What is the physical origin of conjugation? I had a question about the origins of conjugation in conjugated molecules. Take two ethylenes as compared to butadiene. Huckel seemed to argue that conjugation was the result of pi-pi stabilization interactions between the HOMO of the two ethylene molecules. More specifically, he argued that during the creation of Ψ1 and Ψ2, the former is lowered in energy more than the latter is raised in energy, leading to overall stability.

While this explanation satisfies me very much, I have also come across some literature (especially in organic chemistry) citing that π->π* hyperconjugation the main contributor to the stabilization of conjugated molecules. While I do understand this, I "feel" that Huckel's definition is more satisfactory. I am wondering if anyone could give some validity to the latter and if someone could point to the major factor for stabilization in conjugated molecules.

I think you have posed an interesting question. Upon looking through some literature, I have chanced upon the answer to your question in Fleming (2010), p. 30-31. The diagram below was also taken from the discussion in the book.

What you have proposed are two ways of constructing the $\pi$ molecular orbitals of butadiene from the $\pi$ molecular orbitals of two ethene molecules. What is normally discussed is the one shown on the left, whereby we consider the interaction between the two filled MOs of exactly the same energy and a similar interaction between the two unfilled MOs of equivalent energy. Considering the bonding and antibonding combinations allows us to derive the four $\pi$ Mos of butadiene.

You also propose considering $\pi$ to $\pi^*$ conjugation as an alternative way of deriving the butadiene $\pi$ MOs. Diagramfmatically, what you are essentially referring to is shown in the MO diagram on the right. Firstly, this interaction cannot be the main interaction between the two ethene $\pi$ systems simply because it would result in the butadiene $\pi$ MOs comprising of two separate levels of two degenerate MOs each, which is quite different from the picture which we are aware of, which has 4 $\pi$ MOs of different energies. Secondly, note that this interaction is considerably weaker due to the large energy difference between the $\pi$ and $\pi^*$ MOs of an ethene molecule. Recall that for a strong interaction between any two orbitals, one of the important conditions is that the energy difference between them must not be too large. In fact, the strongest interaction is obtained when the energy difference between them is zero. This interaction would contribute to the energies of the $\pi$ MOs in butadiene but the contribution would surely be less than the one that is highlighted in the previous paragraph.

To construct the MOs of the butadiene, it is incomplete to just consider the stronger interactions shown on the left. For completeness, we would have to then mix the resultant MOs obtained from these two interactions (i.e. both left and right diagrams), factoring in the extent of contribution of each interaction. Naturally, we would be curious as to what is the effect of the weaker interaction on the MO energies derived from the stronger interaction. Fleming (2010) describes the effects rather detailedly in the text:

  Mixing these two sets together, and allowing for the greater contribution from the stronger interactions, we get the set of orbitals shown in the figure below. Thus, to take just the filled orbitals, we see that $\psi_1$ is derived by the interaction of $\pi$ with $\pi$ in a bonding sense ($\psi_a$), lowering the energy of $\psi_1$ below that of the $\pi$ orbital, and by the interaction of $\pi$ with $\pi^*$ in a bonding sense ($\psi_w$), also lowering the energy below that of the $\pi$ orbital. Since the former is a strong interaction and the latter weak, the net effect is to lower the energy of $\psi_1$ below the $\pi$ level, but by a little more than the amount ($\beta$ in simple Huckel theory) that a $\pi$ orbital is lowered below the p level in making the $\pi$ bond of ethylene. However, $\psi_2$ is derived from the interaction of $\pi$ with $\pi$ in an antibonding sense ($\psi_b$), raising the energy above that of the $\pi$ orbital, and by the interaction of $\pi^*$ with $\pi$ in a bonding sense ($\psi_x$), lowering it again. Since the former is a strong interaction and the latter weak, the net effect is to raise the energy of $\psi_2$ above the $\pi$ level, but not by as much as a $\pi^*$ orbital is raised above the p level in making the p bond of ethylene. Yet another way of looking at this system is to say that the orbitals $\psi_1$ and $\psi_2$ and the orbitals $\psi_3^*$ and $\psi_4^*$ mutually repel each other.

Does it really matter if we factor in the weaker interaction into our considerations? Yes, it does. If we were to not do that, we would observe that the stabilisation predicted would not be as great as the one which we would predict from just looking at the MO diagram shown on the left. In fact, as Martin writes in the comments, the interaction between orbitals would produce an antibonding MO that is more antibonding than the bonding MO produced is bonding. If we were to not consider the weaker interaction, then we would predict that $\psi_2$ is raised more in energy from the original $\pi$ level than $\psi_1$ is lowered in energy from the original $\pi$ level (i.e. $\ce {E_2}$ is larger than $\ce {E_1}$), and this is clearly not the case.

Of course, as you mentioned, we can also build the $\pi$ MOs of butadiene from first principles via the Huckel method. I am not very well-versed with the mathematics behind the method but I do believe that the method discussed above and the original Huckel method would converge in terms of the results obtained eventually. Hence, they are equally valid approaches to the problem.

Reference

Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Reference ed.). United Kingdom : Wiley, 2010.


Solubility and equilibrium Is equilibrium between dissolution and recrystallization of a salt in an aqueous solution reached only when the solution is saturated; or can this also be established in an unsaturated solution?

Will Le Chatelier's principle apply in this situation (as it is about chemical equilibria)?

I was actually studying purification of salt by common ion effect where salt crystallizes from saturated aqueous solution on addition of common ion. Will the salt still crystallize if solvent still can dissolve more of it in unsaturated solution on addition of little amount of common ion. Since this crystallization is consequence of Le Chateliers principle, will crystallization still occur if solution taken was unsaturated and the common ion added is also in very small amount?

The solubility of a solute in a solvent is determined by the chemical potentials of the solute for the solution and the pure solute. Substances move from high chemical potential to low chemical potential. At the concentration of solubility, these two chemical potentials are equal so there is no net transfer. Below this concentration, the chemical potential of the solute in the solution is lower than that of the pure solute so molecules will transfer from the pure solute to solvent until it is all dissolved or saturation is reached.

When you invoke the common ion effect, you are increasing the chemical potential of the common ion in solution which decreases the solubility. It is possible to have a nearly saturated solution and add enough common ions to reduce the solubility below the original concentration.


What is the purpose of atomisation in atomic absorption spectroscopy? What is the purpose of the atomisation process in atomic absorption spectroscopy?

I understand that solution is evaporated by the flame and then the molecules decompose into atoms.

But why must the molecules become individual atoms before the light is passed through them?

Atomic absorption spectroscopy is a quantitative measure of the elements in the analyte.
We atomise the analyte so that the absorption of light corresponds to the characteristics of the elements.
Otherwise how we can predict the fundamental elements when these are still part of some molecules?
We measure the absorbance and refer to the Beer-Lambert law.


Bonding between amylopectin and amylose I know that amylose and amylopectin are major constituents of starch. I also know the chemical structures of amylose and amylopectin. However, I want to determine how amylose and amylopectin bond together. I think that the OH from the amylose bonds with the HO of the amylopectin to form an oxygen bond and a water. Am I correct, or does the OH chain of amylose bond with the O chain of amylopectin?

Actually you were close, below I have tried to discuss how this occurs:

Glycosidic bonds between monosaccharide units are the basis for the formation of oligosaccharides and polysaccharides such as starch.

It is possible for a sugar hydroxyl group (ROH) bonded to the anomeric carbon (the anomeric carbon is the carbonyl carbon of the open chain form of the sugar and is the one that becomes a chiral centre in the cyclic form) to react with another hydroxyl (R’-OH) to form a glycosidic linkage (R’-OR).

  (A glycosidic linkage is not
  an ether)

Glycosidic linkages can take various forms; the anomeric of one sugar can be bonded to any one of the -OH groups on a second sugar to form an alpha- (as in starch) or beta-glycosidic linkage.

An example of a glycosidic bond (showing formation of maltose):

C-1 of one glucose is linked by a glycosidic bond to the C-4 oxygen of the other glucose
when an ⎯OH (alcohol) of one glucose molecule (right) condenses with the intramolecular hemiacetal of the other glucose molecule (left), with elimination of H2O and formation of a glycosidic bond.

The reversal of this reaction is hydrolysis—attack by H2O on the glycosidic bond. In animal digestion (first phase), it is aided by α (1,4)-glucan 4-glucanohydrolase (salivary α-amylase)

Here is another comparison of the α 1,4 glycosidic linkage and α1,6 glycosidic linkage:

Having discussed about the glycosidic bond, we can refer back to starch:
Starch is composed of two components, α-amylose and amylopectin. α-Amylose is composed of linear chains of D-glucose in α (1,4) glycosidic linkages. The linear linkages in amylopectin are α (1,4), whereas the branch linkages are α (1,6) glycosidic bonds:

Edit

It is also possible for polysaccharide units to form hydrogen bond interactions between neighbouring units, although these are minor and serve to stabilise the helical structure of amylose chains (around  up to 6 residues per turn) but the major bonds are glycosidic bonds

I am sure you now have a clue on how they are bonded. Hope this helps.

References:

Biochemistry (Campbell and Farell)

Lehninger Principles of Biochemistry

Biochemistry (Grisham)


Find atomic number given the group and period I want to find the atomic number of an element in a short way, when I know the exact group and period of that element.

For instance; there is a question in our textbook to find the atomic number of the element which is placed in the 4th group and the 3rd period.

This is the quickest way to do this.

Memorise the following table which contains the atomic number of all the noble gases. Noble gases are placed at the end of every period, so if you remember this table, then using a bit of common sense, we can easily locate the position of any element.

  1st period :  helium - 2

  2nd period :  neon-10

  3rd period :  Argon- 18

  4th period : Krypton -36

  5th period : Xenon -54

  6th period : Radon - 86

Now you are good to go.

Example 1) Say you want to find the atomic number of an element placed in group 4th and period 3.

Neon(10) is the last element of period 2. Come down to the next period ,i.e period 3. We need to reach four blocks to the right to get to group 4. Going each block to the right will increase the atomic number by 1. Therefore, going 4 blocks to the right will give us the atomic number = 10+4= 14.

Example 2) 4th period : 17th group

The last element of period 3 is Argon(18). To get to 4th period and 17th group, we need to come down to period 4 and go 17 blocks to the right. The required atomic number is 18+17=35.

Note : This also requires you to remember that there are 2 elements in the first period, 8 elements each in the second and third period, 18 elements each in third and fourth period.


Dielectric constant and dissolution of ionic salts We say that salts like $\ce{NaCl}$ get easily dissolved in water because the dielectric constant of water is very high. I know the dielectric constant of water is 81 and if a salt is put into water the force between $\ce{Na+}$ and $\ce{Cl-}$ ions gets reduced by 1/81 times, after that the salt can easily be dissolved by simply stirring.

But how can we say that the medium between two very small particles which are bonded together is water? How can water molecules be present between two strongly bonded ions?

The dissolving of a substance into a solvent and the avoidance of contact ion pairing depends on several things. The dissolving relates to the change in Gibbs free energy between the crystaline solid and the solution of the cations and the anions from a salt. A Born-Haber cycle can be used to explain this.

You should be aware of the fact that the Debye-Huckel and extended Debye-Huckel equations for the activity coefficent for ions in solutions have terms in them for the relative dielectric constant of the liquid.

The higher the activity coefficent then the more the ions are able to move freely of each other. The extended Debye-Huckel equation can be modified into the Bromley equation by adding another term for ion-ion interactions.

When considering the effect of changing the relative dielectric constant.

I think that one of the best examples is sodium bromide, in water sodium bromide is a strong electroyte where the cations and anions are well separated. But in acetone where the relative dielectric constant is lower the same salt is a "weak electroyte" where the cations and anions tend to associate with each other.

