make new bond, which is easily done at this temperature.

Now coming to your question, the answer is 'no'. The first law is still safe. You know it wrong. The truth is (from wikipedia) : "the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed".

Give special attention to the word 'total energy', you said the 'thermal energy' in your question. And the total energy (atleast in this case) is conserved in the form of summation of thermal energy and bond energy.


Visible Light Spectrophotometer Recommendations I’m looking into purchasing a cheap uv-vis spectrophotometer but am not affiliated with an institution at the moment given the whole COVID 19 situation. Are there any good recommendations for cheap visible light spectrophotometers that can analyze wavelengths of light on the order of 370nm and are purchasable directly to consumer?

Edit: Budget: $500-700

Disclaimer: I'm not affiliated with any company

However,
Ocean Optics cuvette spectrophotometers are a good choice

This model is used in some undergraduate labs and is good because the cuvette holder is built in (i.e. it doesn't require a fiber optic cable).

As long as the optics don’t get jumbled, the prism detector mechanism work well and reliably for many years, although it says to get it checked each year.

If you're on a tight budget I'd recommend looking at a DIY spectrophotometer. In particular the RasberryPI designs seem relatively simple

The Hackaday website has a blog dedicated to DIY spectrometers.


Which azo compounds are found in DVD-R and CD-R media? Does anyone know which specific azo compounds are used in DVD-R/CD-R recordable media?

Each company in the DVD business has their own proprietary twist on the actual dye they use and they don't go out of their way to reveal the structure of the molecule they use.  Also the actual dye used changes fairly rapidly (several years or so) as technology shifts (e.g. higher density, Blu-ray, etc.).  That said, metal-azo dyes are popular today.  A Mitsubishi patent from a few years ago (WO 2006/104196 A1 (10/03/2006)) says "The recording layer contains a pyridone azo organic dye having a 6-hydroxy-2-pyridone structure as a coupler component and an isoxazole, 1,2,4-triazole or pyrazole as a diazo component, coordinated to a metal, such as nickel, cobalt, iron, zinc, copper or manganese." Here is the structure of 2-pyridone (6-hydroxy would be attached on the other side of the nitrogen) and pyrazole.

Here is a link to a Ciba patent that has more detail. And here is a general structure a 2-pyridone-based metal-azo dye


Why is the ROHF spin density of the propenyl radical not symmetric? I was preparing some material for my students and in order to illustrate the concept of spin density and how the ROHF method gives a purer description of this property than the UHF method. I prepared a simple calculation for the $\ce{CH3}$ and $\ce{CH2CHCH2}$ radicals.

The calculations for both radicals performed properly. Given the symmetry of the molecules, I expected the distributions to be symmetric. That expectation was borne by the results of the calculations for the methyl radical and the UHF method of the allyl radical. However, in the case of the allyl radical the spin density surface is asymmetric as shown in the following picture obtained with GaussView 5.

The canonical MO have the proper symmetry (see below)

Can anybody point to an explanation of these behaviour or a mistake in the calculation?

P.D. Find enclose a copy of the Gaussian input field I used in the calculation.

%chk=c3h5_rohf.chk
# opt=tight rohf/6-31++g(d,p) guess=mix

ROHF

0 2
 C
 H                  1    1.06999999
 C                  1    rCC    2  ACCH
 H                  3    rCH    1  AHCC    2    0.0
 H                  3    rCH    1  AHCC    2  180.0
 C                  1    rCC    2  ACCH    3  180.0
 H                  6    rCH    1  AHCC    3  180.0
 H                  6    rCH    1  AHCC    3    0.0

 rCC 1.40140000
 rCH 1.0700000
 ACCH 120.0
 AHCC 120.0

I believe your wave function is not the one you would expect in ROHF. The reason for this is guess=mix which will try to construct a broken symmetry wavefunction from the HOMO and LUMO. Why the SCF converges to that solution, and why this is stable (-116.460412224 ha), I don't know, and I have currently not the time to investigate (I will try to come back to that), but remove the keyword and you should obtain the symmetric correct solution (-116.457980797 ha). Inspect your output file, and you will notice, that the calculation is run in Cs symmetry, not C2v. (Also HF might not be the best method to show delocalisation.)


Considering the d-d transition how, does tetracyanidonickelate(II) ion exist as a colored complex? As I refer on google, the resources indicated that,

  the d block elements which have unpaired electrons as in their ions, when they are making complexes with ligands, they can absorb different frequencies in the spectrum due to degenerated d-d orbital transition.

But I am confused in understanding how there is a colour for $\ce{[Ni(CN)4]^2-}$ complex.

My question is: In the $\ce{Ni (II)}$ ion there are two unpaired electrons in the d orbitals. But while they are hybridizing there aren't any unpaired electrons, all are paired. In the solution state, the nickel ions do not exist as $\ce{Ni^2+}$ ions, but rather as $\ce{[Ni(CN)4]^2-}$ ions. But $\ce{[Ni(CN)4]^2-}$ has no unpaired electron. So it is unable to absorb energy from the spectrum and transit between d orbitals. (My logic is that there are no more spot in d orbitals to transit between after the complex occur)

But as the image below this complex is yellow in colour.

I can't find the mistake in my logic. I referred to many resources but I wasn't able to find any satisfactory answer.

This is a diagram of the d orbitals of a generic d8 complex in a tetrahedral and square planar configuration. The tetrahedral complex would be expected for pi donor ligands (Cl-, OH-, etc.) where the pairing energy is greater than delta t, while the square planar complex would be expected for strong field pi* acceptor ligands (think CO, NO+, and CN-, which are effectively isoelectronic anyways).

As you can see the frontier orbitals of this complex are all mostly on the metal complex , and I'd expect the excitation of electrons from lower-energy orbitals into these anti bonding orbitals to be the reason for the colors.


From where does sea water get salt? I have learned that in primary classes  that sea water is salty and I had also wondered that, what makes it salty? and my question is, where does sea water gets its salt?

The salt in the oceans came originally from rocks. Rain can dissolve some of the minerals found in rocks, especially rain that is slightly acidic from dissolved $\ce{CO2}$. The chemicals in the minerals are broken down into ions in solution. Some of the ions will tend to precipitate out of solution, or be taken up by living things (for example calcium is removed from water by the shells of molluscs and diatoms). Other ions will tend to build up in the water. Two of the most soluble and common ions are sodium and chloride. Over time these built up in the ocean. The sodium chloride in sea water is what makes the seas salty.


Mixing Different Solutions with Different Wavelengths to Get Different Colors? I recently was told that the reason you can mix colors (blue and yellow make green because wavelengths 605nm + 450nm = green) was based on wavelength (which seems very obvious when you think about it but please humor me).

This is kind of cool but I don't get it - is this based on the wavelength range that's transmitted?  If  605nm + 450nm = green then how would you find that exact wavelength of green? Would it be the average of the two, or is there some constant that I am missing?

You have to distinguish two sorts of colour mixing:

Subtractive mixing is observed when you mix two coloured solutions ot put two colored films on top of each other and shine white light through it. Here, $\mathrm{\color{yellow}{yellow}}$ and $\mathrm{\color{blue}{blue}}$ yield $\mathrm{\color{green}{green}}$.
Additive mixing is observed when you overlay two coloured light sources. In the RGB model, $\mathrm{\color{red}{red}}$ and $\mathrm{\color{green}{green}}$ yield $\mathrm{\color{yellow}{yellow}}$, the combination of $\mathrm{\color{red}{red}}$, $\mathrm{\color{green}{green}}$ and $\mathrm{\color{blue}{blue}}$ yields white.

As far as the wavelengths are concerned: Each "colour" is represented by a wavelength range, rather than an exact value. The fact that there's more than one red or green is intuitively understood - particularly by women ;)


Calculating the pH of a saturated calcium fluoride solution How can I calculate the pH of a saturated solution of calcium fluoride ($\ce{CaF2}$)? I am given the following values:

$$\begin{align}
K_\mathrm{sp}(\ce{CaF2}) &= 3.9 \cdot 10^{-11} \\
K_\mathrm{a}(\ce{HF}) &= 6.8 \cdot 10^{-4} \\
K_\mathrm{w} &= 10^{-14}
\end{align}$$

(The $K_\mathrm{sp}$ and $K_\mathrm a$ values are taken from the appendices of Skoog et al. Fundamentals of Analytical Chemistry, 9th edition.)

In homework land you are right, but not in real life.

The $K_\mathrm b$ of $\ce{Ca^2+}$ ($\mathrm pK_\mathrm b = 2.43$ according to some sources) is within 1 log unit of the $K_\mathrm a$ of $\ce{HF}$.  In other words, near neutral pH, consideration of the $\ce{Ca^2+/CaOH+}$ equilibrium is almost as important as the $\ce{F-/HF}$ equilibrium.


Molar heat capacity and heat capacity Is molar heat capacity and heat capacity

at constant volume both represented by $C_v$?
at constant pressure both represented by $C_p$?

The symbols, which are standardized in ISO/IEC 80000 Quantities and units (14 parts) and recommended by IUPAC in Quantities, Units and Symbols in Physical Chemistry (Green Book), distinguish the various related quantities:

$$\small
\begin{array}{llll}
\hline
\text{Quantity} & \text{Quantity} & \text{Unit} & \text{Unit} \\
\text{name} & \text{symbol} & \text{name} & \text{symbol} \\
\hline
\text{heat capacity} & C & \text{joule per kelvin} & \mathrm{J/K} \\
\text{heat capacity at constant pressure} & C_p & \text{joule per kelvin} & \mathrm{J/K} \\
\text{heat capacity at constant volume} & C_V & \text{joule per kelvin} & \mathrm{J/K} \\
\text{specific heat capacity} & c & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{specific heat capacity at constant pressure} & c_p & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{specific heat capacity at constant volume} & c_V & \text{joule per kilogram kelvin} & \mathrm{J/( kg \cdot K )} \\
\text{molar heat capacity} & C_\mathrm m & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\text{molar heat capacity at constant pressure} & C_{\mathrm m, p} & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\text{molar heat capacity at constant volume} & C_{\mathrm m, V} & \text{joule per mole kelvin} & \mathrm{J/( mol \cdot K )} \\
\hline
\end{array}$$

However, in contrast to symbols for units, symbols for quantities are only recommendations. Unfortunately, quantity symbols are used inconsistently in the literature, sometimes in a sloppy way. In principle, a given symbol can indicate different quantities.


Can we predict with any reliability whether a chemical is carcinogenic in people? I know that some chemical compounds have been known to be carcinogenic in humans.

But can we make a reliable prediction of how likely a substance is to be carcinogenic based on the chemical and physical properties of the substance ? How good will the prediction be ?

