Does aromaticity require a co-planar pi system? It's well established that angular twisting can decrease the degree of delocalization across a $\pi$ system (e.g. biphenyl).

Moreover, there's considerable debate about aromaticity in spherical species like $\ce{C_{60}}$.

So let's ask a naive question for discussion. Does aromatic stabilization require a planar delocalized $\pi$ system?

The $\pi$ system is not obliged to be planar at all; the only limitation is to keep the angular twist between the adjacent atoms low enough so they still can be conjugated. (Biphenyl violates that, so it is out of luck.) The whole structure can be pretty much anything. It can be slightly out-of-plane, like corannulene or circulene; it may have cylindrical shape (see the example below), in which case the Hückel's rule still applies, or it may be all twisted to the topology of a Möbius strip, in which case the said rule is reversed.

Fullerenes are not aromatic, unless you stretch the definition of aromaticity to utter pointlessness. Some of their derivatives are, though. $\ce{C60Cl30}$ comes to mind: its 30 remaining $sp^2$ carbons are grouped in two 6-membered benzene-like rings and one 18-membered belt, which (judging by its equalized bond lengths) is an example of a cylindrical aromatic system.


How do you calculate a unit cell's dimensions from an XRD spectra? My inorganic lab had us do an XRD measurement, but I've never been explained how to interpret the data.

Question:

  Calculate the unit cell dimensions $a$, $b$, and $c$, for $\ce{YBa2Cu3O7}$ from the indexed X-ray powder pattern provided in [the textbook]. Explain why the crystals are nearly tetragonal in terms of the atomic structure of the compound. The following formula is useful:

  $$\sin^2{\theta} = \frac{\lambda^2}{4}\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right)$$
  The three numbers, $hkl$, called Miller indices, indicate the direction of the scattering plane. The Miller indices can be read from the diffractogram provided. For simple reflections of the type $h00$, $0k0$, and $00l$, the value of $h$, $k$, or $l$ corresponds to n in Bragg's law.

Bragg's law:

$$n\lambda = 2d\sin{\theta}$$

The spectrum:

  (This is a scan from my textbook, while working on it I circled some items. Just ignore that.)

My Attempt:

$$\lambda = 0.154~\mathrm{nm}$$

Scouring the internet I found the formula for tetragonal atomic structure: the axes $a = b \neq c$, and the axis angles $\alpha = \beta = \gamma = 90^\circ$.

Solving for $c$:
$$\begin{align}
\sin^2{(22.9^\circ)} &= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right) \\
&=\frac{0.154^2}{4} \left(\frac{0^2}{a^2}+\frac{0^2}{b^2}+\frac{3^2}{c^2}\right) \\
0.151 &= 0.0237\left(\frac{3^2}{c^2}\right) \\
c &= 1.188
\end{align}$$

For $a$ and $b$:

$$\begin{align}
\sin^2{(22.9^\circ)} &= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}\right) \\
&= \frac{0.154^2}{4} \left(\frac{1^2}{a^2} + \frac{0^2}{b^2} + \frac{0^2}{c^2} \right) \\
0.151 &= 0.0237 \left(\frac{1^2}{a^2}\right) \\
a &= 0.156
\end{align}$$

Is this the right approach? If I have the right answer, what should the units be?

You should not really have assumed that the compound was tetragonal. If it was indeed tetragonal, then the peaks corresponding to the $(200)$ and $(020)$ planes would overlap. Instead, you can see that one occurs at $2\theta = 46.7^\circ$ and the other at $2\theta = 47.6^\circ$. The question may be considered a little misleading, but it does hint at this; it says that the compound is "nearly tetragonal", and it gives you the general formula for an orthorhombic crystal, which is one with $a \neq b \neq c \,(\neq a)$ and $\alpha = \beta = \gamma = 90^\circ$.

The next problem is that you are given values of $2\theta$ (which happens to be the convention), but the formula reads $\sin^2{\theta}$. As such, you need to make the appropriate adjustments.

This question is relatively easy because the peaks have already been indexed for you (i.e. the values of $h$, $k$, and $l$ are already found out). Therefore, all that is left is to plug the numbers into the formula you are given. Which peaks you use is entirely up to you, although it is obviously easier to use $(h00)$, $(0k0)$ and $(00l)$ peaks.

The reflections of the $(200)$ plane occur at $2\theta = 47.6^\circ$. So:

$$\begin{align}
\theta &= 23.8^\circ \\
\sin^2{(23.8^\circ)} &= \frac{(1.54~Å)^2}{4} \cdot \frac{2^2}{a^2} \\
a &= 3.81~Å
\end{align}$$

Similarly, you can show using the $(020)$ and $(003)$ peaks that $b = 3.89~Å$ and $c = 11.64~Å$. This is actually pretty good. The reported values of $a$, $b$ and $c$ are $3.8231~Å$, $3.8864~Å$, and $11.6807~Å$ respectively.

Regarding units, if you make sure that you leave your units inside the intermediate workings (as they should be), you should never run into confusion about "what unit should the answer be". In your question, you converted $\lambda$ into units of $\mathrm{nm}$. $\sin^2\theta$, $h$, $k$, and $l$ are all dimensionless. Therefore, your calculated values of $a$, $b$, and $c$ must match the units of $\lambda$ you used, i.e. $\mathrm{nm}$.

In my answer, I used ångströms throughout. It is not wrong to use $\mathrm{nm}$, but as Babounet noted in the comment section, ångströms are more commonly used in crystallography.


Synthesis Golf X: TAK-457 I shall try to revive this!
A full FAQ post has been written on chemistry.meta.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

TAK-457 is a drug (more accurately, an injectable prodrug of TAK-456) that displays antifungal activity:[1,2,3]

Some rules, as usual:

At most one chiral centre may be bought in your starting material(s). [Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.]
Your synthesis must feature a method for construction of either the triazole or tetrazole ring (i.e. do not buy both).
The counterion doesn't matter.
Bonus points for synthesis that can be adapted to produce the other three stereoisomers of the drug (preferably via late stage diversification as opposed to, say, using a different enantiomer of starting material). These stereoisomers were of interest.[4]

References

Ichikawa, T.; Kitazaki, T.; Matsushita, Y.; Yamada, M.; Hayashi, R.; Yamaguchi, M.; Kiyota, Y.; Okonogi, K.; Itoh, K. Optically Active Antifungal Azoles. XII. Synthesis and Antifungal Activity of the Water-Soluble Prodrugs of 1-[(1R,2R)-2-(2,4-Difluorophenyl)-2-hydroxy-1-methyl-3-(1H-1,2,4-triazol-1-yl)propyl]-3-[4-(1H-1-tetrazolyl)phenyl]-2-imidazolidinone. Chem. Pharm. Bull. 2001, 49 (9), 1102–1109. DOI: 10.1248/cpb.49.1102.
Hayashi, R.; Kitamoto, N.; Iizawa, Y.; Ichikawa, T.; Itoh, K.; Kitazaki, T.; Okonogi, K. Efficacy of TAK-457, a novel intravenous triazole, against invasive pulmonary aspergillosis in neutropenic mice. Antimicrob. Agents Chemother. 2002, 46 (2), 283–287. DOI: 10.1128/AAC.46.2.283-287.2002.
Tsuchimori, N.; Hayashi, R.; Kitamoto, N.; Asai, K.; Kitazaki, T.; Iizawa, Y.; Itoh, K.; Okonogi, K. In Vitro and In Vivo Antifungal Activities of TAK-456, a Novel Oral Triazole with a Broad Antifungal Spectrum. Antimicrob. Agents Chemother. 2002, 46 (5), 1388-1393. DOI: 10.1128/AAC.46.5.1388-1393.2002.
Ichikawa, T.; Yamada, M.; Yamaguchi, M.; Kitazaki, T.; Matsushita, Y.; Higashikawa, K.; Itoh, K. Optically Active Antifungal Azoles. XIII. Synthesis of Stereoisomers and Metabolites of 1-[(1R,2R)-2-(2,4-difluorophenyl)-2-hydroxy-1-methyl-3-(1H-1,2,4-triazol-1-yl)propyl]-3-[4-(1H-1-tetrazolyl)phenyl]-2-imidazolidinone (TAK-456). Chem. Pharm. Bull. 2001, 49 (9), 1110–1119. DOI: 10.1248/cpb.49.1110.

Fragment #1 Preparation

1) An exact preparation of an aryl imidazolidinone (54% Yield)1
2) A modified preparation of a tetrazole from an aniline derivative (90% yield with pure aniline)2
3) A modified acetal formation (protocol modifications based on urea) (no reported yield)3

Fragment #2 Preparation

1) An exact preparation of an alpha-bromophenone (Hell-Volhard-Zelinski Reaction) (97% yield --likely prior to purification)4
2) Modified coupling of 1,2,4-triazole and alpha-bromophenone (100% yield with chloroacetone -- likely prior to purification)5
3) Wittig reaction modification (73% yield with phenyl methyl ketone)6
4) A modified triazolium formation (82% yield with methyl iodide)7
5) Stereoselective Jacobsen epoxidation (92% yield with 1-phenylcyclohexene)8

Final Fragment Coupling

1) Deprotonation of the amine with LDA
2) Epoxide ring opening followed by the protonation of the alkoxide and hydrolysis of the acetal to yield the final product

Stereochemical and General Considerations

The use of the Jacobsen epoxide formation allows for the use of the (L, L) Jacobsen catalyst in the final steps of the synthesis to yield the enantiomer of TAK-457. Synthesis of the other stereoisomer would require modification of earlier steps in the reaction.

Longest Linear Sequence: 5 steps
Total Number of Steps: 9 steps (I consider the final two steps as one since they are performed without subsequent purification)

Comment on 1,2,4-Triazole Ring: Yes, I did not prepare the triazole ring directly; however, this seems to be the more efficient route.

Comment on Wittig Reaction: This is the step I am probably the most worried about. The steric bulk from the phenyl group may be enough to slightly favor the desired enantiomer; however, this cannot be determined until further experimentation is performed.

