Calculating ΔH from bomb calorimetry Suppose we carry out a reaction in a bomb calorimeter whose starting temperature is $298.15\ \mathrm K$. Here we assume $\Delta V$ is close enough to zero that we consider the process to be at constant volume.

Based on $\Delta T$ and the calorimeter constant, we determine the internal energy change for the reaction at constant $T$ and $V$, which I'll call $\Delta U_\mathrm r$. (We can assume constant $T$ because we know how much heat flow would be required to return the calorimeter to its starting temperature.)

A typical pchem textbook will then say you can calculate what $\Delta H$ would be if the reaction were instead carried out at the same temperature, but at a constant external pressure of $1\ \mathrm{bar}$, namely $\Delta H_\mathrm r^⦵$, from the $\Delta U$ we determined in the calorimeter, as follows:

$$\Delta H_\mathrm r^⦵=\Delta U_\mathrm r^⦵+\Delta pV=\Delta U_\mathrm r^⦵+p\,\Delta V\approx\Delta U_\mathrm r^⦵+\Delta n_\text{gas}RT$$

But: we didn't determine $\Delta U$ at standard state ($1\ \mathrm{bar}$).  We determined it at constant volume. Thus, strictly speaking, when we do the following calculation using the $\Delta U$ from the calorimeter

$$\Delta H_\mathrm r^⦵\approx\Delta U_\mathrm r^⦵+\Delta nRT$$

aren't we additionally making the approximation that $\Delta U_\mathrm  r=\Delta U_\mathrm r^⦵$? I.e., aren't we ignoring the volume-dependence of $U$? Typically, this is a very good approximation since, for most substances, $U$ is only weakly dependent upon $V$. But it seems it's an approximation nonetheless. If so, it is one that is not (but should be) explicitly acknowledged in these textbooks.

There are other approximations, of course, such as the one shown above, that $p\Delta V\approx\Delta n_\text{gas}RT$ (ignore volume change in liquids and solids, and use ideal gas law to calculate volume change due to change in number of moles of gas), but I'm specifically wondering about this one.

To quote Sime's textbook on Physical Chemistry, experiment 3 on the measurement of heats of combustion of hydrocarbons using bomb calorimetry: "we will assume that $\Delta U = \Delta U_r^o$; that is, we will assume that $\Delta U$ is independent of pressure". There is no further justification of the statement. The Parr Instrument Company provides an introductory guide to bomb calorimetry (document 483M) which has the following to say:

Users interested in high precision calorimetry
will understand that test results obtained in a bomb
calorimeter express calorific values obtained at constant
volume. This differs from the usual conditions
under which fuels are burned at constant pressure
rather than constant volume. If the sample contains
elements which normally would increase in volume
after combustion, a correction must be applied to
account for the heat equivalent of this work. This
value is so small as to be negligible in routine work,
and it usually is disregarded.

The required pressure correction can be computed from thermal expansion and isothermal compressibility coefficients. For instance, for water one estimates that if the bomb pressure rises by $\pu{2 GPa}$ over standard pressure then the energy will increase by $\approx \pu{80 J/mol}$. Compare this correction with the heat of combustion of benzoic acid ($\approx \pu{3221 kJ/mol}$) and (assuming the coefficients for benzoic acid are of the same order of magnitude) it's clear that for condensed substances the correction is (often) negligible.
For gases, based on the assumption of ideal gas behavior, an isothermal PV change for the products will leave $\Delta U$ unchanged, so that in fact $\Delta U_r = \Delta U_r^o$.


Why is the thione tautomeric form of 2-Mercaptopyridine favoured? During some organic chemistry studies (through an old online course material) I ran into a doubt about the equilibrium in tautomeric forms of some heteroaromatics.

First, the context:

On the course webcast, it was said that on the molecule 2-pyridone we have a tautomerism equilibrium like in the image below, where the pyridone form is favored instead of the 2-hydroxipyridine form.

Pyridone Tautomerism Equilibrium

Each tautomer have 2 major contributors for resonance, listed on the image below (for the 2-hydroxipyridine and 2-pyridone respectively):

On the 2-hydroxipyridine resonance forms (left) we have on the left side of the equilibrium an aromatic uncharged molecule and on the right side an aromatic charged molecule. However the most electronegative heteroatom (oxygen) is positively charged and the nitrogen negatively charged. This makes this resonance form a weaker contributor.

On the 2-pyridone resonance forms (right) we have on the left side of the equilibrium a non aromatic (8pi $\ce{e-}$) uncharged molecule and on the right side an aromatic (6pi $\ce{e-}$) charged molecule, where the oxygen is the negatively charged heteroatom, matching the electronegativity character of each heteroatom. This favors the 2-pyridone tautomer.

Another factor mentioned was the dimerization of the molecule, which was geometrically favored in the 2-pyridone form, as seem below:

So far so good, the doubt comes when the tautomers of 2-mercaptopyridine are discussed.

Tautomers equilibrium of 2-mercaptopyridine:

It's known that the thione form (right) is favored. It's not explained as detailed as in the pyridone case why, so I decided to analyse it.

I've drawn the 2 major contributors of resonance on each tautomer. I'll show and discuss them.

First the thiol tautomer form resonance contributors:

On the left we have an uncharged aromatic (6pi $\ce{e-}$) molecule. On the right we have a charged and non aromatic (8pi $\ce{e-}$) molecule, with a positive charge on sulfur and a negative charge on nitrogen.

Now the thione tautomer form resonance contributors:

On the left we have a non aromatic (8pi $\ce{e-}$) uncharged molecule. On the right we have a charged aromatic (6pi $\ce{e-}$) molecule, with a negative charge on sulfur and a positive charge on nitrogen.

I could make the following assumptions after drawing those:

1) Considering only aromaticity, none of the tautomeric forms are favored, since both forms have one aromatic contributor and one non aromatic contributor.

2) Considering the heteroatoms electronegativity, knowing nitrogen is more electronegative than sulfur, the thiol tautomer would have the right charges on its charged resonance contributor, being more stable than the thione tautomer.

3) Considering the geometry for dimerization, the thione tautomer seems to have a better geometry for the hydrogen bonds and would be favored over the thiol tautomer.

Only assumption (3) is in favor of the thione formation, however it's the favored tautomer. Why so?

In the course webcast, the teacher makes it sound like having the negative charge on the sulfur and the positive charge on nitrogen would be the right charge distribution. Wouldn't the opposite be true considering that sulfur is less electronegative than nitrogen?

Here's a youtube link to the webcast mentioned:
Hydroxypyridine-Pyridone Tautomerism

Thank you in advance!

EDIT NOTE:

When I originally wrote the question, I made a mistake while counting the thiol tautomer pi electrons on the resonance form of the right. I originally counted 6pi $\ce{e-}$, but it actually has 8pi $\ce{e-}$ (4 from the carbon-carbon pi-bonds, 2 from the nitrogen lone pair and 1 from the sulfur-carbon double bond), which means the right resonance form isn't aromatic. Thus aromaticity doesn't favor neither of the tautomers. The question still makes sense though, so I'm fixing the mistake and keeping it.

Assumption (1) doesn't hold too much precedence in this case, just like in the case of 2-pyridone, so it is a minor thing to consider. In the case of the assumption (2), electronegativity is only a part of it. Sulfur atom can stabilize the negative charge better than oxygen due to it's size. For the same reason, $\ce{CH3SH}$ is a stronger acid than $\ce{CH3OH}$: $\ce{CH3S-}$ is a more stable conjugate base than $\ce{CH3O-}$. So, although sulfur has a lower electronegativity than oxygen, it is still preferred to have a negative charge on it. Thus, altogether, the thione form is more stable than the thiol.

BTW, thank you for a very detailed question and thorough explanation of your reasoning!


Hybridization of nitrogen in trisilylamine, (SiH3)3N? I want to know the hybridization of the central atom in $\ce{(SiH3)3N}$.

I think it should be $\mathrm{sp^3}$, because $\ce{N}$ is attached to three silicon atoms and one lone pair. But actually it is supposedly $\mathrm{sp^2}$.

How is this so?

Ordinarily and according to Bent’s rule, we would expect nitrogen’s lone pair to be in an $\mathrm s$ orbital and nitrogen using its three $\mathrm p$ orbitals to form three bonds to the three silicon atoms. This configuration would allow for the greatest stabilisation.
However, due to nitrogen’s small size, this perfect world already falls apart for ammonia ($\ce{NH3}$), where nitrogen is bound to three otherwise tiny hydrogen atoms. Because an electronically perfect angle of $90^\circ$ would generate much too much steric strain between the hydrogen atoms, $\mathrm s$ contribution is mixed into the bonding $\mathrm p$ orbitals to a certain extent; for ammonia, this extent happens to be almost perfect $\mathrm{sp^3}$ — results for other amines will vary. This electronic situation is not ideal, however it is clearly better than having $\mathrm{sp^2}$ hybridisation and the lone pair in a $\mathrm p$ type orbital. An $\mathrm{sp^2}$ hybridisation of nitrogen in ammonia can be reached, but only as the transition state of nitrogen inversion.
Carrying on to the compound $\ce{N(SiH3)3}$, we would be inclined to again assume a hybridisation of $\mathrm{sp^3}$ in line with the previous paragraph. However, Beagley and Conrad performed electron diffraction studies on $\ce{N(SiH3)3}$ and found the molecule to be practically planar within experimental error.[1,2] A planar molecule without doubt means that nitrogen is $\mathrm{sp^2}$-configured in $\ce{N(SiH3)3}$.
The question remains why. There must be some kind of stabilising interaction of nitrogen’s remaining $\mathrm p$ orbital with something else to keep that molecule planar. Beagley and Conrad suggest — in line with what was thought at the time — that this be due to π bonds with silicon’s remote $\mathrm d$ orbitals.[1]. Numerous evidence, much of which is collected on this site, speaks the opposite (namely that $\mathrm d$ orbitals do not play any role in the bonding situation of main group metals).
Instead, I think we are dealing with something you may call ‘inverse hyperconjugation’. Remember that $\chi(\ce{Si}) = 1.9$ which is less than hydrogen, meaning that the $\ce{Si-H}$ bonds are polarised towards hydrogen. This in turn means that $\sigma^*_{\ce{Si-H}}$ is a silicon-centred orbital with its primary lobe pointing towards nitrogen. Therefore, nitrogen’s $\mathrm p$ orbital can favourably interact with the antibonding $\sigma_{\ce{Si-H}}^*$ orbital, increasing the $\ce{Si-N}$ bond order and decreasing the $\ce{Si-H}$ bond order. The effects are the same as with hyperconjugation stabilisation of secondary or tertiary carbocations but the electronic demand is reversed. We could attempt to draw the following resonance structures in Lewis formalism to explain this:
$$\ce{H-SiH2-N(SiH3)2 <-> \overset{-}{H}\bond{...}SiH2=\overset{+}{N}(SiH3)2}\tag{1}$$
In this Lewis formalism, the double bond would be generated from a $\mathrm p$ orbital on both silicon and nitrogen.

