Deriving kinetic equations for a complex reaction system I have a system of protein reactions that I've derived differential equations for the kinetics of. I'm using them to run simulations to fit the rate parameters, but the simulations at present are working badly. I'd like to start by checking the system of differential equations. I realise they may be to complex for anyone to write them out here, but you know of software that could be used to check my work, please post it here. (A lot of people face similar problems)

The three monomers are $L$, $A_1$, $A_2$; the chemical equations for the monomers forming dimers are:
$L + A_1 \rightleftarrows^{k_{a1}}_{k_{d1}} LA_1$,
$\qquad$
$L + A_2 \rightleftarrows^{k_{a2}}_{k_{d2}} LA_2$,

and the dimers forming trimers are:

$LA_1 + L\rightleftarrows^{k_{a1}}_{k_{d1}} LA_1L$,
$\qquad$
$LA_1 + A_2\rightleftarrows^{k_{a2}}_{k_{d2}} A_1LA_2$,

$LA_2 + L\rightleftarrows^{k_{a2}}_{k_{d2}} LA_2L$,
$\qquad$
$LA_2 + A_1\rightleftarrows^{k_{a1}}_{k_{d1}} A_1LA_2$,

trimers forming tetramers:

$LA_1L+A_2\rightleftarrows^{k_{a2}}_{k_{d2}} LA_1LA_2$,
$\qquad$
$LA_2L+A_1\rightleftarrows^{k_{a1}}_{k_{d1}} LA_2LA_1$,

$A_1LA_2 + L\rightleftarrows^{k_{a1}}_{k_{d1}} LA_1LA_2$,
$\qquad$
$A_1LA_2 + L\rightleftarrows^{k_{a2}}_{k_{d2}} LA_2LA_1$,

dimers forming tetramers

$LA_1+LA_2\rightleftarrows^{k_{a1}}_{k_{d1}} LA_1LA_2$,
$\qquad$
$LA_1+LA_2\rightleftarrows^{k_{a2}}_{k_{d2}} LA_2LA_1$,

tetramer molecules changing their configuration

$LA_1LA_2 \rightleftarrows^{k_{a4}}_{k_{d4}} B$,
$\qquad$
$LA_1LA_2 \rightleftarrows^{k_{a3}}_{k_{d3}} B$.

Could anyone either point me in the direction of software to derive the system of differential equations, or do the derivation?

In many cases, complex systems of even simple differential equations do not have simple and obvious solutions. Therefore, it could be too difficult and not reasonable to find fully analytical solutions.

Usually, numerical solutions can be easily calculated using iterative algorithms. However, such computer codes can be susceptible to calculation artefacts. Furthermore, the validation and verification of the model and the code can be a difficult issue.

I made a compromise and used an analytical approach when solving systems of differential equations, but I allowed the code to evaluate subexpressions involving constants numerically. Thus, I made numerical calculations to combine all constant parameter values as soon as possible in order to limit the number of parameters in the equations (using excessive digits in order to avoid accumulation of round-off errors).

I found that this approach can be easily and quickly implemented using the commercial software Maple. However, other similar mathematical software could be suitable, too.

I successfully used this approach for the evaluation of radioactive decay of the inventory of a nuclear reactor core in dependence of time after shutdown. The calculation of the radioactive decay of the relevant radionuclides involved more than 100 individual differential equations, which mainly describe chains of simple first-order reactions, e.g.:

$$\frac{\mathrm da_{\text{I-131}}(t)}{\mathrm dt}=-\lambda_{\text{I-131}}\cdot a_{\text{I-131}}(t)+0.778\cdot\lambda_{\text{I-131}}\cdot a_{\text{Te-131m}}(t)+1\cdot\lambda_{\text{I-131}}\cdot a_{\text{Te-131}}(t)$$

> de_a_I131:=diff(a_I131(t),t)=-l_I131*a_I131(t)+0.778*l_I131*a_Te131m(t)+1*l_I131*a_Te131(t);

A large number of further differential equations could be added in order to model transport phenomena during postulated accidents (e.g. loss of coolant).

Differential equations that were dependent on each other and the corresponding initial conditions were grouped into systems of differential equations, e.g.:

> sys_131:={de_a_Xe131m,de_a_I131,de_a_Te131,de_a_Te131m,a_Xe131m(t0)=core_Xe131m,a_I131(t0)=core_I131,a_Te131(t0)=core_Te131,a_Te131m(t0)=core_Te131m};

The systems of differential equations were solved analytically and the subexpressions involving constant parameter values were evaluated numerically.

> sol_131:=evalf(value(dsolve(sys_131,[a_Xe131m(t),a_I131(t),a_Te131(t),a_Te131m(t)])));

The results could be easily rearranged into a suitable form, e.g.

$$\begin{aligned}a_\text{I-131}(t) ={} & -1.6030637832147524320\cdot10^{17}\cdot\mathrm e^{-6.4180294496291232354\cdot10^{-6}\cdot t}\\&-7.9769363886868464215\cdot10^{15}\cdot\mathrm e^{-4.6209812037329687295\cdot10^{-4}\cdot t}\\&+4.7213833147101620896\cdot10^{18}\cdot\mathrm e^{-9.9782796169607015473\cdot10^{-7}\cdot t}\end{aligned}$$

The results could also be directly used for further calculations within the same worksheet.


White snow-like powder below sodium hydroxide solution I have created sodium hydroxide solution with half cell brine electrolysis but, finally I got white snow-like insoluble powder sank below sodium hydroxide.
What should I do keep or remove it?

Sodium hydroxide is easily contaminated with sodium carbonate. The free base continuously reacts with atmospheric carbon dioxide to form the salt.When a relatively concentrated solution of NaOH is prepared, sodium carbonate slowly precipitates out as a white gelatinous mass.
a) How long did you expose it to air?
b) What is the expected concentration of the base?
c) What are your electrodes made of?


Ranking order of nucleophilicity According to me the answer should be (c),(b),(a),(d). As we move left to right in a period the electronegativity increases and hence the tendency to donate electrons will decrease so nitrogen is better electron donor than oxygen. Now to select between three oxygen donating species we can use the steric hindrance concept and hence my answer.

  But the correct answer is (b),(c),(a),(d).

Please tell me what is my mistake.

The first thing to note is that nucleophilicity is a kinetic parameter that cannot be addressed easily with thermodynamic parameters such as electronegativity. Instead, you need to inspect all potential nucleophiles by their kinetic merits.

One easy thing that we can use, even though it is thermodynamic, is the effective charge. The centre of negative charge of the acetate molecule is somewhere in the middle between the two oxygens and that of tosylate is somewhere in the middle between three equivalent oxygens. Also, in both systems the oxygens are bound to very highly oxidised atoms ($\ce{C^{III}}$ and $\ce{S^{IV}}$, respectively), which greatly reduces their electron density and thus their desire to attack nucleophilicly. Finally, in acetate’s case the nucleophilic attack would reduce the effectiveness of the carboxylate resonance as one of the two mesomeric structures becomes disfavoured (charge separation) which adds another grain of salt against nucleophilic attacks.

Tosylate is less nucleophilic than acetate because there are three oxygens than two but they only bear a single negative charge between them.

Now onto cyanide and methanolate. For cyanide, first note that it is both $\ce{C}$-nucleophilic and $\ce{N}$-nucleophilic. Since the HOMO, the highest occupied molecular orbital, has a much larger contribution on the carbon side, the $\ce{C}$-attack is strongly favoured over $\ce{N}$-attack and we can ignore the latter.

We can again use the bond-partners argument here. The oxygen of methanolate is bonded to a $\ce{CH3}$ group which, in general, is electron donating, thus increasing its electron density. The cyanide carbon, however, is bonded to the more electronegative nitrogen which slightly reduces its electron density. Thus, we can infer that methanolate should be a better nucleophile than cyanide. (Note, however, that both are very good nucleophiles, especially when compared to the other two.)


Chaotic melting points of n-alkyl carboxylic acids Is there any trend here at all? This seems very chaotic as a trend.

My guess without looking at any literature:

Higher acidity $\implies$ higher hydrogen bonding $\implies$ higher melting point.

Thus, changing from H and Me where there is virtually no +I effect from the chain, to Et, Pr, Bu where the +I effect of the chain decreases the melting point, as the hydrogen bonds get weaker, as the COOH groups are less acidic.

Further increasing the chain length after hexyl does not really lead to any additional +I effects on the COOH group that is now far away. Thus, the melting point increases as there are additional van der Waals interactions between the longer chains.

Then there seems to be a factor that makes crystals of odd chains more favorable.


How to do complexometry titration of calcium oxalate? My titration of  calcium oxalate with NaEDTA doesn't work. I need to calculate the ion calcium within my sample. My essay is to test anticalculi activity.

Why do I say I have problem? Because the colour keeps turning back to its initial colour (pink) more than twelve times. It's never ending and the titration end point is so premature.

In my preparation, I added HCl 0.5 M to my calcium oxalate so it dissolves. Calcium oxalate has very low solubility.

First, I used EBT indicator and ammonia buffer. When i add the ammonia the solution became foggy and there's a precipitation within the sample.
So I used substitution titration with Mg-EDTA (EBT indicator) = Still happens
Then change the indicator to hydroxynaftol blue and calcon = Still happens (I don't use buffer with this indicator, only NaOH, but when I add NaOH it's became foggy again)

Does it have to be complexometric?

You could acidify the sample and do a redox titration of the oxalate with permanganate or dichromate. And since the amount of oxalate is the same as calcium, bob's your uncle.


Why exactly does molten NaCl explode, when it is poured into water? Why does molten $\ce{NaCl}$ explode, when it is poured into water?
$\ce{NaCl}$ has a high melting point, $1074\ \mathrm{K}$ ($801~\mathrm{^\circ C}$). $\ce{NaCl}$ has a molar mass of $58.44\ \mathrm{g/mol}$, it has specific heat capacity of $36.79\ \mathrm{J/(K\cdot mol)} = 629.53\ \mathrm{J/(K \cdot kg)}$. therefore the $\ce{NaCl}$ at melting point temperature has $491,033\ \mathrm{kJ/kg}$ more thermal energy than $\ce{NaCl}$ in STP conditions ($1\ \mathrm{atm}, 20~\mathrm{^\circ C}$, temperature difference $= 780\ \mathrm{K}$)
According to one amateur video in youtube, the molten salt explodes when it is poured into water (c. 2:15)
What is the exact cause of the explosion?
The author of the video reasons that the phenomenon is purely physical and it is caused by that water heats up, vaporises and expands as a gas inside the glimp of very hot $\ce{NaCl}$ salt. But is it really everything that happens there?
Other possible processes present in such occasion are (this is just a list what comes into my mind):

rapid crystallisation of the $\ce{NaCl}$
chemical reaction between $\ce{Na}$ and water: $$\ce{2 Na (s) + 2 H2O (l) -> 2 NaOH (aq) + H2 (g)}$$ (this causes explosion if $\ce{Na}$ is inserted into $\ce{H2O}$)
reaction between $\ce{Cl}$ and water: $$\ce{Cl2 + H2O -> HOCl + HCl}$$
the rapid solubility of hot $\ce{NaCl}$ into water
thermal decomposition (thermolysis) of $\ce{H2O}$ into either monoatomic or diatomic hydrogen and oxygen, and reactions that follow this.

