dipole–dipole interactions: OCR Sample Paper wrong? I was doing OCR A Level Chemistry A, H432/01 Periodic table, elements and physical Sample Question Paper.

Question 6 was:

The boiling point of hydrogen bromide is –67 ºC.

The boiling point of hydrogen iodide is –34 ºC.

The different boiling points can be explained in terms of the strength of bonds or interactions.

Which bonds or interactions are responsible for the higher boiling point of hydrogen iodide?

I put answer C: permanent dipole–dipole interactions
?
The given answer was D: induced dipole–dipole interactions

[Revision post down-vote]

My learning to date is that:

dipole-dipole interactions could only be one or the other i.e. permanent or induced.
iodine is more electronegative than hydrogen, therefore the interactions are permanent.

Please would someone explain where how the induced interactions are caused?

Permanent Dipoles occur when two atoms bonded to each other have significantly different electronegativities. When the atomic number of an element increases the number of electrons present in them also increases. These electrons are constantly moving around in the atom. At some point, there are more electrons on one side of the atom than the other. This creates an imbalance of charge, creating a small charge of $\delta ^+$ and $\delta ^-$. These small charges induce its neighbors to create additional charges of $\delta ^+$ and $\delta ^-$, thus creating a chain of induced dipoles. These induced dipoles are temporary as the electrons can move, and hence can be reverted. They are, therefore, known as induced dipole forces, or better, known as London Dispersion Forces. In reality, they are not really forces, but rather attraction between molecules.

Naturally, more the number of electrons in the atom, more the amount of London dispersion forces, and hence greater the attraction. The greater the attraction, the more energy you have to put to break the bonds apart. Between the considered hydrogen halides ($\ce{HBr}$ and $\ce{HI}$), iodine has more electrons than bromine, and hence $\ce{HI}$ has a greater attraction force between molecules than $\ce{HBr}$. As a consequence, $\ce{HI}$ has a greater boiling point between them. Appropriate values are listed in following table:

$$\begin{array}{cccccc}
\hline
\ce{H-X} & \text{BP}^\text{Ref.1} & \ce{X}\text{'s EN} & \mu ^\text{Ref.2} & \text{Polarizability }(\pu{10^{-24} cm^3})  & (r_\ce{H-X})^\text{Ref.3} \\ \hline
\ce{HCl} & \pu{-85.1 ^\circ C} & 3.16 & \pu{1.09 D} & 2.63 & \pu{1.27 Å} \\
\ce{HBr} & \pu{-67.1 ^\circ C} & 2.96 & \pu{0.82 D} & 3.61 & \pu{1.42 Å} \\
\ce{HI} & \pu{-35.1 ^\circ C} & 2.66 & \pu{0.45 D} & 5.44 & \pu{1.60 Å} \\
\hline
\end{array}$$

Moreover, a related question you can find here. Orthocresol has given very reliable answer, which included two equations to calculate the strength of permanent dipole interactions ($U_\mathrm{dipole}$) and the strength of dispersion interactions ($U_\mathrm{dispersion}$). The author has stated that:

  If you plug all the numbers in for the hydrogen halides $\ce{HX}$, you will find that the magnitude of the dispersion forces is much larger than the magnitude of the dipole-dipole attractions, i.e. $\left|U_{\mathrm{dispersion}}\right| \gg \left|U_{\mathrm{dipole}}\right|$.

For example, for $\ce{HCl}$, dispersion forces contribute $86\%$ to the intermolecular attractions, and for $\ce{HI}$, they contribute $99\%$.

References:

J. F. Ogilvie, W. R. Rodwell, R. H. Tipping, “Dipole moment functions of the hydrogen halides,” J. Chem. Phys. 1980, 73(10), 5221 – 5229 (https://doi.org/10.1063/1.439950).
N. N. Greenwood, A. Earnshaw, “Chapter 17 – The Halogens: Fluorines, Chlorinee, Bromine, Iodine, and Astatine,” In Chemistry of the Elements; 2nd Edn.; Elsevier: Boston, MA, 1997, pp. 789–887.
K. Fajans, N. Bauer, “Electronic Structure and Stability of Hydrogen Halides and of Complex Ions $\ce{XO4}$,” J. Chem. Phys. 1942, 10(7), 410 – 415 (https://doi.org/10.1063/1.1723742).


Why is methane's molecular formula conventionally "CH₄", while water is "H₂O" (among others)? While revisiting some of my old notes about the Miller-Urey experiment, I stumbled across the "equation"...

  Electricity + $\ce{CH4~/~ NH3~/~H2O~/~CO}$ = Amino Acids

This got me thinking.

Conventionally, why are molecules like $\ce{CH4}$ and $\ce{NH3}$'s molecular formula written differently (in H placement) than others like $\ce{H2O}$ and $\ce{HF}$? Is there a particular reason, or did it just happen to be?

NOTE: I realize the $\ce{H4C}$ or $\ce{H3N}$ make perfect sense, especially when drawing out molecules. It's just that these are rarely used in literature.

The choice of whether to write binary hydrides with the hydrogen first or second depends on whether that hydrogen is considered acidic, with water marking the delineation point.

By convention, if a binary hydride is more acidic than water, then they are written in the form $\ce{H}_{\text{n}}\ce{X}$. If a binary hydride is less acidic than water, they are written as $\ce{YH}_{\text{n}}$. It so happens that this changeover neatly divides the periodic table. Binary hydrides from groups 16 and 17 (VIA and VIIA) are written with the hydrogen atoms first: $\ce{H2O,~H2S,~H2Se,~H2Te,~HF,~HCl,~HBr,~HI}$.

Binary hydrides from groups 1 (IA) through 15 (VIA) are all written with the hydrogen atoms last, regardless of whether the hydride is ionic $(\ce{LiH,~NaH,~KH,~CaH2})$ or molecular $(\ce{BH3,~CH4,~SiH4,~NH3,~PH3})$.

This notation is in keeping with that for more complex compounds. If the compound is an acid, the acidic hydrogen atoms are put at the from of the formula, while nonacidic hydrogen atoms are placed in the main part of the formula. For example, the hydrogen atoms in phosphoric acid $(\ce{H3PO4})$ are considered acidic and the hydrogen atoms in octane $(\ce{C8H18})$ are not. Acetic acid $(\ce{HC2H3O2})$ has one hydrogen atom that is considered acidic and three that are not, this the formula $\ce{HC2H3O2}$ is considered more helpful than $\ce{C2H4O2}$, although both are "correct" by different sets of rules.


What is the "rate factor" or "second-step rate constant" in the reaction rate equation? The reaction rate equation can be written as:

$$r=r_0 K \prod_n a_n$$

where the $a_n$s are the activities of each reactant in the reaction equation (in the case where we find the reaction rate from reactants to products, of course. Else the other way around). These activities are multiplied together. (This is often written as concentrations instead, I have seen, which seems to be an idealization.)

$K$ is the equillibrium constant, $K=\exp\left(\frac{-\Delta G}{RT}\right)$.

I am curious as to what exactly the $r_0$ is? In the wikipedia article it is denoted $k_2$ and called the "rate constant for the second step." The second step here refers to the intermediate step at the actived-complex.

So far so good, but what is this quantity at this intermediate step? Is it a reaction rate from the activated-complex to the final product (in that case, how can this be different from the reaction rate from the reactants to this state)?

The implication is that the reaction mechanism for any particular reaction, if written like this:

$$A + B → C$$

might not proceed as one step. Having a $k_2$ indicates that there is a second step to the mechanism, and potentially even several more. $k_{obs}$ is probably a better name for it, because sometimes you see it written as a simplified "one-step" process, such as in Michaelis-Menten Kinetics, when really, it might be $1$ more step, or it might even be $20$, depending on the reaction (I'm exaggerating a bit, but you get the point).

If we take Michaelis-Menten enzyme kinetics as an example:

$$E+S⇌ES→E+P$$

The first step is an equilibrium in which the concentration of the intermediate is assumed to not change because the intermediate is being produced about as fast as it is being consumed. This would be the Steady State Approximation. In other words:

$\frac{d[ES]}{dt} = 0 = k_1[E][S] - k_{-1}[ES] - k_2[ES]$

Eventually...

$[ES] = \frac{k_1}{k_{-1} + k_2}[E][S]$

The second step is where the $ES$ complex does something in such a way that the enzyme unbinds somehow, and the product is made... somehow. Depending on the reactants and products, the mechanism is different.

Anyways, the point is that $k_2$ describes this second step in this type of reaction (which is what you were wondering---why the "second step"?), hence the $2$.

In any case, $k_2$ corresponds to the reaction step(s) that proceed after an intermediate product is formed on the first step for a reaction mechanism.


Why does the opening of an epoxide occur via an Sn2 like mechanism when using a methanoate ion as a nucleophile and methanol as a solvent? Below is a picture of the mechanism:

I understand the mechanism but what I don't understand is why it's $\mathrm{S_N2}$. The solvent used is polar and protic, there is a tertiary carbon that would be suitable for $\mathrm{S_N1}$ as well. For what reason does this reaction occur exclusively via $\mathrm{S_N2}$?

Quotation from Clayden:

  Without protonation, the epoxide oxygen is a poor leaving group, and leaves only if ‘pushed’
  by a strong nucleophile: the reaction becomes pure SN2. Steric hindrance becomes the controlling
  factor and methoxide attacks only the primary end of the epoxide.


Orthogonality of wavefunctions for different normal modes Vibrational wavefunctions corresponding to different vibrational levels of the same normal mode are orthogonal because they are eigenfunctions of the harmonic oscillator Hamiltonian, which is hermitian.
Displacement vectors corresponding to different normal modes are orthogonal because they are eigenvectors of the Hessian matrix, which is symmetric.

QUESTION: given the above, how can one show that vibrational wavefunctions corresponding to different normal modes are orthogonal?

given the above, how can one show that vibrational wavefunctions
  corresponding to different normal modes are orthogonal?

The Hamiltonian for a system with two (or more) normal modes can be written as a tensor product of (Hermitian) Hamiltonians for each normal mode. For example for two modes (stretching and bending):

$$
H_{\rm{total}} = H_{\rm{stretching}}\otimes H_{\rm{bending}}
$$

The tensor product of two Hermitian matrices is also Hermitian.
Therefore the wavefunctions are orthogonal for the same reason that you figured out on your own for the single mode case.

More generally, all Hamiltonians (Hermitian operators) have orthogonal wavefunctions (eigenfunctions) for distinct energy levels (eigenvalues). So wavefunctions corresponding to different energy levels are orthogonal as a consequence of the Schroedinger equation which is based on a Hermitian object called the Hamiltonian, and is the basis for the existence of all distinct energy states (vibrational or not) in quantum chemistry.


How does sulfuric acid react to heating? How would $\ce{H2SO4}$ react to heat? I assume that it would form toxic sulfur (di/tri)oxide, but I can't find any information on this.
I'm concerned about fire hazards specifically, so assume very intense heating and very low concentration.