Note and reference:
[1]: B. Beagley, A. R. Conrad, Trans. Faraday Soc. 1970, 66, 2740–2744. DOI: 10.1039/TF9706602740.
[2]: Actually, $\angle(\ce{Si-N-Si}) \approx 119.5^\circ < 120^\circ$. The authors state:[1]

The apparant slight deviation from planarity is associated with a shrinkage effect11 on the $\ce{Si\dots Si}$ distance of about $\require{mediawiki-texvc}\pu{0.007 \AA}$ (see [table]). Spectroscopic results12 are entirely in agreement that the molecule is planar.
11 A. Allmenningen, O. Bastiansen and T. Munthe-Kaas, Acta Chem. Scand., 1956, 10, 261. [sic!]
12 E. A. V. Ebsworth, J. R. Hall, M. J. Mackillop, D. C. McKean, N. Sheppard and L. A. Woodward, Spectrochim Acta, 1958, 13, 202. [sic!]


To Debunk the 9/11 conspiracy, what is the science behind melting of the towers architecture? I am wondering what the chemical validity is of jet fuel actually melting the steel beams that held up the twin towers during the September attacks.

Jet Fuel type A (Jet A) burns at approximately 2,230 °C. 0.2% carbon steel has a maximum forging temperature (i.e., the point at which it'll melt) at about 2410 °C, while 1.5% carbon steel is at about 1920 °C. This also doesn't take into account effects that happen with how aerosolized jet fuel will burn, which is, much hotter, and strange effects that'll happen as everything else has it's surface area very spontaneously increased, subjected to a large amount of kinetic energy at once, and then lit on fire.

Not to mention everything else burning at the same time. Finely shredded aluminum is very flammable.

Also: As a final note: The steel didn't actually have to melt, persay. It only had to weaken enough that the mechanical forces of the skyscraper that it was supporting became too great for the softened steel, at which point it'll all collapse in a chain reaction.

TL;DR: Conspiracy theories are BS.


Is it possible to determine the absolute value of entropy of a system? I came across a question (in some assignments I had received) where it was asked:

  Can the absolute value of entropy of a system can be measured?

and to my surprise, the answer said yes. Till now, I knew that entropy is a state function, just like enthalpy and internal energy, and that only change in entropy can be measured.

Am I missing something here?

Well, in practical I would say it is not possible. But there is a way to theoretically approximatly calculate an absolute value for entropy.

According to the third law of thermodynamics the entropy change can be calculated as

$$\mathrm d S = \frac{\delta q}{T}\, .$$

Per definition, a perfect crystal at $\pu{0 K}$ would have an entropy $S=0$.

So when you now heat that crystal and measure the change in entropy, since your initial $S$ was zero, the change is equal to the absolute value.

One problem is though, that the above equation only holds for constant temperatures. So the steps at which you measure the entropy changes have to happen at very (infinitesimal) small temperature changes.


Energy balances for a heated reactor I have a heated mixing reactor, as seen in the figure.
How to set up the energy balance of the heating jacket and the reactor?

The flow rate of the heating fluid $F_\mathrm j$ probably refers to the mass flow rate (e.g. expressed in $\mathrm{kg/s}$) since the indicated flow rates into the heat exchanger and out of the heat exchanger are identical. (The corresponding values for the volume flow rate (e.g. expressed in $\mathrm{m^3/s}$ would be different since the change in temperature causes a difference in density.)

$$F_\mathrm j=\frac mt\tag{1}$$

The change in temperature of the heating fluid $\Delta T$ is given by the different temperatures of heating fluid before and after the heat exchanger:

$$\Delta T=T_\mathrm j-T_\mathrm{ji}\tag{2}$$

The ratio of the amount of heat energy $Q$ (e.g. expressed in $\mathrm{J}$) that is transferred to a system at constant pressure and the resulting increase in temperature $\Delta T$ is given by the heat capacity at constant pressure $C_p$ (e.g. expressed in $\mathrm{J/K}$):

$$C_p=\frac Q{\Delta T}\tag{3}$$

You may know the value of the specific heat capacity at constant pressure $c_p$ (e.g. expressed in $\mathrm{J/(kg\cdot K)}$) for the heating fluid at the average temperature of the heat exchanger, which is defined as heat capacity at constant pressure divided by mass $m$:

$$c_p=\frac{C_p}m\tag{4}$$

Now you can rearrange the definition of specific heat capacity $\text{(3)}$ to solve for the heat $Q$:

$$Q=C_p\cdot\Delta T\tag{5}$$

insert the result for $C_p$ from equation $\text{(4)}$

$$Q=c_p\cdot m\cdot\Delta T\tag{6}$$

the result for $\Delta T$ from equation $\text{(2)}$

$$Q=c_p\cdot m\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{7}$$

and the result for $m$ from equation $\text{(1)}$

$$Q=c_p\cdot F_\mathrm j\cdot t\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{8}$$

Therefore, the thermal power $P$ of the heat exchanger (e.g. expressed in $\mathrm{W=J/s}$) is

$$\begin{align}P&=\frac Qt\\[6pt]
&=c_p\cdot F_\mathrm j\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{9}\end{align}$$

The sign of the values for $Q$ and $P$ depends on whether heat is transferred from the fluid to the reactor or from the reactor to the fluid.

Note that the specific heat capacity $c_p$ generally depends on temperature; therefore, the specific heat capacity of the fluid changes during the passage of the heat exchanger. That is why you should use an appropriate average value. A precise engineering calculation would better be based on the specific enthalpy $h$ of the fluid at the given temperature and pressure before and after the heat exchanger.

Example

The characteristic data of a typical nuclear power plant with a pressurized water reactor are

Coolant mass flow rate $F=18\,800\ \mathrm{kg\ s^{-1}}$
Coolant pressure $p=158\ \mathrm{bar}=15.8\ \mathrm{MPa}$
Reactor pressure vessel inlet temperature $T_\text{in}=291\ \mathrm{^\circ C}$
Reactor pressure vessel outlet temperature $T_\text{out}=326\ \mathrm{^\circ C}$

The specific heat capacity of water at the inlet temperature of $T_\text{in}=291\ \mathrm{^\circ C}$ and a pressure of $p=15.8\ \mathrm{MPa}$ is $c_{p,\text{in}}=5.2553\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$. The corresponding specific heat capacity at the outlet temperature of $T_\text{out}=326\ \mathrm{^\circ C}$ is $c_{p,\text{out}}=6.4388\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$. For simplicity’s sake, we may want to use the specific heat capacity for the average temperature $\tfrac12\left(T_\text{in}+T_\text{out}\right)=308.5\ \mathrm{^\circ C}$, which is $c_p=5.6787\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$.

Using equation $\text(9)$, the thermal power of the reactor can be calculated as

$$\begin{align}P&=c_p\cdot F\cdot\left(T_\text{out}-T_\text{in}\right)\\[6pt]
&=5.6787\ \mathrm{kJ\ kg^{-1}\ K^{-1}}\times18\,800\ \mathrm{kg\ s^{-1}}\times\left(326\ \mathrm{^\circ C}-291\ \mathrm{^\circ C}\right)\\[6pt]
&=3736584.6\ \mathrm{kJ\ s^{-1}}\\[6pt]
&\approx3737\ \mathrm{MW}
\end{align}$$

By way of comparison, the literature value for the reactor core thermal power is $3765\ \mathrm{MW}$.

Note that, in this example, heat is transferred from the reactor to the fluid.


Born-Oppenheimer approximation I'm studying about formulation of Born-Oppenheimer approximation from Atkins, Molecular quantum mechanics, 5th edition. In chapter 8.1 I found the following:

I keep wondering where did this W come from in equation 8.4.? It says later in text that it is responsible for non-adiabatic effects, but I don't see how it just appeared here. If someone is willing to explain, I'd be greatful.

Also, if anyone has tips on where to find more nice and clean text about more general formulation of Born-Oppenheimer approximation, since this one is confined to particular system.

Thanks!

Atkins did a good job on explaining what parametric dependence of the electronic wave function (and electronic energy) on nuclear coordinates is, but failed to mention that in the derivation of the Born-Oppenheimer approximation this dependence is assumed to be continuous and differentiable and that both first and second derivatives of this quantities with respect to nuclear coordinates are in general non-zero.
In particular, for the system described in the text, we have
$$
  \frac{\partial \psi}{\partial Z_j} \neq 0 \, , \quad
  \frac{\partial^2 \psi}{\partial Z_j^2} \neq 0\, .
$$

Now, when the solution of the form (8.3) is substituted into the Schrödinger equation (8.2), the term involving $T_\mathrm{e}$ is trivial: since the nuclear wave function $\chi$ is not a function of electronic coordinates it is just a constant when differentiating with respect to them, so we get
$$
  T_\mathrm{e} (\psi \chi) = \chi T_\mathrm{e} \psi \, .
$$
But the term involving $T_\mathrm{N}$ does not trivially transform in a similar way,
$$
  T_\mathrm{N} (\psi \chi) \neq \psi T_\mathrm{N} \chi
$$
since both $\psi$ and $\chi$ depend on the nuclear coordinates. Rather, applying the product rule twice, we get
\begin{align}
  \frac{\partial^2}{\partial Z_j^2} (\psi \chi)
  &=
  \frac{\partial}{\partial Z_j} \left( \frac{\partial}{\partial Z_j} (\psi \chi) \right) \\
  &=
  \frac{\partial}{\partial Z_j} \left( \psi \frac{\partial \chi}{\partial Z_j} + \chi \frac{\partial \psi}{\partial Z_j} \right) \\
  &=
  \psi \frac{\partial^2 \chi}{\partial Z_j^2}
  +
  2 \frac{\partial \psi}{\partial Z_j} \frac{\partial \chi}{\partial Z_j}
  +
  \chi \frac{\partial^2 \psi}{\partial Z_j^2} \, ,
\end{align}
so that
$$
  T_\mathrm{N} (\psi \chi)
  =
  - \psi \sum\limits_{j=1,2} \frac{\hbar^2}{2 m_j} \frac{\partial^2 \chi}{\partial Z_j^2}
  -
  \sum\limits_{j=1,2} \frac{\hbar^2}{2 m_j} \left(
  2 \frac{\partial \psi}{\partial Z_j} \frac{\partial \chi}{\partial Z_j}
  +
  \chi \frac{\partial^2 \psi}{\partial Z_j^2} \right) \, ,
$$
where the first term is nothing but $\psi T_\mathrm{N} \chi$ and the remainings are designated as $W$.


What is formed when hydrogen peroxide reacts with titanium? I just got night contact lenses, and you are supposed to use a "one-step" cleaning solution to clean it over a period of six hours. At the bottom of the lenses case, there is a "titanium plate neutralizer", which causes the liquid to bubble. What is this gas?

Edit: it was platinum, not titanium

Hydrogen peroxide will damage your eye tissue; that's why there are enzymes in our cells actually to catalyze the conversion of $HOOH$ to water and oxygen gas. The enzymes are actually contained in peroxisomes (name is self-explanatory).* The neutralizer serves the same function of catalyzing the otherwise relatively slow decomposition of $HOOH$ into water and oxygen gas; hydrogen peroxide will decompose in the presence of light and that's why bottles of hydrogen perioxide are opaque - generally black or brown (but not transparent like bottles of which witch hazel or isopropyl alcohol). I have seen platinum used as a catalyst in contact cleaning solutions but I suppose titanium can be used as well.