It would be good if the following methods were compared

Computational chemistry
Crystallography
Bacteria models such as the Ames test
Animal models such as rats and mice
Chemical kinetics experiments
Predictions based on the fact that a substance is from the same class as a known carcinogen or non carcinogen

One of the great problems is that there is no single mechanism by which a chemical carcinogen causes the damage which leads to cancer. There are a range of different mechanisms, I will mention some of them.

One mechanism is alkylation of DNA, the attachment of alkyl groups to DNA will prevent it being read correctly. In some cases the damage can be repared while in other cased the damage will cause the body to misread one base pair as another. There is an interesting problem with electrophilic alkylation agents, if they are too reactive towards water then they will not survive to reach the DNA.

There is an interesting paper in which this idea is discussed E.W. Vogel and M.J.M. Nivard, Mutation Research, 1994, volume 305, pages 13-32 considers it at length. If we consider the ratio of the rate constant for the reaction of an alkylation agent with thiosulfate and acetate, then as the ratio of these reaction rates changes in favour of the thiosulfate reaction something known as the Swain Scott constant (s) also increases. The Swain Scott constant is the ratio between reaction at two different sites (7-alkylguanine/O6-alkylguanine) in DNA. In general the higher the value of the s constant the stronger a carcinogen is. In the paper by A. Barbin and H. Bartsch in Mutation Research, 1989, volume 215, page 95 to 106 a series of directly alkylating carcinogens a good but not perfect relatioship is seen between the s constant and how carconogenic they are in rodents.

So it appears for one class of carcinogen that it is possible by doing some simple chemical kinetics experiments that it is possible to make an assessment of how carcinogenic a substance.

One of the great problems is that many "carcinogens" are in fact "precarcinogens" which require metabolic activation to make them carcinogenic. For example hydrazine reacts with formaldehyde in the liver before being oxidized to form diazomethane which then methylates the liver. This is why acute large hydrazine exposures in humans can cause liver failure.

This brings us to the Ames test, this is a test using bacteria which measures how able a substance is to alter DNA. The idea is that cancer is caused by altering the DNA of normal cells. So something which is able to randomly alter the DNA of cells is able to induce cancer. The great problem with the Ames test is that while it is a good test. There are both false positives and false negatives if it is used as a test for carcinogenic potential.

The normal Ames test protocol includes some sterile extract of rat liver to provide some metabolic activation. This will increase the ability of things like polycyclic aromatic hydrocarbons to alter the DNA of the bacteria.

If we consider an inorganic example, the chromate anion is known to be carcinogenic while chromium(III) is not carcinogenic. To understand this we need to know something about chromium biochemistry.

It is true that chromium(III) can form some very stable complexes which are very slow to react with DNA. But the chromium(III) is unable to reach the DNA in the cell. However chromate is able to pass through the sulfate transport channels in both the cell and nuclear membranes to allow it to reach the nucloplasm. There it is reduced to chromium(III) which then damages the DNA. As a result we can make a judegment about chromate and sulfate.

Sulfate is unable to react in the nucloplasm to form something which will form long lasting complexes with DNA, so even while sulfate is able to reach the DNA by the same route as chromate it is clear that sulfate is unable to act as a carcinogen by the same mechanism as chromate.

Also permanganate or magnanate is a stronger oxidant than chromate, I suspect that as it would be less able to survive the journey through the body to the DNA it is less able to incude cancer. This is likely to be one of the the reasons why permanganate is far less carcinogenic than chromate. Plenty of evidence exists for chromate being carcinogenic but little if any evidence exists for permanaganate.

Also the complexes of manganese with DNA are likely to be shorter lived than those of chromium. This is becuase chromium(III) complexes are slow to exchange ligands while manganese(II) are rather quick to exchange ligands. This is due to the fact that chromium(III) complexes are d3 while manganese(II) are d4.


Why are silyl chlorides more readily hydrolysed than alkyl chlorides? I know that the $\ce{Si-Cl}$ bond has a slightly higher energy, but silyl chlorides are much more readily hydrolyzed compared to alkyl chlorides. I do not fully understand why that is. My thoughts so far are:

There is a better match in energy between the orbitals of the $\ce{Si-Cl}$ bond and water.
Reactions proceed when there is a favorable interaction between HOMO and LUMO.
In this case, the HOMO is the lone pair on the water molecule. The LUMO of a $\ce{Si-Cl}$ bond is closer to it in energy compared to a $\ce{C-Cl}$ bond.

Is this the right argument to take?

Using $\ce{CCl4}$ and $\ce{SiCl4}$ as the basis for comparison, there are several reasons.

The greater electronegativity difference between Si and Cl leads to a larger $\delta^+$ on silicon, making it more electrophilic and susceptible to attack by water.
Larger size of Si means that there is less steric hindrance towards the approach of water.
It is possible that the hydrolysis of $\ce{SiCl4}$ proceeds via an alternative, faster pathway that involves a five-coordinate intermediate (as opposed to $\ce{CCl4}$, in which five-coordinate carbon does not occur).


Which hydrogen of 3-methylcyclohex-2-en-1-one is the least acidic? Which hydrogen of 3-methylcyclohex-2-en-1-one is the least acidic?

According to me it should be 2, as on its removal there won't be any resonance stabilization as would happen in the other cases. In addition to that a negative charge residing on a carbon attached to a double bond would be highly unstable.
But the answer key says that it is 3.

A follow up question: Which is the most acidic hydrogen?

According to me it should be 4 as it leads to a greater number of resonance structures.

You are correct. Position 2 is the least acidic.

Position 3 is not the least acidic hydrogen. It might be the most acidic hydrogen. Position 2 is the least acidic hydrogen of the four marked.

What does position 2 have going for it? Why might it not have been marked as the least acidic? The lone pair that would be left in the conjugate base is sitting in a sp² orbital.

All else equal, sp² hybrid orbital containing $\ce{C-H}$ bonds are more acidic than sp³ hybrid orbital containing $\ce{C-H}$ bonds. The $\mathrm{p}K_\mathrm{a}$ of ethane $\ce{CH3CH3}$ is reported to be $\sim 50$, and the $\mathrm{p}K_\mathrm{a}$ of ethene $\ce{CH2=CH2}$ is reported to be $\sim 44$.

However, all else is not equal. Resonance, especially with a carbonyl group, stabilizes conjugate bases more. Acetone $\ce{CH2COCH3}$ has a $\mathrm{p}K_\mathrm{a}$ of $\sim 19$. Position 2 is not stabilized by resonance. The conjugate bases from positions 1, 3, and 4 are all resonance stabilized through the carbonyl group.

Position 1 has two resonance structures.

Position 3 has three resonance structures.

Position 4 has three resonance structures.

It is between positions 3 and 4 for the most acidic. I chose position three because alkyl groups (the rest of the ring) are electron donating by induction, but alkyl group also stabilize pi bonds, so position 4 is a good candidate for most acidic as well.


Food safe basic polyelectrolytes? Pectin is an example of a food safe polyelectrolyte. It acts as an acid because it is a proton donor and develops negative charges along the polymer.

Can you think of any examples of food safe polyelectrolytes that act as bases and develop positive charges along the polymer?

Pectin is a heteropolysaccharide rich in galacturonic acid:

The carboxylic acid in galacturonic acid gives pectin the ionizable character. Pectin can react with bases to form carboxylate anions along the polysaccharide chain.

There exists a sugar-derivative and polysaccharide with opposite behavior - they react with acids to produce cations along the polysaccharide chain. The modified sugar is glucosamine, and the polysaccharides are chitin and chitosan(http://en.wikipedia.org/wiki/Chitosan). The amine ($\ce{NH2}$) group on glucosamine is basic and will react with acids to become cationic.

Glucosamine is used as a dietary supplement, and chitin and chitosan are found in various foods, like fungi and shellfish.


Ideal Gas Vs Real gas If ideal gas equation is formulated using ideal gas concept then why is the same equation used for real gases os they are not the ideal gases

For real gases there are some contradictions in the assumptions made while defining real gases.
Therefore some corrections were needed in the equation which were given by van der waals in his real gas equation
The van der Waals equation corrects for the volume of, and attractive forces between, gas molecules:

(P+a(n/v)^2)(V-nb) = nRT

There are two corrective factors in van der Waals equation. The first,   , alters the pressure in the ideal gas equation. It accounts for the intermolecular attractive forces between gas molecules. The magnitude of a is indicative of the strength of the intermolecular attractive force.

The factor - nb accounts for the volume occupied by the gas molecules. b has units of L/mol. Since b corresponds to the total volume per mole occupied by gas molecules, it closely corresponds to the volume per mole of the liquid state, whose molecules are closely layered. b is generally much smaller in magnitude than a . The values of a and b generally increase with the size and complexity of the molecule.


Why do people consider caffeine ineffective after its half-life? Assuming you take $\pu{200mg}$ of caffeine, it's half life is roughly 6 hours. So after 6 hours there should still be $\pu{100mg}$ of caffeine in your body.

Then why is it generally said that after six hours caffeine's effect should wear off? Shouldn't it only be diminished since there is still $\pu{100mg}$ of caffeine in your body?

Personal experience: Some years ago, I mowed our lawn on a hot summer day. It took about 2 hours, and I sweated up a storm. To compensate, every 20 minutes or so, I came into the house and swigged down some Pepsi Cola. After I was finished, I noticed that the 2 liter bottle of Pepsi was also finished, and I was tired, so I sat down.
With nothing else going on, I noted that my heart was skipping beats. 1,2,3,skip. 1,2,3,skip. Extra systole. Nothing to get excited about. But never this skippy. So I listened, and checked, and after about 15 minutes, I got 1,2,3,4,skip. 1,2,3,4,skip. Interesting. And about every 15 minutes, my heart would go another beat before the skip. It finally got out to 14 before skipping, then went all crazy: 12, 17, 20, then finally stopped skipping.
It took 2 hours to consume the Pepsi, and about 4 hours to overcome the observable effects of the caffeine. There is certainly a non-linearity involved with the excretion and with the effect of the caffeine on my heart, but yeah, the half life says something, but there is a diminution of effect, not a go/no go.
And I never again drank that much caffeinated beverage all at once.