References

Stabile, P.; Lamonica, A.; Ribecai, A.; Castoldi, D.; Guercio, G.; Curcuruto, O. Mild, Convenient and Versatile Cu-mediated synthesis of N-Aryl-2-imidazolidinones. Tetrahedron Lett. (2010) 51 (24). 3232–3235. DOI:10.1016/j.tetlet.2010.04.064
Kundu, D.; Majee, A.; Hajra, A. Indium Triflate-Catalyzed One-Pot Synthesis of 1-Substituted-1H-1,2,3,4-Tetrazoles Under Solvent-Free Conditions. Tetrahedron Lett. (2009) 50 (22). 2668–2670. DOI:10.1016/j.tetlet.2009.03.131
Shuai, F.; Wang, X.; Zhang, J. Preparation of Dipyridamole Active Pharmaceutical Ingredient. CNIPA Patent 108069972, Nov 16, 2016
Alonso-Alija, Cristina; et al. Preparation of Amino(monocyclic aroyl)pyridinones that Inhibit p38 Map Kinase for use as Antiinflammatory Agents. WIPO Patent 2003076405, Sep. 18, 2003
Sun, Y.; Huang, N.-Y.; Ding, M.-W. Efficient Synthesis of 6-(1H-1,2,4-Triazol-1-yl)-thieno[2,3-d]pyrimidin-4(3H)-ones via an Iminophosphorane. Synth. Commun. (2010) 40 (13). 1985–1991. DOI: 10.1080/00397910903219351
Pratsch, G.; Overman, L. E. Synthesis of 2,5-Diaryl-1,5-dienes from Allylic Bromides Using Visible-Light Photoredox Catalysis. J. Org. Chem. (2015) 80 (22). 11388–11397. DOI:10.1021/acs.joc.5b01962
Myles, L.; Gathergood, N.; Connon, S. J. The Catalytic Versatility of Low Toxicity Dialkyltriazolium Salts: in Situ Modification Facilitates Diametrically Opposed Catalysis Modes in One Pot. Chem. Commun. (2013) 49 (46). 5316. DOI:10.1039/c3cc41588k
Brandes, B. D.; Jacobsen, E. N. Highly Enantioselective, Catalytic Epoxidation of Trisubstituted Olefins. J. Org. Chem. (1994) 59 (16). 4378–4380. DOI:10.1021/jo00095a009


What typically happens to H2 molecules released into atmosphere (with no spark source)? I'm wondering what happens when $\ce{H2}$ leaks from say a transmission pipe in an unenclosed area. No immediate source of ignition, I know it rises more quickly than helium (x2 I believe) and dissipates in concentration quickly but does it react with gases in atmosphere to form methane or other compounds?

Short of ignition and burning causing more $\ce{H2}$ to dissociate what can happen, especially in upper atmosphere with more UV and so on. Methane has reaction in the upper atmosphere giving it a half life of ~7 years for example.

  The covalent radius of a neutral hydrogen atom is 0.0371 nm, smaller
  than that of any other element. Because small atoms can come very
  close to each other, they tend to form strong covalent bonds. As a
  result, the bond dissociation enthalpy for the $\ce{H-H}$ bond is relatively
  large (435 kJ/mol). $\ce{H2}$ therefore tends to be unreactive at room
  temperature. In the presence of a spark, however, a fraction of the $\ce{H2}$ molecules dissociate to form hydrogen atoms that are highly reactive. Source: chemed.chem.purdue.edu

The most common, reactive stuff in the atmosphere are $\ce{-OH}$ radicals (Wikipedia). So I guess they're the natural antagonist for $\ce{H2}$ molecules.
Btw., as soon as it's diluted in the surrounding air, it will no longer rise. Only bulk masses have buoyancy. The g gradient in the atmosphere is much too small to separate gases.
In the end, hydrogen however does diffuse upwards and is ultimately lost into space. Hydrogen and helium are light enough to be able to reach escape velocity in the upper atmosphere.


Major product in nitration My approach was that when the carbon attached was a secondary one, it would have a greater inductive effect compared to the primary one.

Why is the answer 3? Is it due to steric reasons or am I missing some other factor?

Read this first: Why methyl group is 2,4-directing?

From that answer, we understand that methyl groups (or alkyl groups in general) are o/p directing because of hyperconjugation, and not inductive effects! In fact, as you may also know, inductive effects are generally overshadowed by hyperconjugative effects.

The aromatic ring on the left is more activated because it has two alpha hydrogens - implying more number of hyperconjugative structures - as opposed to only one for the right aromatic ring (a small difference, but good enough for elementary level questions). Thus nitration takes place at the para position of the left ring.


Spontaneity of an exothermic reaction when temperature is decreased How is the spontaneity of an exothermic reaction changed when temperature is decreased?
Assuming the entropy is negative.

Using Gibbs free energy equation, $\Delta G=\Delta H-T\Delta S$, $\Delta G$ should become less negative as temperature is decreased meaning the reaction is less spontaneous. However, using Le Chatelier's principle, as temperature decreases the reaction should shift towards the right towards the products, becoming more spontaneous.

What's wrong here?

The key is $\Delta G = -RT \ln K$.

The $T$ specifically is causing your reasoning to be flawed, as it complicates the link between temperature dependence of the equilibrium constant and $\Delta G$.

$-RT \ln K = \Delta H-T\Delta S$

$-R \ln K = \frac{\Delta H}{T}-\Delta S$

In this form you can see that (at least to the extent that $\Delta H$ and $\Delta S$ are temperature independent), the sign of $\Delta H$ (meaning whether the reaction is endothermic or exothermic) determines whether or not temperature change favors the forward or the reverse reaction. You can also see that the sign of $\Delta S$ is irrelavent.

For an exothermic reaction, $\Delta H$ is negative.  If $T$ decreases, $\frac{\Delta H}{T}$ and becomes a larger magnitude negative number, $-R\ln K$ decreases, $R\ln K$ increases therefore $K$ increases and the reaction shifts forward.


Is it okay to heat a test tube for several minutes for evaporation of solvent inside? I have a mixture of water and a small amount of undissolved solute in a test tube. I want to remove the water in it by evaporation method. Can the test tube carries the evaporation process with a Bunsen burner? I don't want to transfer into different glassware to avoid contamination and reduction of the solution when transferring.

There are two issues here:

Heat from a Bunsen burner is pretty strong. You better make sure whatever is left after the evaporation will survive.
If the solvent is organic (and not super halogenated), this is a terrible idea because you run a serious risk of igniting your solvent vapors.


Zero filling leading to increases in resolution (NMR) Zero-filling involves adding data points with zero intensity to the end of an FID. According to this website https://www2.chemistry.msu.edu/facilities/nmr/handouts/DH%20NMR%20Basics.pdf zero filling can take a spectrum with low apparent resolution to one that is much more helpful like so

In this instance, going from bottom to top corresponds with more zero filling. I do not understand how this happens. How does adding data points with zero intensity lead to increases in apparent resolution? I am certain I am missing some crucial point, but I can not pinpoint what that may be. For what its worth, I understand that we are not increasing the resolution of the instrument's measurements, rather we are manipulating the data to view information that would otherwise not be displayed.

The key idea behind zero padding or zero filling is that the "resolution" or the step size of the frequency axis in the Discrete Fourier Transform is dependent on the number of points you have in the time domain. Zero filling is done before doing the Fourier transform. Thus after the FFT the resolution appears to be improved because you have more points in the spectrum and the spacing between each point appears to be decreased.
This is a result of a beautiful theorem (although this is casually mentioned in most textbooks), but never shown as a theorem. I found in Cooley's original paper from 1967 [1, p. 80].
Cooley & Tukey are the key persons who made Discrete Fourier Transform (DFT) possible by computers for lowly mortals like us, otherwise it was an elitist subject among the mathematicians. Here is a snippet of it. It took me several years to find an original paper which showed this theorem. However, math researchers told me this was known in 1754. Full discussion here: History of Integral Transform Theorem
References

Cooley, J.; Lewis, P.; Welch, P. Application of the Fast Fourier Transform to Computation of Fourier Integrals, Fourier Series, and Convolution Integrals. IEEE Transactions on Audio and Electroacoustics 1967, 15 (2), 79–84. https://doi.org/10/bmw8z8.


How does the force constant change when a heavier atom is used in vibrational spectroscopy? (iii) The fundamental band in the IR spectrum of $\ce{^1H^{127}I}$ is found at $\pu{2308 cm^-1}.$ Calculate the force constant for the bond. Assume the molecule can be treated as a harmonic oscillator.
(iv) Calculate how the position of the fundamental band would be expected to differ for molecules of $\ce{^2H^{127}I}.$

I have found the force constant in (iii), but I am struggling with (iv). I don’t know how to calculate the difference. Any insight would be greatly appreciated.

In the harmonic oscillator approximation, the wavenumber of vibration for a diatomic molecule is
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$
Here, $k$ is the force constant of your bond, and $\mu$ is the reduced mass for the two atoms.
$$\mu=\frac{m_1m_2}{m_1+m_2}$$
When you change isotopes, only $\mu$ changes. The force constant depends on the strength of the bond, which depends on the chemical nature of the elements, and can be assumed to stay constant regardless of changing isotopes.
You have to calculate the $\mu$ for the second problem, and use that with the $k$ you got from the first problem to calculate the wavenumber of the fundamental band.


Spin of electrons about it's own axis, release of energy & COLLAPSE of atom? An electron spins about it's own axis. So it has rotational accln. Thus, according  to Maxwell's Electromagnetic Theory , Rotational Kinetic energy must be released. Thus, at some pt. of time , shouldn't the energy of an electron become 0 & then shouldn't the electron move out of  the atom, eventually leading to the atom only having the nucleus .

You are confusing macroscopic spin with the spin angular momentum of electron here. Electron is a point entity and does not posses a macroscopic spin (i.e. it does not spin on its axis).
 It does have angular momentum which is expressed as 'spin angular momentum' which is of-course quantized.

This link   by  John Rennie will be useful for understanding : The electron isn't spinning on it's own axis


How can diesel fuel produce less carbon monoxide but more carbon particulates? My guess at why diesel produces less carbon monoxide is that more heat is required for combustion of diesel fuel, so there is less incomplete combustion. Therefore rather than carbon monoxide being formed and released, the ability to completely combust means that carbon dioxide is often formed instead. Correct me if I'm wrong.

However, a risk of diesel fuel is that more carbon particulates are produced. But surely this is just a contradiction, because carbon particulates are a result of a lack of oxygen supply and incomplete combustion. Surely if there is enough complete combustion to form carbon dioxide over carbon monoxide, then there is a high enough oxygen supply and temperature to stop carbon particulates forming?