*Edit: peroxisomes actually create $HOOH$ upon catabolism of cell by-products but can also break down its own toxic byproduct.


HOMO Odd, LUMO Even: Why does the affected orbital change depending on the carbon? How are carbons numbered? While looking at how substituents affect the absorption spectra of molecules, I came across this:

  Substituents alter the emitted wavelengths of these molecules by affecting the electron densities. Electron Withdrawing Groups (or EWGs) withdraw electron density, thereby making the arrangement more stable. The energy level of the orbital affected subsequently lowers, causing a typically-noticeable red shift in the maximum wavelength. Alternately, Electron Donating Groups (EDGs, for short), add to the electron densities of the molecule. This causes the affected orbital to be less stable, and requires more energy to obtain. The orbital affected depends on the placement of the substituent. A useful mnemonic device in this matter is HOLE: HOMO Odd, LUMO Even. This explains that the affected orbital alternates between odd and even placements.

Why does the orbital affected depend on the placement of the substituent? I thought substituents that caused bathochromic shifts just extended conjugation and hence reduced the HOMO-LUMO gap? Is this odd-even rule only applicable to azulene, or can it be applied to other chromophores?

Also, why are the carbons numbered the way they are? Here's a picture with the numbering.

Shouldn't the carbons with the question mark also be counted?

To start with, I'll address the frontier orbitals. Lets look at what the HOMO and LUMO look like.

Notice the LUMO has no density on carbons $1$ or $3$ and (while it is a little difficult to see in this image) little density on carbons $5$ and $7$. Similarly the HOMO has little to no density on any of the even carbons. So, if we want to affect the energy of these orbitals, we have to add a substituent to a carbon that has electron density from that orbital. The odd-trend is likely not a general rule, but if you know the HOMO and LUMO reside mainly on different carbons, you can assume that selective functionalization of those carbons will affect the orbital of interest. I say "assume" because you should always keep in mind that the orbital picture is an approximation and that once you make a substitution you have to check to make sure you haven't perturbed the system enough where your new compound doesn't have frontier orbitals that look like this anymore.

In regards to the numbering, this is really just a matter of convenience for a fused ring system like this. We don't number these bridging carbons because they can't be functionalized; that is, we can't add a group to these carbons without turning the molecule into something other than azulene. So the numbering starts from next to these bridging carbons and skips over them as you move around the rings.


Two electron integral evaluation over primitive Gaussians without recurrence relations I want to evaluate two-electron integrals over s, p, and d primitive Gaussian functions. The procedure is presented for s-type ones in "Modern Quantum Chemistry" by Szabo and Ostlund, but I need to evaluate these integrals for p and d ones. However, most books, such as "Molecular Electronic-Structure Theory" by Helgaker which I read, evaluate these integrals using recurrence relations. Is there any other method which can evaluate two-electron integrals, without using recurrence relations?

As you mentioned, most implementations such as Obara–Saika (OS), McMurchie–Davidson (MD), Head-Gordon–Pople (HGP), and many others use a series of horizontal and vertical recurrence relations to build up angular momentum and transfer it among each of the four centers. These are analytic solutions, up until the need to solve the Boys function, however we could debate as to whether or not they are closed-form expressions by terminology. They are closed-form in the sense that each part of the recursive expansion does terminate properly with $(ss|ss)$, given in OS equation 44.

In general, any recursive function can be rewritten to be non-recursive using loops of sums and or products, though it may involve considerable effort. In that sense, the recurrence relationships are quite powerful and avoid lots of "boilerplate". If you are interested in a non-recursive expression, there is the earlier Taketa–Huzinaga–O-ohata (THO) paper that, while not very descriptive, includes explicit working expressions for 2-electron and 1-electron (overlap, kinetic energy, and nuclear-electron attraction) integrals. The most relevant expression is

\begin{align*}
\text{ERI} &= \frac{2\pi^{2}}{\gamma_{1}\gamma_{2}} \left( \frac{\pi}{\gamma_{1} + \gamma_{2}} \right)^{1/2} \exp \left\{ - \frac{\alpha_{1}\alpha_{2}\overline{AB}^{2}}{\gamma_{1}} - \frac{\alpha_{3}\alpha_{4}\overline{CD}^{2}}{\gamma_{2}} \right\} \\
&\times \sum_{i_{1}i_{2}r_{1}r_{2}u} B_{i_{1}i_{2}r_{1}r_{2}u}(l_{1}, l_{2}, A_{x}, B_{x}, P_{x}, \gamma_{1} | l_{3}, l_{4}, C_{x}, D_{x}, Q_{x}, \gamma_{2}) \\
&\times \sum_{j_{1}j_{2}s_{1}s_{2}v} B_{j_{1}j_{2}s_{1}s_{2}v}(m_{1}, m_{2}, A_{y}, B_{y}, P_{y}, \gamma_{1} | m_{3}, m_{4}, C_{y}, D_{y}, Q_{y}, \gamma_{2}) \\
&\times \sum_{k_{1}k_{2}t_{1}t_{2}w} B_{k_{1}k_{2}t_{1}t_{2}w}(n_{1}, n_{2}, A_{z}, B_{z}, P_{z}, \gamma_{1} | n_{3}, n_{4}, C_{z}, D_{z}, Q_{z}, \gamma_{2}) F_{\nu}(\overline{PQ}^{2}/4\delta),
\label{tho-2.22-1}\tag{THO eq. 2.22 part 1}
\end{align*}

with

\begin{align*}
&B_{i_{1}i_{2}r_{1}r_{2}u}(l_{1}, l_{2}, A_{x}, B_{x}, P_{x}, \gamma_{1} | l_{3}, l_{4}, C_{x}, D_{x}, Q_{x}, \gamma_{2}) \\
&\quad = (-)^{i_{2}} f_{i_{1}}(l_{1}, l_{2}, \overline{PA_{x}}, \overline{PB_{x}}) f_{i_{2}}(l_{3}, l_{4}, \overline{QC_{x}}, \overline{QD_{x}}) \\
&\qquad \times \frac{i_{1}!i_{2}!}{(4\gamma_{1})^{i_{1}}(4\gamma_{2})^{i_{2}}\delta^{i_{1} + i_{2}}} \cdot \frac{4\gamma_{1})^{r_{1}}(4\gamma_{2})^{r_{2}}\delta^{r_{1} + r_{2}}}{r_{1}!r_{2}!(i_{1}-2r_{1})!(i_{2}-2r_{2})!} \\
&\qquad \times [i_{1} + i_{2} - 2(r_{1} + r_{2})]! \frac{(-)^{u} p_{x}^{i_{1} + i_{2} - 2(r_{1} + r_{2}) - 2u} \delta^{u}}{u! \{ i_{1} + i_{2} - 2(r_{1} + r_{2} - 2u)\}!},
\label{tho-2.22-2}\tag{THO eq. 2.22 part 2}
\end{align*}

and intermediate quantities are

\begin{equation}
\nu = i_{1} + i_{2} + j_{1} + j_{2} + k_{1} + k_{2} - 2 (r_{1} + r_{2} + s_{1} + s_{2} + t_{1} + t_{2}) - u - v - w,
\end{equation}

\begin{align*}
\gamma_{1} &= \alpha_{1} + \alpha_{2},~\gamma_{2}\text{ is similar}\\
\delta &= \frac{1}{4\gamma_{1}} + \frac{1}{4\gamma_{2}} \\
\mathbf{P} &= \frac{\alpha_{1} \mathbf{A} + \alpha_{2} \mathbf{B}}{\alpha_{1} + \alpha_{2}},~\mathbf{Q}\text{ is similar} \\
\mathbf{p} &= \mathbf{Q} - \mathbf{P} \\
\overline{AB} &= A - B
\end{align*}

\begin{equation}
F_{\nu}(t) = \int_{0}^{1} u^{2\nu} \exp(-tu^{2}) \mathop{du},~\text{the Boys function.} \tag{THO eq. 2.11}
\end{equation}

$(l, m, n)$ are the usual polynomial exponents leading to s-, p-, and d-type functions, $(\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D})$ are the usual basis function positions, the $\alpha$ are the basis function exponents, and $f_{j}(l, m, a, b)$ is the coefficient of $x^{j}$ in the expansion of $(x + a)^{l} (x + b)^{m}$.

  In \eqref{tho-2.22-1} the summations with respect to the indices $i_{1}, i_{2}, r_{1}, r_{2}, u$ extend from 0 to $l_{1} + l_{2}$, $l_{3} + l_{4}$, $[i_{1}/2]$, $[i_{2}/2]$ and $[(i_{1} + i_{2})/2 - r_{1} - r_{2}]$ respectively. The ranges of $(j_{1}, j_{2}, s_{1}, s_{2}, v)$ or $(k_{1}, k_{2}, t_{1}, t_{2}, w)$ may be found in the same way.

Regarding the indices used in the summations:

If the index $i_{1}$ runs from 0 to $l_{1} + l_{2}$, and $i_{2}$ runs from 0 to $l_{3} + l_{4}$ we can deduce that $i_{1}$ must cover all possible values of the total angular momentum along the x-component for the bra, and $i_{2}$ goes up to the total angular momentum along the x-component for the ket. The indices $j,k$ match with $m,n$, meaning the y- and z-component.
$r,s,t$ are not to be confused with distances themselves, like how we usually define $r_{A} = |\mathbf{r}_{A}| = |\mathbf{r} - \mathbf{A}|$. Both $r,s,t$ and $u,v,w$ are also related to the three Cartesian coordinates, but in a more complicated manner than $i,j,k$; they appear to originate from the closed-form expressions for the Hermite polynomials.

Their implementation is slightly modified, with the necessary expressions in equations 3.3, 3.4, and 3.5, though the above can be implemented directly as well. A quick search revealed this code that implements all the THO integrals in C++. It is worth commenting that in real packages, the Boys function is precomputed from interpolation tables, rather than from the gamma and incomplete gamma functions.


Why do all inter-halogen compounds contain an even number of atoms? Why do all inter-halogen compounds contain an even number of atoms?

All halogens have one electron in their valence shell. One covalent bond is formed by 2 halogen atoms (sharing one electron each). If in the compound an odd number of atoms were involved, then an odd electron compound (unstable) would be formed. IF4 would contain 1 lone pair and 1 odd electron, hence unfavorable.

Is my answer correct? If you have any better explanation, please share it.

Yes neutral $\ce{IF4}$ would be unstable because it would have one lone pair and one unpaired electron.  The extra lone pair is not a problem because Iodine is large enough to exceed an octet, but the unpaired electron makes it a free radical.


Are the potentials of the electrolytes equal in Galvanic cell? My book defines Electrode Potential for a Galvanic Cell as follows:

A potential difference that develops between the electrode and the electrolyte is called Electrode Potential

$$E= V_{electrode} - V_{electrolyte}\tag1$$
It further defines cell potential as:

The potential difference between the two electrodes of a galvanic cell is called Cell Potential.

Mathematically, I interpret it as
$$E_{Cell} =  V_{electrode,anode} - V_{electrode,cathode}\tag2$$
However my book further mentions that

The cell potential is the difference between the electrode potential of the anode and cathode.