Also, the author fails to explain why no explosion occurs with sodium tetraborate nor sodium carbonate when each of them, in molten state, was poured into the water.

Understand that sodium chloride is not made up of sodium metal and chlorine gas but of sodium ions and chloride ions, held together by ionic interactions. Under these thermal conditions (liquefication), the compound will not decompose into its elements and therefore all reactions you suggested which include elemental chlorine or sodium cannot occur.

Dissolution of sodium chloride in water is neither strongly exothermic nor strongly endothermic, so any effects stemming from the dissolution are neglegible.

Also note that you neglected an important variable in your calculations. As far as I can tell, you only calculate how to arrive at melting-point hot solid sodium chloride. To liquefy, additional melting enthalpy has to be applied, a further reservoir of energy to draw from. This melting enthalpy is of course released upon rapid crystallisation, but you should really subsume it into the heat energy difference altogether.

Finally, thermal decomposition of water is not exactly a process that will happen quickly, as exemplified by (water) steam temperatures that can be huge. Check out this unrelated answer of mine about how much energy is required to heat water to $100~\mathrm{^\circ C}$ and boil it.

So all things considered: The principle contribution to the explosion is the rapid heating of water including boiling, leading to tightly compressed gaseous $\ce{H2O}$ which expands explosively. This concurs with the recrystallisation of $\ce{NaCl}$ which accounts for a non-neglegible portion of the energy added, but which is technically already present in molten $\ce{NaCl}$. All other processes are minor or impossible.


What is the pH of a 5M solution of hydrochloric acid? What is the $\mathrm{pH}$ of a $\pu{5M}$ solution of hydrochloric acid?

So in order to solve this, apparently all you need to do is plug it into the equation:

$$\mathrm{pH} = -\log[\ce{H3O+}]$$

where the concentration of $\ce{H3O+}$ is $\pu{5M}$. But why can we assume this? Don’t we need to create an ice table from $\ce{HCl + H2O-> H3O+ + Cl-}$ and use the $K_\text{a}$ value of $\ce{HCl}$ to solve this?

As explained in pH Paradoxes: Demonstrating That It Is Not True
That pH ≡ -log[H+] Journal of Chemical Education Vol. 83 pages 752-757.

pH = -log (hydrogen ion activity).

and "pH of 7.6 M HCl is about -1.85 (not -0.88)"

So how far off is -log (5) = -0.69 from the real answer?

From this table of activity coefficients 5m HCl has an activity coefficient of 2.38, so activity is 11.9, which yields pH = -1.1.

Unfortunately the table is in terms of molality rather than molarity.  5M HCl is about 5.5m, and activity coefficient is dramaticly increasing with concentration (see table I of the pH paradox reference) so the true pH deviates somewhat further from -0.69, but pH = -1.1 is a reasonable estimate.


Do magnetic fields generated from ion flow through a membrane channel have any physiological relevance? Membrane pores and transporters see millions of ions flow through them per second. This creates a current and therefore a magnetic field. Do cells have any use for these fields (like maybe drawing charged receptors together?) or is it physiologically irrelevant?

For currents in biological systems the magnetic field is minuscule, and likely has no physiological effect.

That said, magnetoencephalography is based on measuring the tiny magnetic fields created when nerves fire. Detection requires an incredibly sensitive detector, a superconducting quantum interference device (SQUID), to detect the location of the currents. A SQUID can detect fields of just a few atoteslas (aT)!

Perhaps one could investigate how much ion flow in electric eel muscle is constrained by the induced magnetic field... It could lead to shocking revelations.


Calculating bond disassociation energy and bond energy for ATP I'd like to calculate Bond Dissociation Energy and Bond Energy for ATP to a similar molecule. What free (at least for academic use) computer programs which natively support and document BDE?

If you don't mind, could you provide a tutorial for using those programs to perform the calculations?

An example program is GAMESS.  It is free. But be warned: the calculation of BDEs is an extremely difficult thing to do.  I am presuming that you are a beginner, so you have to be careful for issues surrounding which BDE:
\begin{align}
\ce{A-B &-> A+ + B-}   \tag{ionic}\\
\ce{A-B &-> A. + .B}  \tag{homolytic}
\end{align}

Most electronic structure theory programs assume gas-phase, absence of solvent.  We know that homolytic dissociation is favored in the gas phase, as charge separation into ions is unfavorable in a Coulombic energy.

That being said, the run-of-the-mill electronic structure theory calculations (e.g., HF, DFT, MP2, CCSD, etc. etc.) will not break the atoms apart in a homolytic manner.  They tend to break apart atoms in the ionic fashion.  But this is a problem, as ions are abhorrent in the gas phase!  Basically, simple electronic structure theory methods over-estimate the bond strength, as the product of the dissociation (ionic) is erroneous.

You must use multiconfigurational methods to get it right (e.g. CASSCF, MCSCF). But these are very, very difficult unless you have had a lot of training.  I did them for years, and they were still hard for me.  But they will give you $\ce{A. + .B}$ as a product!  That is the gas-phase BDE.

You may be able to study closed-shell, closed-shell dissociations without much problem, using the HF and DFT methods.  Examples include dimers, host-guest interactions where the products look like good-old fashioned molecules that can stand by themselves. (Think water dimers, etc.)

Also, beware of Solvent Effects (favoring ionic dissociation $\ce{A-B -> A+ + B-}$) and solvation energy.  In fact, the solvation energy is why $\ce{Na-Cl}$ breaks apart into $\ce{Na+}$ and $\ce{Cl-}$.  In the gas phase, it would be $\ce{Na.}$  and $\ce{Cl.}$ radicals.

In summary, BDEs are very difficult things to compute to qualitative accuracy. Quantitative accuracy is even harder.  You may want to have an expert look over your procedure and results before you trust them completely.


Thermochemistry Problem - How did they get that volume? The chemistry problem in my textbook is : Calculate the work one when 2.0 L of methane gas, $\ce{CH4}$ (g) undergoes combustion in excess oxygen at 0ºC and 1.00 bar. Assume the volume of water formed is negligible.

I know that we should use the work equation $\mathrm{Work=-Pressure \cdot \Delta Volume}$, and that the pressure is 1.00 bar and we must convert this into pascals. All i need is the volume and convert that to $\mathrm{m^3}$. I got the equation $\ce{CH4 + 2O2 -> CO2 + 2H2O}$, but what I don't get is that how do we find the change in volume?

The book says let $\ce{O2}$'s volume = $\mathrm{V_{initial}}$, and that the total volume of gases for initial is $\mathrm{V_{i}}$ + 2.0 L. It says the final volume is 2.0 L of $\ce{CO2}$ (which I get because it should be the same amount of liters for carbons), but then it says that the water's volume is $\mathrm{V_{i}}$ - 4.0 L of $\ce{O2}$.

Can someone please explain the water's volume? How did they get that? Why is then the total final volume $\mathrm{V_f}$= $\mathrm{V_i}$ - 2.0 L. I thought the number of moles for initial should be same as the number of moles according to Gay-Lussac's law, so then the number of volume total should be the same for both sides, but this is clearly not correct. Am I applying the Gay-Lussac's law wrong? Why is the final volume not the same as the initial volume even if the initial and final have the same amount of moles on each side?

I think most of your issues regarding this question will be solved if you actually put the states in your chemical equations, so it becomes: $$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$ It is important to note that all the substances are in the gaseous state except for water which is a liquid. Therefore your application of Gay-Laussac's law is incorrect here as the number of moles of gaseous reactants is not equal to number of moles of gaseous products. Thus the initial and final volume shouldn't be the same. This is also why in the question they have stated to assume that the volume of water is negligible as water is a liquid which has very little volume compared to a gas.

The change in volume can be calculated by noting that 3 moles of gas reacts to form only 1 mole of gas. Since there is 2L of methane, that means that the volume is going to decrease by 4L.

Therefore: $\mathrm{V_f = V_i - 2.0L}$


Where to put the number in organic nomenclature? Should one write 4-hydroxypentan-2-one or 4-hydroxy-2-pentanone?
It has been seen that both have been used frequently.

The IUPAC recommendations on this matter have been changed.

According to the 1979 recommendations, the correct name was still 4-hydroxy-2-pentanone.

According to the 1993 recommendations, however, locants were placed immediately before the part of the name to which they relate. This rule is still included in the 2013 version. Therefore, the correct name is now 4-hydroxypentan-2-one.


Does grignard reagent react with oxygen? Does this reaction actually take place -

$$\ce{C2H5MgCl + O2 ->[H2O] C2H5OH}$$

I never saw this kind of a reaction. I thought that grignard reagent might take a proton from water and just form ethane.

This paper says it does and notes that while the yield of alcohols from aliphatic Grignards is 60-90%, formation of phenols is much less satisfactory. Other methods exist for the formation of phenols from aryl organometallics with much better yields.


Does it make a difference in energy if a reaction is done in many steps or in one step? I am particularly interested in knowing about glucose breakdown. If we artificially burn glucose, will the energy released be the same if it is broken down in steps(like in a living cell)?

You might want to take a look at Hess's law. It states that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps. But that covers only the thermodynamic aspect. In reality, you also have kinetic barriers (activation energy) which add to the energy needed for a reaction to run (or released during a reaction). That's why catalysts are so useful: They break a reaction down into steps (involving different intermediates) with lower activation barriers than the original reaction so that the reaction as a whole can proceed at a lower temperature.


Overall equilibrium expression for competitive equilibria I have a case where I have two reactions that depend on a common reactant. The reactions can be written as:

I) $\ce{A + B <=> AB}$

II) $\ce{A + C <=> AC}$

We can thus write equilibrium expressions for both reactions:

I) $K_\mathrm{I} = \frac{[\ce{AB}]}{[\ce{A}][\ce{B}]}$

II) $K_\mathrm{II} = \frac{[\ce{AC}]}{[\ce{A}][\ce{C}]}$

How can I combine these equations to get an equation that relates the equilibrium concentration of B to that of C?

Secondarily, can this be extended to give an equation that relates the equilibrium concentrations of B and C to the initial concentrations of A, B and C?

I already know the equilibrium and rate constants for both reactions in isolation. If it is relevant, A is a solid with a fixed number of surface sites, to which B and  are adsorbed. B and C are in aqueous solution.

OK, so essentially we have $[A],[B],[C],[AB],\text{ and }[AC]$, and a few equations: three conservation laws and two equilibrium constants.
$$\left\{
\begin{align}
[A]+[AB]+[AC] & = A_0\\
[B]+[AB] & = B_0\\
[C]+[AC] & = C_0\\
{[AB]\over[A][B]}& =K_1 \\
{[AC]\over[A][C]}& =K_2
\end{align}
\right.$$
With 5 unknowns and 5 equations, this should be solvable (because the nature manages to solve it somehow, if not for other reason). There is no guarantee, however, that the solution will be nice. Indeed, if we deal with the system exactly as it stands now, it is equivalent to a certain 4th degree algebraic equation which is better solved numerically.