Control the pressure of Hydrogen in a small container without the use of a pressure regulator I had recently asked this question, and due to the comments and general consensus of safety concerns, I decided to re-ask the question, attacking the problem from another angle.
The Problem
I need to store a relatively large amount of Hydrogen in a small (12 cc) Aluminum container at about 1000 psi. The problem is not the generation of the hydrogen, nor the transfer of the hydrogen into the container at that high pressure, but how to extract the hydrogen out of the container without using a pressure regulator.
I am afraid that the container is too small for a pressure regulator to fit in it, and the whole device has to be small on the outside, so one cannot fit it on externally...
The output pressure I desire would be about 200 psi
An Idea: (I'm an Electrical Engineer, don't laugh!)
Emulate Vapor Pressure
Gasses like CO2, Butane, and Propane have relatively low vapor pressures, allowing them to form a liquid at reasonable pressures. At about 800 psi, CO2 will form a liquid, making it so that any container holding liquid CO2 will let out CO2 at that somewhat constant 800 psi of pressure until there isn't enough CO2 to keep itself in liquid form. Hydrogen, however, will not do this because it's vapor pressure curve is far too sensitive. It would have to be cooled down to some ungodly temperature, which is not an option:

I was wondering if I could somehow emulate this behavior by possible adding something to mix with the hydrogen and change it's vapor pressure. Is this even possible?
Edit:
Flow rate doesn't directly matter, but the more flow the better.

More like a suggestion than a solution.

First, understand what you are trying to achieve.
You are generating hydrogen at some rate and storing it at high pressure. From your previous question I understand that the pressure of your stored $H_{2}$ is only a very general approximation.

Your objective is now to extract $H_{2}$ in a controlled way without using regulator. This sounds absurd.

My complaints:

I am very sure that you do not have a pressure gauge to measure the pressure of your $H_{2}$ container. This is a serious design flaw.
When you will try to extract $H_{2}$ the flow rate will be really high.

Now to the possible solution:

Incorporate atleast one dual valve regulator. This will take care of both issues mentioned above. You will be able to measure container pressure, and  regulate flow-rate.

(But you do not want a such a regulator ! okay. What you can do is to get the smallest one possible and connect the regulator using 1/8 inch or 1/4 inch PTFE tubes and keep it separate. The regulator will then need to have custom input connector. It is not difficult to design one. Regulator to tube can be connected using all metal nut-ferrule connection which are safe upto 2000 psi or more. )

Well, your safety should be prime concern than the size of device I guess. In case oxygen leaks in and temperature at some local point is enough there is a chance of SERIOUS explosion amplified by the high pressure.

Some things about which I am concerned:

Do you evacuate the container before filling $H_{2}$ in ? If not, you are making a mixture of $H_{2}$ and $O_{2}$. Be careful !
The container of $H_{2}$ can handle high pressure or not. Have you tested it ? You can test it with $N_{2}$ or just air before using $H_{2}$.


Can I engrave borosilicate glassware without compromising its structural integrity? Can I use a pen engraver (the kind that's like a tattoo gun with a diamond tip) to engrave my initials on labware without compromising it? I am in a lab space with a shared acid wash, and my glassware keeps getting boosted. I'm worried about e.g. heating a beaker and having it shatter because of the propagation of microfractures potentially caused by the engraver.

Any scratch weakens glass somewhat. However, a vibratory engraver is very likely to shatter the vessel.
You can label glass permanently by etching with warm concentrated $\ce{NaOH}$ or room-tempeerature $\ce{HF}$, perhaps with a stencil. Use great caution with $\ce{NaOH}$, and even more caution with $\ce{HF}$. Both are very caustic, and $\ce{HF}$ can release toxic fumes and cause deep tissue damage.
There are commercial etching creams with HF that might be less likely to splash.
Note that etching weakens glass slightly, but less so then engraving.


How can water be electrolysed if it's not ionic compound? The electrodes have different charges. One is positive ( the cathode ) while one is negative (anode). They will attract the particles in the solution which are charged. Hydrogen to the cathode while oxygen will go to the anode.

I have been taught that water is a covalent bond. Also I have been taught that covalent bonds do no have a charge so how is it possible that this compound is charged if it's covalent?

Pure water contains only small amounts of $\ce{H3O+}$ and $\ce{OH-}$ which are generated by autoprotolysis (the reaction given by Klaus Warzecha in his comment to your question). Reduction of $\ce{H3O+}$ at the cathode produces molecular hydrogen. At the anode, water is in turn oxidized, yielding molecular oxygen and $\ce{H3O+}$ (reference).

$$\mathrm{Cathode:}\ \ce{2H3O+ + 2e- \rightarrow H2 +2H2O}$$

$$\mathrm{Anode:}\ \ce{6H2O \rightarrow O2 + 4H3O+ + 4e-}$$

$$\mathrm{Overall\ reaction:}\ \ce{2H2O \rightarrow 2H2 + O2}$$


Chirality and Optical activity Do all chiral objects rotate the plane of polarization of EM waves? For example my hand is chiral, will it rotate the plane of polarization of EM waves of suitable wavelength?

Of course it will, provided you have a billion of it. (One molecule does not rotate much, after all.) Also, each should come without the rest of the body, or at least without the other hand, otherwise that would make a racemic mixture. (I'll leave the grisly details to the reader's imagination.) Also, each hand should form the same gesture, otherwise the destructive interference will ruin the effect.
Also, light of suitable wavelength would be microwave radiation, and a cursory search seems to indicate that the methods of polarizing it and measuring its polarization are known.
With all these conditions fulfilled, I don't see any difference from the situation with the molecules and ordinary light.


Is this reaction possible? When mixing HCl with K2CO3, the reaction that is usually given is the following:

K2CO3 + 2HCl → 2KCl + CO2 + H2O

I am wondering, however, if I could produce KHCO3 from K2CO3, by using half as many moles HCl. The reaction I imagine could happen would be expressed as follows:

K2CO3 + HCl → KHCO3 + KCl

I've gathered the solubility data for this reaction hereafter:

K2CO3: 112 g/100 mL  (20°C)
HCl: 72,5 g/100ml (20°C)
KHCO3: 22.4 g/100 mL (20°C)
KCl: 34.2 g/100ml (20°C)

From solubility data I see that it wouldn't be extremely practical to conduct as both KHCO3 and KCl are fairly soluble, but I'm wondering if the reaction would occur, or if it would just produce half the amount of KCl + CO2 + H2O through the first reaction (which is generally attributed to these reagents).

If the reaction occurs, can I assume that the first reaction is strongly dominant until I get close to have added as many moles of HCl as the initial moles of K2CO3?

And if it doesn't occur, why does it not?

Note that this question had been asked already, but had received no satisfying answer.

It is interesting to consider that either outcome is possible, depending on the order of addition.

If a solution containing 0.5 moles of HCl is added to a solution of 1 mole of K2CO3, slowly, with mild stirring, cold, the products will be 0.5 KCl + 0.5 KHCO3. Nothing is lost to the atmosphere.

However, if a solution of 1 mole of K2CO3 is added to a solution of 0.5 moles of HCl, it would be possible to get 0.5 KCl + 0.5 K2CO3 with slow addition, rapid stirring, warm/hot. The difference here is that the initial K2CO3 is completely overwhelmed by the HCl and produces KCl + H2CO3, which can be degraded to H2O + CO2. By the time half of the K2CO3 is added, the products are 0.5 KCl + CO2, and the CO2 could be mostly evolved as gas and disappears. The last half of the K2CO3 just comes in and sits there.


Haworth's research of glucose Firstly haworth used CH3I/Ag2O with glucose which replaced 5 H by CH3 forming a acetal which locked the anomer and stopped the equilibirium of ring chain tautomerism. Under dilute acidic conditions acetal group hydrolyze and ethers do not.

So my text (Organic chemistry by L.G. Wade 8th edition) says through these reactions haworth suggested that C5  contains free hydroxyl group and hence proved the pyranose(6 membered cyclic ring) form of glucose. But I fail to understand how did he find that the hydroxyl group is present on the 5th carbon in the chain after hydrolyzation by dilute acid.

I cannot guarantee that the following explanation for the proof of the ring size of the principal cyclic form of glucose is due to Haworth. The link I provided in the comment above explains one method. In "Methylation of Pyranoses" methods are illustrated for the permethylation of the hemiacetal form of glucose and its subsequent hydrolysis to the tetramethoxy hemiacetal. Although the 6-membered ring is shown, a priori the ring size is unknown. This is the issue to be determined. continued below
In "Ring Size of Pyranosides" mild HNO3 oxidation of the tetramethoxy hemiacetal affords a keto carboxylic acid (upper left). Continued oxidation of either enol of the keto group leads to oxidative degradation to an  optically-active dimethoxysuccinic acid (red path) and a meso trimethoxyglutaric acid (green path). These two dicarboxylic acids are only derived from a 6-membered ring of glucose.

Source: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/carbohydrates44.pdf


Is sulfuric acid still hazardous after it dries? One of our housemates decided to pour battery electrolyte (which is 30-40% sulfuric acid) into a motorcycle battery on our back porch next to our laundry washer and dryer.  He did so extremely sloppily, and we don't know exactly what he might have spilled the acid on.

Is the acid safe to touch after it dries, or is it still a safety hazard?

If it isn't safe when dry, what can we do to neutralize the acid?  I know baking soda can neutralize acid, but I'm not really sure how to easily coat an entire porch in baking soda.

My boyfriend suffers from severe, debilitating OCD with a chemical contamination obsession, so this is an absolute nightmare for him.  Thank you in advance for any help you may have!

Sulphuric acid won’t dry — meaning evaporate; what water does when one typically thinks of drying — at any noticeable rate. It will, however, given time penetrate deeper into the material if the material is porous or susceptible to degradation.

Depending on how much sulphuric acid was actually spilt, I suggest either mopping up the area with lots of water (if not much was spilt) or first semi-generously applying dry baking soda to the area. Once the baking soda has stopped generating bubbles immediately on application, you can use normal amounts of water for wiping up.

While I would love to suggest you get the sloppy housemate to do it, I think you might be disinclined because they again will do it sloppily.


Energy absorbed by the gas from a PV-diagram The $PV$-graph for a monoatomic gas is shown in the figure. Find the energy absorbed by the gas during this process.

I’m having trouble with this question because both pressure and volume are changing and I cannot calculate a value for $T$. So far, I ended up with an answer using the ratio of $C_p$ and $C_v$, and got $6.25P_0V_0$. The answer is $8.5P_0V_0$.

I suspect you forgot to take the work done by the gas into account. From the first law of thermodynamics:

$$Q = \Delta U - W \label{eq:1}\tag{1}$$

For the monoatomic gas $U=\frac{3}{2}PV$, so

$$\Delta U = U_2 - U_1 = \frac{3}{2}(P_2V_2 - P_1V_1) = \frac{3}{2}(3P_0 \cdot 2V_0 - 2P_0 \cdot V_0) = 6P_0V_0 \tag{2}$$

On the other hand, since the gas expanded ($V_2 > V_1$), the work is negative and can be found geometrically as the area under the line:

$$W = -(\color{blue}{2P_0(2V_0-V_0)} + \color{orange}{0.5 \cdot (3P_0 - 2P_0)(2V_0 - V_0)}) = -2.5P_0V_0 \tag{3}$$

Now back to \eqref{eq:1}:

$$Q = 6P_0V_0 -(-2.5P_0V_0) = 8.5P_0V_0$$


Spectrochemical Series - Sigma Donor Capability I fully understand and recognise the manner and which the series works.