I interpret this as:
$$E_{Cell} = E_{anode} - E_{cathode}$$
From (1)
$$E_{cell}=(V_{electrode,anode} - V_{electrolyte,anode} )- (V_{electrode,cathode} - V_{electrolyte,cathode})\tag3$$
To keep the above consistent with (2),
$$V_{electrolyte,anode} = V_{electrolyte,cathode}\tag4$$

I find (4) illogical as there is no basis for the two electrolytes to be at same potential.
What is going on here?

There is a potential difference between an electrode and an electrolyte, but it is not directly measurable and the electrolyte potential is rather undefined.
We do not use electrolyte potentials as reference potentials, as reference must be well defined. By mechanical analogy, it would be like defining the relative height wrt a level that is not clearly given. By other words, the conventional electrode potential is NOT the potential difference between the electrode and electrolyte, even if this difference does exist.
By convention, the electrode potential is considered the potential difference between the electrode and the standard hydrogen electrode(SHE) $E - E^{\circ}_{\ce{H2/H+/Pt}}$.
The cell potential is called rather the cell voltage or the cell electrode potential difference. The cell voltage is difference of electrode potentials, which are difference of electrode potentials wrt SHE.
$$U_\mathrm{cell} = E_\mathrm{cathode} - E_\mathrm{anode} \\ = ( E_\mathrm{cathode,absolute} - E^{\circ}_{\ce{H2/H+/Pt}}) - (E_\mathrm{anode,absolute} -  E^{\circ}_{\ce{H2/H+/Pt}})\\
= E_\mathrm{cathode,absolute}  - E_\mathrm{anode,absolute}$$
Note that cathode is the more negative one for electrolytic cells, but more positive one for galvanic cells.
The cell voltage and the potential electrode difference wrt SHE ( = the cell voltage if the other electrode were replaced by SHE ) does not depend on the electrolyte potential.


How do I find change in Gibbs free energy of water formation? How do I find the change in Gibbs free energy of the reaction

$$\ce{H2 + \frac{1}{2}O2 -> H2O}$$

when it is at $\pu{75^\circ C}$ and $\pu{1 atm}$? Do I calculate the Gibbs free energy of each species and then do

$$\sum\Delta G{\mathrm{(products)}} - \sum\Delta G{\mathrm{(reactants)}}?$$

You're using the correct equation, $\Delta G^0_r(T;P)=\sum \nu_i \mu_i^0(T;P)$,
where $\Delta G_r^0(T;P)$ is the standard free energy change for the reaction at $T$ and the pressure $P$.  The task is to calculate each $\mu_i^0$.  Provided the same convention for the standard state pressure $P^0$ and for the zero value of $\mu_i$ is used, then $\Delta G_r^0$ for any reaction will be the same.  Here are some ways to calculate $\mu_i^0(T;P)$ for each species to use in the equation and the data required to calculate them (available in tables). In the following, $P^0$ is the standard state pressure used in the tables and subscript $fi$ denotes the reaction forming the species from the elements in their standard states.  If the tabular values are available at a different pressure than the one you want ($P$), you must add $RT\ln(P/P^0)$ to the values of $\mu_i$ indicated below.

$\mu_i^0 = \Delta G^0_{fi}(T;P^0)$.
$\mu_i^0 = \Delta H^0_{fi}(T) - T\Delta S^0_{fi}$.
$\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T) dT -Ts_i^0(T)$.
$\mu_i^0 = \Delta H^0_{fi}(T) - Ts_i^0(T)$.
$\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T)dT -T\left[s_i^0(298.15) + \int_{298.15}^T \frac{c^0_{pi}}{T}dT\right]$

There are many other ways.  It's an interesting exercise to prove why all the above routes give the same value of $\Delta G^0_r$ for any reaction.


What are the physical properties affected in ZnO when there is a sulfur impurity? I have a question. I work at a brass metal foundry where one of our waste products is $\ce{ZnO}$, which we sell. Because we have been making new alloys which contain sulfur, our $\ce{ZnO}$ has a very small sulfur impurity level and the people that we sell our $\ce{ZnO}$ to have been complaining that our $\ce{ZnO}$ makes their product yellow. Do you think this is due to the sulfur? What are the effects of very little sulfur on tons of $\ce{ZnO}$ dust? thanks!

Of course, if $\ce{ZnO}$ has a very small $\ce{S}$ impurity level, this could be the reason of the yellow colour of products.
As for the effects of very little sulfur on tons of $\ce{ZnO}$ dust, it all depends on the end use of $\ce{ZnO}$:The applications of zinc oxide powder are numerous. Most applications exploit the reactivity of the oxide as a precursor to other zinc compounds. For material science applications, zinc oxide has high refractive index, high thermal conductivity, binding, antibacterial and UV-protection properties. Consequently, it is added into materials and products including plastics, ceramics, glass, cement, rubber, lubricants, paints, ointments, adhesive, sealants, concrete manufacturing, pigments, foods, batteries, ferrites, fire retardants, etc...

Clearly, if $\ce{ZnO}$ is used in rubber industry, sulfur impurities have no harm effect.
If it is used in electrical applications (as a semiconductor), sulfur will affect the electrical conductivity.
If it is used in coating and paints applications, sulfur will affect the the color ....


What elements and/or substances without water are liquid at room temperature? I was thinking about liquids, and I started to wonder theses related questions:

1) Besides mercury, what elements are naturally liquid at room temperature?

2) What naturally found family of substances/mixtures that do not contain  $\ce{H2O}$ are naturally liquid at room temperature?

This question is a bit broad in terms of the sheer amount of chemical compounds and mixtures that are liquid at room temperature. Examples include:

Compounds
Acids, bases, many hydrocarbons (e.g. hexane) and many more

Mixtures
Crude oil, aqua regia and many more

In terms of elements, there are only two that are liquid at room temperature (say about 20 °C or 293 K):

Mercury (as you identified).
Bromine

Francium, cesium, gallium and rubidium are close, with melting points at 300 K, 301.59 K, 303.3 K and 312.46 K respectively.

LennTech provides a list The elements of the periodic table sorted by melting point


HOMO and LUMOs of Carbon based Compounds Why are the HOMO and LUMOs often $\pi$ orbitals in carbon based compounds that contain multiple double bonds or conjugation such as ethene and butadiene?

Why are the HOMO and LUMOs often π  orbitals in carbon based compounds

The heat of hydrogenation of ethylene is -32.5 kcal/m

image source

In other words, ethylene is 32.5 kcal/m less stable (higher in energy) than ethane and this amount of energy (heat) is given off when we hydrogenate ethylene and convert it to ethane.

This tells us that pi bonds are, generally, less stable (higher in energy) than sigma bonds as shown in the following diagram (just forget about the delta bond orbitals, they don't play into this discussion).

image source

This diagram is a pictorial answer to your question.  Since pi bonds are generally higher in energy than sigma bonds, they will become the HOMO in most molecules.

Since bonding and antibonding orbitals are arranged symmetrically about the zero energy point, whatever type of orbital is the HOMO, then the antibonding counterpart will be the LUMO.  If a $\ce{\pi}$ orbital is the HOMO, than a $\pi^{*}%$ (pi star) orbital will be the LUMO.


if a balloon full of warm air was placed in a refrigerator, what is the result? If a balloon full of warm air was placed in a refrigerator, What will happen to the Ballon ?

The balloon will shrink and get smaller as the gas inside it gets cooler, the pressure inside the balloon falls. You might also have seen that a bottle partially filled with warm water kept in a refrigerator squeezes a bit because of the air pressure inside the bottle falls. This result directly follows from the gas laws.


Is water "hardness" dependent more on calcium or magnesium? If one sample of water has a certain concentration (g/L) of calcium ions, and another sample as that same concentration of magnesium ions, will one solution have a greater "hardness than the other"? In other words, does one ion have more influence over water hardness than the other?

The one with the magnesium would be harder (assuming no other hardness-inducing ions) because magnesium has a lower molecular mass, so the molarity of magnesium would be higher. Assuming a 1:1 molar ratio of magnesium:calcium would give the same hardness (i.e. the same $ppm\ \ce{CaCO3}$ rating).


Does firing a proton beam on water make it acidic? Lets say I fire a proton beam upon the surface of water. Lets assume I do the experiment in an air-free environment and that the proton beam doesn't reflect off the surface of water.

Now, protons, which are simply $\ce{H+}$, should enter the bulk of water and probably combine with a $\ce{H2O}$ molecule and form $\ce{H3O+}$. As per the definition of acidity and pH:

$$\ce{pH}=-\log [H^+]$$

If the pH reduces, the solution would become acidic as seen here. But at the same time, the water becomes charged, so I am not sure whether this is applicable.

So, in short, does water become acidic due to a proton beam?

Edit: I do not have access to proton guns and it's certainly not on sale on eBay. I only attempt to theorize on what may happen.

Yes, the concentration of $\ce{H3O+}$ will increase which will lower the pH value of the water. This is theorie based.

In reality you won't be able to measure the change. The mount of protons you need to significantly change the pH will also effect the charge of the water as you mentioned. The charged water will affect your method of measuring AND the protons will leave the water because of it's positive charge.

To neutralize the water you could add elections, which will cause the protons to form hydrogen gas. Or you could add a negative ion like chloride, which would have been the same as adding $\ce{HCl}$ to the water.


Gibbs free energy in standard state vs. equilibrium I have a problem with the definition of the standard Gibbs energy and its connection to the equilibrium constants.

I think, that I've basically understood what the different equation mean but there is one thing, I'm unable to understand:

On the one hand:

One may describe a chemical reaction with $\Delta G=\Delta G^\circ + RT\ln{Q}$. In equilibrium $\Delta G = 0$ and the equation reads $\Delta G^\circ = -RT \ln{K}$.

On the other hand:

The definition of standard state is very clear: pressure = 1 bar and all reactants and products must have activity = 1.

If I consider these two aspects separately, everything seems to be fine. But these two concepts have to be valid at the same time, what leads to $\Delta G^\circ = 0$ (always), since $K=1$ (all activities are per definition = 1).

Therefore, $\Delta G^\circ$ would be always zero. I know that this isn't true, but I don't understand why.

Can anyone explain this to me?

Thanks!

As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$\Delta G=\Delta G^\circ+ RT\ln{1}=\Delta G^\circ$$ On the other hand at equilibrium $Q=K$ and so $$\Delta G=\Delta G^\circ + RT\ln{K}$$ This of course leads to $\Delta G^\circ = -RT\ln{K}$ since at equilibrium $\Delta G=0$.

So you might want to think of it as multiple statements:

For the conversion of reactants to products in their standard states $Q=1$
At equilibrium  $\Delta G=0$
At equilibrium $Q=K$
$\Delta G^\circ$ is the free energy change for conversion of reactants to products in their standard states.

The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.

$K$ and $\Delta G^\circ$ are very much connected, but $\Delta G^\circ$ does not describe the change in free energy from reactants to products at equilibrium. $K$ and $\Delta G^\circ$ are not points on the reaction progress curve. Rather they are related to the slope of that curve.