Sulfuric acid, $\ce{H2SO4}$, having an enthalpy of formation of $-814\ \mathrm{kJ/mol}$, is quite stable and won't decompose easily.
According to A Kinetic Study of the Decomposition of Spent Sulfuric Acids at High Temperature, Dominique Schwartz, Roger Gadiou, Jean-François Brilhac, Gilles Prado, and Ginès Martinez:

The decomposition of $\ce{H2SO4}$ to $\ce{H2O}$ and $\ce{SO3}$ is predominant between $400$ and $700\ \mathrm K$. The formation of a small amount of gaseous sulfuric acid can be observed. Above $673\ \mathrm K$, the equilibrium constant of the reaction R1 becomes higher than 1 and increases rapidly.
$$\ce{H2SO4 <=> H2O + SO3}\tag{R1}$$
The second process is the reduction of sulfur trioxide to $\ce{SO2}$. This endothermic reaction needs a high temperature to take place, the equilibrium constant of the reaction R2 being higher than 1 above $1050\ \mathrm K$.
$$\ce{SO3 <=> SO2 + \dfrac12O2}\tag{R2}$$

So, you only have to worry when the temperature reaches $400\ \mathrm K$, or $127\ \mathrm{^\circ C}$ (false precision).


Naming of a dispiroalkane compound Please help me in naming a dispiro compound:

I propose bispiro[3.0.4.1]decane. Is this correct?

All the special rules you need for naming of such dispiro cycloalkane systems can be found in one chapter of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-24.2.2 Linear polyspiro alicyclic ring systems
Polyspiro parent hydrides consisting of unbranched assemblies of three or more saturated cycloalkane rings are named using the nondetachable prefixes ‘dispiro’, ‘trispiro’, etc., according to the number of spiro atoms present, cited in front of the name of the acyclic hydrocarbon that has the same number of skeletal atoms. The von Baeyer spiro descriptor indicates the number of carbon atoms linking the spiro atoms by arabic numbers that are cited in order, starting at the smaller terminal ring if one is smaller, and proceeding consecutively, always by the shorter path, to the other terminal ring through each spiro atom and then back to the first spiro atom; the numbers are separated by full stops and enclosed in square brackets. The compound is numbered in the order in which the numbers of the von Baeyer spiro descriptor are cited, including spiro atoms when encountered for the first time. Each time a spiro atom is reached for the second time its locant, which has already been assigned, is cited as a superscript number to the number of the preceding linking atoms.
(…)

Following these rules step by step, you will find that the name for the compound that is given in the question is dispiro[3.0.45.14]decane.


Is equilibrium constant at a given temperature derived from Gibbs free energy of reaction valid for doing simple kinetic modelling? I intend to do a kinetic study of simple alcohol catalytic dehydrogenation reactions in the gas phase. I want to start with simple power law kinetics using $K_\mathrm{eq}$ to account for the reversibility of the reactions. Would the equation below, knowing the Gibbs free energy of reaction, be valid?

$$Δ_\mathrm{r}G^\circ = -RT\ln K_\mathrm{eq}$$

I came across a paper that used this equation to calculate $K_\mathrm{eq}$ to build a kinetic model of the ethanol-to-butadiene reaction. However, other papers I read, i.e. on the methanol-to-diethyl ether reaction, have used complex equations derived from empirical observations to calculate $K_\mathrm{eq}.$ Why use one approach over the other?

In general I would say 'no', thermodynamics only tells us about starting and ending points, thus you know the $\Delta G^\text{o}$ for the reaction but nothing about its actual mechanism, (since time does not come into thermodynamics). As $K_\mathrm{eq}$ is a ratio of rate constants these can take on many different values and still have the same ratio. To study the mechanism you will need to monitor species and intermediates vs time as the reaction proceeds.

Before doing any experiments you can only make informed guesses as to what the mechanism might be, even apparently simple reactions, such as between hydrogen and oxygen, is not simple but by experiment found to be a chain reaction.


Do acetals and hemiacetals give 2,4 DNP test? Do acetals and hemiacetals give positive 2,4 DNP test?
The reagent for the reaction is described as:

An aqueous solution of 2,4-dinitrophenyl hydrazine (DNP) is known as Brady's reagent

I recently learnt that acetal don't give positive tollens test whereas hemiacetals did.
Following from the same conclusion I believe that since its an aqueous solution and seems alkaline aswell so according to me acetals should give negative 2,4 DNP Test whereas hemiacetals should give positive.
However I fail to find any literature on this, are my conclusions correct or are there any twists to it!?

The standard mixture for a 2,4-DNP test contains 2,4-DNP and conc. Sulfuric Acid in EtOH/water here. This is sufficient to hydrolyse most acetals and hemi-acetals so they will give a positive test. Cyclic acetals such as dioxolanes are more stable and may not.


Confusion in understanding temperature change in exothermic reactions The question due to which my doubt arose $\ce{->}$

  The following system is in equilibrium:
  $\ce{SO2Cl2 + {Heat} -> SO2 +Cl2}$

  What will happen to the temperature of the system initially if some $\ce{Cl2}$ is added into it at constant volume ?

My attempt at the question : Since if we add more $\ce{Cl2}$, concentration of products will increase which will in turn shift the equilibrium in backward direction, which will in turn release more heat. And since the heat is being released to the surrounding, temperature of the system should decrease.

Though according to my textbook the answer is temperature of the system will increase.

So please do explain where my understanding is wrong.

I think this comes down to what constitutes the system. If we think of the system as the reactants and products in a container, releasing heat will increase the temperature of the system.


Why is phosphorus (v) oxides molecular formula P4O10 and not P2O5 I understand that the name derives from the empirical formula; however, I do not understand why it does not exist as $\ce{P2O5}$. Is there a compound that exists with this as its molecular formula?

There is no clear rule about how to name a compound which is actually existing as a dimer, a trimer... or an octamer, as long as it does not make a difference in a reaction equation, which is usually the case.

It does not matter in an equation if you count sulfur as a single atom entity (S) or as its true form (S8). You will find the same quantity in both cases.

The same stands for P2O5/P4O10: the true form is usually P4O10 (depending on the solvent), but the 2 forms give the same result at the end.


What's special about the purine scaffold? Purine is a remarkable substance, given Nature has chosen it as the scaffold for two nucleobases from DNA/RNA: adenine (A) and guanine (G). Its structure also appears in several other substances of biological interest, by substitution on the fused ring system, like these examples shown in Wikipedia:

The base ring system has 4 nitrogen atoms and 5 carbon atoms. Then I wonder, it's not hard to imagine isomers or at least similar compounds just by changing the positions of the nitrogen atoms around the rings, with a few constraints. The five-membered ring needs at least one N to maintain aromaticity in the pyrrole ring. It's better to have this N at the position where a glycosidic bond is expected, if we want to incorporate it in some kind of nucleic acid. Along the edge connecting the two fused cycles, it's better to have C, to avoid N atoms with four bonds. So we are left with the following template:

We can distribute the remaining 3 N and 3 C atoms among the 6 free positions in $6!/((6-3)! \cdot 3!)  = 20$ different ways.
So my question is, what's special in purine, that caused its scaffold to be picked among all the other possibilities?

It is important to remember that the purine scaffold is one of the most fundamental organic structures in all lifeforms. From an evolutionary standpoint, this means that there must have been a simple, effective way for this scaffold to be prepared in the pre-biotic earth. After all, the first organisms to contain purine scaffolds did not have "access" to complex enzymes, proteins, etc.
Take a look at the molecular formula for adenine, a simple purine derivative. C5H5N5. We could rewrite this as (HCN)5. As the molecular formula alone might lead you to conclude, adenine can be made from only one chemical: hydrogen cyanide. Interestingly, hydrogen cyanide was a very abundant chemical in the earth's primordial atmosphere. Roy et al.1 replicated the atmospheric conditions of the primordial earth and found that substantial quantities of adenine could be produced! The mechanism for this reaction is quite beautiful, and I will include Peter Jacobi's rendition of it below. The mechanism for the formation of adenine also explains why other constitutional isomers are not produced. We must consider the alternating nature involved in these nucleophilic-electrophilic interactions (e.g. two nitrogen atoms will never "decide" to couple).

So, to conclude, the purine substructure is "special" because it not only has the unique ability to hydrogen bond and form a base pair for DNA replication but was also incredibly easy to synthesize in the primordial earth's atmosphere. Although life forms do not use HCN to synthesize the purine substructure anymore, its synthesis is still incredibly simple.
1 Roy, D.; Najafian, K.; Schleyer, P. V. R. Proc. Natl. Acad. Sci USA 2007, 104, 17272-17277.


Why does CaCO3 react with HCl, but not with H2SO4? I have a wonderful reaction of marble chips, $\ce{CaCO3}$, with hydrochloric acid, $\ce{HCl}$, and carbon dioxide was released beautifully (fast, large volume, easy to measure and makes good visual effect too). But there is no reaction between $\ce{CaCO3}$ and $\ce{H2SO4}$. Why not?

Your marble chips react on the surface.

In the case of hydrochloric acid, the resulting salt, calcium chloride, is highly soluble in the acid, dissolves and provides further attack to the (new) surface.

With sulfuric acid, the highly insoluble calcium sulfate is formed on the surface of the marble chip. With other words:

Calcium sulfate acts like a protective layer.


Number of cyclic isomers of C4H6 I faced a question

  How many cyclic isomers are possible for $\ce{C4H6}$?

I and my friend found the following four.

But the answer key says there are 5. So what's the other one?

You can answer such questions using D.U. (Degree of unsaturation)

The formula is $\mathrm{C + 1 - }\frac{H + X - N}{2}$

C = Carbon.
H = Hydrogen.
X = Halogen.
N = Nitrogen.

If you get D.U. to be one, then in the structure there could be:

1 double bond.
1 ring.

Let's take an example of $\ce{C4H8}$ which has D.U. equal to one.

It can have three structures of 1 double bond:

It can have two structures of 1 ring:

Now for D.U. equal to two, the possibilities are:

2 double bond.
2 ring.
1 double bond and 1 ring.
1 triple bond.

Let's take your example of $\ce{C4H6}$ which has D.U. equal to two:

It can have two structures of 2 double bond:

It can have one structure of 2 ring:

It can have four structures of 1 double bond + 1 ring:

It can have two structures of 1 triple bond:

So, the answer is 5 cyclic isomers as you can see above.


HOMO-LUMO and perturbation theory I have some confusion about perturbation theory, and just how it fits into the hard-hard/soft-soft interactions within organic reactions. I have come across a formula for perturbation that is proportional to an overlap constant squared but is inversely proportional to the energy difference of the orbitals in question. This suggests that for orbitals of poor energy match, there is least perturbation, and vice versa for good orbital energy matching. How does this translate into a hard/hard and soft/soft interpretation? I am not looking for a quantitative answer because my chemistry is not advanced enough!