It seems that the mechanism of particulate formation is not yet well known. This article (preview) states some basics about the composition and formation of Diesel Particulate Matter (DPM):

  Diesel particulates form a very complex aerosol system. Despite considerable amount of basic research, neither the formation of PM in the engine cylinder, nor its physical and chemical properties or human health effects are fully understood.
  [...]
  The basic fractions of DPM are elemental carbon, heavy hydrocarbons derived from the fuel and lubricating oil, and hydrated sulfuric acid derived from the fuel sulfur. DPM contains a large portion of the polynuclear aromatic hydrocarbons (PAH) found in diesel exhaust

Also, it seems that the particulates are formed both during combustion and from gas during dilution. During combustion, carbonaceous agglomorates (soot), metallic ash from lubricating oil, and precursors to particle formation during dilution (sulfur oxides and partially burned hydrocarbons from fuel and lubricating oil) are formed. During dilution, materials such as sulfuric acid and other sulfates, as well as heavy hydrocarbons and hydrocarbon derivatives, are for the most part adsorbed onto the carbonaceous agglomorates formed during combustion. Some material may also nucleate by itself, but most nucleation occurs on the agglomorates.

PDF with explanation of ultrafine particle formation mechanisms

Unfortunately, i found it difficult to find any reason why this is such a big problem with diesel and not with gasoline. Perhaps the composition of the diesel is to blame, not temperature or oxygen supply.


Why is there a bulge at the top of my volumetric flask? Sometimes volumetric flasks are designed with a bulge in the neck right below the stopper opening. What is the design intent of that bulge?

From ASTM manual [1, p. 204], emphasis mine:

Volumetric flasks should be diluted to just below the line, mixed thoroughly, then allowed to stand for 4 to 24 h to allow any heat from an acid-water reaction to dissipate. [...] Thorough mixing consists of at least 20 complete inversions of the volumetric flask, allowing enough time on each for the air bubble to completely traverse the length of the vessel. Some borosilicate volumetric flasks contain a bulge in the neck above the line as an aid to proper mixing.

Volumetric pipetts also have a large bulge in the middle, but it serves a different purpose – to keep the $\rm length : volume$ ratio minimal and still maintain quick liquid collection with high precision.
Reference

Dulski, T. R. A Manual for the Chemical Analysis of Metals; ASTM manual series; ASTM: West Conshohocken, PA, 1996. ISBN 978-0-8031-2066-2.


Definition of an intensive variable The IUPAC definition of an intensive variable is:

  Physical quantity whose magnitude is independent of the extent of the system.

Why is it defined for "a system", and not for a homogenous system, as for a heterogenous system the quantity (e.g. density, temperature) might be different for the different extents/parts of the system?

The intensivity of a variable doesn't depend on the homogeneity of the system.

Consider system A, a metal rod of length 2 with a gradient of temperature :

This metal rod has an average temperature, which you could calculate.

Now consider system B, a metal rod of length 4 with the same gradient of temperature :

This metal rod also has an average temperature. Furthermore, it is the same temperature as system A. (if you are unsure about that try to pick a gradient (I arbitrarily chose T = exp(-x) and T = exp(-x/2) for the graphs) and do the calculations yourself)

System B is bigger (it would be twice as massive) but its temperature is the same. Ergo the temperature doesn't depend on the size of the system and is an intensive variable.


What makes something biodegradable? Learning about biodegradable polymers and non-biodegradable polymers. What exactly sets them apart though? Here's the formula from wikipedia for the biodegradable PHB:

and here for polypropylene, it's non-biodegrable counterpart

As far as I understand, biodegradable just means that bacteria will degrade it into other constituent parts. Is it just that bacteria is a picky eater and won't eat certain polymers? But they seem to be made of similar elements, and the bonds between are generally similar. Any help is helpful.

The difference between biodegradable polymers and 'regular' polymers is the bonds that make up the backbone. The backbone is the part of the polymer that extents the brackets. Anything attached to that chain is a side group and is less important.

The two examples you show here are very dissimilar. The polypropylene is only made up of Carbon-Carbon bonds. These bonds are very stable: that holds for chemical and 'biological' stability. The PHB how ever bonds monomers with an Ester functional group (Ester). This group is far more reactive that a C-C bond.

The reactivity of the groups the polymer is made up of, is very, very important. That being said, I cannot claim that all poly-esters are biodegradable. Depending on the groups close to the esters and the enzymes (probably from bacteria) involved, the bond might be cleaved, or might be inaccessible.

Hope this helped!


Regarding logic behind Organic Chemistry Was there any logic or thinking behind the experiments performed by Cannizzaro, Beckmann, Gattermann, Koch etc.?

Did they thoughtfully perform experiments or just add any two reactants and luckily discovered the reactions which are known by their name today?

I meant to ask that in Physics all problems are treated mathematically and logically; But in case of Organic Chemistry, there are lot of exceptions to all statements which are made by Cannizzaro etc. It is said that if a single observation is against a scientific law; the law is abandoned and new perfect law takes its place. But it seems that Organic Chemistry is unaware of this and "perfect" statements are not part of it.

Sometimes it was a lucky accident, sometimes there was a lot of math and logic going in the designing experiments.  The organic chemistry of that era, though, was a lot more applied and empirical than the physics of today.  They usually weren't setting out to find the absolute laws-of-nature governing chemistry, they were just trying to make better dyes, medicines, rubber, etc, and stumbled upon new phenomena along the way.
Further Explanation
Almost all these guys worked in Germany in the late 1800s, when chemical industry was the hot new thing (think of biotechnology in today's world).  There was a lot of money to be made in making better dyes, medicines, rubber, etc, so people would hack/engineer the established methods for making that stuff, and see if the resulting products were any better.
For example:
Chemical A cures headaches really well, but causes nausea.  Chemical B gives a little relief from headaches, but doesn't have any side effects.  I bet I could make chemical C (which no one had ever made before), which has a structure that's sort of a combination of chemicals A and B, and maybe that will cure headaches with no side effects.
==> {made the chemical and did some tests to find out the structure}.
What the heck!  This has a completely different structure than what I was expecting, it must have undergone a second reaction to rearrange itself.
==> {publish paper},{get famous},{people 100 years later question the logic of your experiments} :)


Combustion of carbon with hydrogen Can coal (carbon) be "burned" in a hydrogen atmosphere? I know it would not match the definition of "combustion".

But is the reaction $$\ce{C +2H2 ->CH4}$$ exothermic?

(I'm fairly convinced this is a duplicate, but I didn't find any question like that.)

The formation reaction of CH4 is exothermic. The heat of formation is : -75 kJ/mol


Calculation of equilibrium constant given the reaction temperature I have only a chemical equation, like:
$$\ce{N2 + 3H2 <=> 2NH3}$$
How can I calculate the value of equilibrium constant from only this equation. Provided temperature is $\pu{298K}$. I can use other standard data. I am really stumped on this question.

By definition, the standard equilibrium constant is:

$K = e^{-\Delta_rG/RT}$

Where $\Delta_rG$ is the difference between the Gibbs Energies of the products and reactants.


How can I neutralise H2 gas without burning it? I was thinking of maybe making it bubble into a solution containing something that would react with H2 yielding either a dissolved substance or a precipitate from the reaction.

Maybe a solution with a small amount of H2O2 in it?

Many nuclear power plants use so-called "passive autocatalytic recombiners" for the catalytic oxidization of hydrogen that could be released into the containment in the event of severe accidents. (Note that the affected units of the Fukushima-Daiichi plant did not have such equipment.) The catalytically active materials, typically platinum and/or palladium, are doted on carriers, preferably designed as catalyst plates orientated vertically to support the flow.


Bonding in the nitrate anion Hi so I was studying chemical bonding where i encountered a problem which is stated below.

When we talk about ${NO_3}^-$ we draw its structure as following

But the thing which I dont understand is that in two of the O atoms there are three lone pairs(ie: $6 e^-$) but as we know from O's $e^-$ config. (which is $1s^2 2s^2 2p^4$) there are two unpaired $e^-$ which will take part in bonding as there is no vacant orbitals for $e^- $ to get excited to . Now if in this O atom only one sigma bond is present then technically we are left with $5e^-$ and not $6e^-$ which is shown in the figure.

Similarly in the N atom (whose $e^-$ is $1s^2 2s^2 2p^3$ ) does not have any other vacant orbital to excite its $e^-$ thus showing that it can make only 3 bonds then why is it making 4 bonds( 1 pi and 3 sigma) ?

The problem lies in the electron count. Each bond contains 2 electrons and the pair is counted twice, once for being around each atom in the bond. And don't forget the extra electron to make the ion.

The + and - are from the point of view of the atom nucleus: e.g., consider the nitrogen to share 2 electrons with the oxygen on the right, to give 2 to the oxygen above, and to give 1 to the oxygen on the left. The other electron in that bond is one that makes the anion negative overall.

Then nitrogen gains its octet of 8 electrons, but formally only shares 50% of those electrons, so it started out with 5 but only controls 4 - it now has a positive charge. The right oxygen shares, so no charge. The top oxygen got 2 from nitrogen, so is negative, and the left oxygen got 1 from nitrogen and one from the metal of which this is the anionic part. And then, draw up the other two diagrams that have the "neutral oxygen" in the other positions, add them all and average them. Then each oxygen winds up with a 2/3 negative charge (but the nitrogen is still +1), while the anion is -1 and planar.


Is this heterocyclic boron molecule aromatic? What kind of aromaticity does the following molecule have?

I read somewhere that this is  an aromatic compound, but I believe that in order to be aromatic, it should have continuous resonance. I also think that there should be a break at B that causes no $\pi$ electrons to be shared among the B—B atoms...so I think this is non-aromatic. Is that correct?

That molecule is indeed aromatic.

It satisfies Huckel's rule ($4n+2$) with $n=0$. It's also planar.

Lastly, the boron atom has three electron domains, meaning it can exist with $sp^2$ hybridization. The left-over p-orbital, whilst not actively forming $\pi$  bonds, can still complete a stable $\pi$ molecular orbital.


Z* effective charge and Ionization Energy I'm trying to figure out the patterns for Ionization Energies. I am familiar with the periodic trend, however things become quite different when we hit the 1st I.E. For example, Na has an I.E(1) of 495.8 kJ while its second I.E. rockets up to 4562 kJ while the atoms towards the right are much lower than this. The trend says that the I.E. increase up and right of the periodic table, which is not the case here.

My point is, in order to get a better estimate, would it be safe to say that the Zeff charge and atom size relate to the I.E?

Example: Effective Charges

Mg = 2
Al = 1
S = 4
Si = 2
Na = 1

We see that the strongest pull towards the charged center would be Al and Na in this case. However, having Al being the smaller atom would require more energy to remove the electron from its valence shell.