I think that the formation of contact ion pairs is the first step when going from a perfect solution of ions to a crystaline ionic solid. But the water does more than act as a simple dielectric. What the water molecules will do is to bind to the ions. In particular the water molecules will coordinate as Lewis bases to the cations and then enlarge the size of the cations. This binding of waters can act as a bridge between the cation and the water.


Are there molecules that take the shape of every platonic solid? Methane is the simplest molecule I know of that takes the shape of a platonic solid — in this case, the tetrahedron. Wikipedia has a page on platonic hydrocarbons that includes molecules with tetrahedron, cube, and dodecahedron shapes. There are lots of molecules that have octahedral geometry (e.g. sulfur hexafluoride).

Are there any molecules that form an icosahedron? The closest I've found is $\ce{C_60}$, buckminsterfullerene, which takes the form of an Archimedean solid, the truncated icosahedron.

To be explicit: I am interested in molecules with atoms positioned at the vertices of platonic solids; whether or not there are direct bonds between those atoms is not relevant.

Yes, for each of the 5 platonic solids, there is at least one molecule that takes its shape.

There exist tetrahedron-shaped molecules. Some instances are $\text{CH}_4$ or $\text{CCl}_4$.

There exist octahedron-shaped molecules. Some instances are $\text{SF}_6$ or $\text{Mo(CO)}_6$.

There exist hexahedron-shaped molecules. An instance would be $\text{C}_8\text{H}_8$ AKA "cubane". Perhaps simple cubic lattices (like NaCl) could be considered "hexahedral molecules" as well?

There exist icosahedron-shaped molecules. An instance is elemental boron, $\text{B}_{12}$.

There exist dodecahedron-shaped molecules. An instance is dodecahedrane ($\text{C}_{20}\text{H}_{20}$).


Why are electrolysis cell plates turning a brown color? I have been experimenting with electrolysis for producing oxyhydrogen and using potassium carbonate as an electrolyte.  After several hours of electrolysis, I notice the plates began to tarnish and after 24 hours, are now dark brown.  The plates are made of 316 stainless steel and the electrolyte remains clear.

When I chose to use K2CO3; I thought that it would not decompose during electrolysis but now I suspect that it is decomposing and electroplating the stainless steel plates with carbon. If this is the case; will this have an adverse effect on the plates ability to conduct and produce oxyhydrogen? Also, is CO2 being produced?

Since the carbonate radical is covalently bonded, it probably is not being decomposed electrolytically. Much more likely is that iron in the stainless steel is being removed and oxidized, leaving a thin film of "rust" on the plates.

A few ways you can prove this:

Immerse the browned plates in oxalic acid, which will slowly dissolve iron oxides, but leave carbon untouched.
Substitute platinum electrodes -- admittedly expensive, but Pt is the "gold" standard for electrodes. They should not discolor.
Spray the browned plates with a solution of potassium ferricyanide. If free iron is present, a blue color ("Prussian blue") appear.


Galvanic cells/redox - confused regarding diagram I'm not sure if I got this wrong because I interpreted the cell incorrectly. When they have an arrow pointing in to the tube for $\ce{O2}$, does this imply that $\ce{O2}$ is a reactant? If so then B is definitely wrong, which was my answer.

Is there any other reason why B is wrong? I wasn't sure if it was right or wrong when I did this question.

I think the arrow for $\ce{O2}$ just means there's oxygen gas in that part of the apparatus.

The reason B is wrong is that conversion of hydrogen peroxide to oxygen gas and water is the spontaneous reaction at these conditions. Assuming SLC implies 1 atm of pure oxygen, then everything is under standard conditions so we don't need to bring Nernst into it. The half reactions are as follows with the corresponding potentials.

$$ \ce{H2O2 + 2H+ + 2e- -> 2H2O} (\Delta E^{\circ} = +1.78) $$
$$ \ce{H2O2 -> O2 + 2H+ + 2e-} (\Delta E^{\circ} = -0.7) $$

so the net reaction is
$$ \ce{2H2O2 -> O2 + 2H2O} (\Delta E^{\circ} = +1.08) $$

Since $\Delta E^{\circ} $ is positive, this reaction is spontaneous.

The fact that this reaction doesn't happen means that there's a high kinetic barrier. I think that's what A is getting at, though I have a little bit of an issue with how it's worded. Regardless it's still better than B.


Questions about intramolecular interactions in AMBER forcefield I am trying to run a molecular dynamics simulation with AMBER forcefield. However, it's the first time I use is, and while I know the form of the various intra- and intermolecular energy terms, some details are not clear to me. So, for the AMBER forcefield:

What pairs are excluded in the summation of intramolecular Lennard-Jones interactions? Typical choices are “excluding bonded atoms”, “excluding 1-2 and 1-3 pairs”, “excluding 1-2 and 1-3, applying a factor 1/2 to 1-4 pairs”, and “excluding 1-2, 1-3 and 1-4 pairs”.
For torsion potentials, is the potential scaled by the number of such torsions? (as is done in the DREIDING forcefield, to ensure a constant torsional barrier for the central bond)

As for your first question, this site states that 1-4-interactions are scaled by a factor of 0.5, which I suppose implies 1-2 and 1-3 interactions being excluded.

The answer to your second question is yes, as can be seen here (section TORSIONS). The potential is divided (and thus scaled) by Ntors.


Amorphous to crystalline: Opposite trend? Well my book says that on heating- amorphous solids crystalline at some temperature. As far as I know , amorphous solids have a very disordered structure compared to crystalline solids and on further heating the level of disorder should increase. So why do they become more ordered or crystalline at high temperatures or am I missing something fundamental in here?

What they say in your book is right. Take the case of polyethylene terephthalate (PET) which is a kind of polyester used in water bottles for example. It's an amorphous polymer (remember the bottles are transparent). When it is heated, the strained chains have the possibility to move and reorder themselves and the polymer becomes partially crystalline. This fact can readily be proved experimentally by heating a piece of PET (over its glass transition temperature, around $74 ^0\mathrm{C}$), then quenching the sample. It becomes white and not transparent which means it's partially crystalline.
We can monitor this transformation using a specific instrument called differential scanning calorimeter, where we can notice the formation of an exothermic peak corresponding to crystallization.


How do I calculate the enthalpy change when a gas is being used to heat water? The question I'm trying to answer is:

Calculate the mass of butane gas that would be needed to heat $724\ \mathrm{cm^3}$ of water from an initial temperature of $7.44\ \mathrm{^\circ C}$ to $50.7\ \mathrm{^\circ C}$. The thermochemical equation for the combustion of butane is:
$$\ce{2C4H10(g) + 13O2 -> 8CO2(g) + 10H2O(l)}\quad\Delta H=-5748\ \mathrm{kJ/mol}$$

I know that the equation for enthalpy is
$$Q=m\times c\times\Delta T$$
and I calculated this for water:
$$Q=724\times4.18\times43.26$$
$$Q=130918.6032$$
I don't know where to go from here in order to figure out the mass of butane needed.  Please tell me if I have done something wrong or if you can help me to solve the rest of the problem.

You're on the right track.

\begin{eqnarray}
Q & = & m\cdot c\cdot \Delta T\\
\mathbf{unit\ analysis} \qquad \textrm{kJ} & = & \textrm{kg}\cdot \frac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{K}} \cdot \textrm{K}
\end{eqnarray}

For water, the specific heat capacity $c = 4.18 \large{\frac{\textrm{kJ}}{\textrm{kg}\cdot \textrm{K}}}$. You have that right.

The temperature difference $\Delta T = (50.7 - 7.44) = 43.26\ \textrm{K}$. We're talking about differences, so we can use kelvin as the unit. You have that right too.

Now, what is the mass of $724\ \textrm{cm}^3$ water?

Choose any unit you want, but adjust the outcome of your calculation according to the unit analysis above!

Now you know the kJ you need.

From the combustion equation, you know the amount of heat delivered by $n$ moles of butane. Note the $n$!

Now you can calculate how many moles of butane you will need.

From there, the calculation of the mass is not difficult either.


Isopropyl alcohol forming peroxides I read in many MSDS that Isopropyl alcohol tends to form peroxides (by reacting to air OR light) which may explode when they reach a given concentration.

For example: http://nj.gov/health/eoh/rtkweb/documents/fs/1076.pdf or http://www.hvchemical.com/msds/isal-70.htm as well as many others.

I own a few liters of isopropyl alcohol (a few bottles) and I'm getting concerned.
It came in standard transparent (polypropylene I suppose) bottles with a plastic bag wrapped around it (I suppose to prevent contact with air). I only keep one bottle ready to use, all the others are as when I bought them (with plastic bag wrapped around it), stored in a relatively cool (<= 25°C) dark area of my house.

This is at home and I couldn't find any fire resistant storage cabinet in my area.

Now, I'm not sure if I understand the datasheet correctly, but as far as I understood the problem lies with the opened bottle (which I open and close regularly) and not with the air-tight never opened other bottles, right?
Therefore a simple solution may be to say: after one year I empty the opened bottle, trash it away and buy another one. Should this be enough?

I think this may be an overreaction.  First, MSDS are designed for workers working with industrial quantities in industrial conditions.  Second, the danger comes from concentration of the peroxide from evaporation or, more typically, distillation.  Small quantities of peroxides in solution are not usually an explosion hazard.  Third, I have never seen isopropanol listed in lists of hazardous peroxide prone chemicals as used by lab safety departments (ethers are the typical risk here).  The closest I have seen is sec-butanol, which is the secondary alcohol from butanol, and I've only seen this listed once, in a list that seemed much more cautious than some other lists.  Isopropanol is the secondary alcohol from propanol.  Given your storage, cool and dark, I think you are fine.  Remember that 91% isopropanol is sold in drugstores, though these may contain stabilizers and you larger bottles may not.

But having said all this, Lighthart's brief answer is correct though probably unlikely.  If you DO see any crystals, don't even try to open the bottle (crystals in the threads are the most common source of explosion).  But again, seems unlikely.

If you are still concerned, there are peroxide test strips you can buy that are very easy to use and can tell you what concentration of peroxides are present.


rust caused by salt solution & stainless steel I am not a chemist and I could really use some help for a practical case.
I am fermenting vegetables (cabbage, onions, carrots, etc.). I recently switched from glass and ceramic vessels to stainless steel 304 ones (100L capacity). At the moment, I am only filling the vessels half-way, i.e. 50/60L and using a variable lid system. The vessel is airtight thanks to a plastic band around the lid, which is then placed right on top of the vegetables (it is a stainless steel lid and the plastic band around it is then pumped with air to compress it against the vessel). The lid also has an airlock system in the middle, allowing the co2 generated during the fermentation process to be pushed out. The brine created during the fermentation process also gets pushed out from that airlock (when the vegetables are put in the vessel, about 2% salt is added, so the brine contains salt). The brine ends up covering the lid and I removed it from time to time. It is not overflowing on the plastic band from the outside, but it seems that the brine is also pushed up against the plastic band and gets in between the plastic band and the vessel. After a few days, no more brine gets out.
At the end of the fermentation (after 4-5 weeks), when I removed the lid and emptied the fermentation vessel, I discovered traces of rust on the inside. It is right where the plastic sealed the vessel. It's not only rust, but the surface (or coating) of the stainless steel was even removed in some parts. There is no rust below or above that line (ie where the plastic sealed the vessel).
I am aware that stainless steel 304 is for sure not 100% rust-resistant but I was definitely not expecting such reaction after 1 usage! What a powerful brine...
My question is whether you believe that this reaction is due to the brine pushed up? Probably small amounts of brine end up getting stuck between the plastic and the vessel and thus potentially slightly exposed to air.
If that's the cause, I have two fixes in mind :

put the lid higher in the vessel so that the brine is not pushed up against the plastic/lid. I believe that the CO2 created should be sufficient to push out all the air below the lid.

continue to place the lid right above the vegetables. Once no more brine is pushed out, remove the lid to clean the hedges. Then put the lid back. I believe that the remaining fermentation process should push out the newly added air while not generating sufficient pressure to push new brine out.