What you are referring to as perturbation theory sounds like a very crude approximation to the 'charge-transfer' or CT intermolecular interaction.  In quantum chemistry lingo, the CT interaction is the mixing of occupied orbitals on one molecule/fragment with the virtual orbitals on another fragment (this is the 2-body CT interaction).  This permits the transfer of electron density from one fragment to another (charge-transfer).

Instead of redoing the HF (variational) calculations with the new orbitals included, perturbation theory is instead used to calculate the nth order correction as a result of occ(a)+vir(b) mixing.  Differencing the HOMO on fragment A with the LUMO on fragment is just a way to quickly approximate the more complicated (and expensive) perturbation theory expressions for the CT iteraction.

So this seems to make sense in the context of HSAB theory and Lewis acids where acid/base exchange electron pairs and thus participate in a strong charge-transfer interaction.

*EDIT:  I'd like to add that the orbital energy of the LUMO orbital (in fact, all virtual orbitals) is sensitive to the basis-set size. When a large enough basis-set is used, the LUMO can always be forced to near 0.00.  Orbital energies, at the end of the day, are not nature.


Would singlet oxygen in the engine cause a more efficient combustion? Today someone told me about a new product — a mesh that is made from 5 different metals and when oxygen passes through it, singlet oxygen appears for a short period of time.
This mesh needs to be installed in a car air filter and the "better" oxygen creates a better combustion inside the engine. This is supposed to save fuel, give the engine more power etc.

My questions are:

Is it possible or this is just more one of the advertising lies that stores are trying to sell us?
if an interaction like this is something that can be acquired, will it really cause a better combustion inside the engine?
What are the downsides of interactions like this?
Can a mesh like that be permament without the need to replace it every couple of weeks or months?

This hypermileage scheme is one of the more interesting ones I've come across, so I think it's worth some discussion. I don't know much at all about the engineering aspects of internal combustion engines, so I'll stick to a theoretical analysis.

Let's start by describing the difference between singlet oxygen and triplet oxygen. Both of these are the same substance, molecular (di)oxygen, or $\ce{O2}$. However, they are in different energetic states. In its lowest energy state, a molecule of $\ce{O2}$ contains two unpaired electrons (which is a very interesting and important fact in itself), and this form is called triplet oxygen. This is the regular stuff you breathe all the time.

However, if some energy is given to triplet oxygen, it can be excited into a different state, and the two formerly unpaired electrons now pair up. Since there are now no unpaired electrons, we call this singlet oxygen (technically there is more than one singlet state possible, but for our purposes that can be ignored).

So in summary, singlet oxygen and triplet oxygen are the same molecule, except that singlet oxygen has inherently more energy. So why does this matter? If singlet oxygen has more energy, then one expects it to be more reactive. It is entirely conceivable that a fire supplied with singlet oxygen will burn hotter/more completely than the same fire supplied with triplet oxygen. So that part of the scheme is plausible.

That said, you may ask yourself, if all it takes is some energy to turn triplet oxygen into singlet oxygen, then shouldn't oxygen in any fire be continuously converted from the triplet state to the singlet state, simply by the heat of the roaring flame? This would mean preparing singlet oxygen separately is entirely superfluous! As it turns out, due to the quirks of quantum mechanics it is not easy to excite triplet oxygen into singlet oxygen. Technically speaking, this is called a spin-forbidden process.

However, due to special relativity (!), spin-forbidden processes turn out to not be completely forbidden, just very unlikely. A concept called spin-orbit coupling comes into play. The more spin-orbit coupling in a chemical system, the more likely it is for spin-forbidden process to nevertheless happen. Spin-orbit coupling becomes more and more relevant as you go further down the periodic table. This means heavy atoms such as platinum can favour the transformation of triplet oxygen into singlet oxygen. So again, the necessity of a metal honeycomb as pictured in the scheme is at least conceptually plausible for this singlet oxygen strategy to work.

(Of course, it is possible to produce singlet oxygen through other means, such as the decomposition of some oxygen-containing molecules like hydrogen peroxide and ozone, but doing so would defeat the whole purpose; you'd essentially have to buy a second kind of fuel!)

At this point it seems there is actually something going for this scheme, but here is where so many others like it stumble: the laws of thermodynamics. If it takes energy to transform triplet oxygen into singlet oxygen, then that energy has to come from somewhere. If the energy comes from the metal honeycomb itself (or more precisely, the energy imbued into the metal honeycomb during its manufacturing process), then yes, it is plausible to get increased mileage, but the mesh will lose energy until it is inactive and has to be replaced or reactivated. My bet is that this would cost much more than the meagre fuel savings.

In principle, the energy to make singlet oxygen can come from the extra energy from the more efficient fuel burn, leaving the mesh intact. However, now you're dipping into the very extra energy that was supposed to make the car get more mileage. The detailed energy balance is generally very hard to calculate, and I couldn't begin to tackle all of the real-world variables in play. That said, the answer to whether a hypermileage scheme is cost-effective almost always ranges from a flat-out "no" to "yes in principle, but no in practice". Processes that are cost-effective (precise spark timings, regulation of air-fuel ratio, etc) tend to make it into standard issue vehicles in the first place. I can't help but be reminded of the joke that alternative medicine which works is just called medicine.

Lastly, regardless of energy savings, singlet oxygen may not be great inside the engine. Not only is it more chemically reactive than triplet oxygen, but it can perform reactions which are completely different from triplet oxygen. It could cause more wear of the engine components, or could generate more of some unwanted sideproducts during combustion, such as nitrogen and sulfur oxides. Any singlet oxygen that manages to escape combustion and exit the tailpipe could also pose a significant biological and chemical hazard.


Why don't ions form crystal lattices in water and other polar solvents? Ionic compounds are in the form of crystal lattices, and when they are dissolved in a polar solvent, they are bonded with the corresponding cations/anions of the polar solvent. For example, $Na^+$ bonds with $OH^-$ and $Cl^-$ bonds with $H^+$ when NaOH is dissolved in $H_2O$.

The reason why $NaCl_{(aq)}$ conducts electricity is because $Na^+$ and $Cl^-$ do not form crystal lattices with $OH^-$  and $H^+$  respectively. But why don't they form crystal lattices?

They actually form: ions can be dissolved up till a certain concentration, and beyond that, they form ionic lattices just as you predict.

The easiest way to describe the scenario as an equilibrium between ions-in-water vs ions-in-lattice:

If the ion-water interaction is very strong, the ions can be very stable in water, even more than in an ionic lattice, so ions will be in the solution.
If the ion-water interaction is weak compared to the solid, then the solid phase is much more stable, and only a very low concentration of ions can be in the solution.

Looking at only interaction energies is a very simplistic view as it does not consider entropic factors, yet can describe at least part of the story. Entropy is generally a huge driving force for any kind of dissolution and allows some ions to be in solution even if the solid state energetically more stable.


How does alcohol keep smell and taste flavors in wine? Nowadays alcohol free wine is also sometimes available. Now is seems that this wine has less taste and flavors in it. The cause for this is that alcohol seems to the capability to keep these flavors longer inside the wine. But how can alcohol do this?

First of all, ethanol has its own smell (which we perceive as taste), it also can stimulate pain sensors in the mouth. Therefore alcohol free wine will taste different.

The process of separating alcohol from wine is quite invasive: the water/alcohol is separated from flavour and colour compounds by reverse osmosis, then the water is distilled to remove alcohol, the water is added back, to the concentrated wine flavour. This process is surprisingly good at not removing anything but the alcohol, but it does change the flavour. Other small light volatile compounds are lost by this process. Tannins are also lost in this process, resulting in reds that lack body

Next the presence of alcohol helps some aroma compounds to evaporate. This means that the aroma of the wine changes. The scent chemicals can't get into the air without alcohol to carry them.

Finally, of course, the best wine is sold with alcohol. When you drink non-alcoholic wine, you are usually drinking a lower quality source wine.

source: http://winefolly.com/tutorial/non-alcoholic-wine/


What elements (in the pure form) would harm a human, if eaten? What pure elements, if we eat it in a relatively small dose (around a piece of sugar), can be harmful/lethal for an average human? Is, for example, eating pure carbon bad for the organism?

Is, for example, eating pure carbon bad for the organism?

No, quite the contrary.

Activated charcoal (in the form of powder or tablets) was/is traditionally used to treat diarrea. It is listed on the WHO Model List of Essential Medicines on page 4.


Meaning of mass concentration "units" % m/v In a saline solution that is 5.750% (m/v) potassium iodide $(\ce{KI}),$ there are:
a) 5.750 mL $\ce{KI}$ per 100.0 mL $\ce{H2O}$
b) 575.0 mg $\ce{KI}$ per 10.00 mL of $\ce{KI}$ solution
c) 5.750 g $\ce{KI}$ per 100.0 g $\ce{H2O}$
d) 5.750 g $\ce{KI}$ per 100.0 g of $\ce{KI}$ solution
e) 5.750 g $\ce{KI}$ per 100.0 mL of $\ce{H2O}$

I figured it should be e because for the equation of $(\text{m/v})\%$ is
$$(\text{m/v})\% = \frac{\text{mass solute (g)}}{\text{volume solution (ml)}}$$
and it's $\pu{g/100 mL},$ but it's not the correct answer.

Since the question raised a discussion in the comments, I guess it won't hurt putting up a brief summarizing answer.

Notation % m/v, as you might've guessed, refers to the ratio between the mass and the volume, or, more precisely, mass concentration $ρ_i$ (sometimes denoted as $γ_i$ [1, p. 48] to distinguish from density, see How to distinguish mass concentration and density?):

$$ρ_i = \frac{m_i}{V}$$

where $m_i$ is the mass of solute and $V$ is the volume of solution, suggesting the SI unit $\pu{kg m-3}.$
Where does percent in % m/v come from?
Quoting Wikipedia:

  In biology, the "%" symbol is sometimes incorrectly used to denote mass concentration, also called "mass/volume percentage." A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). The notation is mathematically flawed because the unit "%" can only be used for dimensionless quantities.

So, in order to account for this percentage “unit”, one should normalize the mass concentration to the mass of solute in grams in 100 mL of solution.
This can be done in one's head in most cases, but you must watch out for the units.

In this particular example, only option b satisfy the definition of the % m/v quantity we are expressing since only b lists volume of KI solution.
Let's check the math.
Since $\pu{575.0 mg} = \pu{0.5750 g},$ there is $\pu{0.5750 g}$ $\ce{KI}$ per $\pu{10.00 mL}$ of $\ce{KI}$ solution, and normalizing to $\pu{100 mL}$ we indeed obtain the aforementioned concentration:

$$ρ_\ce{KI} = \frac{\pu{0.5750 g}}{\pu{10.00 mL}} = \frac{\pu{5.750 g}}{\pu{100.0 mL}} = 5.750\%~(\text{m/v})$$

The exercise really just requires to pay more attention and follow the analytical chemistry definitions.

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.


What's the difference between "conversion" in condensation polymerization and in radical polymerization? The conversion in condensation polymerization is defined as the fraction of functional groups reacted,

$$p=\frac{[M]_0-M}{[M]_0}$$

while in radical polymerization it's defined as the fraction of monomer that has reacted.