Now in the case of first Ionization Energy we have:

Mg = 3
Al = 2
S = 5
Si = 3
Na = 2

In this case, Na now has reduced its size due to the fact it jumped from n=3 to n=2 level and Al also reduced in size but is still a bigger atom than Na due to our trend.

My question is, is this approach fairly accurate or should I be looking somewhere else?

Using the common idiom: "full subshells are stable"

It is a little more compact to use the incorrect explanation and correct it than to explain in terms of correct. Also, when you hear this incorrect explanation, you will understand what is meant.

Yes, size is a factor (thus the "up" part- IE is higher in lower periods), but the observed anomaly in second-ionization energies in $\ce{Na}$ and $\ce{Al}$ is better related to removal of an electron from a full subshell.

$\ce{Na}$ has an electron configuration of $\ce{[Ne] 3s^1}.$ The first ionization gives $\ce{Na}$ makes it $\ce{[Ne]}$, the same as a noble gas- full 2p subshell. Because full subshells are stable, it takes a lot of energy to remove the first electron from it. $\ce{Al}$ is analogous; the second ionization means removing an electron from a full subshell.

The same is observed in the first ionization energies (in $kJ/mol$):
$$\ce{Na}:495$$$$\ce{Mg}:737$$$$\ce{Al}:577$$
Ionization of $\ce{Mg}$ removes an electron from a full $\ce{3s}$ subshell, so it is a little higher than the trend.

There are many factors that contribute to exact values, but these are difficult to predict, which is why they taught a simple version in your class.

Correcting the myth

While it is common to refer to full subshells as being more stable, it isn't really what is going on. In reality, the next electron added in destabilized. This occurs because of shielding. Outer electrons feel repulsion from inner electrons, thus outer electrons are easier to remove than inner ones. The saying that "half-filled subshells are stable" is also used. Again, it is that the next added electron is destabilized, but in this case it is because the energy needed to spin pair the two electrons.


What is meant by excess energy in exothermic reaction? I was reading this article on exothermic reactions online: Exothermic reactions

It states that "In exothermic reactions, the products have less enthalpy than the reactants, and as a result, an exothermic reaction is said to have a negative enthalpy of reaction. This means that the energy required to break the bonds in the reactants is less than the energy released when new bonds form in the products. Excess energy from the reaction is released as heat and light."

I'm just curious, is the excess energy from the bonds being formed, or is it because of the fact that products have more potential energy than the reactants, and because of law of convervation of energy, is released as heat energy?

The energy release is because of the total energy of valence electrons in the product bonds is lower  than in reagents.

In fact, the change of the mean total  potential energy of valence electrons is twice as much as the change of their total energy, as the half of that change is compensated by their increased mean total kinetic energy.

As a consequence, the energy released by forming chemical bonds of products is bigger than energy spent on breaking bonds of reagents.


How to properly relate quantum mechanical energy to the classical world The Equation of Interest

The energy equation derived from the "Particle in a 1-D box" problem is as follows:

$E=\dfrac{n^2h^2}{8mL^2}$

where $n$ is the principle quantum number, $h$ is Planck's constant, $m$ is the mass of the particle (e.g. electron), and $L$ is the length of region of space (i.e. box).

The Goal

I would like to illustrate the harmonious nature of this energy equation derived from the quantum mechanical nature of the system and classical mechanics.  In theory we should be able to define a mass and region of space for a clear comparison.

The Example: Electron vs. Car

I want to compute the lowest possible energy of an electron using the equation above.  We define $m=9.11\times10^{-31} \textrm{kg}$ and the region of space as a reasonable value for the diameter of an atom, $L=0.2 \textrm{nm}$.

Plugging this into our equation along with Planck's constant ($h=6.626\times10^{-34} \textrm{ m}^2 \textrm{kg s}^{-1}$) we obtain

$E_{\textrm{electron}} = \dfrac{(6.626\times10^{-34} \textrm{ m}^2 \textrm{kg s}^{-1})^2}{8 *9.11\times10^{-31} \textrm{ kg} * (2\times 10^{-10} \textrm{m})^2} = 1.51 \times 10^{-18} \textrm{ J}$.

Now let us imagine a car with $m=2000 \textrm{ kg}$ and a region of space as $L=1 \textrm{ m}$.  The resulting energy is

$E_{\textrm{car}}=2.74\times10^{-71} \textrm{ J}$.

Now in both cases, the energy is so small that we might as well say its zero (from a classical standpoint anyway).  Of course $E_{\textrm{car}} << E_{\textrm{elecron}}$.

*So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?

We can, however, convert these energies (which are purely kinetic energies) into something which is clearly distinguishable such as velocity by using the equation

$E = \dfrac{1}{2}mv^2$

Alternatively,

$v = \sqrt{\dfrac{2*E}{m}}$

where $m$ is the mass of the object and $v$ is the velocity.  If we plug in our previously determined energies we get the following results...

$v_{\textrm{electron}} = 1.8\times 10^6 \textrm{ m s}^{-1}$

and

$v_{\textrm{car}} = 1.7\times 10^{-37} \textrm{ m s}^{-1}$

Here we see that the velocity of the car is essentially zero (or technically speaking, so slow that it wouldn't ever be observed as moving in billions of lifetimes).  This jives with classical theory.  However, we see that the velocity of the electron is staggering, even though the energy of the electron is 'essentially zero'

The Conclusions

So, in order to properly relate the energy equation derived from the quantum mechanical system of a particle in a 1-D box, one cannot simply compare the raw energies of the particle (i.e. electron) to that of an ordinary object such as a car because both energies result in a number that is essentially zero which defeats the point.  As such, one must relate the energy to something appropriate such as the velocity of the object to properly illustrate how the energy equation still relates to everyday life.

The Question

Are my conclusions correct or can the raw energetics simply be used to make a proper comparison between quantum/classical theory without resorting to recasing the results into a physical property such as velocity?

#not-homework

*So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?

This is wrong. Just because you can't immediately, instinctively comprehend the difference between the $10^{-18}\  \mathrm{J}$ and $10^{-71}\  \mathrm{J}$ and therefore think they are both "indistinguishably small" doesn't mean it's true. The difference is massive. Our brains just haven't evolved to deal with numbers whose logarithms are very far from 0, because for almost all of our existence as a species, it never had to. Perhaps this site may give some very qualitative though fun context on how brains perceive the magnitude of a number.

In order to cope with this limitation of ours, we have developed some strategies. One can be understood as a form of chunking, where we simply modify numbers in a way so they become closer to something we understand. You probably won't be able to immediately give an example of an object that weighs 200,000 ounces or $3\times 10^{-27}$ solar masses, though you may find it quite a bit easier if I say 6 tonnes instead. This is because even if we inherently know the concept of an ounce or a solar mass, we can't really comprehend the magnitudes of 200,000 or $3\times 10^{-27}$ (the latter of course being much worse), and in order to get our heads around it more easily, we instead invent a new unit to measure weights in which we can express the same information using numbers closer to 1. The whole idea of scientific notation for numbers and SI prefixes for units can be understood as a consequence of this.

Another way is the one you use; instead of reframing the units and numbers, reframe the situation. You may feel there is no difference between $10^{-18}\  \mathrm{J}$ and $10^{-71}\  \mathrm{J}$, but the 53 orders of magnitude in difference is about the same as the number of atoms in a human body and the number of atoms in the entire observable Universe! Notice how I simply took the ratio into a slightly more familiar context (you know the Earth is just a small part of the observable Universe, and a human being is puny in comparison to the Earth).

So all of your calculations are correct, but if Homo sapiens were smarter, they would be entirely unnecessary!


Natural Product class I have a quick question: how can I distinguish between terpenoids compounds and polyketides?
For example artemisinin (see image below): it present a continuous carbon chain alternated with $\ce{O2}$, but it is a polycyclic structure as in terpenoid compound.
Which is the main criterion I should consider here?

It’s very hard to tell for some molecules a priori. For most, there are enough indicators.

Polyketides:

will typically have 1,3-functionalisation patterns, so hydroxy groups or keto groups are typically two carbons away from the next one. And most methyl groups also have a nonmethylated carbon in-between. Note that 1,3 can be expanded to 1,5; 1,7; 1,9; …
if cyclic, will often be only monocyclic or any second cycle will typically be something like a ketal (two oxygens bonded to a single carbon; formed by a hydroxy group attacking a ketone often five carbons away).
if cyclic are typically macrolactones, so large cycles (14, 20 or even more carbon atoms) and linked by an ester group.
can contain phenyl groups or other aromatic systems formed by elimination reactions. Often, these will be 3,5-bishydroxyphenyl groups.
typically contain an even number of carbon atoms in the chain (i.e. excluding any methyl groups mentioned above).

On the other hand, terpenoids:

very commonly have a carbon count that is a multiple of five (even often multiples of ten).
can very often be broken down into isoprene (2-methylbutane) units by paper chemistry.
typically have 1,5-methyl functionalisation (although this can change due to Wagner–Meerwein rearrangements in the biosynthesis).
can have 1,1-dimethylated carbons which polyketides cannot.
often form polycyclic structures composed of small all-carbon cycles (small being 5- to 7-membered rings).
rarely contain lactones or lactams (cyclic esters or amides).
are generally less oxygenated.
very rarely contain more than one or two double bonds, never contain phenyl rings to the best of my knowledge.

With that, you have some criteria at hand to attempt a classification. Artemisinin’s tricyclic structure with an all-carbon cycle looks a lot more like a terpenoid than a polyketide.


How to calculate the solubility of CaF2 at different temperatures? At $\pu{25^\circ C}$ I used the free energy table for $G$ for the reaction and got $K_\mathrm{sp}= 10^{-9.02}$.

I now need to find out what the solubility is at $10$, $20$ and $\pu{30^\circ C}$. Could you also show what equation you used and how to work out the solubility by one of these other temperatures?

It depends what you have access to… but if we suppose that you have access to decent thermodynamic tables, here's a hint:

You know how solubility is linked to dissolution reaction (Gibbs) free energy. Now, how does the free energy depend on temperature? In other words, what is $\displaystyle\frac{\partial G}{\partial T}$?


How much does a light heat the air around it? My dad told me to turn off the lights when I'm not using them because it will keep the house cooler. I don't think he is correct that a light can heat up the room. How much of a difference in the temperature would it make? Assume a 60 Watt light bulb running for one hour in a room with approx. 3000ft^3/85m^3 of air. Also assume the air is dry and in constant thermal equilibrium at 1 atm. I think the specific heat capacity of air is about 1 J/g*K, but I am not confident.