Using stoichiometry to split a complex mineral Here is the problem I am stuck with, in particular just the first part of it.

I have calculated the moles $4.492 * 10^{-3} \,\mathrm{mol}$ and I am quizzed with regards to the equation that occurs. First assumption I am making is that the gas released is carbon dioxide.

Is it possible to separate out $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ into simply the $\ce{(Mg_{x}Cu_{1-x})(CO3)}$ part that reacts with the dilute sulfuric acid to produce the carbon dioxide, the moles of which I know?

Is it possible to separate out $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ into simply the $\ce{(Mg_{x}Cu_{1-x})(CO3)}$ part that reacts with the dilute sulfuric acid to produce the carbon dioxide, the moles of which I know?

$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{CO2}$ implies $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$.
$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ is 0.8167 g.
Then, considering the atomic weights of C, O and H, subtract the contribution of these elements from 0.8167 g to determine the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2}$.
Then divide by 2 to get the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{Mg_{x}Cu_{1-x}}$;
Then use the atomic weights of Mg and Cu to solve for $x$.


How to calculate the molarity of a gas? If I have $X$ moles of a gas and I put them in a container at constant volume $V$, will the molarity of the gas then be $X/V$?

Molarity is defined as "moles of solute per volume of solution", which implies that the system is in the liquid state.

As such, molarity is undefined for anything in a gaseous state -- there is no solute/solvent distinction in the gas phase.

You can define the concentration of a gas, though, and that calculation would be exactly the one you've described: placing $X$ moles of a gas in a container of volume $V$ yields a concentration $C=X/V$, with units of, e.g., $\pu{mol\over dm^3}$.

Note that while these units might look the same as the units of molarity, they are different. In this concentration value, the volume is the volume of the container; in a molarity value, the volume is the volume of the solution.


When is donation into an anti-bonding MO stabilising? This has been puzzling me for quite a while and I'm not sure if it could be a misconception on my part. When is interaction of a filled orbital with a antibonding MO stabilising and when is the interaction destabilising? I will now detail a few examples of the "stabilising interactions" and the "destabilising interactions". I will use Clayden's Organic Chemistry (2nd ed.) as my primary source of examples.

Destabilising interactions

Nucleophilic attack at the carbonyl carbon (p. 127)

Here, interaction of the nucleophile's HOMO with the $\ce {C=O \pi ^*}$ LUMO destabilises the $\pi$ bond, resulting in its cleavage. This is thus a destabilising interaction.

Nucleophilic attack in the $\ce {S_N2}$ reaction (p. 340)

Here, again we see the interaction between a filled HOMO and an empty antibonding MO ($\ce {C-X \sigma ^*}$ MO) destabilising the $\ce {C-X}$ bond.

Stabilising interactions

Stabilisation of cis configuration in esters (p. 805, explained in detail here)

Here, we see that the interaction of a filled, nonbonding MO centred on the $\ce {O}$ interacts with the $\ce {C=O \pi ^*}$ LUMO to stabilise the molecule overall.

Unreactivity of dichloromethane (p. 804, see here also)

As can be seen in diagram, interaction of nonbonding electrons centred on $\ce {Cl}$ with the $\ce {C-Cl \sigma ^*}$ stabilises the molecule, as it decreases is reactivity, by raising the level of the new resultant LUMO of the molecule.

Explaining Zaitsev's rule

This is a classic explanation for the increasing stability of alkenes as substitution increases. Interaction of $\ce {C-H \sigma}$ MO with the $\ce {C=C \pi ^*}$ MO increases the stability of the alkene.

I am sure there are likely to be more examples (e.g. various other instances of hyperconjugation) but I believe these examples are sufficient to prove my point. Is this inconsistency in the use of frontier MO theory or am I merely misinterpreting its use?

Update

After pondering over this for a while, I was thinking, perhaps we should make use of an MO diagram to more clearly describe the interaction. Below is a simple, typical diagram (from p. 111, with edits). Based on my understanding, all of the above examples I have presented can in fact be described by this straightforward MO diagram. Of course, the energy gap between the two interacting MOs may be different and thus, the strength of interaction would also differ.

Now, may I alert readers to something:

  In no way does the MO diagram indicate any sort of destabilisation of the molecule.

Such an interaction between a filled MO and an unfilled MO only produces stabilisation, as seen in the diagram. Not only is the filled orbital of a lower energy and hence less likely to interact with LUMOs of other molecules, the empty orbital's energy has also been raised, making it less energetically accessible from the point-of-view of a nucleophile. In other words, the molecule is both less reactive to an electrophile, due to a energetically-lowered filled orbital, and also less reactive towards a nucleophile, due to a energetically-raised empty orbital.

If this is the case, may I question further:

  What is the basis of destabilisation in the first 2 examples I have presented above?

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

The difference between the "stabilising" and "destabilising" examples that you have chosen are that the "stabilising" examples are static molecules, whereas the "destabilising" examples are chemical reactions, in which a lot is happening. We'll look at the "stabilising" cases first, as these are easier to understand.

As with any question involving "stability", it is important to consider what we are measuring stabilisation with respect to. In the example of the anomeric effect (e.g. in esters), we are comparing the true electronic state of the ester against a hypothetical scenario, where the lone pair and σ* do not overlap, but everything else is kept the same. In such cases, it is nearly always true that allowing overlap, or delocalisation, leads to a lower energy. This is captured quite well in the last picture in your question body. (See also: Physical reason for why delocalisation leads to stability?)

In a chemical reaction, as we move along the reaction coordinate (RC), we are creating overlap between two different orbitals, just as in the previous case. Naively we might think that this necessarily leads to stabilisation. But this is not physically sensible: if the reaction proceeds via a transition state (TS), this TS must be higher in energy than the reactants (at infinite separation).[1] The problem is that when we bring two molecules together in space, it is not just overlap that is affected: the energies of every orbital will change as nuclear positions are moved, electron-electron repulsions are different, etc. This means that all the orbital energies change in a way which depends not only on overlap with other orbitals (as was the case before), but also a number of other factors.

Instead of going from reactants at infinite separation to the TS directly along the RC, as would be physically realistic, it may be instructive to consider hypothetically doing it in two stages. In the first step, we bring the reactants together from infinite separation to the TS geometry while forbidding any orbital overlap; secondly, we stay at the TS geometry, but turn on the overlap. In real life, you cannot separate these two things: changing the geometry comes with formation of overlap. But thinking about it this way allows us to link it to the "stabilising" cases talked about before.

Let's use the addition of a nucleophile to a carbonyl group as an example to make this more concrete. In the reactants, there are three key frontier orbitals we need to consider: the nucleophile lone pair ($n$), and the π-type orbitals of the carbonyl group ($\pi$ and $\pi^*$).

In general, as we bring the nucleophile and carbonyl together, mixing of all three orbitals will occur continuously. As the reaction proceeds, we will therefore have three MOs $\{\psi^{(i)}\}$ $(i = 1,2,3)$ which are linear combinations of the three frontier orbitals listed earlier:

$$\psi^{(i)} = c^{(i)}_n\phi_n + c^{(i)}_\pi \phi_\pi + c^{(i)}_{\pi^*}\phi_{\pi^*}$$

For well-separated reactants, the relevant MOs will (for the most part) be the original three frontier orbitals. So, for example, we could have $\psi^{(1)} = \phi_\pi$, $\psi^{(2)} = \phi_n$, $\psi^{(3)} = \phi_{\pi^*}$. But at the end of the reaction, the MOs of the product will be a complicated mixture of the three orbitals. For example, the $\ce{C-Nu}$ $\sigma$ orbital might be made up of something that resembles this:

$$\psi^{(1,\text{new})} \approx \frac{1}{\sqrt{3}}(\phi_n + \phi_\pi + \phi_{\pi^*})$$

The reason why these linear combinations (or mixtures) can form is because there is non-zero overlap between all of these orbitals as the reaction proceeds. If there is no overlap between two orbitals, then they cannot mix. So, the way to forbid any overlap from occurring in the first step is to simply not change the coefficients of the MOs as we bring the reactants closer to each other (towards the TS geometry). Of course, this is not physically possible, which is why it is purely a hypothetical scenario.

Even though the MOs themselves do not change in this first step, their energies do change, because the Hamiltonian is changing (by virtue of the nuclear positions changing). So, the energies of the MOs will change. To throw an additional wrench into the works, it is not just the electronic energy that we must consider, but also the nuclear energy (when we bring nuclei closer to each other, they will repel each other, and this is not captured in MO diagrams which only illustrate electronic energy). As you can imagine, it is very hard to make any real analysis as to what happens in this step.

In any case, once we reach the TS geometry, the energy of the system will definitely have changed in some way. We can now carry out the second step by "turning on" the overlap, i.e. finding the appropriate linear combinations of ${n, \pi, \pi^*}$. Since nothing else here is changing (in particular, no nuclear movement), except for us "allowing" overlap to occur, it is exactly analogous to the "stabilising" case which was discussed earlier. So the second step always leads to a decrease in energy.

The problem is that, from the infinitely-separated starting materials, we have carried out two steps to get to the TS. In the second step, the energy has decreased – a "stabilisation" of sorts. But in the first step, we don't know what has happened. If a TS is to be higher in energy than the reactants, then it follows that there must have been an increase in energy in the first step.[2]

Simple LCAO-MO diagrams don't quite capture the complexity involved in the first step – they can only really properly describe the second step. A better way of drawing it might be to do something like this:

I haven't drawn the orbitals in the TS, but you can imagine that they are starting to morph into what they will be in the products. So $\psi^{(1)}$ will be somewhat intermediate between $\pi_\ce{C=O}$ and $\sigma_\ce{C-Nu}$. Of course, this is purely hypothetical - especially in the middle, the orbitals and their energies have no physical meaning at all. In a real reaction you would go straight from SM to TS, and not via an unphysical intermediate stage as is depicted here.

So, therein lies the difference between the two cases you have described. In the "stabilising" case, the only thing that is happening is a magical "switching on" of overlap. In the "destabilising" case of a chemical reaction, in addition to the overlap (which is stabilising), you also have a glut of other factors which can lower or (more frequently) raise the energy of the system as it move towards the TS. The problem is that we cannot separate the two from each other. Creating overlap comes with all the other strings attached, and so overall as we move along the RC towards the TS, there is (typically) an increase in energy.

Finally, how do we relate this back to the description of nucleophilic addition in Clayden? The wording is definitely somewhat loose (organic chemists frequently do this for ease of use), but it does have a grounding in quantum theory. I'll reproduce the image here and try to relate each phrase:

  Electrons in HOMO begin to interact with LUMO

This refers to the developing overlap that we looked at in step 2, i.e. the formation of MOs which are linear combinations of the HOMO ($n$) and LUMO ($\pi^*$). On its own this seems to be a stabilising effect, but recall that this does not come on its own: we also have to consider the different nuclear positions and its implications on the energy.