Things get somewhat more manageable if we may disregard something. For example, if A is a relatively minor component, so that $[A]<<[B]$, then we may assume $[B]\approx B_0$ (and the same for C). All of a sudden, the problem becomes linear:
$$[AB]=K_1[A]B_0 \\
[AC]=K_2[A]C_0 \\
[A]+K_1[A]B_0+K_2[A]C_0=A_0
$$
which yields
$$
[AB]={K_1B_0\over1+K_1B_0+K_2C_0} \\
[AC]={K_2C_0\over1+K_1B_0+K_2C_0}
$$

Whether or not this approximation is realistic in your particular case is up to you.


Effect of different electron withdrawing/transition state stabilising groups in nucleophillic aromatic substitution I have been asked the following question in my problem sheet:

The only differemce between reaction A and reaction B is the electron withdrawing group that stabilises the anionic transition state changes from $CN$ in reaction A to $NO_2$ in reaction B. I can see no glaringly obvious reason why there should be a different rate. The problem is really (at least in my mind): what is the better electron withdrawing group out of $CN$ and $NO_2$. Is this a reasonable interpretation? If so, what reasoning can I bring to the answer to suggest that one is a better EWG than the other?

Your reasoning is correct.  The relative reaction rates would probably be quite similar if we didn't have the methyl groups on both sides of the substituent X.

In order for the substituent to stabilize the transition state through resonance, the substituent must be coplanar with the aromatic ring.  This isn't a problem for the linear $\ce{CN}$ group, but the $\ce{NO2}$ group with the two oxygens protruding out will not be able to remain planar with the aromatic ring.  The two adjacent methyl groups create steric interference with the planar nitro group; this steric interaction is minimized when the nitro group is twisted out of the aromatic plane.  When the nitro group is twisted out of plane it cannot resonance stabilize the transition state.  Hence the nitro compound reacts more slowly than the cyano compound.

While the underlying reaction is different, the thinking behind this question is identical to that found in this question from yesterday.


Why is the radius of the caesium cation smaller than that of the chloride anion? I am currently studying the structures of ionic compounds and how we can use radius ratios to predict the possible arrangement of ions in the lattice. My book mentions that the cation to anion radius ratio in caesium chloride is about 0.93, and since this number lies between 0.732 and 1, it suggests a cubic arrangement with a coordination number of 8.

But caesium is in the 6th period of the periodic table whereas chlorine is in the 3rd. A caesium cation would be similar to xenon, which is in the 5th period. And a chloride anion would be similar to argon, a period 3 element.This means that there is a difference of two shells between the two atoms. How is the ionic radius of chloride still larger? In general, how many "shells of difference"  are required for a cation to be larger than an anion?

This is something students often forget when comparing radii: shells are not simply additive.

If you move along eight elements to go from lithium to sodium, you are not only filling the second shell and putting one electron into the third, you are also adding a proton to the nucleus with every additional electron. The larger the nuclear charge is, the greater the electrostatic potential the electrons experience. And the greater the potential the stronger the force that draws them towards the nucleus, thus the more compact an orbital will become. Thus, while each new period will enlargen the atomic radius each added proton will reduce it. Therefore, the atoms do not become infinitely larger in size — although a sodium atom will always be larger than an electronically comparable lithium atom.

A similar discussion can be made for charges. Adding an electron to the mix increases the electron-electron repulsion (more negative charge) while removing one decreases it. Thus, with each positive charge a cation will get smaller while an anion will get larger with each negative charge. All of this results in rather complex calculations; one should always remember that an atomic radius cannot be easily guessed unless it is a straightforward up/down or left/right movement in the periodic table.

As for your second question: an anion may be smaller than a cation if the anion has very few shells and the cation very many. Unfortunately, the Wikipedia article on ionic radii does not quote a size of the hydride anion which would otherwise be a strong contestant. Instead, the smallest anion in the tables is $\ce{F-}$ with a crystal ionic radius of $119~\mathrm{pm}$ in hexacoordinate state. Searching for a cation that is larger quickly gives potassium with $152~\mathrm{pm}$ crystal ionic radius. In effective ionic radius terms (the second table), fluoride is mentioned as having $133~\mathrm{pm}$ while $\ce{K+}$ has $138~\mathrm{pm}$. Thus, by both sets of values the cation in $\ce{KF}$ is larger than the anion.

In crystal ionic radii, chloride is smaller than caesium and fracium while the latter two’s effective ionic radii are larger than chloride’s.

As you can see, there is no hard and fast rule. Fluoride to potassium is a period and two elements difference while chloride to eka-francium is four periods and two elements.


Periodic table and p block elements I have two questions:

It's said that in p block there's gradation in physical and chemical properties of elements. What is meant by this statement?
Why are the different blocks named s, p, d and f?

First question

From P.248, Rudiments of Chemistry:

  'p' block elements contain both metal and non-metal and we observe a gradation of properties (such as electropositivity, electronegativity, reducing or oxidising property etc.) as we move from left to right in a period.

Second question

From the Wikipedia article on atomic orbital:

  They are derived from the description by early spectroscopists of certain series of alkali metal spectroscopic lines as sharp, principal, diffuse, and fundamental. Orbitals for ℓ > 3 continue alphabetically, omitting j (g, h, i, k, …)


Any false chemical element in history of chemistry? Was there any false chemical element introduced in history of chemistry? I mean a substance that after a while proved that it was not an element.

The example that comes to mind is didymium, which turned out to be a mixture of praesodymium and neodymium. The term is still used, as far as I know, to refer to glass doped with a mixture of these lanthanides. This is discussed in one of the 'Chemistry in its element' podcast episodes.

As it turns out, wikipedia has a category called 'Misidentified Chemical Elements'. Especially notable are illmenium, dianium and pelopium, which were all likely mixtures of niobium and tantalum, two rather similar elements.


Since the formation of diamond from Graphite is endothermic and the entropy decreases for this process, how is it possible to make diamond? Surely G would always be positive and thus would make it impossible to make diamond. Clearly this is not the case. But how? Would an increase or decrease in temperature aid it's formation? I think a decrease in temperature would favor diamond but wouldn't this contradict le chatilier's principle?

Diamond is significantly denser than graphite.

atomic densities:

$1.14  \times 10^{23}$ cm$^{-3}$ for graphite
$1.77  \times 10^{23}$ cm$^{-3}$ for diamond

This suggests that higher pressure would favor the formation of diamond.  Indeed, as the following phase diagram shows, diamond is the most stable allotrope of solid carbon at high pressures.  Temperature has little effect, as long as the pressure is high, diamond is favored over graphite at all temperatures up to the point of liquefication.


Is heating dilute HCl dangerous? Say dilute $\ce{HCl}$ is used to convert some alkaloid to salt form, then the alkaloid is dried with heat, would gaseous $\ce{HCl}$ be released in the process?  Is there a threshold concentration of the acid, below which,  this would be acceptable from a safety point of view?

Say dilute HCl is used to convert some alkaloid to salt form, then the alkaloid is dried with heat, would gaseous HCl be released in the process?

Likely yes, unless it's the limiting reagent in the reaction in question. Probably best to assume some of it will come off as gas - that's the safest approach, in my opinion.

  Is there a threshold concentration of the acid, below which, this would be acceptable from a safety point of view?

Yes, there are several metrics used by various authorities to determine what "safe" means. You'll have to decide which one (or ones) are relevant to your situation.

Here in the U.S., the Occupational Safety and Health Administration (OSHA) - which is under the Department of Labor - has published this information on hydrochloric acid exposure limits and hazards.

In that linked document, several other agencies such as the National Institute for Occupational Safety and Health (NIOSH) and the American Conference of Governmental Industrial Hygienists (ACGIH) list permissible exposure limits ranging from 3 to 7 mg/m$^{3}$; the Environmental Protection Agency (EPA) given an inhalation reference concentration (RfC) of 2 mg/m$^{3}$, detailed here as the concentration above which (I'm paraphrasing) bad things start to happen.

Good luck on your project and stay safe.


Is alcohol really a solution? A friend recently sent me the following chemistry joke:

As a lover of puns, I'd like to enjoy it, but I think that it is actually false the way it is phrased.

A solution is defined as: "a homogeneous mixture composed of two or more substances" Alcohol alone would not constitute a mixture of multiple substances and would therefore not be a solution.

But in English the word "alcohol" is also commonly used to refer to many recreational drinks containing ethyl alcohol in addition to other types of molecules. Are there any common examples of alcoholic beverages which are solutions in the scientific sense that would make the pun valid, such as liquor, wine, or beer?

Technically, alcohol is the name of a class of organic compounds containing one or several hydroxyl groups.

Colloquially, the term "alcohol" is understood as you have described: A solution (of varying degree of purity) of ethanol and water. Pure ethanol is impossible to create via the traditional method of purification (atmospheric distillation) as the water-ethanol mixture is azeotropic:¹

  An azeotrope [...] is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. This happens because, when an azeotrope is boiled, the vapour has the same proportions of constituents as the unboiled mixture.

  Source: Wikipedia

This is why a chemist would never think "100% pure ethanol" when you mention "alcohol".

If you check for example this vendor page for the purest ethanol (>99.5%) I could find, you see various substances listed as impurities:

<0.2% Water
<0.1% Methanol
Other organic substances, such as isopropanol and acetone

Concerning your second question, I would argue that many distilled spirits (vodka, whisky, and cognac just to name a few) are solutions of ethanol, water, and various flavour compounds.

Some alcoholic beverages, such as unfiltered beer and wine, usually also contain solid particles and would as such be correctly termed "suspensions".

¹ I wonder now whether it would be possible to produce 100% ethanol from another mixture somehow. The problem is probably that every common starting mixture already has some amount of water. But synthesizing ethanol in, say, an ionic liquid under an inert atmosphere and then distilling it away might be an interesting approach.


Mechanism for the formation of butanoic acid from ethylacetoacetate My problem sheet asks me: "Draw a mechanism for the formation of $\ce{EtCH2COOH}$ from the reaction of $\ce{MeCOCH(Et)COOEt}$ with $\ce{NaOH}$ in $\ce{EtOH}$"

This has got me stumped. The mechanisms that I have attempted use the hydroxide ion as the base/nucleophile, is this right? My initial thoughts were to make the enolate by deprotonating between the caronyl groups then doing and aldol into the ketone part of another molecule of ethylacetoacetate but this just gives me a complicated molecule that I can't see reacting further.

I simply cannot find a way to make the product using only the reactants in the question.

I thought about adding solid $\ce{NaOH}$ to ethanol and therefore used $\ce{EtO-}$ ;) :


Lugol's iodine 5 % aqueous solution There is Lugol's 5 % solution with 5 g of pure iodine and 7.5 g of potassium iodide.
In one ml there should be 12.5 mg of iodine in sum.

But there are suggestions telling that two drops of this solution contains 12.5 mg of iodine and the drops I seen used there are no way 1 ml.
Is it a mistake? Why they don't use measurement in ml rather than drops to be precise?