During a period of $\Delta t = 1\ \mathrm h =3600\ \mathrm s$, the light bulb converts electric power of $P = 60\ \mathrm W$ to light and heat. Assuming that none of this energy escapes from the room, the entire power is eventually converted to heat:

$$\begin{align}
Q &= P \cdot \Delta t \\
&= 60\ \mathrm W \times 3600\ \mathrm s \\
&= 216\ \mathrm{kJ}
\end{align}$$

According to REFPROP (NIST Standard Reference Database 23, Version 9.0), the relevant parameter values of air at the assumed initial temperature of $T_0 = 20\ \mathrm{^\circ C}$ and pressure $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$ are a density of $\rho_0 = 1.2046\ \mathrm{kg/m^3}$ and a specific heat capacity at constant pressure of $c_{p,0} = 1.0061\ \mathrm{kJ/(kg\ K)}$.

The initial volume is given as $V_0 = 85\ \mathrm{m^3}$. The corresponding mass is

$$\begin{align}
m &= V_0 \cdot \rho_0 \\
&= 85\ \mathrm{m^3} \times 1.2046\ \mathrm{kg/m^3} \\
&= 102.391\ \mathrm{kg}
\end{align}$$

Ignoring the heat capacity of the walls and inventory, the temperature of the air is approximately increased by

$$\begin{align}
\Delta T &= \frac QC \\
&= \frac Q{m \cdot c_{p,0}} \\
&= \frac{216\ \mathrm{kJ}}{102.391\ \mathrm{kg} \times 1.0061\ \mathrm{kJ/(kg\ K)}} \\
&= 2.1\ \mathrm{K}
\end{align}$$

By way of comparison, a better result may be obtained using the enthalpy of air, since the heat capacity of air is not constant. In this case, however, the difference is insignificant because the temperature change is small. The initial specific enthalpy of air is $h_0 = 419.40\ \mathrm{kJ/kg}$. Thus, the enthalpy is

$$\begin{align}
H_0 &= m \cdot h_0 \\
&= 102.391\ \mathrm{kg} \times 419.40\ \mathrm{kJ/kg} \\
&= 42\,943\ \mathrm{kJ}
\end{align}$$

At constant pressure, the change in enthalpy corresponds to the heat transferred to the air:

$$\begin{align}
H_1 &= H_0 + \Delta H \\
&= H_0 + Q \\
&= 42\,943\ \mathrm{kJ} + 216\ \mathrm{kJ} \\
&= 43\,159\ \mathrm{kJ}
\end{align}$$

Thus, the final specific enthalpy is

$$\begin{align}
h_1 &= \frac{H_1}{m} \\
&= \frac{43\,159\ \mathrm{kJ}}{102.391\ \mathrm{kg}} \\
&= 421.51\ \mathrm{kJ/kg}
\end{align}$$

At a pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, this specific enthalpy corresponds to a temperature of $T = 22.1\ \mathrm{^\circ C}$.

Note that, at the final temperature of $T = 22.1\ \mathrm{^\circ C}$ and the pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, the new density of air amounts to $\rho_1 = 1.1960\ \mathrm{kg/m^3}$. Thus, the new volume is

$$\begin{align}
V_1 &= \frac{m}{\rho_1} \\
&= \frac{102.391\ \mathrm{kg}}{1.1960\ \mathrm{kg/m^3}} \\
&= 85.6\ \mathrm{m^3}
\end{align}$$

Therefore, a small amount of air must escape from the room.

Finally, if the heat from a $60\ \mathrm W$ light bulb really is a problem, you might want to leave the room since the human body generates about $100\ \mathrm W$.


Solubility in carbon tetrachloride Which of these substances is more soluble in $\ce{CCl4}$?

  $\ce{CH3F}$
  $\ce{I2}$
  $\ce{Na2SO4}$
  $\ce{HCl}$

How do I even solve this kind of exercise?

My guess is that $\ce{CCl4}$ is non-polar. Therefore the substance we're gonna need must be non-polar as well. The only non-polar substance is $\ce{I2}$ so that should be the answer.

Is my reasoning fine and, more importantly, is there another way to solve this kind of problem? What if they give me two options that are non-polar?

You are correct in your assumption. For these types of problems it is usually safe to assume like dissolves like and choose the answer with the most similarity in terms of intermolecular bonding. For a general description you can check here.

When you get into questions with two options that have similar intermolecular forces it gets a little hairy. For example, I think a question like, "which is more soluble in carbon tetrachloride: n-Pentane or n-Hexane?" would be fairly challenging to answer without experimental or computational data to compare.


Can caffeine be separated from coffee? Is there a way to extract caffeine from coffee completely?

Yes, this is how decaffeinated coffee is made. If I recall correctly, this is done commercially by dissolving the caffeine in dichloromethane (DCM), or more recently in supercritical carbon dioxide. This switch is probably over concerns regarding chronic industrial exposure to DCM, which is mildly toxic.

From personal experience, I have extracted caffeine from instant coffee using a separatory funnel with a system of water and dichloromethane. Purification by sublimation finished the job.


MacGyvering a sealed tube Is it possible (advisable?) to make a sealed tube by melting and twisting the end of a borosilicate test tube? One issue that I can think of is that the pressure might be too high if I then proceed to heat the tube. I thought that I could maybe connect the tube to a vacuum tube and heat it in the middle. Any experience with this?

Usual way to work with air sensitive compounds is to work on Schlenk-line (search Wiki). When you transfer your substance you can leave it in Schlenk tube, or if it's really sensitive and you need to keep out of the line for longer period of time you have in general three possibilities: (1) use glassware with J. Young valve (available even for NMR tubes); (2) store your standard Schlenk glass in glove box filled with inert gas and (3) the oldest is to melt your glass.

Now back to your question. Usual way is to prepare glass with narrow place on it prior to doing experiment (that's were later you will melt it). By narrow I mean the hole inside should be ~2-3 mm (thinner glass, smaller hole). I used to use Schlenk lines, so on opened end I have standard 14 female cone that was connected to a vacuum line. I used to set vacuum ~0.1 Torr and with propane/butane/oxygen torch one can easily close it. The huge advantage of this old way is that you can prepare your glass even with 3mm thick walls that allows you to work with substances with very low boiling points (in this case to close the glass one need to freeze the stuff with liquid N2 or if not available at least cool it with dry ice bath). Of course later opening is quite tricky and usually also had to be done after freezing of chemicals.

Last part if it's advisable. That one is as usually hard to answer. I have no idea how experienced you are. Judging by question not to much (sorry). So I wouldn't start with anything to reactive in the beginning. Generally try to play first with glass and torch and get a bit used to it. Do it on proper chemical glass (not just bottle glass).
If you are "advanced guy" and you want to make yourself f.e. PfBF2 borane you have to use this technique - any other gives just a miserable yield.


Experimental verification of the field effect March's Advanced Organic Chemistry has the following paragraph

  The other effect operates not
  through bonds, but directly through space or solvent molecules, and is called the field
  effect. It is often very difficult to separate the two kinds of effect, but it has been
  done in a number of cases, generally by taking advantage of the fact that the field effect
  depends on the geometry of the molecule but the inductive effect depends only on the
  nature of the bonds. For example, in isomers 1 and 2 the inductive effect of the chlorine
  atoms on the position of the electrons in the COOH group (and hence on the
  acidity, see Chapter 8) should be the same since the same bonds intervene; but the field
  effect is different because the chlorines are closer in space to the COOH in 1 than they
  are in 2. Thus a comparison of the acidity of 1 and 2 should reveal whether a field
  effect is truly operating. The evidence obtained from such experiments is overwhelming
  that field effects are much more important than inductive effects.

The diagram:

I simply do not understand how 2 is more acidic than 1. Since the field effect is greater in 1 , should not 1 be more acidic?

The inductive effect is transmitted through bonds; the field effect is transmitted through space.

The $\ce{C-Cl}$ bond is polarized due to the electronegativity difference between the two atoms (carbon 2.55; chlorine 3.16), the chlorine end being more negative than the carbon end. We could note this excess of electron density (partial negative charge) around the chlorines in your structures by placing a ${\delta^{-}}$ by the chlorine and a ${\delta^{+}}$ by the carbon.

Now, when the acid ionizes there is also a negative charge associated with the carboxylate group.  When the chlorines are on the same side of the bridge as the carboxylate (syn isomer) the (through space) field effect acts to destabilize the situation because we have placed the partial negative charge around chlorine closer to the negative charge on the carboxylate (placing the negative charges closer together is destabilizing since like charges repel). The destabilization results in a higher activation energy being needed to reach the transition state for ionization.

On the other hand, the anti isomer with the chlorines further from the carboxylate will be less destabilized because the negative charges are further apart.  This means that the anti isomer will have a lower activation energy for ionization compared to the syn isomer.  Therefore, the anti isomer will be more acidic (display a lower pKa) than the syn isomer due to the field effect.


Why isn't CO2 flammable? Given that $\ce{CO2}$ is comprised of carbon and oxygen, why
isn't it a flammable gas?

If I understand what you are saying, then the issue is that the carbon and oxygen have already reacted. Carbon dioxide is thus not flammable because it is the fully oxidized product of the combustion between oxygen and carbon.

"Made of" has multiple meanings. You're confusing the meaning of physical mixture (flammable) with the chemical composition of a compound (not flammable).


Understanding effect of ph on solubility of salt In the following picture why Ksp is not simply S2- ? Why is f included with one species only? Please explain the last part.

What does f represent and why it is used in molar solubility?

Let assume we have the following equilibria:
$$\ce{MX \rightleftharpoons M+ + X-}$$
$$\ce{HX \rightleftharpoons H+ + X-}$$
The expression for their respective equilibrium constants are the following:
$$\mathrm{K_{sp} = [M^+][X-]}$$
$$\mathrm{K_a = \frac{[H^+][X^-]}{[HX]}}$$
We also know the following (let S be solubility):
$$\mathrm{S = [M^+] = [X^-]_{tot}}$$
Now we just need one more equation before we can start solving for S which is a mass balance. Since all the $\mathrm{X^-}$ that is produced either reacts with water to form $\mathrm{HX}$ or remains as $\mathrm{X^-}$, we get the following equation :
$$\mathrm{[X^-]_{tot} = [X^-] + [HX]}$$
Now lets put $\mathrm{[X^-]}$ and $\mathrm{[HX]}$ in terms of $\mathrm{[M^+]}$:
$$\mathrm{[X^-] = \frac{K_{sp}}{[M^+]}}$$
$$\mathrm{[HX] = \frac{[H^+][X^-]}{K_a} = \frac{K_{sp}[H^+]}{K_a[M+]}}$$
Now let plugs those values into our mass balance and replace $\mathrm{[X^-]_{tot}}$ and $\mathrm{[M^+]}$ with S:
$$\mathrm{[X^-]_{tot} = \frac{K_{sp}}{[M^+]} + \frac{K_{sp}[H^+]}{K_a[M+]}}$$
$$\mathrm{S = \frac{K_{sp}}{S} + \frac{K_{sp}[H^+]}{K_a\times S}}$$
$$\mathrm{S^2 = K_{sp} + \frac{K_{sp}[H^+]}{K_a}}$$
$$\mathrm{S = \sqrt{K_{sp} + \frac{K_{sp}[H^+]}{K_a}}}$$
This gives you the exact same expression as the one on the sheet but hopefully this is much clearer for you.