  Filling of $\pi^*$

I think the way to interpret it is that, at the start of the reaction the two filled orbitals $\psi^{(1)}$ and $\psi^{(2)}$ are simply $n$ and $\pi$, so the coefficients $c_{\pi^*}^{(1)}$ and $c_{\pi^*}^{(2)}$ are zero. As the reaction progresses, these coefficients steadily increase, which in a sense corresponds to a population of the $\pi^*$-orbital, even though the $\pi^*$-orbital ceases to exist in its own right once we bring the starting materials together.

  causes $\pi$ bond to break

In a sense we could define a "$\pi$ bond order" to be some kind of difference between the coefficients of the $\pi$ orbital in the filled MOs, and the coefficients of the $\pi^*$ orbital in the filled MOs. At the start these are $1$ and $0$ respectively, for a bond order of $1$. As the reaction proceeds and the orbitals mix, we find that $c_\pi$ decreases in the filled MOs, $c_{\pi^*}$ increases, and at the end of the reaction we have a bond order of $0$. So loosely speaking, it is true that "population of the $\pi^*$ orbital" (i.e. increasing $c_{\pi^*}$ in the filled MOs) leads to the cleavage of the $\pi$-bond.

  Electrons from $\pi$ bond end up as negative charge on oxygen

This is somewhat true again since the filled MOs morph from $\pi_\ce{C=O}$ plus a lone pair to $\sigma_\ce{C-Nu}$ and a lone pair, but it's important to remember that it is not a one-to-one correlation, i.e. the $\pi_\ce{C=O}$ MO does not become the oxygen lone pair all by itself.[3] More properly, the oxygen lone pair is a linear combination of multiple MOs in the starting material.

It's also important to not ascribe too much importance to exactly which electrons go where, as this violates quantum indistinguishability.

Notes

[1] There are barrierless reactions which do not proceed via a transition state, such as $\ce{H + H -> H2}$, and in such cases it is quite true that we always get stabilisation as we move along the RC towards products. However, these reactions don't fully capture the complexity involved in other reactions with TSes and it's better to think of these as being special cases of the other reactions.

[2] Barrierless reactions don't obey this line of thinking, of course. In fact, they don't even have a TS, so there isn't a well-defined geometry for us to distort towards in the first step. You can pick any geometry along the reaction coordinate between reactants and products, and it would probably work.

[3] Under some special circumstances it is possible to have a one-to-one relationship between reactant MOs and product MOs. See e.g. the first correlation diagram in Theoretical basis behind orbital correlation diagram for pericyclic reactions. But note that here there is only one orbital of each symmetry species, so there is no mixing going on and no linear combinations being formed. See also the last section in the answer on "Real pericyclic reactions".


Confused by notation of atomic number Z and mass number A on periodic table of elements I'm totally confused by the different conventions and when to use what. In the Periodic Table I see

$$\ce{^{6}_{12}C}$$

However, in books when talking about isotopes I see

$$\ce{^{12}_{6}C}$$

I don't understand the difference between the two. I know what the numbers mean but I don't understand when to use which convention. Are there names for these two conventions?

Periodic tables of elements (PTEs) are often abused by designers. Books are more trustworthy as long as they are written by scientists. Long story short, the second notation $(\ce{^{12}_{6}C})$ is the correct one.
There is an easy to remember AZE notation: $^A_Z\ce{E}$.

I suspect the PTE you were looking at lists standard (averaged) atomic weights of the elements $A_\mathrm{r}$ rounded to the nearest whole number so it may appear as if those were the mass numbers $A$, probably something like this:

  * The atomic weights listed on this Table of Elements have been rounded to the nearest whole number. As a result, this chart actually displays the mass number of a speciﬁc isotope for each element. An element's complete, unrounded atomic weight can be found on the It's Elemental website: http://education.jlaborgﬁtselemental/

Note that a good periodic table usually includes a legend which deciphers and justifies designer's choice.
Even better example is the Periodic Table of the Elements by NIST.
Despite atomic number $Z$ also located in the upper left corner of the cell, its location is typographically literate as both $Z$ and $A_\mathrm{r}$ have distinct place, different typeface and the legend unambiguously denotes which is which:


Measuring the colloidal nanoparticles stability by dynamic light scattering In this article the ligand exchange reaction is done for thiol coating of gold nanoparticles.In the first step dicholoromethane (DCM) is used as a extraction solvent to remove CTAC(Cetyl trimethylammonium chloride) from the gold nanoparticle's surface.CTAC acts as a stabilizer for gold nanoparticles.However,only incomplete CTAC removal occurs to ensure that there is still enough CTAC present to maintain colloidal stability.The optimum of solvent extraction is determined by dynamic light scattering.
My question is "How can be measured the the optimum solvent extraction by dynamic light scattering?"

Dynamic light scattering is a sizing technique; that is, it provides an indirect measurement of the size of nanoparticles (details may vary depending on experimental set-up).

However, it's important to note that it doesn't provide a direct measurement of the solid particle, but of its effective size when moving in solution. In other words, it measures the hydrodynamic radius of the nanoparticles - which, by the way, is not a single value but an ensemble average, which you usually measure in statistical fashion.

Now, at the nanoparticle-solution interface, you have several layers of material that is bounded to the surface and will move around with the nanoparticle, contributing to the hydrodynamic radius. In this particular paper, you have a CTAB/C monolayer (at least) which will be more or less tightly bounded to the gold particle, and also counter-ions to provide charge balance and several shells of solvation water (progressively less ordered and more weakly bound as you move away from the particle). It's the size of that whole entity that you measure (indirectly) with DLS.

As you can imagine, reducing the concentration of CTAB/C in the solution will change nanoparticle coverage - and will change the size, shape and strength of the layers surrounding the nanoparticle. As with colloidal systems, when synthesising nanoparticles there's a real concern that they can aggregate and precipitate or otherwise change their properties - which is the reason that you produce them in the presence of an stabiliser such as CTAB/C in the first place. If you look at figure 4, the histograms show hydrodynamic size for DLS measurements in different conditions and you can see they show that this aggregation isn't happening: if it did, you'd see far larger objects moving around in the solution. And as the body of the article explains, the small changes in hydrodynamic radius are probably due to the removal of CTAB/C from the nanoparticle surface, which affects solvation shells and changes the DLS-measured radius.


Ether nomenclature: Oxygens on alkyls and alkyls on alkyls Does the ether hold priority over longest chain? If so, my bet would be on:

4-(2-ethylheptanyl)oxy decane

If not, I would say:

4-(octan-2-yl methoxy) decane

Again, if ether takes precedence and not longer chain (which, if you ask me, would be ideal), I'd say:

4-[(1-methyl,2-propylheptan-1-yl)oxy]decane

If not:

4-(ethan-1-yl oxy nonan-4-yl)decane

What is correct?

Names for ethers can be formed by five different methods. Furthermore, various traditional names are retained for use in general nomenclature.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), preferred IUPAC names of ethers $\ce{R-O-R'}$, when $\ce{R}$ and $\ce{R'}$ are both aliphatic groups, are formed by substitutive nomenclature. In substitutive nomenclature, when $\ce{R}$ is different from $\ce{R'}$, $\ce{RH}$ is chosen as parent hydride and $\ce{R'-O-{}}$ is cited as a substituent to it.

  P-63.2.2.1.1 Substituent prefix names for $\ce{R'-O-{}}$ groups are formed by concatenation, i.e., by adding the prefix ‘oxy’ to the substituent prefix name for the group $\ce{R'}$. (…)

For the first example in the question, the name of the parent hydride $\ce{RH}$ is ‘decane’.
The substituent prefix name for the group $\ce{R'-{}}$ is ‘2-ethylheptyl’.
Thus, the substituent prefix names for the $\ce{R'-O-{}}$ group is ‘(2-ethylheptyl)oxy’.
Therefore, the preferred IUPAC name of the compound is ‘4-[(2-ethylheptyl)oxy]decane’.

For the second example in the question, the name of the parent hydride $\ce{RH}$ is ‘decane’.
The substituent prefix name for the group $\ce{R'-{}}$ is ‘3-propyloctan-2-yl’.
Thus, the substituent prefix names for the $\ce{R'-O-{}}$ group is ‘(3-propyloctan-2-yl)oxy’.
Therefore, the preferred IUPAC name of the compound is ‘4-[(3-propyloctan-2-yl)oxy]decane’.


Why are double bonds shorter than single bonds? We're told in chemistry class that the order of bond length is:
triple < double < single

Can anyone explain WHY this is the case? Is it because multiple bonds involve stronger attraction between the two atoms or something else?

In general, in a single bond, there is involvement of sigma bonds only. In double bonds, there is also overlap between pi orbitals. More is the overlap shorter will be the distance between the atoms.


Why can DAP be used to elute a protein in affinity chromatography? A little background: as part of my bioinformatics degree I have to take Protein Chemistry course, but I miss chemistry basics (have CS background), so that's why I'm asking the following question.

We performed protein (Galanthus Nivalis L. agglutinin) purification in class, using affinity chromatography.
The mannose column was loaded with protein (mixed with ammonium sulphate), washed with ammonium sulfate and then the protein was eluted using 20 mM DAP. While I understand the principle of affinity chromatography, I can't really figure out the chemistry of using DAP to elute the protein.

Would appreciate any help.

In this report, diaminopropane is used to elute a lectin from an affinity column. They mention the pH, which is the key to the question.

  The  bound  lectin was  eluted  with  20  mM diaminopropane  containing  0.15M NaCl,  pH ~11.

The diaminopropane is unbuffered, and the high pH weakens the interactions between lectin and carbohydrates.


The effects of acetic acid on porous concrete brick and mortar joints for someone who has no understanding of chemistry I mistakenly applied a patio cleaner containing acetic acid to my porous concrete facing brick thinking it was biocide. The concentrate solution has a ph of 3 and I diluted it to one part concentrate and three parts water. About three hours later, having realised my mistake, I flushed it with water. I repeated this process the following morning and also on the morning after that. The bricks now appear to be dry. I have been told that the pore structure of the type of bricks I have is vital in protecting them from frost attack. I’m very concerned that the acid has damaged the brick and mortar joints and am wondering if the acid has now completely evaporated - one day after the last rinse. Is there anything more I can do to neutralise any residual acid? I would be very grateful for any information. Thank you.

Per Wikipedia on Fired Bricks, to quote their composition;

Normally, bricks contain the following ingredients:[21]
Silica (sand) – 50% to 60% by weight
Alumina (clay) – 20% to 30% by weight
Lime – 2 to 5% by weight
Iron oxide – ≤ 7% by weight
Magnesia – less than 1% by weight

I would expect slowly, with time, that the acetic acid could attack all of the above with the exception of Silica.
Flushing with water would reduce the magnitude of the attack, but I would not categorically claim that the brick structure is completely sound.


Where does the energy come from to lower the temperature of a brine solution? Suppose I have a densely packed bucket of ice chips / cubes made from pure $\ce{H2O}$. It has been at room temperature for days, so the interior temp of the ice is likely close to 0 °C / 32 °F. (I have a working model of this in a clear ice - ice maker in a kitchen that does not have refrigeration; the ice is kept cold solely by the continuous addition of more ice.)

To a measured quantity of that ice (think about an ice bucket to cool wine) if I add:

23% of the weight of the ice in $\ce{NaCl}$ (table salt);
Some water to make a brine solution;
Stir.

The temperature of the resulting solution declines to roughly -22 °C / -8 °F.