A Lugol's iodine solution is an aqueous solution of elemental iodine and potassium iodide. This solution may be prepared in different strengths, typically $2{-}15~\%$, while the strength is determined by the content of elemental iodine (see this table from the Wikipedia article, but note that the first column's title is a bit misleading — the leftmost column lists the different strengths of Lugol's solution, and the other columns list the amount of the each constituent (iodine, iodide, total) in a solution of that strength).

Now, this is in essence an iodine solution, but it is prepared from iodine and potassium iodide due to solubility issues of elemental iodine. See this answer on Chemistry.SE to the question Preparation of iodine solution for a simple and concise explanation.

For the calculation of the amount of iodine and potassium iodide in solution, let us take $5~\%$ Lugol's solution, which is prepared from 5 grams of iodine and 7.5 grams of potassium iodide in $\pu{100 mL}$.

If we have $\pu{5 g}\ (\pu{5000 mg})$ of iodine and $\pu{7.5 g}\ (\pu{7500 mg})$ of potassium iodide in $\pu{100 mL}$, then after dividing by $100$, we have $\pu{50 mg}$ of iodine and $\pu{75 mg}$ of potassium iodide in $\pu{1 mL}$.

Since the commonly accepted volume of one drop is $1/20\ (5~\%)$ of $\pu{1 mL}$, then by multiplying $\pu{50 mg}$ and $\pu{75 mg}$ by $0.05$, we get $\pu{2.5 mg}$ of iodine and $\pu{3.75 mg}$ of potassium iodide in one drop of the solution.

For better clarity, here is the calculation again for I2 content:

$$\frac{\pu{5 g}}{\pu{100 ml}}\cdot \frac{\pu{1000 mg}}{\pu{1 g}} = \frac{\pu{50 mg}}{\pu{1 mL}}  \\ \Longrightarrow  \frac{\pu{50 mg}}{\pu{1 mL}}\cdot \frac{\pu{1  mL}}{\pu{20 drops}} = \frac{\pu{2.5  mg}}{\pu{1 drop}}$$

Do the same for $\pu{7.5 g}$ of potassium iodide to get $\pu{3.75 mg/drop}$.

A word about the actual iodine content of the solution following Jan's comment:

In principle, masses can always be added up (unlike volumes, which should not be added up, due to possible interactions between certain solvents — see here).
Therefore, we can say that the total mass of solutes in one drop is $$\pu{2.5mg}+\pu{3.75 mg} = \pu{6.25 mg}$$ ($\pu{12.5 mg}$ in two drops, per OP question). However, we cannot say that two drops contain $\pu{12.5 mg}$ of iodine, because the solution contains both elemental iodine (I2) and potassium iodide (KI), which are two different constituents. Since KI's role in the solution is to facilitate dissolution of elemental iodine molecules (see description of chemical properties, including equations, here), iodide content is irrelevant and the solution's strength reflects only iodine's content.

In other words, $\pu{12.5 mg}$ is the total quantity of solutes in two drops of Lugol's solution, but this quantity has no practical meaning, since the active ingredient is iodine, and its amount in 2 drops of $5~\%$ Lugol's solution is $\pu{5 mg}$.


Mixing cobalt(III) hydroxide and thallium(III) hydroxide A simple question on solutions is driving me crazy. At this point, I think the textbook's wrong. Here's the question:

  A saturated solution of cobalt(III) hydroxide $(K_\mathrm{sp} = \pu{1.6e-44})$ is added to a saturated solution of thallium(III) hydroxide $(K_\mathrm{sp} = \pu{6.3e-46}).$ What is likely to occur?

The answer given is that $[\ce{OH-}]$ is above saturation levels for both the cobalt and thallium in the solution, so that both will precipitate (thallium(III) hydroxide first, since it has a smaller $K_\mathrm{sp}).$

What I don't understand is how they can make such a statement. We don't know the amounts of each solution originally. Furthermore, wouldn't $[\ce{OH-}]$ be somewhere between the two original $[\ce{OH-}]$ depending on amount of each solution? And if so, wouldn't it not trigger precipitation of the cobalt (III) hydroxide (larger $K_\mathrm{sp},$ larger $[\ce{OH-}]$ at full saturation than $[\ce{OH-}]$ of the mixed solutions)?

Because of this, my answer would be that thallium(III) hydroxide precipitates, but cobalt(III) hydroxide remains dissolved.

This should be a comment, but I decided to post it as an answer, since from my point of view, the present answers fail to shed light on the absolute simplicity of the truth.
We can state from the present answers that there will be no precipitates, but not because the concentration is smaller than an atom per liter, we don't know the volume used since it was not stated.
The simple fact is that the $[\ce{OH^-}]$ concentration stays the same, it will be $10^{-7}$ before and after the mixing of the solutions, no matter how much from each solution we will mix together, since the solubility product is so small and the dominating factor for the $[\ce{OH^-}]$ concentration comes from the autoprotolysis of water. Accepting this little knowledge, no matter what we will never reach the saturation concentration after mixing. The saturation concentration will stay the same always (in this problem), so we will just dilute the initial concentration.
Imagine we mix an equal amount of both solutions, then the before saturated concentration of the substances will be halved. So in the end in this specific case for an example the solution will be half saturated with both hydroxides. That is why there is no precipitation.
Edit: Also there is no sense talking about common ion effect, since this phenomenon arises from the charge balance equation.
$$3[\ce{Tl^{3+}}]+3[\ce{Co^{3+}}]+[\ce{H^+}]=[\ce{OH^-}]$$
As we postulated before $[\ce{H^+}]=[\ce{OH^-}]$, since the autoprotolysis is the domination factor, both of the concentration of the ions will be zero (simplified). In truth the $[\ce{OH^-}]$ concentration will be just slightly higher than the $[\ce{H^+}]$ concentration, but we can neglect this equation since we assumed the only $[\ce{OH^-}]$ concentration comes from autoprotolysis, meaning we don't have to solve a system of equations, just plugging in the value in the solubility product.


What is the oxidation state of nitrogen in nitrate ion and what is its structure? What's the oxidation state of nitrogen in $\ce{NO3-}$ ion. Simple calculation depicts that it's $+5$. But second period elements never show variable oxidation states. And it's structure is so confusing to me. Some people show dative/co-ordinate bonds, whereas others show a positive charge on nitrogen. Which should be considered as correct one?

You are correct that the oxidation state is +5. The second period elements, particularly carbon, nitrogen and oxygen, do show variable oxidation states regularly.

The $\ce{NO3-}$ ion cannot be depicted using a single Lewis structure. However, its resonance structures are shown below:

What this means is that the actual structure of the nitrate ion is somewhere in between the three structures shown below. It is sometimes also depicted like this to show the symmetrical nature of the ion and the distribution of charges across it.

http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements
http://en.wikipedia.org/wiki/Nitrate#Structure


Why do certain groups withdraw/ donate electron to double bond or conjugated system in resonance effect? Why for the Resonance effect certain group of substituents/functional group donate electron from the double bond or to a conjugated system while some withdraw electron from the double bond or to a conjugated system? What causes the groups/substituent to withdraw or donate electron causing +R and -R effect?

Generally, it is observed that if there are lone pair(s) present on the atom of function group directly attached to to the conjugated system, then $+R$ effect is observed.

If the same atom has double bonds or dative bonds with another atom, then $-R$ effect is observed.

Example: –$\ce{OH}$ shows +R due to 2 lone pairs on Oxygen.
— $\ce{NO2}$ shows -R due to one double bonds between Nitrogen and Oxygen, and one dative bond from Nitrogen to another Oxygen.


Books on Chemical Reaction Theory I got interested in Chemical Reaction Theory by studying the lecture notes of Feinberg and related papers for my bachelor thesis and want to dive deeper into that subject.

Are there any books you would recommend for doing so? I would really appreciate it. I'm looking for something which emphasizes the mathematical aspects but also gives good actual chemistry examples or references.

So far I found a book by Peter Érdi & Janos Tóth from 1989 called "Mathematical Models of Chemical Reactions: Theory and Applications of Deterministic and Stochastic Models", which does just that. However I fear it is a little outdated as it seems there has been quite some progress in that field recently. So maybe there isn't even a book out yet as the theory seems relatively fragmented.

Do you know more?

If you are up for some fun and want to learn about unimolecular reactions (e.g., rearrangements decompositions after energy absorption) then you should try Theory of unimolecular and recombination reactions by RG Gilbert, SC Smith. Oxford: Blackwell Scientific Publications (1990).


Calculating reaction products programmatically? I was wondering if there was a known algorithm which could calculate simple reaction products - not anything organic, but things like

$$\ce{NaBr + LiF -> NaF + LiBr}$$

While these are quite easily to be worked out mentally, I cannot think of an easy way in which these reaction products could be predicted. Is there some software which does this (most I have found simply uses databases of known reactions), or even better an algorithm which does it?

You are looking at the most simple form of double displacement reactions of salts with

\[\ce{M^1X^1 + M^2X^2 -> M^1X^2 + M^2X^1}\]

Dissolving these salts leads to complete dissociation $\ce{MX -> M+ + X-}$. From the solution, pairs of anions and cations recombine.

Using a lookup table with elements, you could identify the cation from each string that represents a starting material (=salt). The remaining string is the anion.
Build three lists:

a list of cations cations = ['Na', 'Li']
a list of anions   anions = ['Br', 'F']
a list of the starting materials

With Python, itertools.product(cations, anions) will give a generator object that yields tuples of all possible cation-anion combinations
These tuples can be joined to strings for a possible salts
Put these strings in a list and subtract the list of the starting materials to obtain the possible products

import itertools as it
reactants = ['NaBr', 'LiF']
# splitting the string via lookup table is something I leave up to you
cations = ['Na', 'Li']
anions = ['Br', 'F']

pr = it.product(cations, anions)
pairs = list(pr)
# yields list of tuples [('Na', 'F'), ('Na', Br'), ...]

salts = [''.join(pair) for pair in pairs]
# yields a list of strings by joining the tuples:
# ['NaBr', 'NaF', 'LiBr', 'LiF']

products = [salt for salt in salts if not salt in reactants]
# only gives those salts that are not in the list of reactants


How is the volume kept constant in an NVE MD simulation? The simplest way to conduct a molecular dynamics simulation is to set up some initial conditions and propagate the system in time by solving Newton's equations of motion. When one does not interact with the system at all(no thermostat or barostat) this will be an NVE simulation, since the number of particles, the volume and the total energy is conserved.

My problem is, that I see no reason why(or how) such a simulation without external influences conserves volume. Consider simulating a cluster of water: due to the random motions, every once in a while a molecule on the surface will have sufficient energy to break away from the cluster, and fly away into what is effectively a perfect vacuum. In fact, that is exactly what I would expect from a tiny droplet of water in perfect vacuum: to evaporate and become a gas.

So my question is, how do MD simulations preserve volume?

The answer to this is linked to how you define the “volume” of your system, and the boundary conditions. The simplest case is simply to consider an isolated systems, like a molecule in gas phase. This corresponds to $N$ atoms (or molecules or particles) with no restrictions of their coordinates and no boundary conditions. This is essentially an infinite box. Its volume naturally does not and cannot vary.

Note that the volume of the system is not defined as “the volume actually occupied by the atoms”, such as for example the Connolly volume of your system. Instead, the volume of the system is the boundary of space where the atoms can go.