I cannot find any references for the change in sigma donor capability moving up the series.

For example, comparing $\ce{Cl-}$ and $\ce{NH3}$. Which of the two would be more a more willing sigma donor ?

Does this have something to do with how diffuse the orbitals are ?

The spectrochemical series is an ordering of relative ligand interaction strength with metal ions. It can be used to infer relative $\sigma$-donor strength, if $\pi$-interactions are not possible for those particular ligands. Examples of this type ligand with no $\pi$-interactions include $\ce{NH3}$, $\ce{CH3-}$, and $\ce{H-}$.

Most ligands will exhibit $\pi$-interactions, operating as $\pi$-donors or $\pi$-acceptors, and these interactions dominate the ordering of the spectrochemical series. The $\sigma$-donor interactions are obscured and heavily affected by the $\pi$-interactions. Experimentally all of these effects are important, and it is difficult to separate these interactions from one another.

For example, the $\sigma$-donor interaction of a $\pi$-acceptor ligand is drastically increased from some initial level. $\pi$-backbonding reduces electron density at the metal center, leading to increased $\sigma$-donation (which increases $\pi$-backbonding, and so on). This will make a given $\pi$-acceptor ligand act as a better $\sigma$-donor than would be expected. The converse effect is observed with $\pi$-donor ligands reducing $\sigma$-donor strength.

The complexity of these interactions tempers the question of gauging $\sigma$-donor interactions only. The combined bonding effects influence one another.

Theoretically separating out $\sigma$-donor interactions from all $\pi$-interactions could be possible computationally, but I am unaware of any research on this specific topic. From a very general standpoint, improved metal-ligand orbital overlap will increase $\sigma$-donor strength, which may help estimate these characteristics qualitatively.

Most inorganic textbooks should be able to provide further detail on this topic under sections on ligand field theory.


Mechanism in formation of acetals someone asked a question about this mechanism and there were 2 answers as shown below.

but these two answers seem contradict.
and my questions are
  1. which one is right?
2. does the protonation by H2SO4 and free proton both mean the protonation by H3O+?
3. I know sp2 hybridized C is more electronegative, but at the same time it's losing e- to oxygen through pi bond. is C=O still less polarized than C-O even considering pi bond?
4. in answer #2, it's saying like free proton(H3O+?) is less acidic than hydrogen attached to positively charged oxygen(ROH2+?). how is this right?

Hope somebody can help me understand it !!

The first answer is wrong, and the second answer is somewhat true (alcohols are more polar than carbonyls, but not for this reason), but unhelpful.

  You use protonated ethanol because it is the most likely acid present.

Let me provide an analogy.

Sulfuric acid is a strong acid. When sulfuric acid is dissolved in water, it ionizes (reacts with water)to an extent that only the tiniest fraction of $\ce{H2SO4}$ is left:

$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$

In an aqueous solution, its fair to say that the concentration of hydronium cation outweighs the concentration of free sulfuric acid by quite a lot. If you were drawing an acid-catalyzed mechanism in water, would you use $\ce{H2SO4}$ or would you use $\ce{H3O+}$?

This reaction is done in ethanol as the solvent. The basicity of ethanol is pretty similar to water, and we have a lot of ethanol, so we can expect the sulfuric acid to ionize to a great extent. These acid-base equilibrium shenanigans reaction would occur regardless of whatever else we added to the mix.

$$\ce{CH3CH2OH + H2SO4 -> CH3CH2OH2+ + HSO4-}$$

Now, if we are doing an acid-catalyzed reaction in ethanol using a strong acid, which species should we use as the proton source when drawing a mechanism: $\ce{H2SO4}$ or $\ce{CH3CH2OH2+}$?

Note that this is an example of the "leveling effect". The strongest acid that can exist in your solution is the conjugate acid of your solvent, and the strongest base that can exist in your solvent is the conjugate base of your solvent. Anything stronger reacts with your solvent.


Is the variation in carbonation of soft drinks the variable that affects the effectiveness of the mentos reaction? I went and bought a bottle of Fanta and some Mentos to replicate the famous youtube videos of the big fizzy reaction.

The result was dismal, it weakly foamed up and dribbled over the edge.

I tried again with Pepsi Max and had a spectacular result (that the kids loved).

I wondered if the carbonation of the drinks was what impacted the difference in reaction.

My question is: Is the variation in carbonation of soft drinks the variable that affects the effectiveness of the mentos reaction?

It's my understanding that the two key variables here are carbonation, as you suggested, and viscosity.

People performing this sort of demonstration usually use drinks completely sweetened with artificial sweeteners like aspartame or Sucralose, because they're much less viscous than their full-sugar equivalents, and especially than the high-fructose corn syrup variants.

Common artificial sweeteners are much sweeter than sucrose and glucose [1], so they're added in much smaller quantities and so affect the properties of the liquid less.

The viscosity affects the ability of the liquid to have the rush of bubbles from rapid nucleation required for the fountain effect, and the carbonation affects how much impetus there is behind it.

For an good comparison, you could try three variants of (popular cola product), including a newer one partially sweetened with steviol (look for the ones with green labels: Pepsi Next and Coca-Cola Life are two examples): the wholly artificially sweetened one should jet best, then the part-steviol, part-sugar one, then the full-sugar one.

"Sucralose is about 320 to 1,000 times sweeter than sucrose, three times as sweet as aspartame and twice as sweet as saccharin." -- Wikipedia (I'm not sure how that's measured, but the sucralose figure comes from the FDA approval document cited on the Wikipedia page.)


What makes heat resistant proteins heat resistant? We were doing the chapter Biomolecules in class the other day, and a doubt popped up once we reached the section on proteins.

We were taught that the stuff that keeps proteins together are: Ionic bonds (not always), Covalent bonds (peptide bonds and di-sulfide bridges), van der Waals' forces and hydrophobic interactions.

Now I've heard that certain bacteria are known to survive extremely hot conditions (Inside thermal vents, volcanoes, the like), like Thermus aquaticus, the source of taq polymerase. I'm told they're able to survive in those conditions because their proteins (enzymes and other cell components) show a very high level of thermal tolerance.

I tried reading up more on that, but Wikipedia adopts a very vague, generalized approach to the topic and doesn't specifically mention what imparts this high degree of stability to these proteins. https://en.wikipedia.org/wiki/Thermus_aquaticus

Frankly, if the earlier stated interactions are the only interactions seen in proteins, I don't see how some proteins can possess such high resilience to heat, especially since most of those interactions are fairly weak (and ionic interactions aren't very common).

Sure, maybe an incredibly extensive network of those interactions (say, hydrogen-bonding) would impart a good deal of stability, but I suppose such a protein would be incredibly compact, and twisted in all sorts of wacky shapes to generate the required stability.

I even tried comparing the structures of taq polymerase (First image; which denatures at about $100\ \mathrm{^\circ C}$) and salivary amylase (Second image; which denatures at a little under $40\ \mathrm{^\circ C}$), but to my (untrained) eye, there doesn't seem to be some absolutely fantastic and unique character to taq polymerase going by the structure that is:

So what exactly is it that imparts such high thermal stability to some proteins? Is it just a vastly intricate network of the earlier mentioned interactions, or is it because of something else?

Szilágyi and Závodszky published an article in the journal Structure which analyses a number of different structural parameters of proteins of moderately thermophilic ($45~\mathrm{^\circ C} < \vartheta_\mathrm{opt} < 80~\mathrm{^\circ C}$) and hyperthermophilic ($\vartheta_\mathrm{opt} \approx 100~\mathrm{^\circ C}$) organisms compared to homologous mesophilic organisms.[1] I shall summarise their results in this answer.

They analysed the parameters number of cavities, volume of cavities, surface area of cavities (all relating to cavities on the surface of a given protein), number of hydrogen bonds, number of unsatisfied potential hydrogen-bonding candidates, number of ionic pairs (the ionic pairs were analysed for distances of $4~\mathrm{Å}, 6~\mathrm{Å}$ and $8~\mathrm{Å}$) number of α helices, number of β strands, number of unordered loops and the ratio of polar to apolar amino acids (both of exposed and of buried residues). These were all assigned a $P$-value; this defining the probability of the observed effect occuring if the null hypothesis ($H_0$, ‘this feature does not contribute to heat stability’) were true.

The single most important influence on the stability of both moderately thermophilic and hyperthermophilic proteins proved to be additional ions pairs as shown in table 1 below.

$$\textbf{Table 1: }\text{$P$-values of ion pair distances in both}\\\text{moderately thermophilic and hyperthermophilic proteins}\\
\begin{array}{lcc}\hline
d/\mathrm{Å} & P_{45{-}80} & P_{100}\\ \hline
4 & 0.178\phantom{0} & 0.026\\
6 & 0.024\phantom{0} & 0.016\\
8 & 0.0048 & 0.014\\ \hline \end{array}$$

While close-contact ion pairing is not as important for moderately thermophilic proteins ($P_{45{-}80}$), this quickly changes for less-close contact ion pairings; in hyperthermophiles, all types of ion pairing are significantly important. Other influences were:

polarity of buried areas; while the difference was not as significant for only moderately thermophilic ($P_{45{-}80} = 0.156$) or hyperthermophilic proteins ($P_{100} = 0.101$), over the full data set there was a greater proportion of polar amino acids buried (i.e. not superficial) in thermophilic proteins when compared to mesophilic proteins ($P_\mathrm{full} = 0.047$).
polarity of exposed ares; this mainly stabilised moderately thermophilic proteins ($P_{45{-}80} = 0.024$).
number of cavities; this was mainly a feature of hyperthermophilic proteins ($P_{100} = 0.033$).

Furthermore, the authors point out that the choice of amino acid is somewhat influenced; however, the effects are not as strong. Moderately thermophilic proteins tend to favour arginine over lysine and use less lysine than mesophiles; hyperthermophiles incorporate both more lysine and arginine residues. Tryptophan, glutamate and aspartate are significantly increased in hyperthermophiles; moderate thermophiles incorporate less tryptophan and slightly more aspartate and glutamate than mesophiles. Also, hyperthermophiles exhibit a preference for large unpolar amino acids (leucine and isoleucine) and and have a reduced methionine and asparagine content. Since both latter amino acids decompose more readily at elevated temperatures, their reduction was expected. Cysteine content is either similar or reduced, showing that increased disulphide bridging does not contribute to thermic stability.