Where does the label ‘Dq’ to denote the field split in coordination compounds come from? As established in a previous question, coordination compounds typically have a field split between the $\mathrm{t_{2g}}$ and the $\mathrm{e_g}$ d-orbitals.[1] This energy difference can be explained by the crystal field theory which assumes negative point-charges approaching a complex and destabilising orbitals pointing towards said point charges.

Commonly, the energy difference between the lower and higher orbitals is referred to by the relative value of $\Delta$ or the equally relative value of $10~\mathrm{Dq}$.

The latter is of interest to me. It is clearly used as a (relative) unit, because ligand field stabilisation enthalpies (LFSE’s) are typically given in values of $\mathrm{Dq}$ — e.g. $\ce{[Cr(H2O)6]^3+}$ (hexaaquachromium(III)) has an LFSE of $-12~\mathrm{Dq}$. But what does $\mathrm{Dq}$ mean and what does it derive from?

Note:

[1]: This is explicitly considering the octahedral case. Tetrahedral compounds split the d-orbitals into $\mathrm{t_2}$ and $\mathrm{e}$. This distinction is, however, not relevant to the question.

I will expand my previous comment into an answer.
In the $O_h$ case the metal ion is surrounded by $6$ ligand ions, each with charge $q=-Ze$ located at $\langle\pm a,0,0\rangle$, $\langle0, \pm a,0\rangle$, and $\langle0,0,\pm a\rangle$.
The electrostatic potential of a point charge $q$ is
$$V=\frac{kq}R$$
where
$$k=\frac1{4\pi\epsilon_0}$$
is the Coulomb's Law constant and $\epsilon_0$ is the permittivity of free space and $R=\left\|\vec{r_f}-\vec{r_s}\right\|$ is the distance from the source point $\vec{r_s}$ to the field point $\vec{r_f}$.
Working for the moment with the potential $V_1$ due to the charge at $\langle a,0,0\rangle$,
$$R=\left\|\langle x,y,z\rangle-\langle a,0,0\rangle\right\|=\left\|\langle x-a,y,z\rangle\right\|=\sqrt{(x-a)^2+y^2+z^2}=\sqrt{a^2-2ax+x^2+y^2+z^2}=a\sqrt{1-\frac{2x}a+\frac{r^2}{a^2}}$$
Where $r^2=x^2+y^2+z^2$ is the squared distance from the origin. Then
$$V_1=-\frac{Ze}{4\pi\epsilon_0a}\left(1-\frac{2x}a+\frac{r^2}{a^2}\right)^{-1/2}$$
Now we need to expand $(1+x)^{-1/2}$ to fourth order in $x$. The Taylor series yields
$$(1+x)^{-1/2}=1-\frac12x+\frac38x^2-\frac5{16}x^3+\frac{35}{128}x^4+O\left(x^5\right)$$
We could have obtained this result via algebra alone by assuming that $(1+x)^{-1/2}=1+ax+bx^2+cx^3+dx^4+O\left(x^5\right)$, then squaring and multiplying by $(1+x)$ so that
$$\left(1+ax+bx^2+cx^3+dx^4\right)^2(1+x)=1+O\left(x^5\right)$$
and then finding the unknowns $a$, $b$, $c$, and $d$ from the conditions that the coefficients of $x$, $x^2$, $x^3$, and $x^4$ should be zero. Then we have
$$\begin{align}V_1= & -\frac{Ze}{4\pi\epsilon_0a}\left\{1-\frac12\left[-\frac{2x}a+\frac{r^2}{a^2}\right]+\frac38\left[\frac{4x^2}{a^2}-\frac{4xr^2}{a^3}+\frac{r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac{8x^3}{a^3}+\frac{12x^2r^2}{a^4}\right]+\frac{35}{128}\frac{16x^4}{a^4}\right\}+O\left(\left(\frac ra\right)^5\right)\end{align}$$
This is complicated, but now we simplify by combining the potentials due to all $6$ ligands. We want to find group operations from $O_h$ that map the first ligand to the others, and fortunately the powers of $S_6$, the rotoreflection by angle $\frac{\pi}3$ about the axis through the origin and $\langle1,1,1\rangle$ will do the trick.
Its powers cycle the original ligand point through all the others. We just need to find the mappings of the powers of $x$ by the powers of $S_6$ and sum over all powers of $S_6$:
$$\begin{array}{c|cccccc|c}
f(x,y,z)&Ef&S_6f&S_6^2f&S_6^3f&S_6^4f&S_6^5f&\sum\limits_{i=0}^5S_6^if \\
\hline r^2&r^2&r^2&r^2&r^2&r^2&r^2&6r^2 \\
x&x&-z&y&-x&z&-y&0 \\
x^2&x^2&z^2&y^2&x^2&z^2&y^2&2r^2 \\
x^3&x^3&-z^3&y^3&-x^3&z^3&-y^3&0 \\
x^4&x^4&z^4&y^4&x^4&z^4&y^4&2\left(x^4+y^4+z^4\right)
\end{array}$$
So now
$$\begin{align}V_{O_h}= & \sum\limits_{i=0}^5S_6^iV_1 \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left\{6-\frac12\left[0+\frac{6r^2}{a^2}\right]+\frac38\left[\frac{8r^2}{a^2}-0+\frac{6r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[0+\frac{24r^4}{a^4}\right]+\frac{35}{4}\frac{x^4+y^4+z^4}{a^4}\right\}+O\left(\left(\frac ra\right)^5\right) \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left[6-\frac{21r^4}{4a^4}+\frac{35\left(x^4+y^4+z^4\right)}{4a^4}\right]+O\left(\left(\frac ra\right)^5\right) \\
& =-\frac{6Ze}{4\pi\epsilon_0a}-\frac{35Ze}{16\pi\epsilon_0a^5}\left(x^4+y^4+z^4-\frac35r^4\right)+O\left(\left(\frac ra\right)^5\right)\end{align}$$
All this mathjax is exhausting. I can't finish in one sitting, but at least I have gotten to
$$D=\frac{35Ze}{16\pi\epsilon_0a^5}$$
EDIT: Now that the potential for $O_h$ geometry has been developed, the same will be done for the $T_d$ case where the metal ion is surrounded by $4$ ligand ions, each with charge $q=-Ze$, this time located at
$\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$, $\left\langle\frac a{\sqrt3},-\frac a{\sqrt3},-\frac a{\sqrt3}\right\rangle$, $\left\langle-\frac a{\sqrt3},\frac a{\sqrt3},-\frac a{\sqrt3}\right\rangle$, and $\left\langle-\frac a{\sqrt3},-\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$.
The distance from the source point at $\vec{r_s}=\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$ to the field point at $\vec{r_f}=\langle x,y,z\rangle$ is
$$\begin{align}R= & \left\|\vec{r_f}-\vec{r_s}\right\|=\left\|\left\langle x-\frac a{\sqrt3},y-\frac a{\sqrt3},z-\frac a{\sqrt3}\right\rangle\right\| \\
& =\sqrt{\left(x-\frac a{\sqrt3}\right)^2+\left(y-\frac a{\sqrt3}\right)^2+\left(z-\frac a{\sqrt3}\right)^2} \\
& =\sqrt{a^2-\frac{2a}{\sqrt3}(x+y+z)+x^2+y^2+z^2} \\
& =a\sqrt{1-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}}\end{align}$$
Then the potential due to this charge is
$$\begin{align}V_2 & =\frac{kq}R=-\frac{Ze}{4\pi\epsilon_0a}\left(1-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}\right)^{-1/2} \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left\{1-\frac12\left[-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}\right]+\frac38\left[\frac4{3a^2}(x+y+z)^2-\frac{4(x+y+z)r^2}{\sqrt3a^3}+\frac{r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac8{3\sqrt3a^3}(x+y+z)^3+\frac{4(x+y+z)^2r^2}{a^4}\right]+\frac{35}{128}\left[\frac{16}{9a^4}(x+y+z)^4\right]\right\}+O\left(\left(\frac ra\right)^5\right)\end{align}$$
This time the group operations that map the model point to all the others will be the members of the $D_2$ point group: the identity $E$, and the rotations by $180°$ about the x-axis $C_{2x}$, the y-axis $C_{2y}$, and the z-axis $C_{2z}$.
Again a table for combining the contributions from all point charges:
$$\begin{array}{c|cccc|c}
f(x,y,z)&R_1f=Ef&R_2f=C_{2x}f&R_3f=C_{2y}f&R_4f=C_{2z}f&\sum\limits_{i=1}^4R_if \\
\hline r^2&r^2&r^2&r^2&r^2&4r^2 \\
x&x&x&-x&-x&0 \\
xy&xy&-xy&-xy&xy&0 \\
xyz&xyz&xyz&xyz&xyz&4xyz
\end{array}$$
And the reader can readily verify that also
$$\sum\limits_{i=1}^4R_iy=\sum\limits_{i=1}^4R_iz=\sum\limits_{i=1}^4R_ixz=\sum\limits_{i=1}^4R_iyz=0$$
So the rule is that terms in the expression for $V_2$ with odd powers of any of the coordinates cancels except for $xyz$, which, along with all terms with only even powers of coordinates, adds.
Expanding powers of $(x+y+z)$ and summing over the four sources,
$$\begin{align}V_{T_d} & =\sum\limits_{i=1}^4R_iV_2=-\frac{Ze}{4\pi\epsilon_0a}\left\{4-\frac12\left[0+\frac{4r^2}{a^2}\right]+\frac38\left[\frac4{3a^2}(4r^2)-0+\frac{4r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac8{3\sqrt3a^3}(6)(4xyz)+\frac{4(4r^2)r^2}{a^4}\right]\right. \\
& \left.+\frac{35}{128}\frac{16}{9a^4}\left[4\left(x^4+y^4+z^4\right)+6(4)\left(x^2y^2+x^2z^2+y^2z^2\right)\right]\right\}+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{Ze}{4\pi\epsilon_0a}\left\{4+\frac{3r^4}{2a^4}+\frac{20}{\sqrt3a^3}xyz-\frac{5r^4}{a^4}\right. \\
& \left.+\frac{35}{72a^4}\left[4\left(x^4+y^4+z^4\right)+12\left(r^4-x^4-y^4-z^4\right)\right]\right\}+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{Ze}{4\pi\epsilon_0a}\left[4+\frac{20}{\sqrt3a^3}xyz-\frac{35}{9a^4}\left(x^4+y^4+z^5-\frac35r^4\right)\right]+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{4Ze}{4\pi\epsilon_0a}-\frac{5Ze}{\sqrt3\pi\epsilon_0a^4}xyz+\frac{35Ze}{36\pi\epsilon_0a^5}\left(x^4+y^4+z^4-\frac35r^4\right)+O\left(\left(\frac ra\right)^5\right)\end{align}$$
On comparison with $V_{O_h}$, the first term is the potential due to the net charge of all the ligands uniformly spread over the surface of a sphere of radius $a$.
The second term has no counterpart in $V_{O_h}$ and causes no splitting but does create mixing between the $d_{xy}$ and $p_z$ orbitals and would lower the energy of the former and raise the energy of the latter.
I don't know how significant that effect is, though. The splitting term is $\frac49D\left(x^4+y^4+z^4-\frac35r^4\right)$ and is in fact $-\frac49$ the value of the splitting term for $O_h$ geometry.
Still not done, but it's getting late.