I also came across another possible explanation while browsing full link

The same thing can happen at the water’s surface if the container is only half full. In this case, the steel above the waterline is exposed only to air, and the passive oxide layer is stable. Beneath the surface, the oxide layer is at a different potential and less stable because of the chloride ions. Now the crevice is represented by the waterline: stable area above, less stable but very large area below, and crevice corrosion occurs at the waterline. Usually this type of corrosion manifests as pitting or pinholes. The mechanism described is accelerated by localization, so a pit is most often the result.

Some pictures
1)the vessel
2)the rusted line / removed steel on the inside of the vessel
3)the lid on which the plastic band is then mounted
What do you think ?

Chloride will pit/rust 304 ,not quite as much rust/pits on 316/317. Many things make it worse such as limited air access ( under deposit or crevice corrosion ).Oxygen can help rebuild the passive film. For your long term exposure ceramics and glass containers is the best fix. For metal containers ,much higher alloy is needed such as Incoloy 825.


An ideal gas is undergoing an isothermal expansion against a constant external pressure. Which of the following statements is true? An ideal gas is undergoing an isothermal expansion against a constant external pressure. Which of the following statements is true?

  1) This process is reversible.

  2) This process is irreversible.

  3) This process is exothermic.

  4) The system pressure is always in equilibrium with the external pressure.

  5) The information given is insufficient to describe this process.

(source)

I don't get how the answer can be 2, I thought it would be 5. Since it's expansion, $\Delta S$ of the system must be > 0. Now since it's an expansion, work must be < 0. If work is less than 0, q > 0 due to the process being isothermal. So that means $\Delta S$ of the surroundings < 0 since it is -q/T. That means $\Delta S$ of the universe could be > 0, < 0 (non spontaneous) and = 0 depending on the magnitude of $\Delta S$ of the system and surroundings. As a result, I choose 5 since we don't know the magnitudes of both $\Delta S$.

Why would it be 2?

If the gas is ideal, for an isothermal expansion

$$p=\frac{nRT}{V}=\frac{\textrm{constant}}{V}$$

As the volume of the system changes, so must the pressure.

Now a condition of reversibility is mechanical balance between the system and surroundings, that is, $p=p_{\textrm{ext}}$. However, if a process is performed at constant external pressure and temperature, then as shown above the condition of mechanical balance cannot be satisfied, since $p$ will change during the expansion whereas $p_{\textrm{ext}}$ is constant. Therefore the process must be irreversible.


Are risks (hexavalent chromium) during Chrome plating temporary, or remain with the product? Since some chrome plating production processes generate hexavalent_chromium compounds, which are reportedly highly toxic/carcinogenic, I wonder:

Do the hexavalent chromium substances only occur (temporarily) during the production process (i.e. the chrome plating) or will they remain with the product?

My naïve impression has been that after the chrome plating process is over, the product will have an additionally a more or less thick layer of native Cr. Is this correct?

Should I expect impurities or residuals of toxic forms (e.g. hexvalent chromium) to be with the chrome plated this is possible do primitive test exist to detect product?

If it is possible that the chrome plated object could have
worrisome concentrations/amounts of Cr(IV), is there a primitive test to detect this?

Update
Some internet resources, such as tests/detection procedures for Cr(IV) described here, mention

  The screening method can be used to determine chromium(VI) in
  passivation layers on metal pieces.

Also test kits for Cr(IV) sold online mention in their description to be

  test[s] to detect leachable Cr (VI) ions on virtually any solid surface.

both giving reason to thing that the Cr (VI) might remain in worrisome amounts after the chrome plating. True.

update 2

one of the answers mentioned that "chrome plating", being understood as a general term, does not necessarily imply which material is being plated. If permissible I would like to add that I am most interested in those risks of the base material being copper.

update3

The most upvoted answer has stated little risk if chrome plating procedure is carried out well.

As new information the CHROME PLATINGANDCHROMIC ACID ANODIZING SURVEY strongly suggests that the term chrome plating, can be divided regarding the categories a) Tank process (Hard plating; decorative; chromic acid anodizing) and b) *Tank constituent(s) (Hexavalent chrome; trivalent chrome).

This information to raises the aspect, if the outer passivated layer should be expected to have only Cr(III), irrespective of the used tank constituents.

Multiple questions should be broken in multiple items, but anyway,

If your plating process is well done, and everything is thoroughly cleaned after, the level of Cr(III) or Cr(VI) remaining should be low enough to be inconsequential to health and the environment. You're correct stating that regardin chrome, thre will be only a passivated layer of metallic chromium and not the trivalent or hexavalent.

Please do remember that the main reason that chrome plating makes things resistant to corrosion is that as soon as chromium is exposed to air, it forms a very thin layer of chromium oxide attached to the surface of the metal, preventing further corrosion. So, in fact, there is some Cr(III) on a finished piece of chrome plated material.

We aren't considering the environment were the finished part will be exposed/used. If aggressive enough conditions are present, the chromium layer will be corroded away and will enter the environment as Cr(III)/Cr(VI).

Regarding Cr(VI) detection, there is a "simple" method involving a solution of diphenylcarbazide described at https://www2.mst.dk/udgiv/publications/2009/978-87-7052-987-7/html/kap04_eng.htm

However, one can easily buy test kits for Cr(VI) on Amazon, ready for use, starting at about (February 2019) USD 110.


Making three membered ring using different reactants I am struggling with the above question. It will be the conjugate base of trichloromethane that is involved. The same is true of the carboxylate. However, I have no idea about the reaction mechanisms. Please help.

I think that both reactions involve carbene as common intermediate which then is the species involved in the subsequent cyclopropanation reaction.
These are the two equations which lead to the formation of the carbene:
$$\ce{KOH + CHCl3 ->  :CCl2 + KCl + H2O}$$
$$\ce{Cl3CCO2Na ->  :CCl2 + NaCl + CO2}$$
Subsequently, the alkene undergoes cyclopropanation by dichlorocarbene.
In the second case, heat is required in order to promote decarboxylation, which is entropically favored.


How to calculate the probability of a particular value of an observable being measured A particle on a ring is prepared with a wave function equal to $\frac{1}{\sqrt{\pi}}$ between $\theta=0$ and $\theta=\pi$, and $0$ between $\theta=\pi$ and $\theta=2\pi$. If a measurement of the angular momentum is made, calculate the probability of finding a particular value $l\hbar$. [The angular momentum eigenfunctions are $|l>=(2\pi)^{-1/2}e^{il\theta}]$

I've been reading the Oxford Chemistry Primer on Quantum Mechanics and they don't provide solutions. I've been struggling with this question.

I think I need to calculate the coefficient of expansion for a particular basis eigenfunction, but I'm unsure how to proceed.

$$
\newcommand{\ket}[1]{\,\lvert{#1}\rangle}
\newcommand{\bra}[1]{\langle{#1}\rvert\,}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
\newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
\newcommand{\op}[1]{\hat{#1}}
$$

Solution should be a pretty straightforward application of the Born rule.

In this particular case, we have a self-adjoint operator $\op{L_z}$ with eigenfunctions and eigenvalues satisfying
$$
  \op{L_z} \psi_l(\theta) = l \hbar \psi_l(\theta) \, ,
  \quad \text{where} \quad
  \psi_l(\theta) = \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{\mathrm{i} l \theta}
  \quad \text{and} \quad
  l = 0, \pm 1, \pm 2, \dotsc
$$
such that any wave function $\psi$ can be expanded over orthonormal basis $\{ \psi_l(\theta) \}_{l=-\infty}^{\infty}$ of the corresponding eigenfunctions of $\op{L_z}$ as follows,
$$
  \psi(\theta)
  =
  \sum\limits_{l=-\infty}^{\infty} c_l \psi_l(\theta) \, ,
  \quad \text{where} \quad
  c_l
  =
  \braket{\psi_l}{\psi}
  =
  \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta \, .
$$
And a measurement on a system in arbitrary state $\psi$ can yield any of eigenvalues $l \hbar$ with the probability given as follows
$$
  \Pr(l \hbar) = |c_l|^{2} = c_l^* c_l \, .
$$

Now, since $\psi(\theta)$ is defined to be
$$
  \psi(\theta)
  =
  \begin{cases}
  \frac{1}{\sqrt{\pi}} & \text{if} \quad \theta \in [0, \pi] \\
  0            & \text{if} \quad \theta \in [\pi, 2 \pi]
  \end{cases} \, ,
$$
the coefficient $c_l$ is equal to
$$
  c_l
  =
  \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta
  =
  \int\limits_{0}^{\pi} \psi_l^*(\theta) 1/\sqrt{\pi} \mathrm{d} \theta
  +
  \int\limits_{\pi}^{2 \pi} \psi_l^*(\theta) 0 \mathrm{d} \theta \, ,
$$
where the second term trivially vanishes leading to
$$
  c_l
  =
  \int\limits_{0}^{\pi}
  \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{-\mathrm{i} l \theta} \frac{1}{\sqrt{\pi}}
  \mathrm{d} \theta
  =
  \int\limits_{0}^{\pi}
  \frac{1}{\sqrt{2} \pi} \mathrm{e}^{-\mathrm{i} l \theta}
  \mathrm{d} \theta \, .
$$
And then some simple algebra leads to the final answer, which I leave to the OP, since it is far into the night for me and I'm afraid I will make a lot of stupid mistakes.


How do I find the ground state term symbol for transition metal complexes? I have a problem where I'm trying to match metal complexes with the proper ground state term symbol.  I've tried doing a Google search and looking for a process, but haven't had any luck.  I decided to try the following for complexes that are octahedral ($O_\mathrm{h}$ symmetry):

Determine the d-electron count for the oxidation state of the metal atom
Look at the corresponding Tanabe-Sugano diagram to obtain the appropriate ground state term symbol

Is that considered even somewhat correct?

The other problem I'm having is that I there are certain configurations (e.g. $\ce{Cu^2+}$ has a $\mathrm{d^9}$ configuration) for which there are no Tanabe-Sugano diagrams (at least none that I can find) and so my process doesn't work.  Can someone please point me in the right direction?  Thank you!

To find the ground state term symbol, you should be using symmetry and group theory arguments, you shouldn't have to resort to searching Tanabe-Sugano diagrams to get the answer.

We'll start with octahedral complexes (the general idea can be extended quite easily to tetrahedral or square planar complexes). As requested in the question, I will only cover ground-state term symbols, but this procedure can also be extended easily to excited configurations.

In an octahedral complex, the d orbitals are split into a $\mathrm{t_{2g}}$ set and an $\mathrm{e_g}$ set. You will need the direct product table for the $O_\mathrm{h}$ point group, which you can find here (I will reproduce it here for convenience). You simply need to tack on the g subscripts on everything (since $\mathrm{g \otimes g = g}$, there will never be any ungerade states involved).

$$\begin{array}{c|ccccc} \hline
 & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\ \hline
\mathrm{A_1} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\
\mathrm{A_2} & & \mathrm{A_1} & \mathrm{E} & \mathrm{T_2} & \mathrm{T_1} \\
\mathrm{E} & & & \mathrm{A_1 + [A_2] + E} & \mathrm{T_1 + T_2} & \mathrm{T_1 + T_2} \\
\mathrm{T_1} & & & & \mathrm{A_1 + E + [T_1] + T_2} & \mathrm{A_2 + E + T_1 + T_2} \\
\mathrm{T_2} & & & & & \mathrm{A_1 + E + [T_1] + T_2} \\ \hline
\end{array}$$

I also want to make the disclaimer that there is not necessarily a one-to-one correspondence between electronic configurations and term symbols. Therefore, the configuration $(\mathrm{d}_{xy})^2(\mathrm{d}_{z^2})^1$ does not necessarily lead to one specific term symbol. A term symbol may often be a linear combination of multiple electronic configurations, and vice versa. Nevertheless, I will use the traditional "orbital diagrams" to illustrate points as necessary.