If you take a finite volume for your simulation, for example of cube of size $L$ and volume $V=L^3$. You have to put periodic boundary conditions, for example reflexive (your box is surrounded by a wall, and particles bounce on it) or periodic (look up for periodic boundary conditions in a textbook). Then from the initial positions and velocities of the atoms, Newton's equation allows you to propagate those quantities and determine their evolution in time. But nothing in Newton's law affects the system size $L$ (and thus $V=L^3$) or the boundaries. They will stay fixed and the volume will be constant.

Also, in all cases, because of conservation of matter, $N$ is constant. And if the forces acting on the system are conservative (as are all interatomic forces), then the total energy in the system $E$ is conserved.* Thus, the system samples the $(N, V, E)$ thermodynamic ensemble.

* the momentum of the system is also conserved… and it is not as trivial as it may sound :)


Tautomerism in 3,5‐dimethyl‐4H‐pyrazole How would tautomerism take place in 3,5‐dimethyl‐4H‐pyrazole?

I am aware of nitro–aci nitro and keto–enol, but I haven't seen such a compound before. I recognize that a hydrazone formation was done but nothing more. From what I was able to reason, an alkadiene is formed but that is unstable, so tautomerism shouldn't be possible. How does this compound show tautomerism?

You are lucky, because not only the pyrazole motif, but the complete structure drawn by you is part of the Tautobase database

with a preference for the second structure drawn.
By the way, the values of this database are accessible to the public on GitHub, and since January 2020 implemented as (equally [sub]structure) searchable data file within DataWarrior.  Thus, you may train your chemical eye on 1680 tautomers.
Tautobase: An Open Tautomer Database, Wahl, O.; Sanders, T. J. Chem. Inf. Model. 2020, 60,1085-1089; doi 10.1021/acs.jcim.0c00035


Can Silicon be used as anode for Li-Ion batteries Can Si be used as anode. I assume you could store much more energy in Si since the atom is much larger than C, but then Si seems not to like to bond with itself. Can there be a solution?

Generally, heavier atoms means lower specific stored charge per mass.

Specific energy per mass depends of the cell redox systems, but also on the above.

Silicium, in contrary to carbon, does not form hexagonal planar structure with delocalized electrons like graphite, but prefers the diamond-like structure.

The Li-ion cell anode is formed by lithium-graphite intercalate, for which does not exist Silicium analogue. If it had existed and the cell had been possible to create, the major difference would have been being heavier for the same mAh capacity.

Silicon anodes (thanks @andselisk  for the link) are possible, but not as analogue of the graphene structure  of graphite. The huge problem with them is material swelling in the charge state, limiting the anode life to few cycles. Few percents of silicium together with carbon can be used to increase somewhat the graphite anode capacity.


What are the molecular orbitals of the hypothetical linear H₃⁺ molecule? What would be the wave function of the lowest energy molecular orbital of a hypothetical linear H3+ molecule?
According to the LCAO method, I feel the lowest energy MO will be 1s(A) + 1s(B) + 1s(C).
Where 1s(A) is the wave function of the 1s orbital of one of the Hydrogen atoms.
This has 0 nodes and has the lowest energy.

Isn't this correst?

The lowest energy MO is : $\psi_1 = (1/2)(\phi_A+ \sqrt2\phi_B+\phi_C)$.

Then comes the MO: $\psi_2 = (1/\sqrt2)(\phi_A- \phi_C)$.

The highest energy MO is  $\psi_3 = (1/2)(-\phi_A+ \sqrt2\phi_B-\phi_C)$.

Where  $\phi$ denotes the atomic orbital $1s$ on each hydrogen atom. $A$, $B$ and  $C$ denote hydrogen atoms, where  $B$ is the central one.


What substances form crystal structures? Just getting interested in this stuff. People are saying I should crystallize salt, for example. What substances besides salts form a crystal structure?

Maybe to focus more on the question: what kinds of things can you make crystals from:

Typically a pure element or an inorganic or organic compound can be made in crystalline form. It might be a fine powder or small granular crystals. If grown under controlled conditions, you could get large crystals. If it is a gas or liquid at room temperature, it could perhaps solidify to a crystalline form at lower temperatures.

On the other hand, an impure, indeterminate, or polymeric material, or a mixture of compounds, very seldom will form crystals. Still, there are exceptions. Inorganic salts often crystallize from water solution as "hydrates" - with extra water molecules in the crystal lattice of the salt. Other solvents also sometimes co-crystallize with inorganic or organic substances. Neutral compounds can co-crystallize together as clathrates, such as methane hydrate - a crystalline structure containing both methane and water that forms at high pressures and low temperatures.

Even a complex or high molecular weight substance, such as an enzyme or virus particle can form crystals. In the realm of nanotechnology, new materials are being engineered specifically to aggregate into crystal-type structures.


standard reference for thermodynamic properties? I'm a non-chemist attempting to put together a chemistry paper. Up to now I've got my thermodynamic data from various random pdf files found on the web, together with Wikipedia. Now that the paper is nearly finished I need to replace these with values from a single, respectable source. But I've realised I don't know how to find such a source. I have a few chemistry textbooks on my desk, which tend to have tables in the back, but they're not terribly comprehensive and none of them have all the data I need.

So is there a standard resource for thermodynamic tables? I particularly need the Gibbs energy and enthalpies of formation for gases found in the atmosphere, such as $\ce{O2}$, $\ce{CO2}$, $\ce{CH4}$ and $\ce{H2O(vapour)}$?  In an ideal world this would either be something I can access online or a book that's likely to be found in my institute's rather small library.

A reference that I've used quite a lot (as a student) is SI Chemical Data, which I guess is quite common. It should contain the data you ask for, among other things.

If you are looking for an online source, the NIST-JANAF Thermochemical Tables or NIST Chemistry WebBook should be good places to start. I think both should be respectable enough. The Chemistry WebBook might be a bit more comprehensive, but I suspect that much of the data is from the same source anyway, since NIST is behind both services.

Hope you can use one or the other!


Classification/Nomenclature of aroma and flavour compounds - terpenes, aldehydes, and alcohols, etc I'm collating different types of aroma compounds but the nomenclature and classification structure is confusing. How do you classify flavour compounds that can belong to different groups (e.g. terpenoids and aldehydes)?
Citral (lemony smell), for example, is a mixture of neral and geranial. Citral seems to be generally referred to as a terpene, but is a mixture of aldehydes. This previous post refers to it as an aldehyde: What happens when deodorants or perfumes expire , but is generally referred to as a terepenoid.

Similarly, Citronellal is a terpenoid, but is it not also an aldehyde?

Is there a way to prioritise the classifications of aroma/flavour compounds - i.e. does a terepenoid structure take priority over presence of an aldehyde or alcohol group?

Perhaps by classification/nomenclature you are referring to odour descriptors. If I remember correctly I have already covered a similar question about possibility of defining smell, so I will extract relevant parts here:

Odour descriptors and odour profiles

Odour descriptors are the words that come to mind when smelling a substance.
The more generally understood the words are, the more useful they are as descriptors. In practice, it is easy for any observer, after a little training, to use the standard descriptors of fragrance chemistry. Example of descriptors include musky, camphoraceous  etc

Smelling chemical groups

A fact that has, in our opinion, received too little attention from olfaction researchers is the ability of humans to detect the presence of functional groups with great reliability.

The case of thiols ($\ce{-SH}$) is familiar, but groups ($\ce{NO2}$), aldehydes ($\ce{C=O(H)}$), can be reliably identified once the odour character the functional group character confers is known. When nitriles are used as chemically stable replacement for aldehydes, they impart a metallic character to any smell: cumin nitrile smells like metallic cumin (cuminaldehyde), citronellyl nitrile smells like metallic lemongrass (citronellal), and nonadienylnitrile smells like metallic cucumber (nonadienal). Oximes give a green-camphoraceous character, isonitriles a flat metallic character of great power and unpleasantness, nitro groups a sweet-ethereal character,etc.

Here are some odour categories and their representative molecules, chosen to illustrate
structural diversity:

Musk

  Musk odour descriptors might be “smooth clean, sweet and powdery”. The
  molecules that possess this odour character are exceptionally diverse
  in structure. Macrocyclic musks contain a 15-18 carbon cycle closed
  either by a carbonyl or by a lactone and smell similar but fresher and
  more natural, often with fruity overtones (cyclopentadecanolide,
  ambrettolide). Nitro musks, discovered originally as a byproduct of
  explosives chemistry, smell sweeter and are reminiscent of
  old-fashioned barbershop smells.

Representatives from five chemical classes which yield musk odors. 1 androst-16-en-3a-
ol, a steroid musk. 2: ambrettolide, a macrocyclic musk. 3: Musk Bauer, a nitro musk. 4: Tonalid, a
tetralin musk. 4: Traseolide, a indane musk.

Ambergris

  Originally derived from concretions spat out by whales and aged in the
  sun, ambergris odorants smell nothing like natural ambergris tincture,
  which has a weak animalic marine smell. The smell of ambergris
  odorants was once aptly described to us by a chemist-perfumer as
  “glorified isopropanol”. Ambergris odourants provide an interesting
  combination of very closely related smells with widely different
  structures: amberketal, timberol, karanal and cedramber

Two ambergris odorants, timberol (left) and cedramber (right)

Bitter almonds

This easily-recognized category is interesting  because it includes a small molecule (HCN) which, however, is perceived by a large fraction of observers  to smell as metallic not almond-like to. Benzaldehyde, nitrobenzene,trans-2-hexenal (but see above) are good examples.

Green

  Cut grass, fresh green bean notes with a sharp, almost aggressive
  feel. Diverse compounds possess this descriptor, ranging from classic
  grassy notes of cis-3-hexenol and ligustral, to the cucumber peel of
  nonadienal and the bell-pepper green note of some pyrazines.

Some examples of green odorants. Clockwise from top left cis-3-hexenol, ligustral, nonadienal and ethylmethoxypyrazine.

Camphoraceous

Camphor, cyclooctane, cineole are good
examples of camphoraceous smells, and smell rather similar to each other.

Three camphoraceous odorants: 1,8 cineole, camphor and cyclooctane

None of the above

  Many other categories such as musty, spicy, aldehydic, lactonic,
  indolic, marine, etc exist, each with subdivisions. It must be
  emphasized that the odor categories above are merely convenient
  descriptors and only cover a very small fraction of odor “space”. In
  fact, particularly when one steps out of perfumery materials proper
  into smells noticed by chemists in the course of organic and inorganic
  syntheses the most frequent descriptor appears to be sui generis, i.e.
  a smell associated with nothing in particular

The complexity of structure-odour relations, and the fact that the three dimensional structure of the receptor site is unknown, make it very difficult to apply conventional quantitative structure activity relationships. For detailed SAR and theories please refer to above linked post.

References

Structure-odour relations: a modern perspective: Luca Turin et al


Why do unsaturated hydrocarbons burn with a sooty flame while saturated hydrocarbons burn with a clear flame when burnt in the air? Why do saturated hydrocarbons burn with a clean flame while unsaturated hydrocarbons burn with a sooty flame when burnt in air?