Previous studies and theoretical models[2] have shown that ion pairing is expected to have a marked effect on stability. At room temperature, the formation of a salt bridge leads to an unfavourable reduction of solvation which overcompensates any enthalpy gained. At higher temperatures however, this desolvation penalty is strongly reduced and the stability gain from salt bridges thus increased.

The authors do point out one major caveat concerning hyperthermophilic proteins they studied:[1]

  It should be noted, however, that the number of known structures of proteins from organisms with $T_\mathrm{opt} \approx 100~\mathrm{^\circ C}$ is very small (there are only five such structures in our data set). This means that our conclusions pertaining to proteins from extremely thermophilic organisms might be of limited validity; in fact, it is possible that much of the observed deviations are because of ‘sample bias’ (i.e., the peculiarities of the available protein structures).

References:

[1]: A. Szilágyi, P. Závodszky, Structure 2000, 8, 493. DOI: 10.1016/S0969-2126(00)00133-7.

[2]: A. H. Elcock, J. Mol. Biol. 1998, 284, 489. DOI: 10.1006/jmbi.1998.2159. Reference [42] in [1].


How to rationalise the basicity order of phenylenediamine? Why is the basicity of para- > meta- > aniline > ortho-
for phenylenediamine?
($\mathrm pK_\mathrm a$: ortho: 9.53, meta: 9.12, para: 7.96) I tried deducing it using resonance and inductive effects but got stuck.

As Mithoron pointed out in a comment, we had a similar question convering the cases of 2-methylaniline (o-toluidine) and 2-methylbenzoic acid (o-toluic acid) here.

In the case of 1,2-diaminobenzene (o-phenylenediamine), it is conceivable that hydrogen bonding between the two amino groups reduces the electron density on one of the nitrogen atoms, thereby leading to a slight increase on the other. As a result, o-phenylenediamine is significantly less basic than the m- and p-isomer, but a tad more basic than aniline.


Effect of specific heat capacity on heat loss Assume we want to calculate the heat capacities of two substances with different heat capacity. We heat them both up with a immersion heater delivering a constant rate of energy (constant J/sec), and we stop the immersion heater after 100 J of heat has been added. Would the lower heat capacity substance experience more or less heat loss, and would that translate into more or less error in the calculated specific heat?
I assume that less heat loss would mean less error but I am not sure as to which substance would experience less heat loss.

If anyone stumbles across this later, I am now strongly learning towards more heat loss for lower heat capacity because heat loss is proportional to temperature difference and lower heat capacity will change temperature more easily.

Assume you add a fixed amount of heat to a substance held in an isothermal bath, so that as soon as you start heating it, the substance gives off heat to the surroundings. As heat is added the substance increases in temperature to a maximum. When the heater is turned off, the substance continues to dissipate heat until it is again in equilibrium with the surroundings.
If the rate at which heat is added is much faster than that of heat dissipation and the heat capacity is approximately constant, then the maximum temperature will be approximately
$$T_{\mathrm{max}}=T_{\mathrm{ini}} + \frac{Q}{C_p}$$
where Q is the total amount of heat added. As expected, a substance with a smaller heat capacity will become hotter.
Now consider the rate of heat dissipation. The rate of heat dissipation can be expressed with the heat equation (in one dimension) as
$$\frac{dQ}{dt} = -k \frac{\partial T}{\partial x}$$
where k is the thermal conductivity.
Therefore if k is the same for two substances, then the rate at which heat is released will differ according to the heat capacity, and the lower heat capacity substance will release heat more quickly (since temperature gradients will be greater in that substance).
This can also be argued based on the value of the thermal diffusivity $\alpha$, which is inversely proportional to the heat capacity $c_p$:
$$\alpha = \frac{k}{\rho c_p}$$
As the Wikipedia explains,

In a sense, thermal diffusivity is the measure of thermal inertia.[6] In a substance with high thermal diffusivity, heat moves rapidly through it because the substance conducts heat quickly relative to its volumetric heat capacity or 'thermal bulk'.

Since the substance with smaller $c_p$ will have a larger thermal diffusivity (assuming all else constant) it will also lose heat more quickly.
This can of course be expected to result in greater uncertainty in a measurement of the heat absorbed by the substance.


Buffer formation in weak acid/strong base titration? I'm a little confused about what happens when you titrate a weak acid or base. For example, let's say that you are trying to titrate a weak acid. Based on this question, it seems like the weak acid isn't initially a buffer. But, as you begin to add strong base, conjugate base is created from the reaction between the weak acid and strong base. Does this mean that a buffer is created from the titration?

I think that the above diagram will explain a lot of your questions.

  it seems like the weak acid isn't initially a buffer

That is correct as to form a buffer you need a solution containing a weak acid and its conjugate weak base. So in the example above, initially the solution only contains acetic acid ($\ce{CH3COOH}$, but its conjugate base, acetate $\ce{CH3COO^{-}}$, is not present.

  But, as you begin to add strong base, conjugate base is created from the reaction between the weak acid and strong base

That is absolutely correct. For example, in the above diagram the reaction that is occurring when you titrate the acetic acid with sodium hydroxide is:
$$\ce{CH3COOH + NaOH -> CH3COO- + Na+ + H2O}$$

Now as you begin to titrate the acid, its conjugate base starts to form. This is supported by the fact that the pH doesn't really change when more $\ce{NaOH}$ is added to the solution.

To form an effective buffer solution you need roughly equal concentrations of the acid and conjugate base, hence that is why it says it is a buffer when half of the acid is titrated. However anywhere near that region can be considered as a buffer. Generally a solution is a buffer if the pH is within +/- 1 of its pKa. So for acetic acid, a buffer will have a pH between 3.76 and 5.76.

  Does this mean that a buffer is created from the titration?

Yes, a buffer is created. However is doesn't stay as a buffer for long. As I mentioned above, when the ration of the concentration of acid and the conjugate base becomes too high or too low, it won't be a buffer. That is evident as when more $\ce{NaOH}$ is added to the buffer, the concentration of the conjugate base becomes much greater than the concentration of the acid, resulting in the pH to increase dramatically.


Acid titration of sodium acetate in the presence of sodium ethoxide, is it possible? I have an aqueous solution of ethanol of $30 - 40\text{ V%}$ that contains a few (unknown) $\text{%}$ ethyl acetate (ethyl ethanoate - $\ce{EtOAc}$).
I wonder if I can quantitatively determine the $\ce{EtOAc}$ content by back titration of a small excess of strong $\ce{NaOH}$.
The idea is to add a stoichiometric excess of $\ce{NaOH}$ to a known volume of sample solution, then reflux this mixture to achieve saponification of the $\ce{EtAc}$, acc.:
$$\ce{EtOAc}+\ce{NaOH} \to \ce{NaOAc}+\ce{EtOH}$$
The excess $\ce{NaOH}$ would form sodium ethoxide with the ethanol, acc.:
$$\ce{NaOH}+\ce{EtOH} \to \ce{NaOEt}+\ce{H2O}$$
Since as $\ce{NaOEt}$ is a strong base and $\ce{NaOAc}$ a weak one, I think it should be possible to titrate them sequentially.
Any thoughts would be appreciated.

The pH of 0.1 M NaAc is calculated to be 8.9. https://pubs.acs.org/doi/pdf/10.1021/ed079p29.1
The pH of NaOH will be much higher. The 30-40% of ethanol may have some effects, but in general, you should be able to add about twice as much NaOH as you think will react with all the EtAc, reflux and back titrate the NaOH. As Poutnik comments, there won't be significant NaOEt.
The reaction will look like this:
EtAc + 2 NaOH --> NaAc + EtOH + NaOH
If you can get some thymolphthalein, you could use that as an indicator.

https://en.wikipedia.org/wiki/PH_indicator
Or, you can use pH paper strips if you are satisfied with a cruder analysis.
Possibly a more intricate back titration might work: add about twice as much NaOH as you think will react with all the EtAc in your ethanol, reflux as above, then add the same number of moles of acetic acid as you added moles of NaOH. Then the reaction will look like this:
EtAc + 2 NaOH --> NaAc + EtOH + NaOH;   reflux;   + 2 HAc -->
NaAc + EtOH + NaAc + H$_2$O + HAc
Then back titrate the excess acetic acid with phenolphthalein (common enough - you must have some). If there were no EtAc in your ethanol, the pH of the solution after adding the 2 HAc will be about 8.9; if there is any EtAc in your ethanol, you will have excess acetic acid due to the saponification of the EtAc. And you back titrate that with NaOH. So this is sort of a back to back titration, or a back and forth.


Esterification of racemic acid with optically active alcohol An excess of racemic acid ($\ce{CH3CHClCOOH}$) is allowed to react with (S)-2-methyl-1-butanol to form the ester, $\ce{CH3CHClCOOCH2CH(CH3)CH2CH3}$, and the reaction mixture is carefully distilled. Three fractions are obtained, each of which is optically active. Draw stereochemical formulas of the compound or compounds making up each fraction.

I know that two fractions obtained will be of the diastereomers (R,S) and (S,S), what is the other fraction?

Source: Morison and Boyd

It's a pretty basic concept. Two diastereoisomeric esters will be formed. Now diastereoisomers have different physical and chemical properties just like normal isomers. Hence, obviously, their rate of formation will be different. So, different amount of R and S acid will react

Thus, the third fraction will contain unequal amount of the acids which will show optical activity, and the other two fractions are the two diastereoisomers which are also optically active.


Why is the test for sodium as dihydrogen antimonate so specific? I have been attending lectures of salt analysis in my college. We are at group 6 cations or also known as group zero cations, namely $\ce{NH4^+}$, $\ce{K^+}$, $\ce{Na^+}$, $\ce{Mg^2+}$.
The lecturer wrote an equation:
$$\ce{NaCl + KH2SbO4 -> NaH2SbO4 (s) v  + KCl}$$
Now, I know that the electrochemical series follows the order $\ce{Li}$, $\ce{K}$, $\ce{Ba}$, $\ce{Sr}$, $\ce{Ca}$, $\ce{Na}$, $\dots$.
I would like to know why this reaction occurs so spontaneously that it is included as one of the lab tests. Sodium chloride is completely neutral whereas potassium dihydrogen antimonate is basic.