$\mathrm{d^1}: {}^2T_{\mathrm{2g}}$

The ground state is a $\mathrm{(t_{2g})^1}$ configuration, which corresponds very simply to a $T_\mathrm{2g}$ state. The spin of a single electron is $S = 1/2$, corresponding to a multiplicity of $2S + 1 = 2$.

$\text{High-spin }\mathrm{d^6}: {}^5T_{\mathrm{2g}}$

The high-spin configuration of $\mathrm{(t_{2g})^4(e_g)^2}$ can be considered to be the sum of a $\mathrm{(t_{2g})^3(e_g)^2}$ $\mathrm{d^5}$ configuration (in black), and a $\mathrm{(t_{2g})^1}$ $\mathrm{d^1}$ configuration (in blue).

The five spin-up electrons are symmetrically distributed amongst the five d-orbitals, and as a consequence, these five electrons together have a spherically symmetric electron density. You can prove this quite easily by looking up the wavefunctions of the five d-orbitals, and evaluating the function:

$$\rho(\mathrm{d^5}) = |\mathrm{d}_{xy}|^2 + |\mathrm{d}_{xz}|^2 + |\mathrm{d}_{yz}|^2 + |\mathrm{d}_{x^2-y^2}|^2 + |\mathrm{d}_{z^2}|^2$$

In fact, it is probably much easier to add up the complex spherical harmonics (add up the square modulus of the five functions with $l = 2$). The radial part $R(r)$ can be factored out (and ignored) since it only depends on the quantum numbers $n$ and $l$ (and is therefore constant for all five 3d orbitals).

Whichever way you do it, you would find that the resulting electron density $\rho$ would only be dependent on $r$, and no longer on $\theta$ and $\phi$. Such spherically symmetric spatial components transform as the totally symmetric irreducible representation, which in our case is $\mathrm{A_{1g}}$. (This is always the case, but if you need further evidence, it is provided in the tables of descent in symmetry.) Since $\mathrm{A_{1g}} \otimes \Gamma = \Gamma$ (where $\Gamma$ can be any generic irrep), we can safely discard the $\mathrm{A_{1g}}$ component, and focus on the blue component...

...which we already found in the previous section to be $\mathrm{T_{2g}}$. Therefore, the spatial component of a high-spin $\mathrm{d^6}$ configuration is also $\mathrm{T_{2g}}$.

The spin, however, is different. This time, there are four unpaired electrons, such that $S = 2$ and $2S + 1 = 5$.

$\text{Low-spin }\mathrm{d^6}: {}^1\!A_\mathrm{1g}$

In fact, there is nearly no group theory here to be discussed. This is an example of a closed-shell configuration: all electrons are paired up and there is no unequal population of degenerate orbitals. Such configurations always lead to totally symmetric term symbols, i.e. a spatial part of $A_\mathrm{1g}$. Since there are no unpaired electrons, $S = 0$ and $2S + 1 = 1$.

$\mathrm{d^9}: {}^2\!E_\mathrm{g}$

This is a good point in time to introduce what is known as the hole formalism. The idea is that a $\mathrm{d^9}$ configuration is the same as a $\mathrm{d^{10}}$ configuration, minus one electron. A fully filled d subshell is spherically symmetric, meaning that the $\mathrm{d^{10}}$ configuration has a $A_{\mathrm{1g}}$ term symbol. Therefore, if the $\mathrm{d^{9}}$ configuration transforms as $\Gamma_{(9)}$, and the "missing" $\mathrm{d^{1}}$ transforms as $\Gamma_{(1)}$, then:

$$\Gamma_{(9)} \otimes \Gamma_{(1)} = \mathrm{A_{1g}}$$

There is a result in group theory that states that the totally symmetric irrep ($\mathrm{A_{1g}}$ in our case) only occurs along the main diagonal of the direct product table. In other words, the above equation can only be fulfilled if

$$\Gamma_{(9)} = \Gamma_{(1)}$$

Therefore, our task is simply to find what the term symbol of the "missing" electron is. The missing electron is known as a hole.

In our case, in order to minimise the energy of the $\mathrm{d^9}$ electron configuration, we would want to put the hole in the higher-energy $\mathrm{e_g}$ orbital. As a result, our task becomes to find the term symbol for a $\mathrm{(e_g)^1}$ hole configuration, which is simply $E_\mathrm{g}$.

The total spin of a $\mathrm{d^9}$ configuration is easily seen to be $S = 1/2$, leading to a doublet state again.

$\mathrm{d^2}: {}^3T_{\mathrm{1g}}$

The ground state is a $\mathrm{(t_{2g})^2}$ configuration. Clearly, the green configuration is more favourable than the red one (Hund's first rule). The question then becomes how you obtain the possible term symbols for a $\mathrm{(t_{2g})^2}$ configuration, and which of them correspond to the green configuration and which to the red. The different possibilities are given by the direct product

$$\mathrm{T_{2g} \otimes T_{2g} = A_{1g} + E_g + [T_{1g}] + T_{2g}}$$

The square brackets are used to indicate spatial states that are antisymmetrised, i.e. are antisymmetric under the permutation of two electrons. Due to the spin-statistics theorem, the total electronic wavefunction, which is a product of spin and spatial parts, must be antisymmetric. To satisfy this, you have two possible choices:

$$\begin{array}{cc|c} \hline
\text{Spatial} & \text{Spin} & \text{Overall} \\ \hline
\color{green}{\text{AS } (\mathrm{T_{1g}})} & \color{green}{\text{S (“triplet")}} & \color{green}{\text{AS}} \\
\color{red}{\text{S } (\mathrm{A_{1g}, E_g, T_{2g}})} & \color{red}{\text{AS (“singlet")}} & \color{red}{\text{AS}} \\ \hline
\end{array}$$

The antisymmetric spatial wavefunction is more stable than the symmetric spatial wavefunctions, due to exchange energy. Consequently, the ground-state term symbol has a $T_{\mathrm{1g}}$ component.

As can be seen from the above table, this necessitates that the spins of the electrons are parallel, i.e. $S = 1$ and $2S + 1 = 3$. We can therefore associate the $T_\mathrm{1g}$ state with the green electronic configuration, with two spin-up electrons in different orbitals.

$\mathrm{d^3}: {}^4\!A_\mathrm{2g}$

The $\mathrm{(t_{2g})^3}$ configuration here is simply a spherically symmetric $\mathrm{d^5}$ configuration, minus an $\mathrm{(e_g)^2}$ hole configuration. To find the symmetry of the two holes, we take

$$\mathrm{E_g \otimes E_g = A_{1g} + [A_{2g}] + E_g}$$

Since the two holes are in different orbitals, their spatial component has to be antisymmetrised. Therefore, it is really the $A_\mathrm{2g}$ state that we want here.

As for the multiplicity, $S = 3/2$ gives $2S + 1 = 4$.

Now that all the ground work has been done,

we can whizz through the remaining electronic configurations - how exciting! It's a good idea to try working these out yourself before looking at the answer, to see if you truly understand the concepts above. I'm also not going to comment on the spin multiplicities any more - hopefully you can look at the diagram, calculate $S$, and work out $2S + 1$ yourself.

$\text{High-spin }\mathrm{d^4}: {}^5\!E_\mathrm{g}$

Using the hole formalism, you simply subtract one $\mathrm{e_g}$ electron from a spherically symmetric $\mathrm{d^5}$ configuration, which straightforwardly gives the term symbol above.

$\text{Low-spin }\mathrm{d^4}: {}^3T_\mathrm{1g}$

The black $\mathrm{d^3}$ configuration has symmetry $\mathrm{A_{2g}}$, and the lone blue d-electron has symmetry $\mathrm{T_{2g}}$. The direct product gives the symmetry for the low-spin $\mathrm{d^4}$ case:

$$\mathrm{A_{1g} \otimes T_{2g} = T_{1g}}$$

$\text{High-spin }\mathrm{d^5}: {}^6\!A_\mathrm{1g}$

We already established earlier that this is spherically symmetric and therefore transforms as the totally symmetric irrep $\mathrm{A_{1g}}$.

$\text{Low-spin }\mathrm{d^5}: {}^2T_\mathrm{2g}$

This is best treated as a $\mathrm{t_{2g}}$ hole in a totally symmetric low-spin $\mathrm{d^6}$ configuration, which gives rise to a $T_\mathrm{2g}$ state.

$\text{High-spin }\mathrm{d^7}: {}^4T_\mathrm{1g}$

Treating this as a sum of a totally symmetric $\mathrm{d^5}$ configuration and a $\mathrm{d^2}$ configuration, we can tell that this has the same spatial symmetry as the original $\mathrm{d^2}$ configuration, i.e. it is a $T_\mathrm{1g}$ state.

$\text{Low-spin }\mathrm{d^7}: {}^2\!E_\mathrm{g}$

Noticing that this is the sum of the totally symmetric low-spin $\mathrm{d^6}$ and one $\mathrm{e_g}$ electron makes this easy.

$\mathrm{d^8}: {}^3\!A_\mathrm{2g}$

This is, again, a sum of the totally symmetric low-spin $\mathrm{d^6}$ configuration and the $\mathrm{d^3}$ configuration, which has $\mathrm{A_{2g}}$ symmetry.

For extra practice, why not try doing this using the hole formalism instead? You should reach the same answer!


How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure? How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure? Is there some kind of mathematical proof for it? Is there some way I can understand this intuitively?

I have been trying to think about it using this picture but I cannot seem to get anywhere.

To be precise, if in a close packed structure (ccp or fcc) there are $n$ atoms or ions then the number of octahedral voids and tetrahedral voids will be $n$ and $2n$ respectively.

For example, there are 8 tetrahedral voids per unit cell of fcc structure $(Z_{\text{eff}}=4)$. If you divide the FCC unit cell into 8 small cubes then each small cube has 1 tetrahedral void located at its own body center. Thus, total number of TVs in a unit cell $= 8 = 2  \times  Z_\text{eff}$

Now, again consider a CCP or FCC unit cell. The body centre of the cube is not occupied but it is surrounded by 6 atoms (4 in same plane, 1 above and 1 below). On joining these face centers, an octahedral void is formed. Thus, number of OV at body centre of the cube is 1.

Besides the body center, there is one OV at the center of each of the 12 edges. It is surrounded by 6 atoms, three belonging to the same cell unit (two on the corners and once on face center), and the three belonging to two adjacent unit cells.

Since, each edge of the cube is shared between four adjacent unit cells only one fourth of each void belongs to a particular unit.

Therefore, OV at body centre = 1

OV at the 12 edges of the cube and shared between four unit cells  $= 12\times\frac{1}{4}=3$

Total number of OVs $= 1 + 3 = 4 = Z_\text{eff}$


How to get equation of state from fugacity Given that the fugacity of a substance is $f=Pe^{bT}$, how do I obtain the equation of state of the substance?
This is my approach, but I have not made good headway:
$$RT d (\ln f) = VdP \\
d(\ln f) = \frac{V}{RT} dP \\
bdT = \left( \frac{V}{RT} - \frac{1}{P}\right)dP$$
But what next? I see that
$$\left( \frac{V}{RT} - \frac{1}{P}\right)dP = \frac{(Z-1)}{P}dP$$
But I don't see how I can reduce this further.
My end goal is to find derivatives of pressure wrt temperature keeping volume constant. How do I go about this problem?

That first equation is supposed to be at constant T.  You can rewrite it as $$RTd\ln{(f/P)}=RT(z-1)\frac{dP}{P}$$or $$\ln{\frac{f}{P}}=bT=\int_0^P{(z-1)\frac{dP'}{P'}}$$So if z behaves like z = 1+aPT, this becomes:
$$bT=aPT$$ or b = aP


Doubts on Diffusion/ Effusion of gases at different pressures Let's start with a question:

  "One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of an unknown compound of Xenon with fluorine at 1.6 atm  takes 57 s to diffuse through the same hole. Calculate the molecular formula of the compound."