My chemistry book says that it happens due to the amount of carbon being more than the amount of hydrogen in unsaturated hydrocarbons. But isn't that also the case with saturated hydrocarbons?

Your observation that unsaturated hydrocarbons burn with a sooty flame is not always true and depends on the circumstances. There are more factors at play than the nature of the hydrocarbon that is burning.

The most important reason why flames are sometimes smoky is that the combustion is incomplete. This means that not enough oxygen was present to fully oxidise all the carbon to carbon dioxide or carbon monoxide gases (and the amount of CO will vary with the ratio as well as, given enough oxygen, it should also be completely combusted to carbon dioxide). If there is not enough oxygen in the flame there are many competing side reactions that generate sooty resides (the reactions are pretty complex). Some pyrolysis reactions of hydrocarbons may occur before combustion itself (if the mixing in the flame is less than perfect) and these may contribute to the generation of sooty products which may be harder to burn than the original hydrocarbon.

Incomplete combustion is as much a function of the environment of burning as it is the thing that burns. Some environments don't enable enough oxygen to get into the gas mixture to drive complete combustion and the reactions can easily leave a lot of sooty emissions. A candle, for example, burns with a smoky flame despite being made from mostly saturated long chain hydrocarbons. It does so because a candle wick doesn't do a good job of mixing air with the vapours of the burning hydrocarbon (this is deliberate as the prime role of candles is to provide light and this comes from incandescent particles of soot in the flame: a pure flame burning efficiently would provide little light by itself).

Even well controlled environment such as engines can't completely avoid side reactions. But Diesel engines (which use mostly saturated hydrocarbons) produce more sooty particles than petrol (gasoline) engines though petrol engines have far more unsaturated fuels. The exact reasons are complex and depend on the precise environment inside the engine.

In well controlled environment where the oxygen-fuel ration is set carefully most hydrocarbons burn cleanly with little soot. Gas-based cookers (using mostly saturated methane, ethane, propane or butane) are very clean because they control the gas mix very precisely (use the wrong burner and they may well be sooty!). Welding and cutting torches can happily use the highly unsaturated acetylene with a clean, non-sooty, flame (the oxygen-acetylene flame is one of the hottest achievable) because they control the gas mix very carefully leaving little room for side reactions.

Uncontrolled flames might well show your general observation that unsaturated fuels burn more sootily, but this is as much a product of the environment as it is the product of the hydrocarbon. It is hard, without a carefully controlled experiment, to tell whether the general observation is true at all.


Reduction of an Allylic Enoate to afford allylic alcohol in presence of Benzyl Acetal Groups I have a question that requires finding a reagent for the reduction of an enoate to the corresponding allylic alcohol. Preferably done in one step too with an acid workup at the end too.

What types of reducing agents would work?

Would Lithal be too strong and reduce the acetal groups or would it be okay?

Your benzyl acetal groups should be stable to all varieties of hydride reducing agents.
For the enoate reduction I suggest Dibal-H or LiBH4/Et2O.

I don't have references for this as I no longer have Scifinder access, and Google is not finding me any references that are not behind paywalls.


What are the important implications to the field of chemistry, if any, of the recently confirmed hexavalent carbon species? I’m struggling to understanding the true nature and implications of recent confirmations of the existence of stable species containing carbon atoms bound six (in another case 5) other carbon atoms. Is this really a breakthrough of science or one of semantics and scientific media hype? I've had a hard time understanding what I've read on this subject.

From Moritz Malischewski, Prof. Dr. K. Seppelt. Angew. Chem. Int. Ed. 2017, 56, 368-370, an open, online article:

  Herein we report the crystal structure of $\ce{C6(CH3)6^{2+} (SbF6^−)2⋅HSO3F}$,
  which revealed a non-classical structure for this labile species
  (Figure 1).

Figure 1.
Molecular structure of $\ce{C6(CH3)6^{2+} in C6(CH3)6^{2+} (SbF6^−)2⋅HSO3F}$, ellipsoids are shown at 50 % probability, C grey, H white; counteranions and co-crystallized $\ce{HSO3F}$ omitted for clarity.

Further discussion of this matter was also given in C&EN: "Six bonds to carbon: Confirmed", C&EN, Volume 94 Issue 49, p. 13  (abstract).

So, I understand how the octet rule is not actually violated. But then, can these molecules really contain carbons with 6 primarily covalently bound carbon neighbors? In what orbitals do all these electrons lie? Should we not be surprised as similar hypervalency occurs elsewhere on the periodic table?  Any clarification at all of these issues would be greatly appreciated.

The carbon is not  hexavalent, it is hexacoordinated. A covalent bond does not equal to a total of two electrons between the bonding partners and the nature of  the chemical bond may lie somewhere between totally covalent and totally ionic.
Examples of this include boranes, with their three-centre-two-electron bonds. But we don't need to stop there; any π coordination has fewer electrons in the bonding orbitals than double the bonding partners.

We struggle understanding the bonding in simple molecules like carbon monoxide, while more complex molecules like many organic molecules have all similar bonding situations which are quite easy to grasp. The concept of aromaticity is still not fully understood and while we believe it is as simple as drawing a MO diagram, it most certainly is not. However, molecular orbitals certainly enhance our understanding.

Chemist in general are interested in unusual bonding situations, since they challenge our understanding of the bonding and the chemistry of such molecules itself.
One of the prominent examples is the 2-norbonyl cation, the structure of which remained a mystery for a long time since it didn't fit within the common constraint of organic chemistry. Such ions are nowadays usually referred to as non-classical ions. Their bonding is different from the more common two-electron-two-centre bonds we expect in organic (and inorganic) molecules. In a first approximation they can be described with resonance (see here: What is resonance, and are resonance structures real?), but the understanding comes with a few misconceptions. A MO description involves multi-centre bonds and typically bond orders less than one.
Another interesting example of unusual bonding are fluxional molecules like bullvalene. (See also here: What is the conformer distribution in monosubstituted fluoro bullvalene?) Similarly the bonding situation in these molecules is quite fluid, which allow them to change shape in such a way and at room temperature we obtain a single signal in the proton NMR (Addison Ault. J. Chem. Educ. 2001, 78 (7), 924-927.).

As such the hexamethylbenzene dication is another representative of unusual bonding situations. While it has been prepared more than forty years ago, it took until now to confirm the actual structure. It is special since it contains a pyramidal coordinated carbon exclusively bonded to other carbons, and it marks the highest observed coordination number for carbon so far. The authors actually state the general motivation behind such approaches in the first two sentences:

  The tetravalency of carbon and the hexagonal-planar ring structure of benzene are fundamental axioms of organic chemistry, and were developed 150 years ago by Kekulé. Chemists have long been fascinated by finding exceptions from these rules.

Let's go a little deeper, and look at the questions you are asking step by step:

  So, I understand how the octet rule is not actually violated. But then, can these molecules really contain carbons with 6 primarily covalently bound carbon neighbors?

The covalent-to-ionic character of a bonding orbital is completely independent of whether it is occupied or not. I believe there still is no unilaterally accepted criterion for what constitutes a covalent bond and what does an ionic bond.
As previously stated, a bond does quite often deviate from the traditional Lewis-concept. This is to be expected as the Lewis concept is quite crude and cannot account for many chemical phenomena.
We can analyse the electron density with the Quantum Theory of Atoms in Molecules, which defines us a set of parameters on which we can judge whether a bond is predominantly covalent of ionic. First and foremost we can obtain bond paths, i.e. these are paths that connect nuclei in a way that the value of the electron density is maximum. On this path we will find a bond critical point, where the electron density is minimal. The Laplacian at this point gives us a measure whether the bond is predominantly covalent (negative) or ionic (positive). In the picture below areas of positive Laplacian are marked with solid lines, i.e. charge depletion, and negative values with dashed lines, i.e. charge accumulation. The location of the plane is given in the 3D model. The hexa-coordinated carbon is C10.

        rho/a.u.    Lapl./a.u.
BCP44   0.259       -0.811
BCP47   0.294       -0.121
BCP53   0.157        0.226
BCP61   0.270       -0.882
BCP64   0.288       -0.117

From the pure visual inspection we can see that the area between C10 and C12 clearly is a predominantly covalent bond, which is obvious from the dashed lines. This is also supported by the numbers.
Visual inspection of C10 to C4 would suggest the same. However, the bond critical point lies outside of the charge accumulation area. This might well be an issue with the methodology. I have reproduced the calculation from the paper only on their geometry. Therefore the optimum MP2 density might be slightly different and completely describes the bond as covalent. Also note that there are no symmetry restrictions, which could also lead to slight variations. From what we can see, we can derive that there is a significant covalent contribution though.
The values of the electron density (rho) also show that the pyramidal bonds are only about half as strong as the bonds in the 1-ethylium-1-ylidene or the five-membered ring.
(Fun fact: The AIM analysis shows no cage critical point. From a technical point of view, the five-membered ring is no ring. The whole molecule is rather a goblet (of fire, lol).)

  In what orbitals do all these electrons lie?

Since it is a cation of benzene, the atomic orbitals involved in bonding are the same as in regular benzene. The total number of molecular orbitals that are formed are one fewer than in the neutral case. Obviously due to the different geometric arrangement they combine also differently.
The canonical orbitals are actually quite pretty and reproduce the (almost) symmetry quite nicely as some of the orbitals are nearly degenerate (for example HOMO and HOMO-1, or LUMO and LUMO+1), but have a look for yourself:

  Should we not be surprised as similar hypervalency occurs elsewhere on the periodic table?

Well, no. Hypercoordinate bonding is one of the most common patterns for anything outside of carbon chemistry.

A few words of advice: It has nothing to do with hypervalency. The term is an ancient relic from the times where people believed that d-orbitals are necessary to describe molecules that exceed the coordination number dictated by Lewis theory. The gold book writes on hypervalency:

  The ability of an atom in a molecular entity to expand its valence shell beyond the limits of the Lewis octet rule. Hypervalent compounds are common for the second and subsequent row elements in groups 15–18 of the periodic table. A description of the hypervalent bonding implies a transfer of the electrons from the central (hypervalent) atom to the nonbonding molecular orbitals which it forms with (usually more electronegative) ligands.

Note that this orbital expansion is not necessary when considering MO theory. As such, the term valency, and especially hypervalency, should be avoided at all cost.
It is important to know that according to the gold book, valence is an absolute property of an element, it does not change:

  The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

Carbon is tetravalent. Always. (Seriously, from now on never use *valen{ce .. cy .. t} again, please.)

  Is this really a breakthrough of science or one of semantics and scientific media hype?

Of course it is. It is always something incredible when we refine our understanding of the universe, challenge our own theories. Is it as ground breaking as the discovery of electricity; probably not. However, with such advances in synthetic chemistry we can obtain more robust data to develop our theoretical models. We can in general expand our knowledge of bonding, think outside the box.
With advances like this we achieve the complete break down of our theories; without such we would probably still be using Lewis structures (for everything, and that would be ridiculous).


Discrepancies in calculating free energy values listed in Stumm and Morgan I am hoping someone can help me with a clarifying a calculation on a fundamental thermodynamic understanding of reactions. Below is Table 2.5 from Stumm and Morgan's Aquatic Chemistry.