I know that the $\ce{NaCl}$ depressed the freezing point of the water and that is why it can be liquid at this temperature. I suspect that $\ce{NaCl}$ is forming ions in the solution.

My question is: Where does the thermodynamic energy come from to lower a mass of ice and water to a significantly lower temperature?

The minimum temperature reached should be −21 °C at 29 % NaCl. Initially suppose that the ice,  water and water vapour are at 0 °C in a thermally insulated container. Say point A in the figure and some salt is now added.
However, the mixture is now not at equilibrium, and some ice melts and dissolves some salt. The resulting solution is now too concentrated to be in equilibrium with the ice so more ice melts, diluting the solution and so more ice melts. At the same time the temperature of the whole system spontaneously falls until after sufficient salt is added a temperature of −21 °C is reached.
This is the eutetic point (point E) and equilibrium has been reached. The temperature falls because for a phase change to occur the ice uses the heat of fusion it needs to melt by drawing that much heat from its surroundings.
Many other salts can be used instead of NaCl, such as ammonium chloride and calcium chloride.

Figure 1. Crude phase diagram of the ice/water/salt mixture.


Chemicals that Decrease fertility Is there any known substance that can decrease fertility of a male?
Less specifically what about one time chemical castrations (ignoring a perfect sterilization)?
I heard about them before once but wasn't sure really.

Numerous chemicals are linked with decreasing male fertility.
In fact, studies suggest that unhealthy lifestyles and environmental factors (e.g. plasticisers in the water system) may be causing this decrease in fertility.
See: http://www.telegraph.co.uk/men/active/mens-health/10695991/Why-are-mens-sperm-rates-falling.html etc
Obviously conclusive studies are limited at this stage.

But if you want to be more specific, I recall that chloroform is known to cause miscarriages and sterility in mice. Googling that to find a citation led me to this article:
http://www.ncbi.nlm.nih.gov/pubmed/10969727
Which found evidence of an otherwise non-toxic antifertility agent for male rats.

To be honest, I would be surprised to find studies which have tested a anti-fertility drug on men. Such a thing should not make it past an ethics board (given the ability to prevent impregnation by far less drastic measures, e.g. condoms).

If however you were talking about the sort of chemical castration they did to someone like Alan Turing, they were not designed to sterilise him (heck, his 'crime' was homosexuality and children were not a likely outcome of that), but to alter his hormones and personality. Anyway, modern drugs might be something like Cyproterone acetate or Depot medroxyprogesterone acetate - though that is also used as a birth control drug for women.


Can dust or fingerprints interfere in ultra-violet spectroscopy My question pertains generally to any form of spectroscopy, but I'll focus on UV spectroscopy. I am trying to get information on carrying out spectroscopy, but most of it is theoretical. Wikipedia is completely useless. Textbooks generally go into more detail about the underlying sub atomic processes being carried out. However, they do not provide too much details about how spectroscopy works in practice. So my question is this

Will something like a fingerprint, or a speck of dust, or a label on the beaker (not even sure what material the beaker would be made of), or even air for that matter interfere with the absorption spectrum?

You can deal with this through background correction of your spectra, which is a very common practice in spectrometry and implemented in any good spectrometry software.

In the most general sense, background correction consists of recording the spectrum of the spectrometer path without the analyte, and then subtracting that background spectrum from the spectrum of the analyte. In that sense, background correction is a form of difference spectroscopy. The spectrum you obtain from this ideally contains only the peaks attributable to the analyte.

A concrete example is subtracting the absorption spectrum of a solvent in a cuvette from an analyte. You would run a cuvette with solvent but without the analyte and use that as your background, followed by a cuvette with the analyte. Subtracting the first spectrum from the second would yield the signal attributable only to the analyte, assuming that the cuvettes, solvents and atmosphere between the two runs are negligibly different.
Obviously, there are a lot of potential confounds here that need to be accounted for:

Cuvettes have to be made of the same material, have the same dimensions (path length is important!) Cuvettes and such are often designed to be maximally transparent to radiation in the region of the spectrum of interest to maximise sensitivity. This is why they sell uber-expensive quartz cuvettes for UV/vis.
Solvents/substrates have to be of identical composition between runs. You also don't want a solvent or substrate that absorbs strongly in the spectral region of interest for similar reasons to the above.
Cuvettes have to be homogeneously clean - if one is dirty, greasy or scuffed and scatters light more than the other, it may ruin the results.
The atmosphere has to be the same between runs - I've heard stories about quantitative IR spectroscopy of $\ce{C=O}$ stretches being thrown out by an increase in $\ce{CO2}$ due to people breathing (how rude!). You can understand that researchers who look at atmospheric $\ce{CO2}$ concentration need very careful sampling and analysis protocols for reasons similar to this.


Electron affinity of nitrogen and exchange energy When studying the effects of electron filling in chemistry, I was studying why energy is needed to add an electron to a nitrogen atom i.e. electron affinity.

Nitrogen already has an electron structure of $\ce{1s^2 2s^2 2p^3}$ and using my knowledge of exchange energy, the energy released as exchange energy is

$$\frac{3 \times 2}{2} = 3K $$

But adding an electron means there is 1 electron  $\ce{1s^2 2s^2 2p^4}$ with opposite spin to the rest of the other electrons so the total exchange energy is still

$$\frac{3 \times 2}{2} + \frac{1 \times 0}{2} = 3K $$

so why is it unfavourable i.e. endothermic for nitrogen to became a nitrogen anion $N^-$?

$ \rm N^{-}$ has the outer electron structure of $ \rm 2s^2 2p^4$. This means we have coulombic repulsion in one of the p orbitals which means adding a single electron will be endothermic. This accounts for the decrease in ionization energy between N and O.

When you get to $\rm N^{3-} $ we have 6K units of exchange energy which is enough to offset the coulombic repulsion.

You see this effect when considering the electron configurations of the transition elements.


Optimal CFM for kerosene to burn? I am going to be making a kerosene forge, and I need a refresher on calculus-based chemistry.

Given a certain distribution of alkanes, what is the amount of air (21% O$_2$) needed (in cubic feet per minute [CFM]) at 21C to burn kerosene at a rate of X mL/s?

The distribution of alkanes is unknown but from Wikipedia I've got that the molecules have no more than 16 and no less than 6 carbon atoms. The density is $\mathrm{0.795~g~cm^3}$. If any more info or help is needed, I'll be glad to pitch in where I can.

This is something of a standard reaction stochiometry question, although the situation is a bit more involved than the ones you normally see in textbooks.

The first step is to work out what the chemical reaction we're talking about actually is. One complication is that "kerosene" doesn't name a single molecular species, but is instead a mix of hydrocarbons. But the hydrocarbons are similar enough that we can probably approximate the mixture as a "typical" hydrocarbon of roughly the right type and mass. Wikipedia approximates kerosene as dodecane, and even gives a combustion reaction (saving us the effort of balancing one):

$$\ce{2 C12H26(l) + 37 O2(g) → 24 CO2(g) + 26 H2O(g)}$$

Now that we have stoichiometry, we just need to convert the quantities of reactants involved to numbers of molecules (moles). We're dealing with rates of consumption for both reactant, so let's just run the numbers for a fixed period of time, say 1 minute.

We have an amount of kerosene we want to burn ($\ce{X\; mL*s^{-1} * 60s}$) and a conversion from volume to mass (the density), and all we need to get the "moles of kerosene" would be the molecular weight of kerosene. Since we're approximating kerosene as dodecane, use the number for dodecane ($170.34\;  \ce{g*mol^{−1}}$).

Using the knowledge from the balanced chemical reaction equation that 2 moles of dodecane require 37 moles of oxygen to combust completely, we can determine how many moles of oxygen we need to burn the given quantity of kerosene.

Now we need to convert moles of oxygen to a volume of air. If we assume oxygen and air is an ideal gas (a safe assumption under normal conditions), we can use the ideal gas law to interconvert moles ($n$) into volume ($V$): $PV = nRT$, where $P$ is the pressure, $T$ is the temperature, and $R$ is the ideal gas constant (about $8.314\;\ce{L*kPa*K^{−1}*mol^{−1}}$). Or, if we're dealing with standard temperature and pressure (about 1 atmosphere at sea level and about room temperature), then 1 mol of ideal gas takes up about $22.4\;\ce{L}$. So we can find out what volume of oxygen we need.

But we're not dealing with pure oxygen, so we have to divide by 0.21 to get the volume of air which would contain that volume of oxygen. (This works because the 21% is a volumetric percentage.) The final little bit is to convert the number of liters of air used per minute into cubic feet used per minute.

So there's really no calculus involved - it's just lining up the conversion factors and multiplying through. The tricky bit is probably figuring out the chemical reaction formula you want to use (e.g. the assumption of dodecane). It also helps to know that things are easiest if you do most of the calculations in SI (metric) units, and only convert to non-SI at the end.

One final caveat is to keep in mind that we assumed that kerosene was well represented as dodecane. If your kerosene is substantially different from dodecane, the reaction will have to be adjusted. We also likely assumed that the input air was at standard temperature and pressure, and was dry (the 21% value is for dry air). Warmer/cooler air, air at altitude, and humid air will all change the calculation. Also keep in mind that the calculation is for combustion under ideal conditions - incomplete mixing and non-ideal combustion issues might mean you need more air in order to ensure complete combustion.


What is the complete mechanism of reaction for sodium benzoate and ascorbic acid? What are the complete steps of reaction from sodium benzoate and ascorbic acid to benzene? I can't seem to visualize it, and, much less, find anything online regarding the actual steps of the reaction.

Additionally: how does it make use of, say Cu(II) or Fe(III) to complete the reaction? It seems that the reaction can only proceed with these present.

From Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst:

$\ce {Cu^{2+} + H_2Asc \rightarrow Cu^+ + HAsc^.}$

$\ce {Cu^+ + O_2 \rightarrow Cu^{2+} + O_2^-}$

$\ce {2O_2^- + 2H^+ \rightarrow H_2O_2 + O_2}$

$\ce {Cu^+ + H_2O_2 \rightarrow Cu^{2+} + OH^- + OH^.}$

Then the hydroxyl radical reacts with benzonic acid as opposed to benzoate.


How to calculate the required concentration for sodium sulfite to remove oxygen as an oxygen scavenger from a solution? If I know the amount of oxygen in such a solution, how can I know the required concentration of sodium sulfite to be used and react as oxygen scavenger?

From the equation
\[\ce{2SO3^{2-} + O2 -> 2SO4^{2-} } \]
you could conclude that you need (at least) 2 moles of sodium sulfite $(\ce{Na2SO3})$ for each mole of oxygen in solution.


When a thin lead sheet is hammered, what happens to internal energy When a thin lead sheet is hammered, what happens to internal energy, does it increase, decrease or remains the same or first increases and then decreases?

When you hammer a sheet of lead, you are adding energy in form of work into it, increasing it's internal energy. If this experiment were to be done in complete vacuum, it'd retain most of the energy, however if this experiment were to be on air, it'd lose a lot of energy in the form of sound.


How does the degree of saturation influence the colour of the flame? If the colour of flame produced by the atom of an element is the property of that element then why do unsaturated and saturated carbon compounds give different colours of flame even though both contain the element carbon and hydrogen?