It is a question where pressures of the gases are not similar. Could be solved quite easily, if we calculate the rates of diffusion ($r_{N_2}$ and $r_{un}$) of each gas separately as no. of moles of the respective gas diffused per unit time, find the ratio of the rates and then equate this ratio, with another ratio found using pressures and molar masses of the given gases, i.e, with
$$\frac{r_{N_2}}{r_{un}}= \frac{p_1}{p_2}\sqrt{\frac{M_{un}}{M_{N_2}}}$$, ($M_{un}$ and $M_{N_2}$ are molar masses of the unknown gas and nitrogen gas respectively) and finally find the no. of fluorine atoms attached with xenon from the value of molar mass of the unknown gas.

It's fine that way. But we also know that, volumes of a gas diffused per unit time gives us rate of diffusion too. And in the question above, the volumes (in case they were given instead of moles) would have been different because of different pressures. Consequently, the first ratio (the ratio involving volume and time) would have been different compared to the ratio involving moles and time. And that would have affected the final value of molar mass and thus messed up our solution.
So my question is, what would be the actual cause of such a situation? And how to fare in those situations?

"But we also know that, volumes of a gas diffused per unit time gives us rate of diffusion too. And in the question above, the volumes (in case they were given instead of moles) would have been different because of different pressures."

If you wanted to measure the effused gas in terms of volume, you had to somehow calculate this volume from the number of moles effused. This volume has nothing to do with the volume or the pressure inside the vessel from which the gas effused.

[edit]

From the number of particles that escaped to the right, you could calculate a volume, given an arbitrary pressure (e.g. standard pressure). With this you could express the rate of diffusion in units of volume per time unit.


Is there any way to chemically check sweetness level in any drink or water or tea etc? I was wondering if there is a test or way to find out how sweet any drink is? For example, if I mix sugar in water, how can I test its level or number of sweetness?

As asked by the OP, I think the answer to the question has to be no. If it was known that a drink contained only glucose as a sweetener, then it would presumably be possible to measure accurately the glucose concentration in the drink. But maybe other substances in the drink would be interferences in the measurement. Following on, as @Alchimista commented, perhaps a Scoville-type sweetness test could be used. But the possible interferences problem complicates this as well. In addition, perceptions of sweetness may also be affected by seemingly innocuous substances in the drink, e.g., I find that a dash of table salt makes watermelon taste sweeter and I have no clue how that might arise.

Then there is the matter of inorganic substances. The old fashioned name for lead acetate was ‘sugar of lead’. Poisonous though it is, it earned its deadly sweet name. As well, some salts of beryllium ('glucine'; all berylium salts are highly poisonous) and yttrium are sweet tasting: see wiki. So it is not just various organic substances that can taste sweet and who knows what else is sweet tasting: no sane chemist (nowadays) tastes chemical substances.

My guess is that there are quite a few chemicals that would taste sweet, but we will probably never know what they are. Then how would sweetness levels of two lead acetate aqueous solutions be compared? I don’t see how it can be assumed, a priori, that sweetness is directly proportional to lead ion concentration and no one should do such an human subjects experiment anyway. And what about a drink containing glucose and lead acetate? Even if the lead acetate did not interfere with the quantification of the glucose concentration, the drink would be sweeter than expected from just the glucose measurement result.


What is the meaning of scores when comparing reaction classes in the KEGG database With respect to related R class in KEGG website:
http://www.genome.jp/tools-bin/relclass?RC00064   What is the biological or statistical relevance of the score? How is it computed? And does it signify the bonding strength of how probably it is for a pair of compounds to be an R pair?
Note: This may be a silly question to all chemists, but not for me, i am a computer science student , and my research field in bioinformatics and so i have to study both biology and bit of chemistry, so please just dont close the question or be harsh on me, I have very little knowledge in chemistry and also googling doesn't help much. I asked this question in Biology forum, they closed the question and asked me to ask it here

And does [the score] signify the bonding strength of how probably it is for a pair of compounds to be an R pair?

No, it has nothing to do with that. It is not about how probably a particular reaction is.

When comparing RCs (reaction classes), the scores in the table of related reaction classes are just a measure on how similar these reactions are.

RC00064 apparently is an N-acylation or the formation of a peptide bond. RC00197 with a score of 1.0 is a N-formylation and considered very similar.

Change the cutoff to a lower value, such as 0.4, and compare the results at the end of the table with RC00064. You'll find dissimilar reaction classes like RC02518, the oxidation of a selenol to a selenic acid.


Effect of isotopes on H-Bonding The following question was asked in one of the assignments my teacher has given.

Acetone ($\ce{Me2CO^16}$) on treatment with $\ce{H2O^18}$ gives a mixture of $\ce{Me2CO^16}$ and $\ce{Me2CO^18}$, the
latter being in slight excess. This may be explained by
(A) hydration of acetone is an equilibrium process
(B) $\ce{C – O^18}$ bond is slightly stronger than $\ce{C – O^16}$
(C) $\ce{Me2CO^18}$ forms stronger hydrogen bonds than $\ce{Me2CO^16}$ with water
(D)hydration of acetone is irreversible

I have doubt about option (C). According to me, (C) should be correct, but it is given incorrect. Why the $\ce{^18O...H}$ is weaker than $\ce{^16O...H}$ bonding?

Hydration of acetone is an equilibrium process.
$$\ce{(CH3)2C^16O + H2^18O <=> (CH3)2C^18O + H2^16O }$$
The equilibrium constant of this reaction will depend on how the bond strengths of the $\ce{O-H}$ bond in water and the $\ce{O=C}$ bond in acetone is affected by isotope exchange. The isotope-dependent strength of hydrogen bonds might be a secondary effect, as well.
The ratio of heavy to light acetone will not only be dependent on the equilibrium constant, but also on the ratio of acetone to heavy water initially present.
So none of the statements are sufficient to explain the observation completely. Statements (a) and (b) are correct. Statement (c) is not directly related to the ratio of the species. You also have to consider the isotope effect on the strength of hydrogen bonds with water (as one acetone changes from light to heavy, one water changes from heavy to light).


How can an excited helium dimer be formed? Why does a collision between an excited helium molecule ($\ce{He^*}$) and a ground-state helium atom create $\ce{He2^*}$?

The electronic configurations are $\mathrm{(1s)^2}$ for $\ce{He}$ and $\mathrm{(1s)^1(2s)^1}$ for $\ce{He^*}$.

Put this information in an energy diagram and see what the molecular orbitals look like, especially how they are filled. (Further hint: Filled-unfilled orbital interactions are generally favorable.)

I have attached a graphic:

This leads to a bond order of $\frac{1}{2}(2-1+1) = 1$, and therefore $\ce{He2^*}$ is stable with respect to $\ce{He + He^*}$. The excited dimer is known as an excimer.

As an aside: The relative energy levels of the molecular orbitals are very hard to get right out of the blue, so take them with a grain of salt. As was kindly pointed out by Martin, the excitation of helium will initially form a singlet excited configuration, depicted above. This should be able to relax to a triplet excited state.


Both calcium hydroxide and ethanol have a hydroxide group, then why isn't ethanol ionic? I am confused on what is a hydroxide exactly is. For example, ethanol has a hydroxide group so does this mean it contains hydroxide ion? Does this make it ionic? I know it is a covalent molecule but why isn't it ionic?

Calcium hydroxide has $\ce{OH-}$ but how is this different than the $\ce{OH}$ group in ethanol? Is it covalent due to the fact it does not dissociate into ions when in aqueous solution? I know I'm missing something here.

No, it is not ionic.
Your doubts are somehow legitimate if you are not aware of the concept of electronegativity. As a rule of thumb if two atoms have similar electronegativity is harder to have ionic bonds. You can use Pauling scale of electronegativity to have some insights on it.

A difference of electronegativity of over 1.7 is likely to be ionic,
and a difference of less than 1.7 is likely to be covalent.
Source: Atkins, Peter; Loretta Jones (1997). Chemistry: Molecules, Matter and Change. New York: W. H. Freeman & Co. pp. 294–295. ISBN 0-7167-3107-X.

This means you will often deal with ionic compounds made of one ion with a strong electronegativity and one with a weak electronegativity. In fact, ions are composed of anions and cations. With a lot of approximations, we can calculate the electronegativity of a group averaging the electronegativities of the components for $\ce{-OH}~\dfrac{3.44 + 2.82}2 = 3.13$.
In your example, $\ce{C2H5OH}$ ethanol the bond is with $\ce{C}$ ( electronegativity 2.55)  so the difference is 0.58 hence covalent. If the bond is with $\ce{Ca}$ (1.00) the difference would be 2.12 which is greater than 1.7 hence is categorized as ionic.
Of course is not always easy to determine if it's ionic or not so in complex cases we can say that it has an ionic character o covalent character.


Why isn't dry CaCl₂ used to dry HCl gas instead of concentrated sulfuric acid? In our chemistry books, it is said that concentrated sulphuric acid is used to dry $\ce{HCl}$ gas.

But, concentrated sulphuric acid is a strong oxidising agent so, why isn't $\ce{HCl}$ gas oxidised to $\ce{H2O}$ and chlorine gas?

Why dry $\ce{CaCl2}$ (Calcium Chloride) isn't used as drying agent?

[Note: $\ce{CaCl2}$ have no reaction with $\ce{HCl}$ as they both have chloride ion in common]

There are several reasons:

Efficiency
When a stream of wet hydrogen chloride is run through a tube filled with coarse calcium chloride "rocks", interaction with the drying agent takes place at at its surface.
When the same wet $\ce{HCl}$ ist passed through sulfuric acid, the contact area is given by the surface area of the bubbles. Moreover, bubbles mean mixing. The trapped water is distributed over the drying agent.
Safety
Wet calcium chloride will clog, which might result in a pressure buildup in the drying tube.


Are halogens a further classification of non-metals, or are they another group themselves? I've searched many pages in the internet, and many just classify halogens
(Grp 7A) as non-metals. However, my teacher classified them as another group in the periodic table, and when asked which groups are classified under non-metals, she written the correct answer to be group 3A-6A. I'm very confused as I thought non-metals include halogens?

Thanks in advance!!:)

Halogens are merely just a subset of non-metals. Consider this periodic table:

Clearly halogens do come under the fluoroscent colored elements, which signifies non-metals. There are several other groups, which have their own unique name. For instance, the oxygen group members are called chalcogens, and the nitrogen group members are called pnictogens.

However, there is something special about halogens though. They are infact the most non-metallic elements known, considering their great tendency to gain electrons. This also makes the halogen group the only group other than inert gases to have only non metals in them (ignoring Ununseptium of course).

In short, halogens are a subset of non-metals and not an entire class by themselves. But they are, the most important subset of non metals.


Interpreting a Formula Tattoo (Bipolar?) Found This Online. Something to do with bipolar disorder?

This is lamotrigine, a sodium channel blocker prescribed against epilepsy.  Apparently it's also prescribed for bipolar disorder.


Converting fractional coordinates into cartesian coordinates for crystallography So let's say for a given molecule I have the volume of a unit cell, the shape of the unit cell (monoclinic, cubic, etc), the cell angles alpha, beta, and gamma, the cell lengths a,b, and c, and a list of atoms with their corresponding fractional units. How do I convert these fractional coordinates of the atoms into cartesian coordinates for any given cell shape? Orthogonal or not?