  Table 2.5. Influence of Pressure and Temperature on the Energies of Water Ionization a
  $$
\begin{array}{rrr}
\hline
p (\pu{bars}) &\log K_\mathrm{w} &\Delta G_\mathrm{w}^\circ (\pu{kJ mol-1})\\
\hline
    &T = \pu{298.17 K} &\\
1 &-14.00 & 79.94\\
200 & -13.92 & 79.48\\
400 & -13.84 & 79.02\\
600 & -13.77 & 78.63\\
800 & -13.70 & 78.22\\
1000 & -13.63 & 77.83\\
\hline
T (\pu{^\circ C}) &\log K_\mathrm{w} &\Delta G_\mathrm{w}^\circ (\pu{kJ mol-1})\\
\hline
&p = \pu{1 bar} &\\
0 & -14.93 & 85.25\\
10 & -14.53 & 82.97\\
20 & -14.17 & 80.91\\
30 & -13.83 & 78.97\\
50 & -13.26 & 75.71\\
\hline
\end{array}
$$
  aData from Harned and Owen (1958).

I'd like to focus on the second portion of the table, where temperature is varied. $K_\mathrm{w}$ and $\Delta G_\mathrm{w}^\circ$ values are listed, but using the given $K_\mathrm{w}$ values, I calculate different $\Delta G_\mathrm{w}^\circ$ values than what the table offers. Here is a sample calculation for $\pu{50^\circ C}$:

$$K_\mathrm{w} = 10^{-13.26} = 5.49\cdot10^{-14}$$
$$\Delta G_\mathrm{w}^\circ = -RT\ln K$$
$$\Delta G_\mathrm{w}^\circ = -\pu{8.314e-3 kJ mol-1 K-1} \cdot \pu{323.15 K} \ln (\pu{5.49e-14}) = \pu{82.03 kJ mol-1}$$

The table lists $\Delta G_\mathrm{w}^\circ = \pu{75.71 kJ mol-1}$

Not only is this answer substantially different, but the value is heading in the opposite direction of where my calculations would head (more positive vs more negative). Unfortunately, I don't have access to the original source (Harned and Owen).

Can someone please help me to understand my error?

TL;DR: It appears that OP is right and has estimated standard Gibbs free energy correctly. The latter book accurately fetched data from the original source, but the calculations were carried out based on a false premise that $\Delta G^\circ$ refers to $\pu{25^\circ C}$ exclusively.

First, a general disclaimer. The uppercase circle ($^\circ$) generally denotes standard state. According to IUPAC, temperature is not declared within, but many authors attribute (negligently) $\pu{298.15 K}$ to Gibbs standard free energy $\Delta G^\circ$. Also check out What is the difference between ∆G and ∆G°?

For the given equilibrium one can use

$$\Delta G_T^\circ = -2.303RT\log K_T, \tag{1}$$

and at $\pu{298.15 K}$ this equation can be rewritten as follows, assuming $[\Delta G_{298.15}^\circ] = \pu{kJ mol-1}$:

$$\Delta G_{298.15}^\circ = -5.71\log K_{298.15}. \tag{2}$$

Now to the references. I didn't find the 3rd edition of The Physical Chemistry of Electrolyte Solutions by Harned and Owen (1958), but I do have an earlier version from 1943 [1]. In chapter 15 "The Ionization and Thermodynamic Properties of Weak Electrolytes" [1, p. 485], table 15-1-1 accumulates values for $K_\mathrm{w}$ determined electrometrically:

  In subsequent investigations, the cells without liquid junction, were employed in certain combinations for the determination of the ionization constant itself, as well as $m_\ce{H}m_\ce{OH}$ and $\gamma_\ce{H}\gamma_\ce{OH}/a_\ce{H2O}$ salt solutions.
  \begin{align}
&\ce{H2 | HX($m$) | AgX-Ag} \tag{I}\\
&\ce{H2 | MOH($m_1$), MX($m_2$) | AgX-Ag} \tag{II}\\
&\ce{H2 | HX($m_1$), MX($m_2$) | AgX-Ag} \tag{III}
\end{align}

  $$-\log K'_\mathrm{w} \equiv \frac{(\ce{E_{II} - \ce{E_{III}}})F}{2.303RT} -
 \log\frac{(m_\ce{H}m_\ce{X})_\mathrm{III}}{(m_\ce{X}/m_\ce{OH})_\mathrm{II}} + \\ + \frac{3.629 \times 10^6}{(DT)^{3/2}}\sqrt{\mu} = -\log K_\mathrm{w} + f(\mu) \tag{15-1-10}$$

So, Stumm and Morgan [2, p. 57] correctly adapted the data from table 15-1-1:

\begin{array}{rr|rr}
T (\pu{^\circ C}) &K_\mathrm{w} \times 10^{-14} {^{[1]}} &\log K_\mathrm{w}{^\text{calcd. [1]}} &\log K_\mathrm{w} {^{[2]}}\\
\hline
0 & 0.1139 &-14.9435 &-14.93\\
10 & 0.292 &-14.5346 &-14.53\\
20 & 0.6809 &-14.1669 &-14.17\\
30 & 1.469 &-13.8330 &-13.83\\
50 & 5.474 &-13.2617 &-13.26
\end{array}

But, what Stumm and Morgan did is they calculated the standard free energy exclusively at $\pu{25^\circ C}$ using the $K_\mathrm{w}$ values for other temperatures; disregarding the fact that equation (2) is only valid for $K_{298.15}$, but it was applied to all values as follows:

$$\Delta G_{298.15}^\circ {^{[2]}} = -5.71\log K_T, \tag{2a}$$

which is $\color{red}{\text{incorrect}}$. The latter column is populated with appropriate values calculated with (1), as it was done by OP:

\begin{array}{rr|rr}
T (\pu{^\circ C}) &\log K_\mathrm{w}{^\text{calcd. [1]}} &\color{red}{\Delta G_{298.15}^\circ {^{[2], (2a)}} (\pu{kJ mol-1})} &\Delta G_T^\circ {^{(1)}} (\pu{kJ mol-1})\\
\hline
0 & 0.1139 &\color{red}{85.29} &78.14\\
10 & 0.292 &\color{red}{82.96} &78.79\\
20 & 0.6809 &\color{red}{80.86} &79.50\\
30 & 1.469 &\color{red}{78.95} &80.28\\
50 & 5.474 &\color{red}{75.69} &82.04
\end{array}

Bibliography

Harned, H. S.; Owen, B. B. The Physical Chemistry of Electrolyte Solutions; American Chemical Society Monograph Series; Reinhold Publishing Corporation: New York, 1943; Vol. 95.
Stumm, W.; Morgan, J. J. Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters, 3rd ed.; Environmental science and technology; Wiley: New York, 1996.


Making 1,3,5-Tribromobenzene from Phenol The problem was to make 1,3,5-Tribromobenzene from Phenol.

I thought to do it the following way:

React phenol with $\ce{Br2/H2O}$.
Then add $\ce{Zn}$ dust to get the product.

Now I am in doubt whether the second step would even work.

Are there any better method of doing this ?

First react phenol with ammonia in presence of zinc chloride to obtain aniline.

Then react aniline with bromine water to form 1,3,5-tribromoaniline

Then react 1,3,5-tribromoaniline  with sodium nitrite and hydrochloric acid at 273 K to form 1,3,5 tribromobenzenediazonium chloride.

Then react the diazonium salt with ethanol/phosphorous acid and water in presence of copper to form  1,3,5 tribromonbenzene.

Detail explanation about the reaction, conditions  and the mechanism is given here by actually conducting the experiment.


Why does methane have 4-different bond dissociation energies? I will elaborate on my question here.
In my Lecture Notes it is mentioned that for methane, there are 4 different $\text{C-H}$ bond energies. I plotted a quick graph of the bond energies to analyse the trend.

Here the vertical axis represents the bond dissociation energy while the horizontal axis represents the bond dissociation number. Looking at the graph, there is no clear trend so I decided to turn my attention to the structure.
$\text{CH$_4$}$ has the following structure

It is from here that I am unable to proceed. Here are my questions.

Am I barking up the wrong tree by looking at the structure and how do I account for the changing bond energies
Is there a VSEPR structure for molecules like $\text{CH$_3$}$ $\text{CH$_2$}$ $\text{CH}$? If so how do I figure it out.
Why am I even learning about bond energies in the first place? Is there a way I can link this to a bigger picture? I find this topic very interesting and I want to tie up my knowledge neatly here.

Bond energies are important, but you have to start somewhere. Actually, I'm surprised that the bond energies for the four dissociations are so similar. But that's a good start.
Now let's get into a bit more interesting compound: ethane. Very interesting!

The table is only part of the information on ethane given in an open access article (Ref 1). Note that the first C-H bond dissociation requires 422 kJ/mol, while the second requires 150 to 465 kJ/mol, depending on which bond you are breaking, because of what product you are making. Each line in the table is worth several paragraphs of explanation.
I admire your desire to tie up your knowledge neatly here - get a lot of string, because it's a very big package!
Ref 1. https://pubs.acs.org/doi/pdf/10.1021/acs.jpca.5b01346   Branko Ruscic, J. Phys. Chem. A 2015, 119, 28, 7810–7837


What are the different types of catalysts used in catalytic hydrogenation of alkynes? This is a question from Org. Chem. by Solomons, Fryhle & Snyder, adapted for JEE by M.S. Chouhan :

In the first step, I expected that the reaction would stop at alkene formation via syn addition of $\ce{H_2}$ since Lindlar catalyst is poisobed and heterogeneous. What are some other other poisoned catalysts that stop at alkene?
In the second step, I expected syn addition of $\ce{D_2}$ since $\ce{Pd-C}$ is a heterogeneous catalyst, forming product (a). But the answer given is (c) — does this mean that some heterogeneous catalyts can give anti addition, or is it just an error on the adapting author's part?

The "Lindlar catalyst" (Pd/CaCO3) is probably the most common way of effecting stereoselective reduction of alkynes to (Z)-alkenes. It is fairly accessible and the reaction is easy to run. But it is not the only way: there are other catalyst systems. One of the more notable ones is probably the Brown catalyst "P2-Ni" (Wikipedia).
There is a nice review on alkyne semihydrogenation here: Chem. Rev. 2013, 113 (3), 1313–1350, which covers both heterogeneous and homogeneous methods, as well as methods involving other hydrogen sources apart from H2 gas itself. I recommend it to interested readers and especially anybody considering carrying out such a reduction.

The Lindlar hydrogenation of an alkyne forms (Z)-alkenes stereoselectively. Syn hydrogenation of the (Z)-alkene gives product (c), not product (a).

Note that in the fully reduced alkane, there is free rotation about the central C–C bond. So, having both D atoms pointing "out of the plane" is irrelevant, as you can easily rotate to a conformation in which one D atom is pointing in the plane. However, the configuration of the product cannot be changed, and that is the difference between options (a) and (c).