Why is it not the same thing?

The reason they are defined differently is based on the differing mechanisms:

All of the monomer in step-growth (of which condensation is a subset) polymerization are reactive, and they are consumed very quickly, forming (mostly) dimers which then go on to react further.  At this point the average length of chains is 2-- not really a polymer, but also relatively monodisperse.  However this reaction is going to continue, and you'll get long chains after some time.  But the monomer itself is gone.

In addition polymerization (of which radical is a subset) there are very few reactive intermediates, and many monomers get added per chain. After 50% of the monomer is converted in addition polymerization, you have some really long chains, swimming around in the remaining monomer (half the original amount).  This contrasts the other situation.

Basically, you can't measure monomer concentration to reasonably monitor the progress of a step-growth polymerization, while you can do such for addition polymerization.  Conversely, you can generally monitor functional groups to monitor both kinds of polymerization.


Are there any magnetic particles that are safe to inhale? In medicine, there is an (old?) experiment where a subject is asked to inhale radioactive Xenon gas, and radiation counters are placed at different positions near the lungs. Using this experiment, it has been demonstrated that the (alveolar?) volume of the lungs is highest in the bottom parts, and smallest in the top parts.

I don't know if this experiment is still used today, since inhaling radioactive gas has the obvious radiation related risks.

I was wondering if there are some magnetic particles that would be safe to inhale? I know that for example helium does not diffuse into blood, and it does not permanently bind into anything. However, I think it does not have any suitable magnetic properties. Let's assume the detector is very very sensitive to any magnetic fields (so, a super-conducting quantum interference device).

$\ce{O_2}$ is paramagnetic.

source: http://www.chem1.com/acad/webtext/chembond/CB-images/liquidoxygeninmagnet.jpg

The stable Xenon isotope Xe-129 has been used in MRI studies:

Hyperpolarized Xenon MR Imaging of the Brain

  A radically new magnetic resonance imaging (MRI) technique, using hyperpolarized 3He and 129Xe, is being developed to produce high-contrast images of important body tissues that have resisted conventional MRI techniques. Hyperpolarization of the noble gas nuclei is achieved by spin exchange with rubidium atoms that have been optically pumped into one electronic spin state with the use of circularly polarized light from a high-power diode laser system. Hyperpolarization can increase the detectability of 3He and 129Xe up to a hundred thousand times. This technology is showing dramatic results for diagnostic imaging of the lungs, brain, and other parts of the body. Most notably, we have achieved high-resolution dynamic images of the lungs and airways using 3He, and measured cerebral perfusion in the brain using 129Xe.

See video of lung being imaged: http://www.spl.harvard.edu/archive/HypX/about.html


Why does decreasing the distance between the electrodes in an electrolytic cell increase the rate of electroplating? I did an experiment on the effect of the distance between the electrodes on the rate of electroplating in an electrolytic cell (copper electrodes with copper sulfate electrolyte).

As I expected, when the electrodes were closer, the current was greater and the rate of copper plating was also faster. But why does less distance increase the current? My initial thought was that less distance to travel=ions arriving at cathode quicker=greater current. That doesn't seem right though, because different ions are depositing and liberating electrons and hence not really flowing through (right?). I'm probably overthinking this, but any help is appreciated.

It's a bit of an oversimplification to explain this with a simple invocation of Ohm's law. The thing to remember about electrochemistry is that electron transfer only happens at every close distances to the surface (<10 nm) and it's the potential difference between the electrode and this tiny region at the surface that is important. You might be applying 10 V between the two electrodes, but if you have a lot of solution resistance between them, the voltage near the surface may only be 1 V as the other 9 V are dissipated as heat in the electrolyte. Thus, by moving the electrodes closer, the resistance goes down and the voltage the electrode "sees" is higher.

The trouble with this type of two-electrode cell is that many factors are being conflated. The current that is measured (the rate of electroplating) is dependant on what is happening at both electrodes as well as the resistance of the bulk solution. As the electrode voltage increases, the current usually increases too as increasing the voltage makes it easier for electrons to transfer between the electrode and the redox-active species in solution. Eventually, however, the rate will not be limited by electron transfer, but rather by how fast copper ions can move from the bulk of the solution to the active region near the electrode and a "limiting current" is reached. You can see this below as the voltage at an electrode is swept to higher values:
 (Source: Own work)

At low voltage, there isn't enough voltage for any electron transfer to occur, but as it increases along the x-axis, we see an increase in current until it reaches a peak and the current is limited by diffusion to the electrode. (the reason it goes down after this point rather than levelling out has to do with the time-scale of the experiment, but it's not important for this)

Because the system is so complicated, when we're interested in examining the reactions that are happening at just one electrode without worrying about the other electrode or the solution resistance, an alternate setup is used. If we use a feedback amplifier (known as a potentiostat) to control the voltage applied between the two active electrodes using a third (reference) electrode to measure the voltage close to the surface of the electrode of interest, we can decouple the effects of the solution resistance and other electrode.

A sidenote:

  My initial thought was that less distance to travel=ions arriving at cathode quicker=greater current. That doesn't seem right though, because different ions are depositing and liberating electrons and hence not really flowing through (right?).

You were actually onto something interesting here. You're right in that it takes a while for copper ions from one electrode to make their way to the other so most of the copper being deposited is coming from what was already in the solution (at least on relatively short time scales), but what would happen if we moved the electrodes very close to one another? Eventually we would reach a situation where copper could dissolve from one electrode and be almost immediately deposited on the other and we would not see the limiting current behaviour as above.


Why are MD (Molecular Dynamics) simulations performed at room temperature for the purposes of studying proteins? All of the MD simulations which are cited in the literature seem to be performed at room temperature (~300 Kelvins), while enzymes are usually bioactive at body temperature (~310 Kelvins).

I am wondering why not just use 310 K for the MD simulation? I suspect it is because most water models are designed specifically for room temperature (ie. TIP3P waters).

Another reason why people don't bother to use 310K versus 300K is because any time you are using MD, you have an approximate force field describing the dynamics of the system which means you have an approximate phase diagram for the system. When comparing MD and experiment, it is a reasonable question if you really want to compare properties at the same temperature or compare properties at "equivalent" points on the phase diagram. As an example, some models of water which are very widely used have a phase diagram which is drastically different from the real phase diagram of water (i.e. they boil at 400K and other weird things) and you will see studies which describe the phase diagram of DFT water or TIP4P water so that you can make comparisons based on the temperature and on the phase diagram.

So, this is to say, if you submit a paper where you do MD at 300K when studying an enzyme in solution, it is unlikely you will get complaints that this is a biomolecule so it should be at 310K because even comparing to properties of the actual enzyme at 310K isn't a direct comparison since you aren't at the same locations on the respective phase diagrams. What could be more relevant is to make comparisons to the real world across a range of temperatures so that you can see if the trends are the same rather than just a single point.

Another less physical reason is that people have already been doing simulations at 300K, so you can't really get complaints about a simulation at 300K since everyone has already been using 300K. This is actually very common in science.


How to calculate the emission wavelength of a substance? A light photon (700nm) gets absorbed by a substance*, this substance than emits the light photo.

How to find out the wavelength emitted by this light photon?

*substance can be anything, I've just used the beryl mineral as an example

There is no general way.

For a particular case, there is need a good quantum chemistry model of energy levels, and a knowledge of character of the emission. It could be an elastic excitation, releasing the same wavelength. It can be fluorescence or phosphorescence with longer wavelength.

For very simple cases, like a hydrogen atom, it is easy and googleable. See the Rydberg constant :

  $$\frac{1}{\lambda} = Ry \cdot {1\over h} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{m_\text{e} e^4}{8 \varepsilon_0^2 h^3} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
  where $n_1$  and $n_2$ are any two different positive integers (1, 2, 3, ...), and $\lambda$ is the wavelength (in vacuum) of the emitted or absorbed light.

  $$\frac{1}{\lambda} =  R_M\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

  where $R_M = R_\infty/(1+m_{\text{e}}/M)$ and $M$ is the total mass of the nucleus. This formula comes from substituting the  of the electron.

The more complex the real scenario is, the less are usable theoretical models and chemists must rely either on experimental data, either on empirical models not supported by underlying theoretical principle.

There is an example of mercury melting point. Experimentally, it is a piece of cake, some $\pu{-38.8 ^{\circ}C}$. Theoretically, physical chemists waited more then 50 years until there were recently powerful enough computers available. It took combination of extensive relativistic quantum chemistry computations and random calculations to reach result within $\pu{\pm 1 ^{\circ}C}$ error. The prior non-relativistic calculations predicted the melting point with the error more than $\pu{100 ^{\circ}C}$ !!


Why is water evaporation not more commonly used to cool buildings Water has a very high heat of vaporization (2260 kJ per kg). Why not just gradually sprinkle water on your roof to keep the building cool in opposed to using air conditioners?

It wouldn't take much water to keep the building a few degrees cooler for a long time. Air conditioners use a lot of electricity which is bad for the environment. I read some places ban water vaporization cooling systems because of droughts and because tap water is government subsidized but that seems silly.

As a society it would be cheaper and more environmentally friendly to use water. The water will simply evaporate and return as rain which it will do anyway regardless. The energy air conditioning uses puts more greenhouse gases  in the environment which (allegedly) causes more global warming and puts more chemicals in rain water.

"This question seems rhetorically posed: how can you justify calling your solution better and cheaper without having run the numbers? – Aesin 1 hour ago"

A central air conditioning system for a house takes about 3500 watts.

http://michaelbluejay.com/electricity/howmuch.html

1 watt = 1 J/sec so that's about 1,260,0000 J/hour

A gallon of water weighs about 3.7854118 Kg

so at 2260 kJ per kg you can extract 6,780,000 J of energy per gallon of water which is more energy than the air conditioning system even uses and the air conditioning system doesn't extract the amount of heat energy from a building than the energy it uses.

If you run the numbers even with a gallon of water the energy efficiency of water usage is much superior than that of an air conditioning system.

The high heat of vaporization of water is how sweat keeps you cool. Ancient civilizations used to pour water on their tents or tepees or whatever to keep themselves cool. It's cheap and efficient. The superiority of this method when opposed to air conditioning, in dry areas at least, is very obvious. I have yet to see one good reason why it's not used more often.


In Goodman and Gilman what does R,S-citalopram refer to? In Goodman & Gilman's The Pharmacological Basis of Therapeutics, 12e R,S-citalopram is referred to. The problem is that from what I am aware of citalopram only has one stereocentre, and hence I am forced to ask were the author having a stroke when they added the comma and really meant the racemic mixture of R and S stereoisomers?

Figure 1: Screenshot of the Relevant Page in Goodman and Gilman

Figure 2: Citalopram

(The latter has the stereocentral carbon highlighted)

It seems fairly clear that the authors intended to refer to the racemic mixture there. You're right that a comma would normally be used to separate the locants of distinct stereocenters when multiple ones are present, but then the individual chiral carbons would have to be numbered, which obviously isn't the case here.