You have to use a matrix to convert. This is derived in most textbooks on crystallography, such as McKie & McKie 'Essentials of Crystallography'.
The matrix is
$$ M=\begin{bmatrix} a & 0 & \\b\cos(\gamma)& b\sin(\gamma) & 0\\c\cos(\beta) & cn_2 & c\sqrt{\sin^2(\beta)-n_2^2} \end{bmatrix}$$
where $$n_2=\frac{\cos(\alpha)-\cos(\gamma)\cos(\beta)}{\sin(\gamma)}$$
and $a,b,c $ are the unit cell dimensions and $\alpha,\beta,\gamma$ the angles in radians. The determinant of $M$ is the cell volume.
If $V_1$ is the vector of fractional cell coordinates $x/a,y/b,z/c$ etc. the matrix multiplication calculation $\hat d_{12}=(V_1-V_2)\cdot M $ and then $L_{12}=\sqrt{ \hat d_{12}\cdot \hat d_{12}}$ gives the bond length atom 1 to 2. The angle $\theta$ between two bonds 12 and 23 is given by a dot product $\cos(\theta)=\hat d_{12}\cdot \hat d_{23}/(L_{12}L_{23})$
An example calculation in python is shown below for a triclinic crystal. The @ is matrix multiplication.
import numpy as np
a = 7.55                # cell parameters
b = 4.99
c = 12.50
alpha = 122.5*np.pi/180
beta  = (95+18/60)*np.pi/180
gama  = (118+54/60)*np.pi/180

V1 = np.array( [-0.2812 , -0.0628 , 0.1928 ] ) # x/a, y/a, z/a
V2 = np.array( [-0.2308 , -0.0972 , 0.2931 ] )
V3 = np.array( [-0.3639 , -0.1913 , 0.3521 ] )

n2 = (np.cos(alpha)-np.cos(gama)*np.cos(beta))/np.sin(gama)
M  = np.array([[a,0,0],[b*np.cos(gama),b*np.sin(gama),0],
     [c*np.cos(beta),c*n2,c*np.sqrt(np.sin(beta)**2-n2**2)]])

dcm1 = (V1-V2) @ M             # row x matrix
L12  =  np.sqrt(dcm1 @ dcm1)   # sqrt(dot product)
L12


Why can't the conjugate base of benzoic acid be stabilized by conjugation with the aromatic ring? All the carbon and oxygen atoms are sp2 hybridised and can have a p orbital in the correct plane yet only conjugation of the COO- group occurs. I can't think why. I can't draw resonance structures for the negative charge being delocalised around the ring but if you look at it from the orbital perspective (sp2 hybridisation) it seems clear that this conjugation should occur.

Here are some of the resonance structures we can draw for the carboxylate anion of benzoic acid.

You're correct that we can't delocalize the negative charge from the oxygen into the benzene ring.  Even though there is continuous p-orbital overlap between the oxygen and benzene carbons, there is just no way to push the electrons so that a negative charge winds up in the benzene ring. However, as the resonance structures show, we can move electrons the other way and place a second negative charge on the oxygens and leave a positive charge delocalized around the ortho and para positions in the benzene ring.  This is why the carboxylate anion of benzoic acid will not react with electrophiles (electrophilic aromatic substitution) at those positions in the aromatic ring.


Vacuum at different altitude I have tried to find this searching the internet and I am not sure if I understand it. I want to turn water from solid to gas without going into the liquid phase - Sublimation.

The altitude above sea level is 1500 meter about 5000 feet. I have small 1 stage vacuum pump and a cheap vacuum gauge. The pump is rated to 5Pa (at sea level). Atmospheric pressure is around 85kPa. The vacuum gauge reads 20 inHg but what is that in Pascal, if I convert it using Google or the calculator it give me a value of 67kPa.

Two questions:
1. Can sublimation occur at an altitude of 1500 meter above sea level and atmospheric pressure of around 85kPa

How does one convert vacuum inHg into pascal

I am not chemist nor a chemist student and just like experimenting with stuff

Because you are manipulating the pressure using a vacuum pump, the altitude change should not matter. If you look at the phase diagram for water in the attached link, you can see that if you keep your system below 0C, say at -10C, than you will cross the solid-vapor equilibrium line and your ice will start to vaporize when the vacuum pump decreases the pressure to ~200 pa.

https://en.wikipedia.org/wiki/Phase_diagram#/media/File:Phase_diagram_of_water.svg

To convert from pa to in Hg, I suggest searching for the conversion factor. From google: 1 in Hg = 3386.39 pa. So if you have a reading of say, 2 in Hg on your gauge then: $ 2 inHg*(3386.9 pa)/(1 in Hg) = 6772.78 pa $


What are the units of 85000 Molecular Weight of a polymer I have a PVA bottle, which says that the molecular weight (Mw) of PVA is between 85000 - 124000.

I know that this means that there are repeated chains of the PVA monomer, but what are the units to the molecular weight (of PVA). Is it $\frac{\mathrm{g}}{\mathrm{mol}}$? I am having trouble grasping this concept. Or does 85000 - 124000 mean the number of monomers in the chain of this PVA?

First of all $M_w$ is not the number of monomers in a chain of PVA. The number of monomers in a chain is called the degree of polymerization.
$\overline {M_w}$ is the weight average molecular weight. It is a way of determining the molecular mass of a polymer. As polymer molecules have different chain lengths, the average molecular mass will depend on the method of averaging. For example, we can use the number average molecular weight which is based on different type of averaging.

As for the unit, it's as you mentioned: g/mol. In biochemistry, they use the dalton (Da) as a mass unit for biopolymers (1 $N_a.$Da= g/mol, where $N_a.$ is the Avogadro number). So, when we say a protein has a molecular weight of 64000 g·mol$^{−1}$, it has  also a mass of 64 kDa.

P.S. To be rigorous, we have to say ** molecular mass** instead of ** molecular weight**. But this was an old way of naming things.


Why there is a temperature change in adiabatic process? I am learning the topic thermodynamics and I came to know that in an adiabatic process the heat of the system remains constant. So is it a hard and fast rule that temperature will  change in an adiabatic process?

Remember adiabatic process implies process in which no heat exchange takes place between the system and the surrounding i.e. neither heat is supplied nor heat is loosed.

Temperature of gases changes only when internal energy of the gases changes.

Net heat of the system constant  means dq = 0

dq= du+dw, to maintain dq = 0, you can have many possible set of values of du and dw. Like whatever amount work you do, say dw, to maintain addition of du and dw always as zero, du = -dw. So any amount of work you do , du becomes negative of that work you perform to maintain net heat of the system as constant (i.e. dq as zero).
So even while maintaining addition of du and dw as zero, you can have infinite possible values of du and dw ( i.e. do any aribitary work and du becomes the negative of the work you performed and hence addition becomes always zero)

Temperature of the gas is related to du by du = nCvΔT. Hence for all those infinite possible values of du, you will have infinite possible temperatures, still maintaining constant net heat of the system .i.e dq as zero.

So in an adiabatic process, if you performed work, then du has to change to a value which is negative of the amount of work you performed (otherwise you won't be able to maintain their addition to be zero) , and as du has to change, hence temperature also changes.

So every time you perform work in an adiabatic container, temperature  has to change.


what contributes to the temperature of a vacuum what is the temperature of Vacuum since temperature of a system is related to the average of the molecular kinetic energy and there is no molecule in a vacuum? i know there could be radiations but i want to know how much the contribution is?

It really depends on how the vacuum is defined. If you, for instance, pump all the air out of a steel container, you will have a vacuum in there. Yet, there will be photons constantly radiated off of the walls. This interplay of photons will be in thermal equilibrium with the walls and therefore will have a contribution to the temperature. For instance, if you would place a thermometer inside this container, the material of the thermometer would be hit by the photons of the container wall and therefore the thermometer would give a temperature equal to the temperature of the wall. As a matter of fact, even in the deepest of deep space, there is some radiation to be found which gives a temperature of about 3K. (This radiation is left over from the Big Bang)

Nevertheless, if you make a true vacuum inside some container of which the walls are at 0 K that cannot be penetrated by any kind of radiation or matter, then the temperature therein is 0 K.


What are analytical expressions for the isothermal change of Gibbs energy? The change of Gibbs energy at constant temperature and species numbers, $\Delta G$, is given by an integral $\int_{p_1}^{p_2}V\,{\mathrm d}p$. For the ideal gas law, $p\,V=n\,RT$, this comes down to
$$\int_{p_1}^{p_2}\frac{1}{p}\,{\mathrm d}p=\ln\frac{p_2}{p_1}.$$
We find this logarithm in many formulas in chemistry.

I tried and failed to find a general formula for $\Delta G$ for the Van der Waals gas (see this question on the physics board).

It makes me wonder:
Is there even another case of a chemical situation, other than the ideal gas, where $\int_{p_1}^{p_2}V(p)\,{\mathrm d}p$ or $\int_{V_1}^{V_2}p(V)\,{\mathrm d}V$ is known analytically?

My questions are mostly motivated by the desire to understand the functional dependencies of the chemical potential $\mu(T)$, that is essentially given by the Gibbs energy. For the ideal gas and any constant $c$, we see that a state change from e.g. the pressure $c\,p_1$ to another pressure $c\,p_2$ doesn't actually affect the Gibbs energy. The constant factors out in $\frac{1}{p}\,{\mathrm d}p$, resp. $\ln\frac{p_2}{p_1}$. However, this is a mere feature of the gas law with $V\propto \frac{1}{p}$, i.e. it likely comes from the ideal gas law being a model of particles without interaction with each other.

With regard to the Van der Waals equation, have you considered integrating by parts:  $V\mathrm dp=\mathrm d(pV)-p\mathrm dV$.  That can certainly be integrated analytically.

The answer will be in terms of $V$, but that’s not a big limitation.


Finding [A] when looking for the velocity constant Some reaction of first order on $[A]$ has the data:

  What is the velocity constant for this reaction? The units are
  $s^{-1}$.

Well, since the units are $s^{-1}$, then the reaction must be of global order one. Meaning that $A$ is the only reagent, yes?

Anyway, the velocity equation is

$$V = k \cdot [A]^1$$

Let's pick the first experiment...

$$\left (-\frac{0.40 - 1.60}{10} \right ) = k \cdot [A]$$

Huh. I don't actually know how to find $[A]$. Should I add $1.6 + 0.40$? Subtract them maybe? Or pick just one?

I tried all three of them and still couldn't get the answer. The options were

$0.030$
$3.1x10^{-3}$
$0.013$
$3.0$
$0.14$

For a first-order reaction, the half-life is constant and is given by the equation

$$t_{1/2} = \frac{\ln 2}{k}$$

In $10~\mathrm{s}$, two half-lives have passed (the concentration of $\ce{A}$ drops to a quarter), so $t_{1/2} = 5~\mathrm{s}$. Accordingly,

$$\begin{align}
k &= \frac{\ln 2}{5~\mathrm{s}} \\
&= 0.1386~\mathrm{s^{-1}}
\end{align}$$


Does a reaction have to have a rate determining step? I am a bit confused about the concept of the rate determining step. From what I understand, a step in a reaction is the RDS if it meets the following requirements:

It is the slowest step in a reaction
It causes a bottleneck in the progress of the reaction

Therefore, if there is a catalyst, it means the slowest step in the reaction will no longer be the slowest step because the reaction can take place faster with a different mechanism involving the catalyst. Would it therefore be of any significance to say that a particular step in this reaction, say for example the step with the highest activation energy is the RDS, or is there no point in saying there is a rate determining step, ie, that it does not exist? Is it possible to say that the concept of the RDS has no significance in some reactions?

Any kind of chemical reaction can be modelled as a (complex) network of elementary reaction steps. For instance, if we model the conversion of R to P via the intermediates $\ce{I^1}$, $\ce{I^2}$, and $\ce{I^3}$, we have the following kinetic network:

$$\ce{R <=> I^1 <=> I^2 <=> I^3 <=> P}$$

Note that on the basis of microscopic reversibility, all these reactions can go both ways and the boundary conditions of the system (pressure, temperature and chemical composition) determine in which way and at what rate the reactions occur.