EDIT:To get the $q$ part of $Dq$ we are going to have to do some integrals. I am going to use the notation
$$\oint f(x,y,z)d^2\Omega$$ to represent the integral of $f(x,y,z)$ over the surface of the unit sphere.
One integral we know right off the bat is
$$\oint 1d^2\Omega=4\pi$$
using the mensuration formula from geometry for the area of a sphere.
Actually, if we didn't know this formula, it would be no problem since if the answer were an unknown, say $\xi$, all the $\xi$'s would cancel out in the final results.
If the integrand is a monomial with an odd power of $z$, then
$$\begin{align}\oint x^ay^bz^{2c+1}d^2\Omega & =\oint \sigma_{hz}\left(x^ay^bz^{2c+1}\right)d^2\Omega=(-1)^{2c+1}\oint x^ay^bz^{2c+1}d^2\Omega \\
& =\frac{1+(-1)^{2c+1}}2\oint x^ay^bz^{2c+1}d^2\Omega=0\end{align}$$
Where $\sigma_{hz}$ is a reflection though the $xy$-plane. Clearly any proper rotation of the integrand about an axis through the origin doesn't affect the value of the integral because it just corresponds to viewing the same function from a different angle.
Also inversion of the integrand through the origin doesn't change the integral and any improper rotation including reflection can be written as inversion times a proper rotation. Thus it can be seen that if
$$\oint x^ay^bz^cd^2\Omega \ne 0$$
Then $a$, $b$, and $c$ must all be even. Also we can rotate coordinates so that we only need compute
$$\oint x^{2a}y^{2b}z^{2c}d^2\Omega$$
for $a \ge b \ge c$.
We can calculate
$$\oint r^2d^2\Omega=\oint(x^2+y^2+z^2)d^2\Omega=3\oint x^2d^2\Omega=4\pi$$
where we have rotated individual terms and then used $r^2=1$ on the surface of the unit sphere. Now,
$$\oint x^2d^2\Omega=\frac{4\pi}3=\oint x^2r^2d^2\Omega=\oint \left(x^4+x^2y^2+x^2z^2\right)d^2\Omega=\oint x^4d^2\Omega+2\oint x^2y^2d^2\Omega$$
Rotating by $\frac{\pi}4$ about the $z$-axis,
$$\begin{align}\oint x^4d^2\Omega & =\oint \left(\frac{x+y}{\sqrt2}\right)^4d^2\Omega=\frac14\oint\left(x^4+4x^3y+6x^2y^2+4xy^3+y^4\right)d^2\Omega \\
& =\frac12\oint x^4d^2\Omega+\frac32\oint x^2y^2d^2\Omega\end{align}$$
This gives us two eqautions in two unknowns and we can solve using Excel or Matlab or whatever to get
$$\oint x^4d^2\Omega=\frac{4\pi}5\text{ and }\oint x^2y^2d^2\Omega=\frac{4\pi}{15}$$
Again we have
$$\oint x^2d^2\Omega=\frac{4\pi}3=\oint x^2r^4d^2\Omega=\oint x^6d^2\Omega+6\oint x^4y^2d^2\Omega+2\oint x^2y^2z^2d^2\Omega$$
$$\oint x^4d^2\Omega=\frac{4\pi}5=\oint x^4r^2d^2\Omega=\oint x^6d^2\Omega+2\oint x^4y^2d^2\Omega$$
$$\oint x^6d^2\Omega=\oint \left(\frac{x+y}{\sqrt2}\right)^6d^2\Omega=\frac14\oint x^6d^2\Omega+\frac{15}4\oint x^4y^2d^2\Omega$$
Solving the last $3$ equations in $3$ unknowns we get
$$\oint x^6d^2\Omega=\frac{4\pi}7\text{, }\oint x^4y^2d^2\Omega=\frac{4\pi}{35}\text{, and }\oint x^2y^2z^2d^2\Omega=\frac{4\pi}{105}$$
Almost there :)
$$\begin{align}\oint x^2d^2\Omega & =\frac{4\pi}3=\oint x^2r^6d^2\Omega \\
& =\oint x^8d^2\Omega+8\oint x^6y^2d^2\Omega+6\oint x^4y^4d^2\Omega+12\oint x^4y^2z^2d^2\Omega\end{align}$$
$$\begin{align}\oint x^4d^2\Omega & =\frac{4\pi}5=\oint x^4r^4d^2\Omega \\
& =\oint x^8d^2\Omega+4\oint x^6y^2d^2\Omega+2\oint x^4y^4d^2\Omega+2\oint x^4y^2z^2d^2\Omega\end{align}$$
$$\oint x^6d^2\Omega=\frac{4\pi}7=\oint x^6r^2d^2\Omega=\oint x^8d^2\Omega+2\oint x^6y^2d^2\Omega$$
$$\oint x^8d^2\Omega=\oint \left(\frac{x+y}{\sqrt2}\right)^8d^2\Omega=\frac18\oint x^8d^2\Omega+\frac72\oint x^6y^2d^2\Omega+\frac{35}8\oint x^4y^4d^2\Omega$$
Solving these $4$ equations in $4$ unknowns,
$$\oint x^8d^2\Omega=\frac{4\pi}{9}\text{, }\oint x^6y^2d^2\Omega=\frac{4\pi}{63}\text{, }\oint x^4y^4d^2\Omega=\frac{4\pi}{105}\text{, and }\oint x^4y^2z^2d^2\Omega=\frac{4\pi}{315}$$
With these integrals in hand we want to normalize the polynomials $xy$ and $x^2-y^2$.
$$\oint (xy)^2d^2\Omega=\oint x^2y^2d^2\Omega=\frac{4\pi}{15}$$
So a normalized spherical harmonic that transforms as the $xy$ partner of the $T_{2g}$ representation of $O_h$ and as the $z$ partner the $T_2$ representation of $T_d$ is
$$Y_{xy}^2(x,y,z)=\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}$$
And we have
$$\psi_{xy}(x,y,z)=Y_{xy}^2(x,y,z)f_{xy}(r)$$
To find the electrostatic potential energy of an electron with such a wave function in the $O_h$ potential we integrate over its charge density to get
$$\begin{align}U_{O_h,xy} & =\int V_{O_h}(x,y,z)\rho_{xy}(x,y,z)d^3volume \\
& =\int\left[-\frac{6Ze}{4\pi\epsilon_0a}-D\left(x^4+y^4+z^4-\frac35r^4\right)\right]\times \\
& (-e)\left[\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}f_{xy}(r)\right]^*\left[\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}f_{xy}(r)\right]d^3volume \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}\oint\left|\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}\right|^2d^2\Omega\int_0^{\infty}\left|f_{xy}(r)\right|^2r^2dr \\
& +De\left(\frac{15}{4\pi}\right)\oint \left(\frac{xy}{r^2}\right)^2\frac{x^4+y^4+z^4-\frac35r^4}{r^4}\int_0^{\infty}r^4\left|f_{xy}(r)\right|^2r^2dr \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}(1)(1)+\frac{15De}{4\pi}\left(\frac{4\pi}{63}+\frac{4\pi}{63}+\frac{4\pi}{315}-\frac35\frac{4\pi}{15}\right)\left\langle r^4\right\rangle \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\end{align}$$
Where we have used the separate normalizations of the angular part of the wave function $Y_{xy}^2(x,y,z)$ and the radial part $f_{xy}(r)$ and the average value of $r^4$ for the radial distribution, $\left\langle r^4\right\rangle$.
Now we calculate
$$\oint \left(x^2-y^2\right)^2d^2\Omega=2\oint x^4d^2\Omega-2\oint x^2y^2d^2\Omega=2\left(\frac{4\pi}{5}\right)-2\left(\frac{4\pi}{15}\right)=\frac{16\pi}{15}$$
So a normalized spherical harmonic that transforms as the $x^2-y^2$ partner of the $E_g$ representation of $O_h$ and the $E$ representation of $T_d$ is
$$Y_{x^2-y^2}^2(x,y,z)=\left(\frac{15}{16\pi}\right)^{1/2}\left(x^2-y^2\right)$$
We could also have obtained this normalization by rotating $Y_{xy}^2(x,y,z)$ by $-90°$ about the $z$-axis.
Now we can get the electrostatic potential energy of an electron with wave function
$$\psi_{x^2-y^2}(x,y,z)=Y_{x^2-y^2}^2(x,y,z)f_{x^2-y^2}(r)$$
in the $O_h$ potential as
$$\begin{align}U_{O_h,x^2-y^2} & =\int V_{O_h}(x,y,z)\rho_{x^2-y^2}(x,y,z)d^3volume \\
& =\int\left[-\frac{6Ze}{4\pi\epsilon_0a}-D\left(x^4+y^4+z^4-\frac35r^4\right)\right]\times \\
& (-e)\left[\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}f_{x^2-y^2}(r)\right]^*\left[\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}f_{x^2-y^2}(r)\right]d^3volume \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}\oint\left|\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}\right|^2d^2\Omega\int_0^{\infty}\left|f_{x^2-y^2}(r)\right|^2r^2dr \\
& +De\left(\frac{15}{16\pi}\right)\oint \left(\frac{x^2-y^2}{r^2}\right)^2\frac{x^4+y^4+z^4-\frac35r^4}{r^4}\int_0^{\infty}r^4\left|f_{x^2-y^2}(r)\right|^2r^2dr \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}(1)(1)+\frac{15De}{16\pi}\left((2)\frac{4\pi}{9}+(4)\frac{4\pi}{105}-(4)\frac{4\pi}{63}-(2)\frac{4\pi}{315}-\frac65\frac{4\pi}{5}+\frac65\frac{4\pi}{15}\right)\left\langle r^4\right\rangle \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\end{align}$$
Where we have assumed the average value of $r^4$ is the same as for $\psi_{xy}(x,y,z)$ because they both started out as $d$ orbitals and we haven't taken into account higher-order effects.
Since there are $3$ $T_{2g}$ orbitals and $2$ $E_g$ orbitals, the average potential energy over all orbitals is
$$U_{O_h,ave}=\frac15\left[3\left(\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\right)+2\left(\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\right)\right]=\frac{6Ze^2}{4\pi\epsilon_0a}$$
The splitting in potential energies of the orbitals is
$$\begin{align}U_{O_h,x^2-y^2}-U_{O_h,xy} & =\left(\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\right)-\left(\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\right) \\
& =\frac{4De}{21}\left\langle r^4\right\rangle\equiv10Dq\end{align}$$
Then we get
$$q=\frac{2e}{105}\left\langle r^4\right\rangle$$
and so
$$U_{O_h,x^2-y^2}=\left(\frac{6Ze^2}{4\pi\epsilon_0a}\right)+6Dq$$and$$U_{O_h,xy}=\left(\frac{6Ze^2}{4\pi\epsilon_0a}\right)-4Dq$$
This derivation was a little bloody because I chose to do everything with only algebra.
There is a much more elegant exposition possible that requires higher level math.
Is it worth adding to the end of this answer or posting as another answer?