How does the calorimeter affect heat released by the system (q)? Here are the 2 questions I am pondering about

How would the calculated quantity of heat (q) released by a system differ if the calorimeter's heat capacity was taken into account?

My thought process is that the calculated quantity of heat (q) released by a system would increase if the calorimeter's heat capacity was taken into account, b/c the calorimeter would aid in absorbing the heat along with the surroundings, so change in temperature would increase ($q=m c \Delta T$)

How would quantity of heat (q) released by a system differ if the calorimeter was not a good insulator?

Here, I am not sure how to word the answer, but if the calorimeter wasnt a good insulator, heat released by a system would be lost to the environment, and thus less heat released would be calculated via $q=m c \Delta T$ due to a smaller change in $\Delta T$.

Let's take a numerical example. $0.2432$ $g$ magnesium metal ($0.01$ mol) is introduced into a calorimeter containing $50~ g$ HCl $1$ M. The chemical reaction will be $$\ce{Mg + 2 H^+ ->  Mg^{2+} + H2}$$ The composition of the solution is changing between the beginning and the end of the reaction. There is an excess of HCl.  But let's admit in first approximation that the specific heat of the solution does not change and remains as in water ($C_w = 4.185~ J g^{-1} K^{-1}$). The tabulated enthalpy change of the reaction is - $462~ kJ/mol$. The total amount of heat $Q$ produced by the reaction of $0.01 mol$ Mg is : $$Q\ce{= 0.01 mol·462 kJ/mol = 4.620 kJ = 4620 J}$$
If the calorimeter was ideal, without any weight, and with the already mentioned approximation, the change of temperature should be : $$\Delta T = \frac{Q}{C_w·m} =  \frac{4620~ J}{4.185~J g^{-1} K^{-1}·50~ g} = \frac{4620~ J}{209.25~ J K^{-1}} = 22.07 K$$
Now suppose that the calorimeter is made of copper and weighs $m_{Cu} = 20~ g$. The heat capacity of copper is $C_{Cu} = 0.385~ J g^{-1} K^{-1}$. The change of temperature will be reduced and becomes $$\Delta T = \frac{Q}{C_w·m + C_{Cu}·m_{Cu}} =  \frac{4620~ J}{(209.25 ~J K^{-1})~ + (0.385 ~J g^{-1} K^{-1}·20~ g)} = 21.3 K$$
And now if the calorimeter is not well insulated, the final temperature change will be smaller than $21.3 K$.
Sorry for my poor management of the italicized letters. Any help would be appreciated.


What visual observations can confirm that a reactant is the limiting reagent? I have this lab question for the lab called Copper Collection Stoichiometry, where we choose an amount of the limiting reagent (iron) for a reaction between copper (II) sulfate and iron. We are to dissolve copper (II) sulfate in water, making it blue, then add iron, and after filtrating, we should have copper left.

How do we know if iron was the limiting reagent in terms of the visual observations? We made it so that iron was the limiting reagent, but how do we know visually if it is?

I understand that in order for iron to be a limiting reagent, it should've reacted with the solution and all completely turned into copper after the experiment. However how do we know if all the iron actually turned into copper? There is no way that we can visually count every solid flake to make sure there's no iron. I also understand that if iron was the limiting reagent, then there should be still copper (II) sulfate left in the solution, as the reaction didn't completely occur. Is that the answer?

The colour of the filtrate after filtering out the solid is still lightly blue in colour. This indicates that Copper (II) sulfate still is present in the solution, meaning that there was not enough iron to completely react with all the copper (II) sulfate, visually confirming it to be the limiting reagent.

A solution containing only Iron (II) sulfate is colourless, so if the solution is blue in colour, then it shows that Copper (II) sulfate still is present.

Credit to Ivan Neretin for helping me.


Why do we use hydrochloric in the test for sulfate ions but nitric acid for halide ions? Recently, I have learnt about testing for ions in qualitative analysis (BBC Bitesize).

The link above which I used for reference mentions that for the test for sulfate anions, hydrochloric acid $(\ce{HCl})$ is used to ensure that there is no presence of carbonate ions.  For the test for halide ions $(\ce{F-},$ $\ce{Cl-},$ $\ce{Br-},$ $\ce{I-},$ etc), nitric acid $(\ce{HNO3})$ is used instead, for the same purpose.

I would like to know why the acids used are different, since they serve the same purpose.

Update: $HCl$ cannot be used for removal of carbonates in halides since $HCl$ contains the $Cl-$ halide.  However, as to why nitric acid cannot be universally applied to all, I am still unsure.

The link above which I used for reference mentions that for the test
  for sulfate anions, hydrochloric acid (HCl) is used to ensure that
  there is no presence of carbonate ions. For the test for halide ions
  (F−, Cl−, Br−, I−, etc), nitric acid (HNO3) is used instead, for the
  same purpose.

Good question, but this is not a universal practice. In reality, the choice of acid depends on the actual sample. Qualitative tests, like these, have become merely academic exercises due the to advances in instrumental analysis these days. As you said, one cannot add HCl to destroy carbonates in a sample which is to be tested for halide ions. Dilute nitric is the only choice, because you would not use sulfuric acid in the case of halides because (i) your test sample may contain ions which will precipitate with sulfate (Ba, Ag, Pb, Ca etc.). Concentrated nitric acid would be a poor choice as well because it will oxidize iodide to free iodine.

An excellent book, which is far better than any online resource, is Vogel's Qualitative Inorganic Analysis. It should be available in any good library and it will give an in depth view of qualitative analysis.


Effect on kinetic energy of a body with change in temperature Why doesn’t the energy of a body doubles when temperature in degrees Celsius is doubled?
I know that it doubles in case of kelvin scale.

You might find it interesting to read up on the differences between the different temperature scales. The kelvin scale effectively measures the vibration (average kinetic energy of molecules) so it makes sense that a doubling in the kelvin means a doubling in kinetic energy. On the other hand the Celsius scale is defined by the melting and boiling points of water (quite arbitrary).


Relative Solubilities of Salts in Acidic Solution I stumbled across this problem which I don't understand the answer to:
Which solid is much more soluble in 1 M $\ce{HCl}$ than in $\ce{H2O}$?
(A) $\ce{CaHPO_4}$
(B) $\ce{CaCl_2}$
(C) $\ce{BaBr2}$
(D) $\ce{BaSO4}$

Chloride, Bromide, and Sulfate all correspond to strong acids, while phosphate corresponds to a weak acid.

With 1M H+ available, phosphate will be essentially fully protonated, while the others will still be anions.


What is the pH of kvass? I know that it is less than 7.0, but would be nice to know exact value of pH of natural fermented kvass (not soda-like counterfeit). Another aspect is to know if kvass is more gentle than Cola (pH=2.5) ?

According to this paper$^{[1]}$, the kvas in their study started out at a pH of 6.1 prior to fermentation, then became stable at about a pH of 3.8 after 12 hours.  This is of course significantly less acidic than a cola with a pH of 2.5 (and some colas can have a pH closer to 2).

1) QUALITY CHANGES OF NATURALLY FERMENTED KVASS DURING PRODUCTION STAGES, Department of Food Technology, Faculty of Food Technology, Latvia University of Agriculture, Liela Street2, Jelgava, Latvia


Is energy required to form bonds [phase change] My question is if any energy is required to form bonds, for instance when there is a phase change?

If I am correct, energy might be required in the beginning, to make the reaction start and then release a bigger amount of energy than it was putted in. However, in terms of molecular behaviour in a phase change, I think there should not be any energy required to make the molecules to form bonds, if they are placed in a system which has a lower temperature than the molecules. Would they not lose the kinetic energy after a specific time?

Bond formation always lowers the energy of the system (or bond formation is a consequence of lower energy, take your pick.)

Indeed, you may have to add energy, because presumably certain bonds must break in order to rearrange the atoms.

Adding energy won't guarantee formation of much higher energy isomers, because the atoms have a large amount of kinetic energy, and so they can just as easily turn around and go back whence they came.

Sometimes the kinetic energy can be dissipated by solvent, collisions with inert gasses, etc. and then you may end up with a measurable quantity of the higher energy species.


How is the 'rate law' and 'order' for the ozone decomposition reaction can be defined? How is the 'rate law' and 'order' for the following reaction can be defined?

$$\ce{2O3 -> 3O2}$$

"The expression that correlates the rate of reaction with concentration of reactants is known as rate law for that reaction."
So the rate law only include the concentration of reactants, not products.
But the rate law for this reaction is:

$$r = k[\ce{O3}]^2[\ce{O2}]^{-1}$$

where $k$ is rate constant.

"Order of a reaction is the number of reactant molecules whose concentration determines the rate expression."

Order for this reaction w.r.t. ozone is 2 and that w.r.t. oxygen is -1. But why we are considering oxygen, being a product of this irreversible reaction. This also means that the rate of this reaction will decrease with the increase in concentration of oxygen.

Now it seems that, this is a reversible reaction. But chemical kinetics is studied for only irreversible reaction?

Chemical kinetic describes both reversible and irreversible reactions.

The reaction scheme and rate expression do not refer to the same thing. The reaction scheme is an overall one whereas the rate equation refers to a particular multi-step reaction. Two examples are given below:

Ozone can be destroyed by UV light in the $240$ to $\pu{320 nm}$ region with the scheme $(\ce{M}$ is an inert gas needed to take away excess energy):

$$
\begin{align}
\ce{O3 &->[$hν$] O2 + O} \\
\ce{O3 + O + M &-> O2 + O2 + M} \\
\hline
\ce{2 O3 &-> 3 O2}
\end{align}
$$

Similarly, oxygen molecules can react with short wavelength UV as

$$
\begin{align}
\ce{O2 &->[$hν$][$(<\pu{242 nm})$] O + O} & \\
\ce{O + O2 + M &-> O3 + M} &|\cdot 2 \\
\hline
\ce{3 O2 &-> 2 O3}
\end{align}
$$

which also gives a similar net reaction. This is the Chapman mechanism and is very important in atmospheric chemistry.


Does a patina on a knife prevent rust? Before I begin, I think it's worth mentioning that I know little about chemistry and even less about knives.

My question is this, having seen many people talk about how patinas work and the best ways of applying them and which ones are best etc., It seemed to me like people just assumed that they work, presumably because they heard someone else do it, and them from someone else and so forth but that doesn't always make it true. If anyone has any idea If there is any reasoning behind this and if there any flaws behind the logic in it that would be greatly appreciated.

I don't exactly want to ruin or change my new knife for no reason. I assume people might think that this works similar to the thin oxide coating on aluminium, but my guess is that the corrosion acts like it would on any metal and slowly "eat away" at it.

So, simply, does it work?

In this case, the use of the word patina is a little misleading. Patina refers to the oxidation film on bronze or aged polish. I assume that you mean that the knife is coated with a substance.