It has no (direct) relation to the electrochemical series nor to acido-basic properties.
It is not a true chemical reaction, rather just crystallization of the least soluble salt from the available ions.
There are no distinguished molecules $\ce{NaCl}$, $\ce{KH2SbO4}$, $\ce{NaH2SbO4}$, $\ce{KCl}$ in the solution, but mutually independent hydrated ions $\ce{Na+}$,  $\ce{Cl-}$, $\ce{K+}$, $\ce{H2SbO4-}$ instead (plus some minor hydrolysis products). Which salt precipitates first depends on their solubility products. $\ce{NaH2SbO4}$ has smaller solubility product than $\ce{KH2SbO4}$.
Similar things would happen if you mix solutions of potassium  chloride and sodium nitrate ( chlorate, perchlorate ). If original solutions are concentrated enough, and especially if you cool the mixture below ambient temperature, there will precipitate the respective potassium salt with the oxoanion.
Another case is the $\ce{NaCl}$ solution  with dissolved $\ce{NH3}$ And $\ce{ CO2}$, with ions $\ce{Na+, NH4+, Cl^-, HCO3-}$, leading to precipitation of $\ce{NaHCO3}$ and solution of $\ce{NH4Cl}$. It is used commercially in the Solvay process
producing sodium carbonate from sodium chloride.


Forming the resonance structures for the nitrite anion Yesterday I learned about resonance structures and came across this problem.

The resonance structures are formed when there is a delocalized π electron cloud. I understand that for benzene, there are two resonance structures. For making π bonds I know that there are three main reasons:

The two orbitals should overlap each other.
There should only be one electron for each orbital.
The two orbitals should be in one level.

In benzene there are six $\mathrm{p}_z$ unhybridized orbitals, therefore they can share their electrons with each other (delocalize the π bond electrons). So benzene has 2 resonance structures.

But I don't understand how this works for the nitrite anion, $\ce{NO2-}$.

In $\ce{NO2-}$ there is also π electron delocalization, so there are 2 resonance structures. Now I am going to say how I think the delocalization is happening.

When the first structure changes into the second structure, nitrogen has to make a π bond with the right side oxygen with each of the $\mathrm{p}_z$ orbitals before, but now that π bond's two electrons take to the right side oxygen, and then there are 2 electrons in the right side $\mathrm{p}_z$ orbital and no electrons in the nitrogen $\mathrm{p}_z$ orbital. But the left side oxygen (in the first structure, $\ce{O-}$) has 2 electrons in its $\mathrm{p}_z$ orbital. Then the nitrogen $\mathrm{p}_z$ and oxygen $\mathrm{p}_z$ overlap each other and make a π bond.

I have said before "There should only be one electron for each orbital." But, according to my understanding above, the π bond is made by $\mathrm{p}_z$ with 2 electrons (oxygen) and $\mathrm{p}_z$ with no electrons (nitrogen) overlapping each other.

So, is there any mistake in my imagination?

Yes. I think that there is a mistake in your imagination. Nitrogen is evidently $sp^2$-hybridised in this case. Nitrogen has 5 valence electrons. Two electrons are used to make 2 $\sigma$ bonds with two hybrid orbitals. Another two electrons become the lone pair and reside also in a hybrid orbital. This leaves us with one electron in one unhybridised $p$ orbital. This $p$ orbital is then used to make the $\pi$ bond to either oxygen atom, contributing one electron to the $\pi$ bond while the oxygen atom also contributes one electron from its $p$ orbital.


Geminal coupling constant of formaldehyde protons in NMR I read in a book that the protons in formaldehyde have the largest known positive geminal coupling constant (41 Hz). I thought only couplings between non-equivalent protons can be seen in the spectrum. Since formaldehyde has C2v symmetry and (if I remember correctly) coupling constants cannot be calculated from theory accurately, how come this coupling is apparent in the spectrum?

This is an extrapolation of an experimentally measured coupling constant. It is possible to determine the coupling constant, $J_{\ce{H-H}}$ based on the $J_{\ce{H-D}}$, where the following relationship is valid:

$$ \left |J_{\ce{H-H}} \right | = (\gamma_\ce{H}/\gamma_\ce{D}) \cdot \left |J_{\ce{H-D}}\right |$$

Shapiro, Kopchick and Ebersole reported the coupling constant for $J_{\ce{H-D}}$ in formaldehyde-d1 in 1963 (The Journal of Chemical Physics 39, 3154 (1963); doi: 10.1063/1.1734163), and calculated the $J_{\ce{H-H}}$ from this. Changes in coupling constants due to the influence of isotopically-modified electron density are apparently very modest for H/D, and so this relationship appears to hold well.

There is no experimental method to measure coupling between magnetically equivalent spins. In order to break the redundancy of the energy transitions, one must remove the equivalence.

You can actually check this isotope relationship very easily with common NMR solvents and the $\ce{C-H}$ and $\ce{C-D}$ coupling if you run carbon experiments ($\gamma_\ce{H}/\gamma_\ce{D}=6.51439$).

$$\begin{array}{ccc} \hline
\text{Solvent} & J_{\ce{C-D}} & J_{\ce{C-H}}\\[1.1em] \hline
\ce{CHCl3} & 32 & 208\\
\ce{DMSO} & 21 & 137 \\ \hline \end{array}$$


Why is the outermost part of the Bunsen burner flame not the hottest region? Why is the blue flame the hottest on a Bunsen burner?

I thought the outermost part of the flame must be the hottest region because of complete combustion of gases taking place there. Where am I getting wrong?

When you have the intake valve (at the bottom of the bunsen burner) opened, the gas can mix with incoming oxygen. By the time is gets to the top of the tube, you've got a nice continuously-flowing source of well-mixed gas and oxygen. That allows combustion to occur very quickly because the methane and oxygen molecules don't have to travel very far to react, increasing the reaction rate, and therefore the rate at which heat is released.

In contrast, when the intake valve is closed (or when it's open and you're considering the outer non-luminous flame), the oxygen for the combustion reaction comes from the surrounding air, which is not thoroughly mixed with the gas at the time of ignition. That results in a slower reaction, and therefore a slower rate of heat being released.

Periodic videos just did a nice thermal imaging of a bunsen burner with this explanation: https://www.youtube.com/watch?v=in9BGz958jg


Superscript open circle notation in formula I may be only a physicist but I'm familiar with the superscripted dot to indicate a radical.  In a paper I'm reading the authors have also used a superscripted open circle.  I can't find any reference online or in my physical chemistry textbook (Atkins). Here's an example:

As you can see they've done this a few times, and also used the dot to indicate a radical.  This is in a published paper "Photocatalytic degradation pathway of methylene blue in water", Houas et al., Applied Catalysis B: Environmental (2001).  I'm starting to wonder (given the OH◦ in the text and OH• in the equation) if it's a typo, but I'd be surprised if something so glaring could get through review or even proofreading by coauthors.

So does the open circle have a meaning of its own?

According to The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers 9.2.5.12

  Indicate a free radical by placing either a centered dot or a superscript dot, signifying the unshared electron, next to the chemical symbol or compound formula

This convention is also followed by the ACS.

So it would be better if the article used the solid symbol throughout, but the solid and open symbols mean the same thing in the article.


Shifting of axial and equatorial protons I know this NMR represents cyclohexanol and I have figured out what each peak represents. If the 3 peaks at 1.9, 1.7,1.5 all represent equatorial protons as they are shifted downfield compared to the axial. How does the peak at 1.3 represent 5 axial protons. Shouldn't they have split signals just like the equatorial protons, due to the electronegativity of the oxygen an its deshielding effect. Or is there a reason why axial protons aren't affected by the electronegativity of the Oxygen.

The distinction between axis and equatorial hydrogens is not going to be visible in an NMR spectrum for cyclohexanol. The conformations of the molecule interconvert too quickly at room temperature to make a distinction in the proton NMR spectrum.

So, in practice, you will only see 4 distinct C-H bond types in the molecule and the splits that will be visible on your spectrum depend on the resolution (which is not particularly high on the spectrum you have). The splits cause by neighbouring hydrogens for the groups between shifts of 1 and 2 overlap even in high resolution spectra and it isn't clear whether what you are seeing this jumble of peaks as 4 separate environments because of the low resolution or whether one of the peaks is an impurity.

Some software packages for structure drawing also generate theoretical NMR plots which can help interpret such spectra.


Surface Chemistry - Dipole Moment of Surfactant with Water and Phases of a Monomolecular Layer Recently, I have been trying to study surface chemistry and I have come across questions that I do not understand how to answer. I understand what a dipole moment is, defined as a charge separation over a defined distance, and I have a fair understanding of surfactants and other areas of surface chemistry such as Langmuir Isotherms. But I am struggling to answer such questions:

How would surfactants with a dipole moment in contact with water be characterised?

Would this mean that most likely the ionic head of a surfactant interacts with either the partially-positive hydrogen or partially-negative oxygen moiety of a water molecule? If so, what would be the proper way to characterise this phenomenon?

In what phase of the monomolecular layer are the sizes of surfactants determined?

I am puzzled by this question, as I do not understand what exactly is meant by phase. I have spent relentless time trying to search for monomolecular layers and their phases but I have come out empty-handed. I see that phase can refer to distinct boundaries of physical states, such as those between gasses and liquids, but I do not know how this could be used to find the size of molecules. I know that in order to find the area that each surfactant molecule covers at an interface the Gibbs Adsorption Equation is utilised, however it uses concentration of surfactant molecules plotted against surface tension.

I would be much obliged if anyone could provide assistance to aid me in understanding this subject of surface chemistry better.

Wow, that is a question!

Surfactants (aka tensioactive compound) have a dipolar moment at the air-water interface but that is not enough to characterize their behavior. There is a cohesive force between their lipophilic tails (due to Van der Walls force) and there is also a cohesive force between their polar head but only because water of metal ions can make a bridge between identically charged moieties.
I will address only your main question which seems to be "how to know the size of molecules in a monolayer"? Well, a surfactant floats at the surface of water. So you can make an apparatus filled with water and with a movable barrier which will move across the surface to slowly reduce the overall surface available for a tensioactive compound. We also know how to measure variations in the "tensioactive power" (whatever its name is) which depends on the number of molecules at the surface.

Using such a device, we can create a monolayer of a lipophilic compound, reduce the whole area (which we call a "compression") and detect the resulting effect (which correlates to the number of molecules at the surface). Whenever the available surface becomes too small for all the molecules to be nicely arranged at the surface (i.e. the "collapse").


How do I get Mass spectrometry graph, history, and Infra-red spectroscopy graph for Secnidazole? My chemistry teacher wants me to create an academic poster for a recently-discovered biologically important organic molecule with chirality that is relatively small and is also recent.