The flames that you see when you burn carbon-containing compounds have colour for a completely different reason that if you add ionic compounds (e.g. sodium chloride) to a flame. Ionic compounds won’t combust, their bonds will merely be broken and the metallic atom then has an electron that can be excited and that can re-relax under photon emission.

For carbon, we are actually talking about proper combustion, i.e. a reaction of the carbon compound with oxygen to give carbon dioxide and water. This comes in two flavours: complete and incomplete.

In complete combustion, all carbon atoms are fully oxidised to carbon dioxide due to sufficient oxygen amounts. We can write a nice equation e.g. for methane:

$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}\tag{1}$$

Complete combustion is often shown with methane as it often tends to combust completely.
Incomplete combustion, as the name says, is incomplete, i.e. some carbon atoms won’t make it all the way to carbon dioxide. It is impossible to give an exact equation for incomplete combustion, but one possible one could be the following:

$$\ce{2 CH4 + 3 O2 -> CO2 + C + 4 H2O}\tag{2}$$

Incomplete combustion is commonly shown with unsaturated compounds because they have a greater tendency to display it.

The key factor in incomplete combustion is the fact that elemental carbon (soot) is generated; it is that soot whose electronic transitions cause the yellow colour of a flame. This is not a feature of unsaturated versus saturated hydrocarbons: as the equation above shows, methane can burn both in a complete and in an incomplete manner depending on how much oxygen is added.

Likewise, the most common example for incomplete combustion, ethyne, can be combusted completely if only enough oxygen is added, see this question.


Why does NF3 have a smaller bond angle than NH3? I've already read many answers about the reason why $\ce{NF3}$  has a smaller bond angle than $\ce{NH3}$ , but I can't seem to understand them. Here's my understanding of the situation:

$\ce{NH3}$: Here N is more electronegative than H so a large electron cloud is crowded over N. This will push the bond pairs $\ce{N-H}$ away from the central atom. So, the angle of $\ce{H-N-H}$ will decrease.
$\ce{NF3}$: Here F is more electronegative than N, so the lone pair cloud over N is scattered into the $\ce{N-F}$ bonds. Thus the smaller electron cloud over central atom is unable to push the $\ce{N-F}$ bonds away from itself as much as it did previously. So, bond angle should be higher than previous case.

This can be argued on the basis of Bent's rule; concisely stated

  Atomic s character concentrates in orbitals directed toward electropositive substituents

What follows below is a crude explanation. Before that, I'll note that we concern ourselves with the hybridisation of the orbitals at the central atom.

Since s orbitals are lower in energy than p orbitals, they are better at stabilising electrons. Which is why molecules like to use them (s orbitals) in bonds where there is "more" electron density to stabilise.

Thus, Bent's rule can just as easily be formulated as follows:  p character concentrates in orbitals directed towards electronegative substituents; (the central atom doesn't have to waste its low energy s orbitals, the orbitals on the electronegative atom, can (possibly) take care of it.

What has all of this got to do with bond angles? Well, more s character leads to larger bond angles. You can think of how s orbitals are spherical in shape, and a large s character would lead to more "spherical" hybrid orbitals, and that would lead to larger bond angles.

Going back to your question, we are supposed to compare $\ce{NF3}$ and $\ce{NH3}$.

Fluorine is more electronegative that hydrogen, and the $\ce{N-F}$ bond would have greater p character than the $\ce{N-H}$ bond, (or $\ce{N-H}$ bond has greater s character, whichever you prefer). Thus $\ce{H-N-H}$ angle is greater in $\ce{NH3}$ than it is in $\ce{NF3}$

P.S I'm sure you can learn more about what I just tried to explain (rather poorly) by reading the Wikipedia article, and/or a textbook and/or using the search function on this site to find related questions/answers.


Can dissolving be a chemical change? My textbook says that $\ce{SO2}$ dissolving in water produces sulfurous acid, but that seems to be a chemical change. But $\ce{NaCl}$ dissolving in water is not a chemical change. Please explain.

Your book is glossing over some details.

There are three things occurring when you dissolve non-metal oxides in water (e.g., $\ce{SO2}$, $\ce{CO2}$, $\ce{NO2}$, etc.):

The gas simply dissolves in the water. You are correct that this step is a physical change.  How much dissolves depends on the partial pressure above the water and the Henry's law constant.
Once dissolved in water, some of the non-metal oxide will react with water to form an acid (sulfuric/sulfurous acid, carbonic acid, nitric/nitrous acid, etc.) and this is the chemical change that your book is referring to. There will be an equilibrium constant that will dictate the relative concentration of dissolved gas to the corresponding acid that is formed.
The acid (usually weak) will dissociate a little bit.  Again, the exact amount of dissociation will depend on the equilibrium constant, $K_\mathrm{a}$.

Using your example, we would have:

\begin{align}
\ce{SO2(g) &<=> SO2(aq)}\\
\ce{SO2(aq) + H2O(l) &<=> H2SO3(aq)}\\
\ce{H2SO3(aq) &<=> HSO3^- (aq) + H+(aq)}
\end{align}

Similar things happen when metal salts dissolve in water (although not $\ce{Na^+}$ or the other group I and II metals). The metal cations will react with water and form a weak acid.


Will a container made of ice hold an alcoholic beverage that is below freezing in an environment that is also below freezing? This very specific question that came after saw I a travel episode where they served alcoholic beverages in ice glasses in an ice bar. I wondered if the drink contents would stay in the glass (of ice) without the glass (of ice) melting. I reason that the drink would be cold enough to not melt the ice thus allowing it to stay frozen.

Depends on the temperature and alcohol content. At 0°C, the container will definitively melt.

Below, your container will melt from the inside until the liquor is diluted so much that its freezing point is raised to the current temperature. (Or the other way round, water ice precipitates from your booze, concentrating it until its freezing point has dropped to the temperature of the surrounding.)

You can easily calculate the equillibrium alcohol content for a given temperature, https://en.wikipedia.org/wiki/Freezing-point_depression#Calculation
(The result will be slightly off, because this calculation is derived for infinite dilution.)


Application of the Woodward-Hoffman rules to a [14+2] cycloaddition (follow up) This is a follow up question to another question I made earlier.

I put some of this information into my own notes just so I can get my head around this.

So just to clarify. Does the Woodward Hoffman rule given below only apply to ground state reactions, thus the rule cannot be applied to the above reaction scheme because one of the components is in an excited state?

A ground-state pericyclic change is symmetry-allowed when the total number of (4n+2)s and (4n)a components is odd.

EDIT: (1) Correct labeling of the suprafacial components, (2) Removal of orbital phases, (3) Stereochemistry situated at the correct stereocentres.

For reactions in the excited state, a modified statement of the Woodward Hoffman rule applies:

  A pericyclic change in the first excited state is symmetry allowed when the total number of (4q+2)s and (4r)a components is even.

The only difference is that we're now looking for the total to be even rather than odd.


Carcinogens, how do they work? The easiest carcinogenic thing for me to grasp is radiation, as it directly messes with DNA. Then it seems there are other compounds that simply mimic hormones, but these shouldn't necessarily cause cancer, right? Then there are things that can accelerate mitosis or mess with other processes, which can inadvertently mess with DNA. So my question is, in general, how many different flavors of carcinogens are there? I know there are a lot, but can someone list a good few? Maybe some examples? I'm most interested in how plastics and things like smoke cause cancer. I think a few examples would be best in aiding my understanding.
Thank you!
Edit: I also just realized this is the Chemistry Stack Exchange, I think I am more interested in the chemistry side of this question, but it might need to be migrated.

There are at least two different radiationless pathways to wreak havoc on the DNA:

intercalaction: compounds like benzo[a]yrene or ethidium molecules slide between DNA strands and cause transcription errors
alkylation: compounds like N-Nitroso-N-methylurea may transfer alkyl groups to nucleobases

But the list is by no means complete.


Is the transformation of pi bonds to sigma bonds always energetically favorable? Is the move from pi bonds to sigma bonds always favorable?

My professor claims so, except in the case of conjugated pi bonds. I can see how pi bonds going to sigma bonds might be favorable; I know that double bonds are actually not twice as strong as single bonds because of poorer overlap ... but are there any exceptions to this guideline that the reduction of pi bonds to sigma bonds is favorable except in the case of conjugated systems?

EDIT: what about in the case of organic chemistry? It certainly seems that hydrogenation (a way of transforming pi bonds to sigma bonds) is generally a favorable move.

When thinking about questions along these lines, I always try to remember the bent-bond description of ethylene.  In all aspects related to physical reality (structure, electron distribution, etc.) the bent bond model pictured below on the right is equivalent to the pi model shown on the left (at least within the scope of all measurements made to date).

The "nice" thing about the bent bond model is that it casts ethylene as a highly-strained, two-membered ring.  When wouldn't releasing this strain and forming sigma bonds be an energy lowering process.  Aside from conjugated examples, I think that transformation of a strained, ethylenic system to a saturated, sigma-bonded system would always be energy lowering.

EDIT: Further Thoughts

Instead of analyzing hydrogenations or other reactions where the starting material and product have a different number of atoms, let's consider the following isomerization reaction.

Thermodynamically, cyclohexane is the most stable $\ce{C6H12}$ isomer in this series by a significant margin.  All we've done is convert a C-C pi bond into a C-C sigma bond and convert 3 $\ce{C(sp^2)-H}$ bonds into 3 $\ce{C(sp^3)-H}$ bonds.  Which (at least as I read it) was the OP's premise.  So I would say that, in carbon-hydrogen systems, pi bonds generally prefer to transform into sigma bonds, as pi bonds are inherently destablilized due to poor overlap (pi model) or strain (bent bond model).

That said, I retract what I wrote above that carbon-carbon pi to sigma transformations would "always" be energy lowering.  Consider the photolysis of benzene to produce dewar benzene, prismane and benzvalene.  These conversions are examples where one or more pi bonds have been converted to one or more sigma bonds and the reaction is not energetically (thermodynamically) favorable.

The products are much less stable (due to the strain in the small rings that have been created) than the starting benzene.  Because of these relative stabilities, the forward reaction does not occur thermally (in fact, thermally, each of those 3 products will revert to benzene), but by pumping in excess energy photochemically, the transformation can be brought about.

To summarize, I would maintain that generally, in model, carbon-hydrogen systems, ethylenic linkages are destabilized (due to poor overlap in the pi model, or strain in the bent bond model) and will thermodynamically prefer to exist as thermodynamically more-stable saturated linkages.


Common Elemental Termolecular Reactions Are there any common elemental (single-step) reactions with three reactants?

Everything I've found either says that they're very rare, or can be broken down into two bimolecular steps with combined reaction rates far greater than the termolecular step.

What reactions require an indivisible termolecular step?

Termolecular reactions almost never happen.  The chance of three molecules "colliding" at the same instant is so small as to be ignorable.

On the other hand there are "almost" termolecular reactions when one molecule hits another and "sticks" to it, and then a third molecule hits the two that are stuck together.  This most often happens when one of the molecules is part of a solid.  But this is really two sequential bimolecular reactions.