First, when steel oxidizes, the iron oxide that forms does not stick well and will flake off the surface. This would not be useful for a knife since the edge will become dull quickly.

Second, if the knife is made with stainless steel, it does not need to be coated to prevent rust, only keep reasonably dry. Some stainless steel will actually rust since it was not properly formulated. Very few knifes are made of other metals since stainless steel is reasonably cheap.


Do more double bonds cause a bigger number of stereoisomers in cycloalkenes? If trans double bonds cause cycloalkenes to be chiral, does a second trans or maybe cis double bond induce another chiral plane and hence four stereoisomers are possible for the given cyclodiene?

The question is based upon these previous questions:
Are all higher cycloalkenes chiral?
Why is trans-cyclooctene chiral?

If trans double bonds cause cycloalkenes to be chiral

Keep in mind that a trans double bond in a cycloalkene does not necessarily produce a chiral cycloalkene.  Only when the chain connecting the ends of the double bond is short enough so that the trans double bond is twisted will a chiral conformation exist.

So in large trans-cycloalkenes the double bond is not twisted and cannot produce chirality.

Remember too, that even if the trans double bond is twisted and a chiral conformation results, the molecule may still be optically inactive if the chiral conformation rapidly interconverts with its enantiomeric conformation. This latter point was discussed in one of your earlier questions where it was pointed out that

  At room temperature, trans-cyclooctene is the only trans-cycloalkene
  where the two enantiomeric conformations are stable enough (e.g. the
  barrier to reach the planar conformation is high enough) to be
  isolated.  If we lower the temperature to 0°C, then we can isolate the
  enatiomers of trans-cyclononene.

Back to your question.

  does a second trans or maybe cis double bond induce another chiral
  plane

The key here is that, as mentioned above, if the second double bond is also twisted, then there will be the potential for an increase in the number of chiral stereoisomers (of course inserting a double bond will always increase the number of stereoisomers because we have additional cis and trans double bond possibilities, but I think your question is focused on chiral stereoisomers, not geometric stereoisomers).

Look at the above figure on the left, of a trans double bond in a small ring.  We are sighting down a trans double bond. To alleviate ring strain, the double bond if forced to twist.  It is this twisting out of plane that infers chirality upon the double bond.  Its insertion into the ring will produce an additional chiral plane and the R or S chirality of the new double bond will  need to be specified in the name of the compound.

Note the figure on the right of a cis double bond in a small ring. To alleviate strain here, the $\ce{C_{ring}-C=C}$ angle is pinched down; there is no out of plane twisting motion, the plane of the double bond still exists.  Therefore, in the cis double bond case, there is no chirality associated with the double bond; no additional chiral isomers will be produced.

  and hence four stereoisomer are possible for the given cyclodiene?

Not always.  Just as in more common compounds with 2 chiral centers, meso stereoisomers are possible.  As an example, if the known E,E-cycloocta-1,5-diene
adopted a conformation as pictured below, then it has a plane of symmetry and would be an achiral meso compound where one trans double bond has the R configuration and the other double bond the S configuration.

(image source)


Is methoxide a better nucleophile than hydroxide? What is more nucleophilic, $\ce{OH-}$ or $\ce{CH3O-}$?
When I look it from the viewpoint of steric hindrance, the answer seems to be $\ce{OH-}$, but when I see it from the viewpoint of extra electron density provided by hyperconjugation of $\ce{CH3}$, the answer seems to be $\ce{CH3O-}$.

Alkoxides tend to be stronger bases than the hydroxide ion. Adjacent atoms in an alkane chain will donate electron density to oxygen, allowing it to be more polarizable. An increase in polarizability will allow oxygen to more easily donate electrons.

The oxygen in a hydroxide ion does not experience the inductive effect. Therefore its valence electrons do not "reach" out as far.

Methoxide is a relatively unhindered base, as is hydroxide. But as explained above, methoxide experiences induction whereas hydroxide does not. So one would expect methoxide to be a better nucleophile than hydroxide because it is a stronger base than hydroxide and still remains unhindered.


Slight variation in NMR integrals of CH protons vs. CH2 or CH3 While acquiring some NMRs I find that there is a consistent pattern in that protons attached to sp2 carbons tend to have slightly smaller integrals than protons attached to sp3 carbons:

(peaks in aromatic region are from a para-disubstituted benzene ring; peak at 4.92 ppm is from a trisubstituted alkene; 4.50 is a NH; the remainder are aliphatic CH2's or CH3's.)

I calibrated the total sum of integrals to 21; one can see from the spectrum that the aromatic and olefinic peaks are slightly smaller than integral values, and the others are slightly larger. Obviously it doesn't hinder the analysis of the spectrum, but it was an interesting trend (the same was observed in multiple spectra).

Am I reading too much into this, or is there a reason behind it? I strongly suspect there is something - perhaps related to the marginally slower relaxation of these protons, since there are no geminal protons (only vicinal protons)?

If it is of relevance - a 60° pulse is used (pulse program zg60). The number of scans ns is 16, and the relaxation delay d1 is 1 second. I'm happy to provide any other necessary acquisition parameters, or to provide the full structure of the compound if it's necessary.

P/S I found something in Findeisen & Berger's 50 and More Essential NMR Experiments, in which they mention that the integral of an aromatic proton in strychnine is smaller than expected as it has the longest relaxation time. Even though the authors used ns = 1, too short a delay between the receiver gain adjust and the scan was used, which led to incomplete relaxation and hence a smaller integral.

That's to be expected: Less protons as spatially close neighbours, less dipolar coupling, slower relaxation.

For a relaxation/recycle delay of $\pu{1 s}$ and $T_1$ of perhaps $\pu{3 s}$, the Ernst angle is already only $44^\circ$, so your measurement is far from optimal anyway. Use a shorter pulse or longer recycle delay.

At one second, your are anyway dangerously close to generating echoes which overlay the signal from further scans, distorting the accumulated spectrum.


Grignard reagent and acidic hydrogen Why does the Grignard reagent not react as a base in case of ketones even though the α-hydrogen atom in case of ketones is acidic? Grignard reagents do react with ethyne to give hydrocarbons, so why not here?

The only thing I can put my finger on is electrophilicity of the carbon.

Grignard reagent is a hard base. The HSAB theory states that soft acids react faster and form stronger bonds with soft bases, whereas hard acids react faster and form stronger bonds with hard bases, all other factors being equal (Wikipedia). Thus, Grignard reagent (a hard base) would preferably attack the harder position when it is available and in offer, like in the situation with its reaction with ketone carbonyl carbon even with the presence of $\alpha$-$\ce{H}$s. Keep in mind that the Grignard reagent probably react with $\alpha$-$\ce{H}$s to certain extent, but when the carbonyl group is readily accessible, that reaction proceeds to much faster rate. Yes, Grignard reagents do react with ethyne to give hydrocarbons, but it is a different story if the molecule contains a carbonyl group.


Are pi bond like like interactions possible when substituents in a compound are not directly connected? identify the more acidic compound (and the answer is the III is more acidic than II ).

My reasoning: Consider II compound, we have oxygen attached to the carbon forming a sigma and a $\pi$ bond, this oxygen has two filled p orbitals. The oxygen in the ortho position to it (in II), also has two filled p orbitals. My theory is that as we have filled p orbitals, they can form a $\pi$ bond type interactions (*). After forming the two $\pi$ bonds, the compound is stable.
Now discussing III, applying the same idea as above, there is a larger distance for the  $\pi$ overlap and hence I think $\pi$ overlap is less stable compared to the case in II.
Hence, we can say that III is more acidic (**) than II, as the pi overlap of the p orbitals of the two oxygen atoms are farther and long in III, and II is more stable (less acidic) as the pi overlap is over a small distance.
I reached the correct answer but I am not sure if my reasoning for it is correct because I haven't seen directly similar explanations in textbooks, however, I felt this idea was hinted at.

*: Not a $ \pi$ bond in the strong sense of the word
**: Less stable

First, let's get to the root of this problem, then I'll double back to your question. We are asked which of the two compounds is most acidic. As a side note there are various definitions of acidity, but note that broadly these theories describe acidity as linked to propensity to accept electron pairs in some form rather than general instability.
Considering neither II nor III have any immediately acidic functionalities (alcohol, carboxylic acid, sulfonic acid), we identify the alpha carbons of the carbonyl functionalities in each molecules as the most acidic hydrogens with pka ~20-25 (since all other hydrogens are alkyl (pKa ~50) and cannot form a stable conjugate base upon deprotonation).
With that in mind, the real difference between II and III is the identity of the carbonyl group. Comparing II to III we see II is an ester, with the carbonyl center bonded to an alkoxyl substituent and an alkyl substituent. Comparatively, III is a ketone, with its carbonyl center bound to to alkyl substituents. Recognizing that O lone pairs can hybridize to conjugate with the carbonyl center were C atoms cannot, we predict III to be more acidic.
The reasoning for this can be approached from a couple of angles, so here's two:

O lone pairs present in II destabilize the enol formed by deprotonation, making it more likely to donate electrons (I.E has more basic character than III), thus making its conjugate acid form less acidic (since weaker acids have conjugate bases that are stronger bases and vice versa)

O lone pairs can form an additional bond to the carbonyl carbon in a octet-satisfied resonance contributor, which greatly reduces the contribution of the analogous hyperconjugative resonance contributor with dissociation of H+. Note that this hyperconjugative contributor is a rather poor description in a thermodynamic sense, but it does play an effective stand in for a mechanism with a basic species.

Now to your actual question: You are correct with your analysis of the pi system of II, but I would note that your description is using atomic orbitals which can become very confusing in the world of ochem. In the hand-wavy hybridization terms, each oxygen would adopt an sp2 bonding orbital hybridization with unhybridized p orbitals participating in the pi-system. However, these pi-interactions can only occur if sequential atoms have unhybridized p orbitals.
This is due to the fact that the stability of pi systems arises from overlap of p orbitals. When two p orbitals are close enough to overlap, their wavefunctions can recombine into  relatively lower energy (bonding) and relatively higher energy (antibonding) wavefunctions which become orbitals. This same recombination can occur for any number of p-orbitals, with the sole condition being that overlap actually occurs.
This can only have a fully stabilizing effect when atoms are ~1.20-1.5 angstroms apart as indicated by spectroscopic bond lengths. Overlap is feasible between single-bonded substituents (Like C-O in II), but impossible once the continuous system of unhybridized p orbitals is broken.
Thus while the O in II can directly conjugate its electrons with the carbonyl pi system by "rehybridizing" from the general expected sp3 hybridization to stabilize the system, in III there are no adjacent unhybridized p orbitals to interact with the carbonyl pi system. The only electronic interactions present between the substituent (cyclic ether) and the carbonyl system would be inductive/field effects.
While one could argue that a conformer of III could possibly involve delocalization of O p electrons into the carbonyl system, I would respond by noting that such a conformer

Would have to have an unhybridized O p-orbital in order to be involved with pi interactions
Would have to have a torsional angle Θ between the oxygen and the carbonyl group of Θ=/=0 in order for overlap to actually occur (Distance between orbitals r<~1.5).