To answer your question: there exists a formal way of identifying the rate-determining step. If we look at the overall reaction rate $r$; then we can define a concept called degree of rate control:

$$
\chi_{i} = \left( \frac{\partial \ln r}{\partial \ln k} \right)_{k_{i \ne j}, K_{i}}
$$

Here, $\chi_{i}$ is the degree of rate control coefficient that indicates the extend to which an elementary reaction steps controls the overall rate and $k$ is the rate constant of the particular elementary reaction step you want to calculate the DRC coefficient for. Importantly, in this calculation you keep the rate constants for all other reaction steps and for the equilibrium constant of the elementary reaction step you are looking at constant. (in other words, you only change the height of the reaction barrier). When you find that this coefficient is 1 and all other elementary reaction steps have a coefficient of 0; in that case this elementary reaction step is the rate-determining step.

Interestingly, when investigating a lot of chemical systems, it turns out that only roughly half of these systems have a rate-determining step. A lot of the systems (especially in catalysis) have multiple rate-controlling steps and even rate-inhibiting steps (where the DRC coefficient is less than zero).

I will not go too much in-depth, but there exists a complete mathematical treatment and derivation of this theory that shows that the sum of all coefficients has to be unity. This means that a chemical system should always have either a single rate controlling step (the RDS) or multiple rate-controlling steps and possible one or more rate-inhibiting steps.

To conclude: Not every overall chemical reaction has an RDS, but they should have one or more rate-determining steps.

Notable references:
The above theory was proposed by Campbell and co-authors in this research paper: J. Am. Chem. Soc. 2009, 131 (23), 8077–8082. DOI:10.1021/ja9000097.


Advanced Level Chemistry Book Recommendation I have a test comming up on Chemistry in Advanced Level and I sincerly ask for a book recommendation for chemistry. I already got the book chemistry by Ramond Chang. And I am pretty near to the exam. What I really want is a book with Lesson and Questions related to that specific lesson book.

Here are a few recommendations for A-level chemistry, made by fellow students such as yourself. TSR is a good student resource for A-level prep since there is are masses of students talking about the same topics.

(Zumdahl is a good general chemistry textbook as well, with quite the number of practice problems. However it is a full-fledged textbook you probably don't need a few days before an exam.)


Product of oxidation of alkanes I'm trying to find the product of the oxidation of iso-butane using KMnO4, i know that alkanes undergo step wise oxidation with strong oxidising agents to finally give carboxylic acids.. With straight chain carbons, its very straightforward - ethane becomes acetic acid etc. But with branched chain alkanes I don't know how to proceed. Can someone tell me if there's a general rule for oxidation of alkanes- to get the product?

Potassium permanganate, $\ce{KMnO4}$, is the most common and probably the most applicable oxidizing agent discussed in organic chemistry teaching classes. $\ce{KMnO4}$ is a very strong oxidizing agent, and is capable of oxidizing a wide range of organic molecules. Usually, it will oxidize organic compounds to appropriate carboxylic acids, but, the final products can vary on the conditions used. If you googled enough to find reduction potentials of $\ce{KMnO4}$, you may find following conditions, which are the most used reduction conditions of $\ce{KMnO4}$ (Wikipedia):
\begin{align}
\ce{MnO4- + H+ + e- &-> HMnO4^-} & \mathrm{E^o} &= \pu{0.90V}\tag{1}\label{1}\\
\ce{MnO4- + 8H+ + 5e- &-> Mn^2+ + 4H2O} & \mathrm{E^o} &= \pu{1.51V}\tag{2}\label{2}\\ \ce{MnO4- + 2H2O + 3e- &-> MnO2(s) + 4OH-} & \mathrm{E^o} &= \pu{1.69V}\tag{3}\label{3}\\
\ce{MnO4- + 4H+ + 3e- &-> MnO2(s) + 2H2O} & \mathrm{E^o} &=\pu{1.7V}\tag{4}\label{4}
\end{align}
Other manganese containing compounds other than $\ce{KMnO4}$, which can also be used as oxidizing reagents are listed below:
\begin{align}
\ce{MnO2(s) + 4H+ + e- &-> Mn^3+ + 2H2O} & \mathrm{E^o} &= \pu{0.95V}\tag{5}\label{5}\\
\ce{MnO2(s) + 4H+ + 2e- &-> Mn^2+ + 2H2O} & \mathrm{E^o} &= \pu{1.23V}\tag{6}\label{6}\\
\ce{HMnO4- + 3H+ + 2e- &-> MnO2(s) + 2H2O} & \mathrm{E^o} &= \pu{2.09V}\tag{7}\label{7}\\
\end{align}

Yet, as most of the viewers already commented, I also believe $\ce{KMnO4}$ does not oxidize alkanes, except in following situations:

$\ce{KMnO4}$ oxidizes alkane carbon atoms, which contain sufficiently weak $\ce{C-H}$ bonds such as those in benzylic positions ($\ce{C-H}$ bonds in the $\alpha$-positions of substituted aromatic rings). These oxidizable alkyl benzene substituents can be -$\ce{CH3}$, or any $1^\mathrm{o}$-alkyl groups such as -$\ce{CH2CH3}$, or $2^\mathrm{o}$-alkyl groups such as -$\ce{CH(CH3)2}$. All of them will be oxidized to carboxylic acid group, -$\ce{CO2H}$ (See Scheme 1 below):

It is also important to note that when the benzylic carbon is a $3^\mathrm{o}$-$\ce{C}$ (e.g., corresponding alkyl groups such as $-\ce{C(CH3)3}$ group in tert-butylbenzene, see the example given in RHS of Scheme 1), the oxidation reaction does not happen.

In addition to benzylic carbons, $\ce{KMnO4}$ also oxidizes weak $\ce{C-C}$ bonds such as $\pi$-bonds in alkenes and alkynes (Scheme 2):

$\ce{KMnO4}$ also oxidizes $\sigma~\ce{C-C}$ bonds such as that in glycol-moiety (e.g., $\ce{HO-C-C-OH}$) and $\ce{C^{\mathrm{sp^2}}-O}$ bonds such as those in phenolic compounds (e.g., it oxidizes phenol to 1,4-benzoquinone, commonly known as para-quinone).

At last, as mentioned above, conditions of the reaction can be adjusted such a way that desired compound can be achied in high yields. For example, see following example (Ref.1):

Note that no benzylic carbon or -$\ce{OCH2}$- was oxidized here.

Late Addition:

Although tertiary $\ce{C-H}$ bond in isobutane (preferred IUPAC name: 2-methylpropane; $\ce{CH3CH(CH3)CH3}$) is relatively weaker than othe alkane $\ce{C-H}$ bonds, I’m highly skeptical of the statement of “Oxidizing agents such as potassium permanganate readily oxidize a tertiary hydrogen atom to a hydroxyl group, e.g., isobutane is oxidized to t-butanol.” (vide infra). To my knowledge, the literature evidence for that type of oxidation reactions does not exist. Nevertheless, 2-methylpropane (isobutane) can be oxidized to 2-methylprop-2-ol (t-butanol) by reacting supercritical-phase isobutane with air in the presence of a catalyst such as $\ce{Pd/C}$ at high temperature, Instead of using $\ce{KMnO4}$ (Ref.2):

  Abstract: tert-2-Butyl alcohol can be synthesized efficiently on $\ce{SiO2 –TiO2}$ or $\ce{Pd–C}$ catalyst if air is directly introduced to the supercritical-phase isobutane.

tert-Butanol (TBA) is the initial material in methyl tert-butyl ether (MTBE) production and methyl methacrylate (MMA) synthesis.  MTBE is high-octane-number gasoline additive and MMA is a resin material for plastic synthesis (e.g., PMMA or poly-methyl methacrylate). Thus, one would have wondered why scientists did not attempted to synthesize TBA in a such an easy way like using $\ce{KMnO4}$.

However, in a different note, almost exclusive $\ce{C-H}$ bond activation has been observed in the reaction of linear and branched alkanes with $\ce{CeO2+}$ (Ref.3)

Reference:

A. Abiko, J. C Roberts, T. Takemasa, S. Masamune, “$\ce{KMnO4}$ revisited: Oxidation of aldehydes to carboxylic acids in the tert-butyl alcohol - aqueous $\ce{NaH2PO4}$ system,” Tetrahedron Letters 1986, 27(38), 4537-4540 (https://doi.org/10.1016/S0040-4039(00)84997-7)
L. Fan, Y. Nakayama, K. Fujimoto, “Air oxidation of supercritical phase isobutane to tert-butyl alcohol,”  Chem. Commun. 1997, (13), 1179-1180 (DOI:10.1039/A702308A).
A. Gunay, K. H. Theopold, “$\ce{C-H}$ Bond Activations by Metal Oxo Compounds,”  Chem. Rev. 2010, 110(2), 1060–1081 (DOI: 10.1021/cr900269x).


Which is a better oxidising agent: concentrated nitric acid or dilute nitric acid? I was reading the Wikipedia page on nitric acid and saw the reaction of concentrated $\ce{HNO_3}$ and dilute $\ce{HNO_3}$ with $\ce{Cu}$.
It is mentioned that:

Copper reacts with dilute nitric acid at ambient temperatures with a 3:8 stoichiometry:
$$\quad\qquad\qquad\qquad\ce{3 Cu + 8 HNO_3 -> 3 Cu^{2+} + 2 NO + 4 H_2O + 6 NO_3^−}$$
With more concentrated nitric acid, nitrogen dioxide is produced directly in a reaction with 1:4 stoichiometry:
$$\ce{Cu + 4 H^+ + 2 NO_3^− -> Cu^{2+} + 2 NO_2 + 2 H_2O}$$

My question is: If concentrated nitric acid is a better oxidising agent, then why does it oxidises only 2 moles of $\ce{Cu}$ per 8 moles of nitric acid against the 3:8 ratio in dilute one. Why is there change of -1 in oxidation state of $\ce{N}$ in concentrated solution as opposed to -3 in dilute solution?
Does this indicate dilute $\ce{HNO3}$ is better oxidising agent than concentrated one?

I think your assignments for the equations need to be inspected closely. Let's look at those equation with their redox potentials:

In dilute conditions (say, ratio of $\ce{HNO3}$ to $\ce{Cu}$ is 8:3):
$$
\begin{align}
\ce{NO3-(aq) + 4H3O+(aq) + 3e- &<=> NO(g) + 6H2O(l)} &E^\circ = \pu{0.957 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &E^\circ = \pu{-0.342 V}\\
\ce{2NO3-(aq) + 8H3O+(aq) + 3Cu (s) &-> 3Cu^2+(aq) + 2NO(g) + 12H2O(l)}&E^\circ_{\text{rxn}} = \pu{0.615 V}
\end{align}$$

In concentration conditions (say, ratio of $\ce{HNO3}$ to $\ce{Cu}$ is 4:1):
$$
\begin{align}
\ce{2NO3-(aq) + 4H3O+(aq) + 2e- &<=> N2O4(g) + 6H2O(l)} &&E^\circ = \pu{0.803 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &&E^\circ = \pu{-0.342 V}\\
\ce{2NO3-(aq) + 4H3O+(aq) + Cu (s) &-> Cu^2+(aq) + N2O4(g) + 6H2O(l)}&&E^\circ_{\text{rxn}} = \pu{0.461 V}
\end{align}$$

Because  this concentrated reaction is exotermic, the colorless $\ce{N2O4}$, gas decomposes to reddish brown $\ce{NO2}$ gas according to $\ce{N2O4 (g) <=> 2NO2 (g)},~\Delta H^\circ = \pu{+57.2 kJ mol^{-1}}$

Based on $E^\circ_{\text{rxn}}$ values, one can assume that the concentrated reaction is less spontaneous. However, some dissolved $\ce{N2O4}$ can still undergo following redox reaction which is the most spontaneous reaction among three:

$$
\begin{align}
\ce{N2O4(aq) + 4H3O+(aq) + 4e- &<=> 2NO(g) + 6H2O(l)} &E^\circ = \pu{1.035 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &E^\circ = \pu{-0.342 V}\\
\ce{N2O4(aq) + 4H3O+(aq) + 2Cu(s) &-> 2Cu^2+(aq) + 2NO(g) + 6H2O(l)}&E^\circ_{\text{rxn}} = \pu{0.693 V}
\end{align}$$

If we can consider the loss of $\ce{N2O4}$ to $\ce{NO2}$ is negligible, then total oxidation of $\ce{Cu}$ with both conc. $\ce{HNO3}$ and $\ce{N2O4}$ can be written as:
$$
\begin{align}
\ce{2NO3-(aq) + 8H3O+(aq) + 3Cu (s) &-> 3Cu^2+(aq) + 2NO(g) + 12H2O(l)}&E^\circ_{\text{rxn}} = \pu{1.154 V}
\end{align}$$

Although it is identical reaction to that in dilute conditions, the reaction proceeds much more favorably (compare $E^\circ_{\text{rxn}}$ values in both conditions).