Ring currents of anti-aromatic and aromatic systems Why are ring currents associated with aromatic systems diamagnetic while those of anti-aromatic systems are paramagnetic?
Does it have anything to do with the number of delocalised electrons 4n+2 and 4n , respectively?

Yes it has to do with the number of electrons. Consider a simple model of the electron on a ring, then the electron has to follow a circular path and its wavelength has to 'fit' exactly  round the ring. Thus the wavelength has to satisfy the condition $L=q\lambda$ where $q$ is a quantum number $q=0, \pm 1, \pm 2 ...$ characterising each eigenstate. The energy in this simple model comes to $E=(hq)^2/(2mL^2)$ ($h$ is the planck constant, $m$ the electron mass). Now in this model filling up the energy levels according to the aufbau principle, (Pauli principle & Hund's rules) leads to 2 electrons in $q=0$ (which is singly degenerate) and 2 in $q=+1$ , and $q=-1$ etc which are all doubly degenerate in energy.

Thus $4n+2 ~~ \pi$ electrons fill up all levels with two electrons in each level. The $4n~~ \pi$ contains only a single electron each. This is because it takes less energy to add one electron to each of the $q$ levels rather than pairing them up. The sketch shows this.

The unpaired electrons are shown in the $4n$ column,  however, the levels are also shown as being split in energy. This is due to the Jahn-Teller effect. This states that a slight interaction between degenerate levels can cause the degeneracy to be split and so lower the energy. The interaction could be due to slightly different bond lengths and the energy levels are split by only a small amount relative to the gap to other levels.

Although not shown in the diagram, (which is thus slightly inaccurate) both spins will occupy the same lower level so in this sense, [4n]annulenes do not really exhibit molecular paramagnetism. The electrons interact with the magnetic field $B_0$ and this leads to a mixing of levels and this causes a paramagnetic contribution to the shielding constant. This is larger than the diamagnetic contribution for  lower orbitals. Thus in [4n]annulenes we speak illustratively of a 'paramagnetic ring current' and molecules showing this effect are called paratropic and those with no ring current atropic.


store alcohol tinctured herbs in polypropylene for extended time Can I store alcohol tinctured medicinal herbs in the 2 oz. food grade polypropylene cups  (PP 5) for about a week?  I would like to have several "doses" ready to go because they are taken several times a day, and I would like to have several days ready to use at all times.  Thanks

The alcohols should not be an issue, it is used for storing often.  Some oils will affect it.

Here is a random guide that includes a few common oils.


How can I obtain oxygen gas from plagioclase I would like to know if there is a possibility to decompose plagioclase to obtain oxygen gas to be used in a combustion reaction?

I know that oxygen represents $47\%$ of the weight of 1 mole of plagioclase (albite $\ce{NaAlSi3O8}$ – anorthite $\ce{CaAl2Si2O8}$), which equals to $\pu{130 g}$ out of $\pu{277 g}$, so it will be very nice if I could decompose plagioclase and obtain oxygen gas, but I don't know if that is possible or not?

Yes, I collect oxygen isotope data from plagioclase using the laser flourination method. To liberate the oxygen from the plagioclase, crystals are reacted with purified $\ce{BrF5}$. Then the plagioclase is hit with a $\ce{CO2}$ laser. Gasses are sent through a couple of cryogenic (liquid nitrogen) traps to filter, and then through a mercury diffusion pump to remove that nasty flourine.


Are there any early industrial age methods capable of determining diamond or graphite's crystalline structure? I'm writing a fictional story with early industrial age technology where a chemist discovers not only that diamond and graphite are made of the same substance, but that they differ by crystalline structure (cubic vs hexagonal).  However, I'd like this to be a realistic depiction if possible and the chemist does not have access to electron tunneling microscopy or advanced spectrographic methods.  The farthest spectroscopy has gotten is Fraunhofer's spectroscope and John Herschel's evaporogram, with visible light and the beginnings of an understanding of infrared and ultraviolet.  Would 1850's level spectroscopy or other technology of the time be able to demonstrate the crystal structures of carbon (specifically graphite and diamond) without a 1900's understanding of atomic structure?

Would 1850's level spectroscopy or other technology of the time be able to demonstrate the crystal structures of carbon (specifically graphite and diamond) without a 1900's understanding of atomic structure?

Yes, using visible light and petrographic microscopes. The field of optical mineralogy was mature enough by the mid-1800's for one to determine the crystal structure of minerals. In particular, the work of Sir William Nicol is worth investigating (thin but useful Wikipedia article).

I will add that there's a pretty simple way to determine whether or not the crystal structure of a mineral is the same or different than that of another mineral - the so-called crystal habit.

Finally, diamond exhibits conchoidal fracture, and graphite does not. This might be the easiest way to tell that the two minerals have different crystal structures.


How to understand this form of writing the solution: (some salt • n H₂O)? For example — "high purity chloride salt of Zinc $(\ce{ZnCl2.2H2O})$" or "various concentrations of $\ce{FeCl2.4H2O}$". What does the number before $\ce{H2O}$ mean?

The salt's crystal lattice's repeating unit is constituted of n molecules of salt and m molecules of water. Such salts are called hydrates. Wikipedia even has a nice picture of a hydrated vs non hydrated salt.

You can have anhydrous ferrous chloride, as well as ferrous chloride tetrahydrate. The number indicates how many water molecules are present in the crystal lattice's unit cell.

EDIT

This has just crossed my mind: be careful when preparing solutions of metal complexes to make sure you know what compound you're working with. One gram of anhydrous salt contains more equivalents than a gram of hydrated salt. This information should be obvious on the bottle. If not, ask a lab tech if they can identify the hydrated/anhydrous salt by memory.


Is it possible to have a dispersion interaction with no inductive interactions? In symmetry-adapted perturbation theory, the interaction energy of a non-covalently bound pair of molecules (it could be done to arbitrary order but is usually just pairs) is built up from a perturbative expansion of the interaction potential between two monomers.

The first-order perturbation drops out a term which we call induction. This has components associated with the induction on monomer B by monomer A and vice versa. The second-order perturbation drops out a term which we call dispersion (this is what people usually refer to when they say van der Waal's force (although that's not technically correct...)).

My question is, can that second-order perturbation term--dispersion-- have a non-zero contribution to the total energy while the first-order induction terms do not contribute?

When I imagine this situation physically, it doesn't make much sense because it means that the presence of monomer A near monomer B does not perturb the eigenstates of monomer B (beyond electrostatics, exchange, etc.) and vice versa, but somehow there is a mutual dispersion effect.

So basically what I'm asking is if it is mathematically possible to have a non-zero dispersion term while simultaneously having zero contributions from the first-order induction terms in the interaction potential expansion?

Also, if it is not possible, it seems as if we ought to be able to write the dispersion energy in terms of the induction energies, but clearly we can't because dispersion is a second-order perturbation. If the effects are related, however, can we approximate dispersion using induction? I imagine this to be similar to how in Moller-Plesset perturbation theory you can approximate the fourth-order perturbation using the second-order perturbation.

I don't know too much about the details of what I'm asking about, so if I've said something quite stupid please let me know.

The first-order correction in the interaction operator is an integral over the frozen reference densities, so in a product basis $|i_A,j_B\rangle$ that would read
$$ E^{(1)} = \langle 00|V|00 \rangle $$
and as such is the electrostatic interaction. This is the same as the Coulomb interaction of the two non-interacting reference densities.

Induction is defined as the  polarization of one monomer due to the reference density of the other monomer, so must involve integrals over the virtual orbitals of the monomer being polarized in order to capture its polarization. So induction is a second-order correction and has the form
$$ E^{(2)}_{ind} = \sum_{j\neq 0} \frac{|\langle 00|V|0j \rangle|^2}{E_{00}^{(0)}-E_{0j}^{(0)}} + \sum_{i\neq 0}\frac{|\langle 00|V|i0 \rangle|^2}{E_{00}^{(0)}-E_{i0}^{(0)}}$$
where the first term represents the polarization of monomer $B$ due to the reference density of monomer $A$ and the second term is the reverse process.

What sets the dispersion energy apart from the induction energy is that it is the simultaneous polarization of both monomers, and so the integrals run over the virtual orbitals of both monomers as in
$$
E^{(2)}_{disp} = \sum_{i\neq 0} \sum_{j\neq 0}\frac{|\langle 00|V|ij \rangle|^2}{E_{00}^{(0)}-E_{ij}^{(0)}}
$$
Mathematically, the dispersion and induction terms are both second-order correction terms in the energy expansion, but they have been grouped in a physically-intuitive way to give two different types of interactions. There are systems where dispersion is the dominant interaction, such as rare gas dimers.

EDIT:

An example of a dispersion-dominated bound system is the neon dimer. Below is the SAPT decomposition of the binding curve at the HF-SAPT(2)/aug-cc-pVDZ level.

At the equilibrium distance, dispersion dominates the binding interaction but there is a non-zero induction contribution. I am inclined to say that although the ratio of dispersion and induction may become large, you can never truly have zero induction. Some naive mathematical reasoning follows.

We need $ \langle 00|\frac{1}{r_{12}}|0i\rangle, i\neq 0$ to be zero in a Gaussian product basis. At a long-enough distance we can expand the Coulomb operator in a multipole expansion to give something like
$$
\frac{1}{r_{12}} = \alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)+...
$$
in a one-dimensional model picture. I've also folded all the coefficients into the constants $\alpha,\beta,...,etc.$ The important thing here is the symmetry of the integrals. So, lets take $i=1$ and see what happens...
$$
\langle 00|\alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)|01\rangle = \langle 00|\alpha x_1x_2|01\rangle + \langle 00|\beta x_1x_2^2|01\rangle-\langle 00|\beta x_1^2x_2|01\rangle+ ...
$$
The first two integrals are zero because $\langle 0|x_1|0\rangle=0$. The third integral is non-zero and survives in the sum. This can happen any time the operator in monomer $A$ is an even function. As $i$ gets larger, higher-order terms in the expansion contribute to the induction energy. So mathematically, there is a tiny but non-zero induction contribution.


Software platforms with integration of multi vendor mass spectrometer device data? I know about ACD/Spectrus Processor which is a software platform to integrate analytical chemistry data from multiple mass spectrometer device vendors (e.g. Agilent Technologies, Bruker, Waters) to allow data processing. Are there comparable software platforms out there which enable the integration of MS/chromatography data of different vendors for data processing?

BTW: I know that questions of this kind usually fits better to Software Recommnedations. However cause it's specific to the chemical domain I think it's more suitable to ask here.

Some of the MS vendors made their software work with other vendors as well. E.g. ThermoFisher's Chromeleon, Waters' Empower seems to have Data Converters for this. If you already have software bought from your vendors it may make sense to research their cross-vendor capabilities.

There are also pure software companies that build vendor-agnostic tools:

As you mentioned - ACD/Spectrus which has a wide range of vendors supported.
Virscidian's Analytical Studio.
Elsci's Peaksel (disclaimer: I'm one of the developers).

The last two have good support for handling libraries (24, 96, 384, 1536 or more samples per experiment).

There are also open-source packages and libraries:

OpenChrom, if you don't like something you can contribute or just change it to your own needs.
MZmine2. There's also a repo with MZmine3, but I don't know what's the status there.