However, this would mean that the oxygen's unhybridized p orbital and the carbonyl pi system would have different symmetry elements, meaning such a pi-interaction would be symmetry-forbidden
That last bit might be jumping the shark a little, but I hope this helps!


Calculating empirical formula of vitamin C Combustion of 0.2000 g of vitamin C gives 0.2998 g $\ce{CO2}$ and 0.819 g of water. What is the empirical formula of vitamin C?

My work:

I calculated moles of products and found the mole ratio but it's not coming to be a whole number.

You are using the wrong number for the grams of water, it should be 0.0819.  Go back and redo your calculations with that change.  Now you should get the correct ratio of the elements in the molecular formula for ascorbic acid.  If you don't, report back and show your math.


Struggling to understand wavevector K for deriving energy levels in reciprocal space I don't really understand why, when deriving wavefunctions for arrays of atoms, the wavevector k is introduced and the general idea surrounding momentum space. please help!!

The Hamiltonian for a particle in a periodic array of atoms is invariant (doesn't change) under translation by any vector of the form $n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3$, where the vectors $\vec{a}_1,\vec{a}_2,\vec{a}_3$ for the unit cell and $n_1,n_2,n_3$ are any integers. The set of all such translations form a group.

When a Hamiltonian is invariant under all elements of some group, it follows that its eigenfunctions can be written as functions of definite symmetry with respect to that group. A function of definite symmetry with respect to the group described above turns out to be characterized by a "wavevector" $\vec{k}$. This is completely analogous to situation for the hydrogen atom, where spherical symmetry implies $[L^2,H]=0$ which in turn implies that total angular momentum is conserved and that our eigenfunctions possess a quantum number $l$ (along with $n$ and $m$) that quantifies the wavefunction's total angular momentum. To make the analogy clear:

Hamiltonian of hydrogen atom commutes with arbitrary rotations $\to$ eigenfunctions of Hamiltonian have quantum number $l$
Hamiltonian of particle in periodic potential commutes with arbitrary translations $n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3$ $\to$ eigenfunctions of Hamiltonian have wavevector $\vec{k}$.

A function $\phi_{\vec{k}}$ of wavevector $\vec{k}$ happens to have a special behavior under group operations:
$$
\phi_{\vec{k}}(\vec{x} + n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3) = e^{i\vec{k}\cdot(n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3)} \phi_{\vec{k}}(\vec{x})
$$
At this point we usually note that if two wavevectors $\vec{k}$ and $\vec{k}'$ differ by some $m_1 \vec{b_1} + m_2 \vec{b_2} + m_1 \vec{b_2}$, where
$$
\vec{b}_i = 2 \pi \frac{ \sum_{j,k=1}^3 \epsilon_{ijk} \vec{a}_j \times \vec{a}_k}{\sum_{m,n=1}^3 \epsilon_{imn} \vec{a}_i \cdot \left( \vec{a}_m \times \vec{a}_n \right)}
$$
and $m_1,m_2,m_3$ are integers, then
$$
\left( \vec{k'} - \vec{k} \right) \cdot(n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3) = 2 \pi \left( n_1 m_1 + n_2 m_2 + n_3 m_3 \right)
$$
so that the wave vectors have equivalent symmetry properties. This allows us to restrict our consideration of wavevectors to those of the form $c_1 \vec{b}_1 + c_2 \vec{b}_2 + c_3 \vec{b}_3$, where $ 0 \leq c_j \lt 1 $. More commonly we equivalently restrict the possible wavevectors to lie in the Brillouin zone.

As to why the concept is introduced, the answer is that recognizing that our energy eigenfunctions can be classified by some symmetry often enormously simplifies problem solving. Using again the hydrogen atom as an example, recognizing the Hamiltonian's spherical symmetry allowed us to break up a 3-dimensional Schrodinger equation into three 1-dimensional differential equations, which is an enormous simplification. For a periodic pontential, knowledge that our eigenfunctions have a wavevector $\vec{k}$ allows us to find the eigenfunctions by solving a differential equation over a single unit cell (as opposed to all space).


Determine stable oxidation state from Ionization energy Question:
Given the first, second and third ionization energies of an element (7 eV, 12.5 eV and 42.5 eV respectively), find the most stable oxidation state (or charge on the resultant ion) of that element.

I think that the element has to ionize with oxygen to reach a certain oxidation state, since oxygen has a 6 valence electrons, the element has to lose 2 electrons to attain stability. So the second ionization energy will the give the most stable oxidation state. So my answer is +2.
Now that happens to be the correct answer but can someone verify my reasoning?

You don't necessarily need to have your element binding with oxygen to be oxidized, per se. Unless it is, of course, specified that we're talking about a reaction with oxygen. So, don't misguide yourself on focusing on the oxygen here :) After all, if we, for instance, think about burning of potassium, it is being oxidized with oxygen, however, it still only makes $\ce{K+}$ and not $\ce{K^2+}$!
The trick in this question is to look for the values, or a magnitude, of those ionization energies in question. To remove the first electron, you need to provide $\pu{7 eV}$, to remove the second one, you need to provide $\pu{12.5 eV}$. So the difference between 7 and 12.5 is not that big if we compare it with the difference between $\pu{12.5 eV}$ and $\pu{42.5 eV}$ that is required to remove the third electron. If explaining it in simple terms, the first to energy "bumps" are easy to "climb", while the last one is way high and you really need to provide a sufficient amount of energy to get that high. Thus, removing the first two electrons is relatively easy, while removing the third one is not.
This gives you the most stable oxidation state for this element to be +2. If we perform a reaction of this element (which looks like $\ce{Sr}$) with any other electronegative element, say chlorine, it would still have a charge of +2.


Yellow color of NaCl My book says:

  $\ce{NaCl}$ shows metal excess defect, which results into the formation of an F-center in the crystal. This F-center is responsible for the yellow color of $\ce{NaCl}$.

How does an F-center provide color to the solid?

An F-center (or any color center) is a point defect in the material that generates localized electronic states. When these states are in the band gap, they will result in strong peaks in optical absorption. An F-center in an alkali-halide material is an electron bound to a negative ion vacancy.  A brief overview may be found in Ashcroft and Mermin's Solid State Physics, Chapter 30.


Synthetic route to hydroxychloroquine This drug hydroxychloroquine has become a new burning topic of discussion as a chemistry enthusiast:

I know that in an industrial preparation we need to account fpr a lot of factors like temperature, solvent, expenditure etc. I tried this, which did not seem good enough:

It seems that the nucleophilic aromatic substitution is not appropriate and the starting products chosen may not be commercially available. Can someone suggest a better yielding and more economical pathway?

This took a bit of thinking about but here is a viable route from commercially available bulk materials.

Start with nitroethane, deprotonate ($\ce{t-BuOK}$ in THF would be my choice, but $\ce{NaOEt/EtOH}$ should do it) and react with the commercially available dioxane protected 3-bromopropanal. Reduce the nitro group - alkyl nitro groups are tricky to reduce, I've found Hg/Al amalgam to be pretty good but others may have different conditions. There's a recent review of reductions$\ce{^{[1]}}$. The product of this will do the SNAr with 4,7-dichloroquinoline, I would try $\ce{K2CO3/MeCN}$ for this, but here are plenty of other possible conditions. This is  a good point to purify, then reveal the masked aldehyde by acid hydrolysis and do a reductive amination with the commercially available N-Ethylethanolamine to give the desired product.

Reference

Manuel Orlandi, Davide Brenna, Reentje Harms, Sonja Jost, and Maurizio Benaglia, Organic Process Research & Development 2018 22 (4), 430-445
DOI: 10.1021/acs.oprd.6b00205


Thermodynamic statements Question 13 from [1, p. 4.2]:

  Which of the following statements is true?

  (a) An ideal gas always obeys the equation: $PV^γ = \mathrm{constant}$ in adiabatic process.

  (b) An ideal gas always obeys the equation: $PV^x = \mathrm{constant}$ in
  polytropic process.

  (c) In a polytropic process, the heat capacity of the system remains same
  throughout.

  (d) In all the cyclic process, $w_\mathrm{net}$ by the system is non-zero.

To me (a), (b) and (c) appear to be correct, but the solution manual says only the (c) is right.

References

Neeraj, K. Advanced Problem in Physical Chemistry for Competitive Examinations, 2nd ed.; Pearson India Education Services Pvt. Ltd: Dehli, Chennai, 2015. ISBN 978-93-325-4373-7.

(a) is not true, because, in addition to being adiabatic, the process must be reversible.

(b) is not true, because the gas must be assumed to have constant heat capacity for this equation to apply


Understanding a chemical reaction from kegg I came across a reaction that looks as follows:

In this reaction the polyphosphate is common in both the sides of the reaction. Does this act as enzyme? But from the website it seems that the enzyme is ADP-specific glucokinase(in the enzyme section click 2.7.1.63). Then what is the function of polyphosphate here?

The $n$ of the polyphosphate should be decreased by 1 on the reactant side relative to the product side:

$\ce{(P)_{n} + Glc -> G 6P + (P)_{n-1} }$

Polyphosphate, like ATP, contains several phosphoanhydride bonds.  These are "high energy" bonds that can be used for biosynthesis.  But to do so one of these bonds needs to be broken, and that is exactly what happens when one of the end phosphates on the phosphoanhydride chain is peeled off and trasferred to glucose.


How to make approximation of rotational partition function of diatomic linear molecules? Using the rigid rotor approximation to the level energies, and such other appropriate assumptions, we can approximate rotational partition function, $Q_{\mathrm{rot}}$, of linear molecules as follows:
\begin{eqnarray}
  Q_{\mathrm{rot}} & = & \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{E_J}{kT}\right)}\\
  & \simeq & \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{hB_0J(J+1)}{kT}\right)} \\
\end{eqnarray}
In the book, Microwave Molecular Spectra (Gordy & Cook, 1984), Chapter 3, Equation 3.64, the authors made approximation as:
$$Q_{rot}\simeq\frac{kT}{hB_0}+\frac{1}{3}+\frac{1}{15}\left(\frac{hB_0}{kT}\right)+\frac{4}{315}\left(\frac{hB_0}{kT}\right)^2$$
Also, there is another approximation made by McDowell, R. S. 1988, Journal of Chemical Physics, 88, 356:
$$Q_{rot}\simeq\frac{kT}{hB_0}\exp\left(\frac{hB_0}{3kT}\right)$$

And we can see from the picture that $Q_{\mathrm{rot}}\simeq\frac{kT}{hB_0}+\frac{1}{3}$ is also a good approximation. Then I can guess how to derive McDowell's approximation, but I cannot derive Gordy & Cook's approximation.