Late edition:

I have tried to avoid criticizing literature values on this subject, but my answer so far arose questions from concerned people. Thus, I'd try to do my best to handle it as delicate as possible.

To my knowledge, there shouldn't be a question about whether conc. $\ce{HNO3}$ is better oxidizing agent than dilute $\ce{HNO3}$. The oxidizing reagent here is the $\ce{NO3-}$ ion, as my equations described above. For example, any $\ce{NO3-}$ ion (say $\ce{KNO3}$) will oxidize $\ce{Cu}$ in an acid medium (say, $\ce{HCl}$). In using dilute or concentrated $\ce{HNO3}$, we conveniently provide both $\ce{NO3-}$ ions and $\ce{H3O+}$ for our redox reaction. This reaction is highly based on reaction conditions. I think nobody knows exact process, but everybody speculate by the results. That's the reason you are seeing too many different equations in different sources. I'm using the comprehensive table in: http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/electrochemical%20series.pdf, which I believe more realistic. Now, consider following 2 questions:

  The question 1: If concentrated nitric acid is a better oxidizing agent, then why does it oxidizes only 2 moles of Cu per 8 moles of nitric acid against the 3:8 ratio in dilute one?

Answer: Conc. $\ce{HNO3}$ has more $\ce{NO3-}$ available to react. When available, two moles of it react with 1 mole of $\ce{Cu}$ to give more stable product, $\ce{N2O4}$ ($\Delta H^\circ_f = \pu{+9.16 kJ/mol}$) compared to $\ce{NO}$ ($\Delta H^\circ_f = \pu{+90.25 kJ/mol}$).

  The question 2: Why is there change of -1 in oxidation state of $\ce{N}$ in concentrated solution as opposed to -3 in dilute solution?

Answer: Look at the last series of reactions involving $\ce{N2O4}$ given in my original answer. Eventual change of oxidation state of $\ce{N}$ is same in both cases. This reaction is not just a matter of power. I don't think anybody studied this process in details yet. I'd like to make my point by giving some known redox processes of of $\ce{NO3-}$.

$$
\begin{align}
\ce{NO3-(aq) + 3H3O+(aq) + 2e- &<=> HNO2(aq) + 4H2O(l)} &E^\circ = \pu{0.934 V}\\
\ce{NO3-(aq) + 4H3O+(aq) + 3e- &<=> NO(g) + 6H2O(l)}&E^\circ = \pu{0.957 V}
\end{align}$$

Both processes are under relatively similar conditions (say, $\pu{2M}$ vs $\pu{3M}~\ce{HNO3}$). In process (1), the oxidation state of $\ce{N}$ changes from +5 to +3 while in (2), it was +5 to +2. Which process dominate when they were subjected to oxidize appropriate amount of $\ce{Cu}$?


Pressure at which graphite and diamond are in equilibrium I came across this problem :

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.
The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $\pu{2e-6 m3 mol-1}$.
If C(graphite) is converted to C(diamond) isothermally at $\pu{T = 298 K}$, the pressure at which C(graphite) is in equilibrium with C(diamond), is
(A) $\pu{14501 bar}$
(B) $\pu{58001 bar}$
(C) $\pu{1450 bar}$
(D) $\pu{29001 bar}$

I applied : $$\Delta G_{(P,T)} =\Delta_f G^{\circ}+\int_{p_1}^{p_2}Vdp$$.
Since the system is at equilibrium,$$ \Delta_f G^{\circ}= -\int_{p_1}^{p_2}Vdp$$
Now I am stuck. I have not been given any relation between Pressure and Volume. Is there any assumption i have to make to solve this integral ?

For each phase (graphite or diamond) you can write

$$ \mu = \mu^\circ+\int_{P^\circ}^{P} V_mdP $$
at constant T.

We are asked to find the pressure $P=P_{eq}$ at which carbon coexists in the two phases (the Gibbs free energy is equal in both) so that

$$ \mu(diamond) = \mu(graphite)$$

This leads to

$$\Delta \mu^{\circ}=-\int_{P^\circ}^{P_{eq}}\Delta V_mdP$$

and, since $\Delta \mu^{\circ}= \Delta_f G_m^{\circ}$, ultimately to the expression you provided:

$$\Delta_f G_m^{\circ}=-\int_{P^\circ}^{P_{eq}}\Delta V_mdP$$

The approximation you are allowed to make at this point is that the solids are incompressible, such that their $V_m$ are constant with change of pressure. It follows that

$$\Delta_f G_m^{\circ}=-\Delta V_m\int_{P^\circ}^{P_{eq}}dP=-\Delta V_m(P_{eq}-P^\circ)$$

which can be solved for $P_{eq}$:

$$P_{eq}=-\frac{\Delta_f G_m^{\circ}}{\Delta V_m}+P^\circ$$


What is non PV work? How is it related to Gibbs energy? Recently, I was trying to understand more about Gibbs free energy and got stuck at this point. Some books write that if $\Delta G=0$, then the reaction is reversible while some other sources say that a reaction is reversible when  $\Delta G$ = Non-$PV$ work.
I am not able clarify things here and am unable to understand the core concept of non $PV$ work.

At constant temperature and pressure, the maximum non expansion work is given by the change in the Gibbs Free Energy.

$$dW_{\text{max}}= dG$$

Proof:

At constant temperature and pressure,

from the definition of enthalpy: $H= U+pV$
first law of thermodynamics: $dU = dq+dw$

$\implies dH = dq+ dw + d(pV)$

Now,

$dG = dH - TdS -SdT \\\implies dG = dq + dw + d(pV)- TdS - SdT$

For isothermal process $dT =0$

$\implies dG = dq+dw+ d(pV)+ -TdS$

When the change is reversible,

$dq = dq_{\text{rev}}= TdS$ and $dw= dw_{\text{rev}}$

$\implies dG = dw_{\text{rev}}+d(pV)$

The work consists of expansion work, which for a reversible change is given by
−pdV, and possibly some other kind of work (for instance, the electrical work of
pushing electrons through a circuit or of raising a column of liquid); let's denote it by $dw_{\text{add},\text{rev}}$

$\implies dG= dw_{\text{add}, \text{rev}}+ Vdp$

For isobaric process, $dp =0$

$\implies dG = dw_{\text{add}, \text{rev}}$

You must be aware the the maximum work is obtained when the process is reversible, if not you may refer to this answer.

$\implies dG = dw_{\text{add}, \text{max}}$

  Some books write that if $\Delta G=0$, then the reaction is reversible while some other sources say that a reaction is reversible when  $\Delta G$ = Non-$PV$ work.

$\Delta S_{\text{total}}= \dfrac{-\Delta G_{\text{system}}}{T}$

Thus, for a reversible process $\Delta G_{\text{sys}}= 0$

And I have also proved to you above that $\Delta G_{\text{sys}}=$ maximum non expansion (non PV) work.

  I am not able clarify things here and am unable to understand the core concept of non $PV$ work.

PV work is only the work done during the change in volume. There can be different types of works that a system can do during a process, for instance: electrical work, work against gravity, work against friction, magnetic work, shaft work, etc. This is the concept of non expansion or non PV work.

Source:

Physical Chemistry by Atkins (8th ed. )
Wikipedia article on work in thermodynamics


NGP in Finkelstein reaction The question is to find the product in the following reaction-

In the answer given, they have replaced Cl with I and inverted the configuration at that carbon(Finkelstein reaction).
But $\text{I}^-$ being a good base, wouldn't it first take away O's acidic H and then $\text{O}^-$ will show Neighboring Group Participation attacking the carbon with chlorine and finally resulting in retention of configuration?

You are thinking too far ahead for a simple reaction. First of all, $\ce{I-}$ is not a very strong base to abstract a protone from alcohol. $\ce{I-}$ is the conjugate base of very strong acid, $\ce{HI}$. Yet, $\ce{I-}$ is a very good nucleophile. Therefore, the condition given ($\ce{NaI}$ in anhydrous acetone; $\ce{NaI}$ dissolves in anhydrous acetone and therefore exists as $\ce{I-}$ in solution) is suitable only for nucleophilic substitution reaction and also favored $\mathrm{S_N2}$ mechanism. Since the byproduct form ($\ce{NaCl}$) is not soluble in acetone, the reaction favors only the forward reaction:
$$\ce{R-Cl + Na+ + I- -> R-I + NaCl_{(s)}}$$
Since it is a $\mathrm{S_N2}$ reaction, inversion is the result ($(1R,2S)$ to $(1R,2R)$ with closed to $100\% e.e.$).
However, note that there is a neighboring group participation (NGP) possible in 1,2-halohydrins in certain reactions (Ref.1). As OP suggested, if NGP is contributed here (without strong base I doubt it though), the product from NGP is going to be the retention product with $(1R,2S)$ configuration. Thus, total $\%e.e.$ of final product is $\lt 100\% e.e.$.
References:

T. A. Geissman, Richard I. Akawie, "Rearrangements of Halomagnesium Derivatives of Halohydrins," J. Am. Chem. Soc. 1951, 73(5), 1993–1998 (https://doi.org/10.1021/ja01149a029).


When you increase the temperature such that change in Gibb's free energy becomes zero, does Keq = 1 at that temperature? $$\Delta G = \Delta H - T \Delta S$$

say for example, both $\Delta H$ and $\Delta S$ are positive. So assuming $\Delta H$ and $\Delta S$ do not change or change minimally with temperature, if you increase the temperature until $\Delta G = 0$, is the equilibrium that you reach an equilibrium in which reactants and products are in standard conditions i.e. $K_\mathrm{eq} = 1$?

Yes and no.

$K = 1$ is true, and it follows directly from the equation $\Delta G^\circ = -RT \ln K$.

However, there is no reason for it to be "an equilibrium in which reactants and products are in standard conditions". $K = 1$ does not necessarily mean that the concentrations (or activities if you prefer) of all reactants and products are also equal to $1~\mathrm{mol~dm^{-3}}$ (or $1$, in the case of activities). Remember that $K$ is a fraction. For the reaction:

$$\nu_{r1}\ce{R1} + \nu_{r2}\ce{R2}\cdots \longrightarrow \nu_{p1}\ce{P1} + \nu_{p2}\ce{P2} \cdots$$

we have:

$$K = \frac{[\ce{P1}]^{\nu_{p1}}[\ce{P2}]^{\nu_{p2}}\cdots}{[\ce{R_1}]^{\nu_{r1}}[\ce{R_2}]^{\nu_{r2}}\cdots}$$

Just like how the fractions $\displaystyle \frac{2}{2}$, $\displaystyle \frac{4}{4}$, or $\displaystyle \frac{4000\hbar\pi e}{4000\hbar\pi e}$ are all equal to $1$, the individual concentrations $[\ce{P1}], [\ce{P2}], [\ce{R1}], [\ce{R2}]\cdots$ might not be equal to $1~\mathrm{mol~dm^{-3}}$, but the result will still be $1$.