Secnidazole (also called Solosec) seems to fulfill these requirements fairly well, however I have no clue on how I would be able to obtain the mass spec or infra-red spec of it, as they doesn't seem to be listed online anywhere. As well as this, there doesn't seem to be a listing for what organisation created it or anything either.

How do I either get a rough approximation of the results, or find a legitimate set? Bonus points for directing me to the history of the molecule.

https://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi (or search sdbs database, it should be the first link) is a useful database that contains a variety of spectral data for a large number of compounds. If you search for Secnidazole, they appear to have MS and IR data for this compound.

If you have access to journal articles, you can also do a literature search for this compound. It is likely that spectral data for this compound will be present in any paper where this molecule was synthesized (either in the main article or its supporting information).

Edit: I just did a brief literature search through Reaxys - most of the papers seemed centered around clinical trials rather than organic synthesis, although I will admit I didn't look over any of them in any great detail.  I'll leave that to you.

Chemical companies such as Sigma Aldrich or Fisher Scientific also sometimes supply spectroscopic data for the compounds they sell.

If you have access to ChemDraw, you can simulate $^1$H and $^{13}$C NMR data for any compound you draw.  I'm sure other spectral simulation programs exist, but don't have much experience in that area.


what are reduction oxidation reactions? in the picture, the product is a neutral material, but in the illustration of the molecular structure, the compound is shown in its ions, why we can't show it in its ions even in the equation or if we can't, why in this molecular structure it is shown as ionic although it is neutral?

Your questions involve a lot of chemical concepts.

An oxydo-reduction is reaction where electrons are exchanged between species including or not molecular rearrangement.

You totally can write down any salt in the following way, emphasizing ion charges:

$\mathrm{NaCl}_{(s)} \equiv \mathrm{Na^+Cl^-}_{(s)}$

As long as you specify the phase, because when writing down an ionic form the chemist might assume that it is solvated into water. To underline it you may write the following:

$\mathrm{Na^+Cl^-}_{(aq)}$

Further more solid generally have numerous possible distinct phases, therefore you may need to specify which crystalline structure the salt have.

The following equation is explicit:

$2\mathrm{Na}^{(0)}_{(s)} + \mathrm{Cl_{2(g)}} = 2\mathrm{Na^+Cl^-}_{(s)}$

And underlines that an electron is withdrawn from metallic sodium in order to produce chlorine. Anyway, this equation is global balance this it hides many details of the complete and complex chemical process, especially the formation of the crystalline structure. And it does not provide any information on conditions of reaction.


Multiple-quantum coherences in NMR Why are multiple-quantum coherences (coherences with changes in magnetic quantum numbers, $|\Delta m| \neq 1$) unobservable in regular NMR experiments?

I'm not sure what you mean by 'regular' NMR experiments. NMR experiments which observe multi quantum signal are still 'regular' experiments, so I'll assume you asking why you wouldn't see DQ signal in a basic 1D 1H spectrum, for instance. I will attempt to explain without the use of product operator formalism, and scary looking maths.

In its simplest terms, the frequency of a double quantum coherence is about twice the normal observation frequency, and so cannot be observed without transformation back to single quantum coherence. Similarly, zero quantum coherence is typically less than a few kilohertz, and so is also not observable. If you set your receiver to the correct frequency, this transition would be observable, although significantly lower in intensity due to the lower probability of this transition path.

In the case of an isolated, uncoupled spin, there are only two energy states, and the transition between these two states (the single quantum transition) is the only possible transition. For the case of coupled nuclei, the state of spin I is affected by the spins state of nucleus S. The diagram above shows the typical spin transitions for a spin system, IS. The most common transition is a single quantum transition, where only a single spin changes state, and this generates magnetisation in the transverse plane. When both spins change state simultaneously, you can get either zero quantum (the spins flip to opposite states) or double quantum (spins flip the same way) transitions. The frequency of the double quantum transition is roughly twice that of a single quantum transition. In the vector model, double quantum transitions have no magnetisation in the transverse plane (only transverse magnetisation can be observed), but can be transferred to single quantum transition through the process of generation of SQC(first pulse), evolution (linked to couplings) and generation of MQC (second pulse) in a multi quantum experiment. This is typical of a DQF/TDF COSY type experiment, but equally utilised in studies of dipolar couplings of small organics in liquid crystal media (where quantum orders of >5 are typically observed)


Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions? Free chlorine ($\ce{Cl}$) in the stratosphere can deplete ozone ($\ce{O3}$)  as follows:
$$\ce{Cl + O3 -> ClO + O2}$$

The chlorine atoms can then react with oxygen and return to the beginning of the cycle:
$$\ce{ClO + O -> Cl + O2}$$

The EPA estimates the ozone-depleting potential of a single chlorine atom:

  It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere.

From the equations above, it seems that each chlorine atom returns to its original form. Why, after about 100,000 cycles, does it lose its ozone-depleting potential?

As a side question, how is this estimated?

Since the stratosphere is nowhere near a closed system, a chlorine atom will eventually leave it. Look at the phrasing again:

  It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere.

It is not stated, that it looses its potential, it just leaves the region, where there is sufficient ozone concentration.

In a closed system, the catalytic cycle would continue until the ozone concentration is small enough, that the chance of chlorine atoms recombining is bigger.

Thanks to Nicolau's comment I decided to further look into the issue. Also qwersjc gave some valuable insights. For further reading I suggest, the following sources:

http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1995/ (overview)
Nobel Lecture by S. Rowland (Free Access)
Stratospheric Chlorine: a Possible Sink for Ozone. R. S. Stolarski, R. J. Cicerone; Canadian Journal of Chemistry, 1974, 52(8): 1610-1615. (Open access)
Original Paper by M. J. Molina and F. S. Rowland; Nature 249, 810 - 812 (28 June 1974) (Might not be free), reprint on ozone.unep.org

The Ozon Cycle

The ozone cycle is what protects us from quite dangerous UV-B radiation ($M$ is an inert Air molecule like $\ce{N2}$):
$$
\ce{O2 ->[h\nu] O + O} \\
\ce{O + O2 ->[<M>] O3}\\
\ce{O3 ->[h\nu] O + O2}
$$

The Troublemakers

The main reason for chlorine atoms in the stratosphere is man made. Chlorofluorocarbons were very common in many products. These molecules are colourless and water-insoluble. Diffusion taking its course the end up in the stratosphere, where they may undergo dissociation.
$$\ce{CCl3F ->[h\nu] Cl + CCl2F}$$
So the main problem is, that free chlorine radicals are produced in the stratosphere. Also these CFCs have a lifetime in the stratosphere of up to 150 years.

There are some natural sources (of close to insignificant importance) that also produces free Chloride Radicals, i.e. volcanic eruptions. Those may transport Hydrochloride into the stratosphere, where it can react with the Hydroxide radical:
$$\ce{HCl + OH -> Cl + H2O}$$

Removal of Chlorine

There are several steps involved, that lead to the removal of Chlorine from the stratosphere. Basically it will stay there for quite a while and cause significant damage. The stratosphere is very dry and water-soluble compounds, like HCl, will not be washed out quickly. In order to leave the stratosphere, the chlorine atoms or compounds therefore have to be near the troposphere.

Probably the most important reaction is with methane forming Hydrochloride, which may the be washed out by water (causing at least a part of acid rain nevertheless).
$$\ce{Cl + CH4 -> HCl + CH3}$$ The likeliness of this event is about 1000 times smaller than colliding with another ozone.

However eventually, after about two years (a number I picked up on the way), a chlorine will leave the stratosphere. In the troposphere is more water present, so Hydrochloric acid will be formed and washed out.

  In our estimates, termination of the Cl--ClO chain results from a downward diffusion of the longer lived species in the chain (HCl, ClO) and eventual removal by tropospheric process. From: Nature 249, 810-812 (28 June 1974).

There are also a number of reactions that decrease the potential of Chlorine radicals.
$$\ce{ClO + NO2 -> ClONO2}\\
\ce{ClO + O2H -> ClOH + O2}\\
...$$

However these also form reservoir molecules that might diffuse back to the stratosphere.

The calculations behind this are rather complex and use a wide variety of empiric data along with complex models.

I think there is still a lot of research going on in this area, considering more molecules like $\ce{CO2}$ and other greenhouse gases, as well as global climate change.

The Nobel lecture (1995) closes with the following remarks and we can only hope that this holds true:

  However, with the limitations just expressed, the threat of extensive further stratospheric ozone depletion during the 21st century appears to be under control.


Is it safe to lyophilize/Freeze-dry a chelated compound? I have a compound that consists of a metal ion chelated by a linear organic structure with assorted moieties in aqueous medium. I'd like to dehydrate this mixture fully to get solid product, but heating is out of question due to unwanted repercussions on aforementioned "assorted moieties". If I lyophilize the compound, will it remain chelated or is there a risk of the metal ion unbinding?

I lyophilized my sample and redissolved a portion of it and confirmed chelation by verifying the presence of the absorption peak of the Lanthanide I was chelating and using Beer's Law.

Thanks for your help.

Lyophilization is regarded safer as normal drying because it is performed at considerably lower temperature; however in order to achieve this, you have to lower the pressure above the condensed phase, too.  Naturally volatile compounds (including some organic compounds solid at room temperature) may sublime under these conditions, too; which may be hold for hours, even days to remove the water gradually.  This simultaneous process may lower the yield of isolated product.

Other light organic molecules (like methanol, ethanol, THF) equally may leave the vessel, too.  The complexation may be affected if leaving molecules were involved as part of the no-further specified chelate, too.


How to determine the dimensions of a bixin molecule? I need the dimension of a bixin molecule.  I do not know where to look or how to calculate this information.

I understand that this molecule has several rotational axes and conformation, so there is no uniform size to be determined. But I am interested in the standard method for the determination of the molecule's dimensions.

I am doing a series of refractive and scattering experiments in a concentrated bixin medium. I need to determine the most probable cross section area normal to its principal axes.

There is a paper on bolaamphiphiles (1) where it is stated that measured surface area corresponding to the area of a bixin molecule lying on the water surface with most methyl groups pointing outside is approx. $\pu{1.0 nm^2/molecule}$ at zero pressure. If the methyl groups were parallel to the water surface, a molecular area of approx. $\pu{1.7 nm^2/molecule}$ would be occupied.

Also, there is a linear size of a single molecule stated as the following:

  Bixin therefore constitutes a valuable starting material for various bolaamphiphiles which can be tied within lipid membranes. Its molecular length is about $\pu{3.1 nm}$ in the all-trans configuration.

I think it might be worth for OP to have a look at the full-text (1) paper as it also discusses various methods of characterization and altering of bolaamphiphiles.