What I have got are as follows:

Since $$n\exp(-x(n+1)n)+\int_{(n+1)n}^{(m+1)m}{\exp(-xy)}\,\mathrm dy+(m+1)\exp(-x(m+1)m) \\
\leq\sum_{J=n}^{m}{(2J+1)\exp(-x(J+1)J)} \\
\leq\sum_{i=n^2}^{m^2+2m}{\exp(-xi)}$$ hence, $$\int_{0}^{\infty}{\exp(-xy)}\,\mathrm dy=\frac{1}{x} \\
\leq\sum_{J=0}^{\infty}{(2J+1)\exp(-x(J+1)J)} \\
\leq\sum_{i=0}^{\infty}{\exp(-xi)}=\frac{1}{1-\exp(-x)}$$

i.e., $$\frac{kT}{hB_0}<Q_{\mathrm{rot}}<\frac{1}{1-\exp\left(-\frac{hB_0}{kT}\right)}$$

While the latter one approximates $\frac{1}{1-\exp\left(-\frac{hB_0}{kT}\right)}\approx \frac{kT}{hB_0}+\frac{1}{2}+\frac{1}{12}\frac{hB_0}{kT}-\frac{1}{720}\left(\frac{hB_0}{kT}\right)^3$ when $T$ is rather high.

Is there anyone knowing how to derive Gordy & Cook's approximation? Please tell me and I appreciate it very much.

Besides, another little question, are such approximations still useful as nowadays computers are greatly powerful?

Sorry for the late answer. I just discovered this question.

Preludium

Firstly, I think you might have an error or some non-standard notation in your formula for $Q_{\mathrm{rot}}$. The rotational energy levels are given by

\begin{align}
  E_{\mathrm{rot}} = \frac{\hbar^2}{2 I} J ( J + 1 ) \ ,
\end{align}

where $I$ is the moment of inertia.
Then the rotational constant $B_{0} = \frac{\hbar^2}{2 I}$ (in units of energy) is introduced, so

\begin{align}
  E_{\mathrm{rot}} = B_{0} J ( J + 1 ) \ .
\end{align}

Thus

\begin{align}
  Q_{\mathrm{rot}} = \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{B_0 J(J+1)}{kT}\right)} \ ,
\end{align}

whereas your exponent contains another $h$ which should not be there. At least that's the standard notation I am used to. For convenience I'll also introduce the rotational temperature $\Theta_{\mathrm{rot}} = \frac{B_{0}}{k}$, so that

\begin{align}
  Q_{\mathrm{rot}} = \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right)} \ .
\end{align}

Main Answer

Ok, now that that's sorted out, let's start the real work. It is actually a pretty standard approximation that is used here: They use the Euler-Maclaurin formula

\begin{align}
  \sum_{n = a}^{b} f(n) &= \int_{a}^{b} f(n) \, \mathrm{d} n + \frac{1}{2} \bigl( f(b) + f(a) \bigr) + \sum_{i=1}^{\infty} (-1)^{i} \frac{b_{2i}}{(2i)!} \Bigl( f^{(2i - 1)}(a) - f^{(2i - 1)}(b) \Bigr)
\end{align}

where $f^{(k)}(a)$ is the $k$th derivative of $f$ evaluated at $a$. The $b_{2i}$s are the absolute values of the even Bernoulli numbers, $b_{2} = \frac{1}{6}$, $b_{4} = \frac{1}{30}$, $b_{6} = \frac{1}{42}$, $...$.
It provides a way to approximate the summation of a function over a discrete variable by an analogous integration over a continuous variable.

In the special case at hand,

\begin{align}
  f(J) = (2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right) \ ,
\end{align}

where the summation limits are $a=0$ and $b=\infty$ and where $f(\infty) = f^{\prime}(\infty) = \ldots = 0$ this formula boils down to

\begin{align}
  Q_{\mathrm{rot}} = \sum_{J = 0}^{\infty} f(J) &= \int_{0}^{\infty} f(J) \, \mathrm{d} J + \frac{1}{2} f(0) + \sum_{i=1}^{\infty} (-1)^{i} \frac{b_{2i}}{(2i)!} f^{(2i - 1)}(0)
\end{align}

The first term is what is usually called the classical or the high-temperature limit. It can be easily calculated by making the substitution $K = J(J + 1)$ $\Rightarrow$ $\mathrm{d}J = \frac{\mathrm{d}K}{2J + 1}$. So

\begin{align}
  \int_{0}^{\infty} f(J) \, \mathrm{d} J &= \int_{0}^{\infty} (2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right) \, \mathrm{d} J \\
  &= \int_{0}^{\infty} \exp\left(-\frac{\Theta_{\mathrm{rot}} K}{T}\right) \, \mathrm{d} K \\
  &= \frac{T}{\Theta_{\mathrm{rot}}}
\end{align}

Now, only the derivatives $f^{(k)}(0)$ need to be sorted out. Those that contain terms below the third order in $\frac{\Theta_{\mathrm{rot}}}{T}$ can be calculated to be:

\begin{align}
  f(0) &= 1 \\
  f^{(1)}(0) &= 2 - \frac{\Theta_{\mathrm{rot}}}{T} \\
  f^{(3)}(0) &= - 12 \frac{\Theta_{\mathrm{rot}}}{T} + 12 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \\
  f^{(5)}(0) &= 120 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - 180 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + 30 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{5} \ .
\end{align}

Doing this by hand is very tedious. Best you use something like Mathematica. Plugging all this into the Euler-MacLaurin formula and dropping all terms above second order in $\frac{\Theta_{\mathrm{rot}}}{T}$ you get exactly the result of Gordy & Cook

\begin{align}
  Q_{\mathrm{rot}} &\approx \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{2} \cdot 1 + (-1) \frac{1}{6 \cdot 2!} \left( 2 - \frac{\Theta_{\mathrm{rot}}}{T} \right) \\
 &\quad + (-1)^{2} \frac{1}{30 \cdot 4!} \left( - 12 \frac{\Theta_{\mathrm{rot}}}{T} + 12 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \\
  &\quad + (-1)^{3} \frac{1}{42 \cdot 6!} \left(120 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - 180 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + 30 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{5} \right) \\
  &= \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{2} - \frac{1}{6} + \frac{1}{12} \frac{\Theta_{\mathrm{rot}}}{T} - \frac{1}{60} \frac{\Theta_{\mathrm{rot}}}{T} \\
  &\quad + \frac{1}{60} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \frac{1}{252} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \\
  &= \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right)
\end{align}

As a side note: If you have a homonuclear diatomic molecule you need to introduce the so called symmetry number $\sigma$ which represents the number of indistinguishable orientations a molecule can have. It originates from symmetry constrains: for homonuclear diatomic molecules there is a two-fold axis of symmetry perpendicular to the internuclear axis which gives rise to two indistinguishable orientations, so the above formula for $Q_{\mathrm{rot}}$ overcounts the number of available quantum states
by a factor of two because of the inherent indistinguishability of the nuclear pair. Thus, $Q_{\mathrm{rot}}$ needs to be divided by $\sigma = 2$. For heteronuclear diatomic molecules you simply have $\sigma = 1$.
With the introduction of $\sigma$ the equation for $Q_{\mathrm{rot}}$ becomes:

\begin{align}
  Q_{\mathrm{rot}} &\approx \frac{1}{\sigma} \left( \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \right) \\
  &= \frac{T}{\sigma \Theta_{\mathrm{rot}}} \left(1  + \frac{1}{3} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{1}{15} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} \right) \right)
\end{align}


I drink city water, city has lead pipe, how is this safe? 6 years ago, I needed to replace the main water shut off valve to my house.  This required the city to turn off the water at the street. Incompetent city employee drives key through water pipe. Plumbers fix pipe, I realize that pipe is lead.

Yearly, the city (90% of which was built before 1950) sends out a water quality report, which reports undetectable amounts of lead.  That is interesting since probably 90% of the city has lead pipe.

Several questions :

How is lead pipe safe from leaching lead into water?
Do cities usually treat the water to prevent lead from leaching into the water?
What is the form of harmful lead (e.g. is it in aqueous solution, lead oxide, or forming  ligands with some other molecule)?

In light of the recent events in Flint Michigan, this issue has come to the forefront (see this article from the American Chemical Society's Chemical  & Engineering News and see this article). To summarize:

Active Measures to prevent lead corrosion:

Maintain high pH. This makes lead carbonates less soluble in water.
Add orthophosphates which create relatively insoluble layer on the inside of the pipes.

Without these treatments, the addition of chloride disinfectants aggravates the corrosion problem by forming lead (II) chlorides (which is water soluble).

As mentioned by Jaroslav, the time that water sits in the system also plays a factor.


Can decarboxylation be done under carbon dioxide? From Why is the decarboxylation of 2-hydroxybenzoic acid (and related compounds) done 'under nitrogen'? I confirmed that nitrogen is used to exclude oxygen - in order to prevent oxidation of the material.

Can $\ce{CO_2}$ be substituted for nitrogen, or will the presence of carbon dioxide slow down the decarboxylation reaction (which is causing carbon dioxide to be released)?

Because nitrogen is the standard inert gas in lab practice. If you are "rich" You can use argon, but carbon dioxide is seldom used, because it won't work on alkaline solutions.

  or will the presence of carbon dioxide slow down the reaction

First, what you presumably mean is influencing the equilibrium unfavourably, don't you? You should learn to discern kinetics and equilibrium.

Answer: not really. Whether you have a carbon dioxide atmosphere or not will shift the decomposition temperature not more than some degrees.


What is the expected number of rings obtained in the following ring formation process? Refluxing, say, one mole of 1,5-dibromo-pentane with excess ammonia in ethanol can result in the formation of rings. (e.g. $1\lambda^2$-piperidine)

However, it is also possible for the 1,5-dibromo-pentane molecules to be connected together, forming longer chains without forming rings. The longer chains can be reacted again to form larger rings. So in theory, we can have rings that are arbitrarily large.
This is about the rings' size distribution
Question: What is the expected number of rings (large and small) after the reaction finishes? How many of the rings are $1\lambda^2$-piperidine? Generally, is there a mathematical distribution that describes the expected number of rings of any given size?

There is no simple answer here because the outcome will be heavily dependent on the concentration and on the rate of addition.

The most salient point in your question to me is: "excess ammonia."

You can ask what the competing reactions are.

1,5-dibromopentane reacts with ammonia to form 1-amino-5-bromopentane
1-amino-5-bromopentane reacts with itself to form piperidine
1-amino-5-bromopentane reacts with ammonia to form 1,5-diaminopentane
1-amino-5-bromopentane reacts with another straight chain alkyl amine to elongate the chain
1,5-dibromopentane reacts with piperidine to form N-(5-bromopentyl)piperidine
and others...

The main problem I see is that in excess ammonia, unless you perform a slow addition of ammonia, the concentration may be such that the ammonia will over add via (3). Then you're going to get any product. The intermediates are also reactive, so the reactant concentration has to be relatively low as well, otherwise, you may see (4) or (5) or other side reactions. You may also need to run the reaction at lower temperature to disfavor reaction of the product with any alkyl bromide intermediate or reactant.

These issues need to resolved before you even begin to think about distribution of ring sizes because you might not even form any rings.