A few years later a crystal structure has been published (2). There is not much to add, as it seems like there is no contradiction with the data obtained with reflection spectroscopy (1). Here are the notable linear dimensions of cis-bixin and overall geometry in 2 projections:

(1) Fuhrhop, J. H.; Krull, M.; Schulz, A.; Moebius, D. Langmuir 1990, 6 (2), 497–505. DOI: 10.1021/la00092a034
(2) Kelly, D. R.; Edwards, A. A.; Parkinson, J. A.; Olovsson, G.; Trotter, J.; Jones, S.; Malik, K. M. A.; Hursthouse, M. B.; Hibbs D. E. 1996, J. Chem. Res., 446, 2640


What is wrong with the Aufbau principle that results in so many exceptions? From what I have observed in the periodic table is that many transition elements do not follow the Aufbau principle. And it's not just the elements like $\ce{Cu}$ and $\ce{Cr}$ that differ from predicted configurations because half-filled and completely filled d-orbitals are more stable, their are many odd exceptions like Nb ($\mathrm{[Kr]}4d^4\,5s^1$), Pd ($\mathrm{[Kr]}4d^{10}\,5s^0$). What reasons lead to such configurations?

It's basically because you can't just add up electronic energies using a shell-by-shell, electron-by-electron approach.  As soon as you have two or more electrons in close quarters, they interact with each other in ways that cannot be described by simple electrostatic laws because of the quantum mechanical nature of the electrons.  The comments (specifically Mithoron's) refer to other questions whose answers explain this much better than I can.  Please look at those.


When are Kp and Kc values equal? The picture shows the correct answers to this question.

I'm not sure why these answers are correct. $K_p$ only counts with gases, while $K_c$ only counts with aqueous solution + gases. I assumed that any reactions with only gases, or a mix of gases and solids, will have an equal $K_p$ and $K_c$ value, but that is clearly not the case.

Disclaimer: The following information has been extracted from my own Chemistry lecture notes which was built on contributions from many other general chemistry textbooks.  It is a stereotypical example of relating $K_c$ and $K_p$.

$K_c$ and $K_p$ relationship

For a simple reaction involving gas phase substances

$$\ce{aA(g) <=> bB(g)}$$

The equilibrium constant can be written in terms of molar concentrations or partial pressures as

$$K_c = \frac{[\ce{B}]^b}{[\ce{A}]^a} \qquad\text{or}\qquad K_p = \frac{(P_\ce{B})^b}{(P_\ce{A})^a}$$

Assuming that the gas behaves ideally (a common and convenient assumption for general cases), we can write the ideal gas law for each gas such that

$$ PV = nRT \quad\to\quad P_\ce{A}V = n_\ce{A}RT \quad\to\quad P_\ce{A} = \left(\frac{n_\ce{A}}{V}\right)RT$$

$$ PV = nRT \quad\to\quad P_\ce{B}V = n_\ce{B}RT \quad\to\quad P_\ce{B} = \left(\frac{n_\ce{B}}{V}\right)RT$$

We note that
$$ \frac{n_A}{V} \equiv M $$

where $M$ is simply molarity (moles/unit volume).  Since $[\ce{X}] = M_x$, we get

$$P_\ce{A} = [\ce{A}]RT \qquad\text{and}\qquad P_\ce{B} = [\ce{B}]RT$$

Substitute these terms into the $K_p$ equation to give

$$K_p = \frac{(P_\ce{B})^b}{(P_\ce{A})^a} = \frac{([\ce{B}]RT)^b}{([\ce{A}]RT)^a}$$

and recognize that

$$\frac{[\ce{B}]^b}{[\ce{A}]^a} = K_c$$

Plug this relationship into the $K_p$ expression to give

$$K_p = K_c (RT)^{b-a} \quad\to\quad K_p = K_c (RT)^{\Delta n}$$

Where $\Delta n$ is simply the difference of the sum of the stoichiometric coefficients of the products and reactants from a balanced chemical equation like so

$$\Delta n = \sum (\mathrm{products}) - \sum (\mathrm{reactants})$$

It should be clear from this that the only way $K_c = K_p$ within the context of an ideal gas is when $\Delta n = 0$.

$$\begin{align*}
K_p = K_c(RT)^{\Delta n} \quad\to\quad K_p &= K_c(RT)^{0} \\
&= K_c (1) \\
&= K_c
\end{align*}$$

By visual inspection of your image, it is clear why these are the correct answers based upon the derived relationship offered here.


Molecular Orbitals (only pi system) of Allyl system using group theory I have just started reading group theory and their applications in the MO theory. I tried to get the equations of the pi molecular orbitals of allyl system using group theory, but I am stuck at the last part.

This is what I did so far—

Determined the point group of the allyl system to be $\ce{C_{2v}}$.
Used the 3 p orbitals as basis.
Found out the reducible representation. Reduction gives $\ce{A2 + 2B1}$

I am stuck at this point. I named the three p orbitals of C atoms as $\ce{\psi1, \psi2, \psi3}$. However, when I try to use the 'projection operator method' using the $\ce{\psi1}$ as generation vector, it gives me $\ce{\psi1 +\;\psi3}$ and $\ce{\psi1 - \psi3}$ (unnormalized). I don't know how to proceed.
 [image copied from Master Organic Chemistry webpage ]

The calculation below is not exactly the same as yours but I had it to hand and it is very similar. The bit you specifically want is towards the end. In your case you add will vectors along the $\pi$ orbitals.

What projection operators  operators do is to extract the symmetry adapted functions from the basis functions used to form them and so produce a linear combination L of the basis functions. Many basis functions can be chosen, but the easiest to use are usually vectors pointing along bonds or those representing a p$\pi$ orbital. To form a linear combination, a 'victim' vector is chosen and operated on with each symmetry operation in the point group, $C_2\,, \sigma(xz)$, etc. to find what vector this turns into. This vector's name (the basis function) is then multiplied with the character from the point group corresponding to that symmetry operation and symmetry species. Finally, the names are added together by moving from one symmetry operation to the next along the top row of the point group. We expect the result to be the sum or difference of the displacement vectors for a molecular vibration, such as $2v_1 + v_2 - v_3 \cdots$ and so forth. The same formula or algorithm is used if the combinations of atomic orbitals that make up molecular orbitals are sought.
The equation to make a linear combination is

$$ L_M=\frac{d}{h}\sum_{j=1\cdots h}c_jS_j(v)$$

where $M$ is the symmetry species label, $A_g, B_{3g}$, and so forth, $h$ is the order of the group, $d$ the dimension of the irreducible representation, and the sum is over all the classes. Because the resulting vector will be normalized d/h can be ignored. One further point is important. In the $C_{3v}$ point group, for example, there is a heading $2C_3$ as there are two members in this class, and this must be split in the summation into $C_3^+$ and $C_3^-$. This is because moving a vector $120$ degrees to the right, say, will turn it into a different vector than turning 120 degrees to the left, similarly, any mirror planes must be separated out. The character for symmetry species $M$ is $c_j$ and $S_j(t)$ is the effect that operator $S$ in column $j$ of the point group has on our victim vector $v$. The result of this operation is to produce a vector whose name is recorded. For example, if $S$ is the identity then the result is $E_1(v) = v$, other operations may leave $v$ unchanged or change it into another vector. Finally, note that if the reduced representation of the vibrations or molecular orbitals contains two or more symmetry species of the same type, e.g. $2A_1$, then two or more different victim vectors will have to be chosen to obtain all the linear combinations.

In H$_2$O, SO$_2$%, and other triatomics with $C_{2v}$ symmetry, it is clear that two of the vibrational modes stretch the bonds and one changes the bond angle. To work out the normal mode vectors, some representative vectors are placed along the bonds as in the figure. You can choose where to put these but physically realistic choices will usually make the calculation simpler.

In $C_{2v}$, the normal mode vibrations have been found to comprise the symmetry species $2A_1$ and $B_1$. The $L_M$ formula is shown as a table with $v_1$ as the victim vector. The table also shows the characters for each symmetry species.

$$\begin {array}{l|llll|l}
C_{2v}  & E & C_2 & \sigma(x,z) & \sigma'(y,z) & L_m \;(d=4,h=4)\\
\hline
A_1& 1&1&1&1\\
L_{A1} & v_1 & v_2& v_1& v_2&  2(v_1+v_2)\\
\hline
A_2& 1&1&-1&-1\\
L_{A2} & v_1 & v_2& -v_1& -v_2&  0\\
\hline
B_1& 1&-1&1&-1\\
L_{B1} & v_1 & -v_2& v_1& -v_2&  2(v_1-v_2)\\
\hline
B_2& 1&-1&-1&1\\
L_{B2} & v_1 & -v_2& -v_1& v_2&  0\\
\hline
\end{array}$$

Notice that what the vector changes into is (obviously) the same for each symmetry species as this is determined by the symmetry operations. Only the value of the character changes in front of each term The final $L_M$ is not multiplied by $d/h$ because the resulting vectors are instead normalized giving, $L_{A1} = (v_1 + v_2)/ \sqrt{2}$ and $L_{B1} = (v_1 - v_2)/\sqrt{2}$. Note also that $v_3$ and $v_4$ do not enter into this table, they are at $90$ degrees  to $v_1$ and $v_2$ and no operation in this point group can inter-convert them.

There are two $A_1$ species produced from the reducible representation but only one has been found so far. This is because the vectors $v_1$ and $v_2$ cannot produce a bend, which is the other normal mode. Using vector $v_3$ as the victim this species appears. As this mode has $A_1$ symmetry all the characters $c_j$ in the $L_M$ formula are $1$, making the calculation easy and the combination is shown below.

$$\begin {array}{l|llll|l}
C_{2v}  & E & C_2 & \sigma(x,z) & \sigma'(y,z) & L_m \\
\hline
L_{A1} & v_3 & v_4 & v_3 & v_4 &  2(v_3+v_4) \\
\hline
\end{array}$$

which describes the bending mode shown in the figure in terms of its unit vectors. When normalized this mode is $(v_3 + v_4)/\sqrt{2}$. (As the centre of mass cannot move during a vibration, there being no force to make it do so, therefore the O atom has to move a small distance away from the H atoms. Vectors
could be added onto the O atom to show this and $L_M$ terms re-calculated.

In the case of degenerate symmetry species as occurs in the $C_{3v}$ and most other point groups the results of the summation $L_M$ is to produce two or more vector sums that are not orthogonal. Either these have to be made orthogonal, by choosing another victim for the degenerate term and then using the Gram-Schmidt method to make the two vectors orthogonal. R. Carter 'Molecular Symmetry and Group Theory' and A. Vincent 'Molecular Symmetry and Group Theory' both give method by which to do this.