Why do the splittings in mirrored COSY crosspeaks differ? I have a COSY spectrum of a mixture of butanal and butyl bromide and I don't understand why the cross-peaks have various/different splitting on the other sides of the diagonal. I think it may have something to do with polarisation transfer but I'm not sure.

Let's focus for now on the two circled crosspeaks. In a COSY sequence, the frequency in the indirect dimension (the vertical axis) corresponds to the spin which the magnetisation is on during the $t_1$ period. The frequency in the direct dimension (horizontal axis) corresponds to the spin which you measure the magnetisation of during $t_2$.

It turns out that these crosspeaks are from butanal, so let's also colour-code the protons in butanal as follows: we have $\ce{C\color{red}{H}_3-C\color{blue}{H}_2-C\color{green}{H}_2-CHO}$.

The red crosspeak occurs at ~1.65 ppm in the indirect dimension, and ~0.95 ppm in the direct dimension. This means that the magnetisation during $t_1$ was on the blue $\ce{C\color{blue}{H}_2}$ proton, and was subsequently transferred to the red
$\ce{C\color{red}{H}_3}$ proton during acquisition (in $t_2$). The coupling between these two sets of protons is referred to as an active coupling; it is the coupling that is responsible for generating the crosspeak.

Conversely, the the blue crosspeak arises due to transfer from $\ce{C\color{red}{H}_3}$ (during $t_1$) to $\ce{C\color{blue}{H}_2}$ (during $t_2$). So, it also has an active coupling between the same nuclei. The difference here is that the final $\ce{C\color{blue}{H}_2}$ protons are also coupled to the green $\ce{C\color{green}{H}_2}$ protons! This is what is called a passive coupling: one that does not generate the crosspeak, but is still reflected in the multiplet structure as a splitting.

The differences in the crosspeak structure arise because of differences in passive couplings to other nuclei.

Exactly how this is manifested is not easy to explain in this case, because there are multiple equivalent nuclei. But if you are keen, read on. (I'm also not going to present any quantum mechanical analysis, because that would really take way too long.)

In fact, the red crosspeak also has one passive coupling – the $\ce{C\color{red}{H}_3}$ group is coupled to not just one, but two protons adjacent to it, and the crosspeak resembles a case where one of the two protons is an active coupling partner and the other is passive.

We first deal with passive couplings. These lead to in-phase splittings, i.e. peaks which are both positive and are separated by the coupling constant, approximately 7 Hz in our case. For simplicity's sake we will stick to just the direct dimension for now. Because we only have one effective passive coupling, this leads to a doublet with a peak separation of ~7 Hz.

(The indirect dimension, broadly speaking, has similar behaviour in that it is also affected by passive and active couplings, but the exact source of these couplings can differ. Also, it is sometimes harder to observe multiplet structure in the indirect dimension because of different resolutions in the two dimensions.)

Active couplings, on the other hand, lead to antiphase splitting, where peaks have opposite signs. So, our doublet is further split into two doublets: one positive and one negative, and these two doublets themselves are 7 Hz apart. We find that the middle peak exactly cancels itself out, and we are left with one positive and one negative peak separated by exactly twice the coupling constant. This is what is observed in the red crosspeak.

On the other hand, the blue $\ce{C\color{red}{H}_2}$ protons have one active coupling (to one of the $\ce{C\color{red}{H}_3}$ protons), but the other two $\ce{C\color{red}{H}_3}$ protons are effectively passive couplings. On top of that, it has two further passive couplings to the $\ce{C\color{green}{H}_2}$ protons.

The four passive couplings lead to the formation of a $1:4:6:4:1$ quintet (we're going to assume that all the couplings have similar size, which is quite reasonable). The single active coupling causes an antiphase splitting of this quintet, which is easier to express mathematically:

1   4   6   4   1
   -1  -4  -6  -4  -1
-----------------------
1   3   2  -2  -3  -1

(imagine that there is 7 Hz separation between every column of numbers). This $1:3:2:-2:-3:-1$ pattern exactly matches how the blue crosspeak looks like.

For more information, see e.g. Claridge, High-Resolution NMR Techniques in Organic Chemistry (3rd ed.), Section 6.1.5.2.


Why is law of mass action only applicable for elementary reaction? Why can't we apply law of mass action for non-elementary reaction? Why does is vary with the rate law in case of non-elementary reactions?

The Law of Mass action states:

The rate of a reaction is directly proportional to the concentration of reactants raised to powers equal to their stoichometric coefficients

This is only valid for elementary reactions since they happen in a single step. As a result, such simple relations hold good.

But what about the more complicated ones. Certain reactions, like the oxidation of ethylbenzene to benzoic acid using hot $\ce{KMnO4}$ don't happen in a single magical step. They involve lots of reactive intermediates and transition states, so it's basically a bunch of several elementary reactions. You can't consider the whole reaction as an elementary reaction.

But you can formulate a rate equation for non elementary reactions by considering them as a bunch of elementary ones. However such formulations involve knowing the mechanism of the reaction in question, along with other data.


Autoionization of water Consider as solution of pure water at $\pu{25 ^\circ{}C}$ with a $K_{\mathrm{w}} =  10^{-14}$.
If we are to add an acid to the mix we would observe an increase in hydronium and an equal decrease in hydroxide content.
Yet, I fail to see how the hydroxide ion concentration can decrease with an increase in hydronium concentration.
Lets add $\ce{HCl}$ to our solution: $$\ce{HCl + H2O -> Cl- + H3O+}$$
Concentration $[\ce{H3O+}]$ increases,  concentration $[\ce{OH-}]$ decreases in order to fulfill the $K_{\mathrm{w}}$ constant of $10^{-14}$.
Through what reaction exactly then will the hydroxide content decrease (by turning into water I assume)? Where does $\ce{OH-}$ gets an $\ce{H+}$ from? Not from $\ce{H3O+}$, I assume seeing that every molecule of a strong acid is said to turn into an hydronium ion 1:1.
Please enlighten me.

Maybe a numerical example will help you. Let's start from pure water, with the following concentrations :$\ce{[H+] = [OH-] = 10^{-7} M}$. Now we will suppose you add $\ce{10^{-7}} mol$ $\ce{HCl}$ in one liter of this water. Suddenly, the concentration of $\ce{H+}$ should double. This is not possible in the long run. A part $a$ of these supplementary $\ce{H+}$ ions must react with the same amount $a$ of the $\ce{OH-}$ ions to form water. The final molar concentrations of both ions are given by  : $$\ce{[H+] = 2.000·10^{-7}} - a$$  $$\ce{[OH-] = 1.000·10^{-7} } - a$$   This amount $a$ may be obtained by solving the equation : $$\ce{ (2.000·10^{-7}} - a)\ce{(1.000 10^{-7}} - a)= 1.000·10^{-14}$$ The solution is : $a = 0.382·10^{-7} mol/L$. So about $38.2$% of the initial $\ce{OH-}$ ions have been destroyed by the addition of the new acidic ions. And the numerical values of the final concentrations are $$\ce{[H+] = 2.00·10^{-7}}- a = 1.618·10^{-7} mol/L$$$$\ce{[OH-] = 1.00 10^{-7} } - a = 0.618· 10^{-7} mol/L$$ And of course : $\ce{[Cl-] = 1.000·10^{-7}$ $ mol/L}$. If you want you may check the product of the two final concentrations $\ce{[H+]}$ and $\ce{[OH-]}$. It is equal to ${1.000·10^{-14} mol^2L^{-2}}$.


Is it correct to say that dehydration isn't a subset of condensation? To my understanding, dehydration is defined as a subset of a condensation reaction. A condensation defined as being two molecules joining together (in the case of loss of water, this is referred to as a dehydration).

I also thought however, if a molecule of water leaves without a condensation necessarily occurring, that this is also a dehydration. In this case, how could the dehydration be a subset of condensation?

I'm not familiar with any such molecule off the top of my head, but some easily protonated alcohol that is resonance stabilized when water leaves? Would this leaving group of water not be a dehydration, without a condensation, since no addition would occur thereafter?

Alkanols are indeed a good example. Think in the E2 and E1 reactions of primary and tertiary alkanols  in the presence of sulfuric acid at elevated temperatures, yielding alkenes.


Frost diagram and acidic/basic conditions. The Frost diagram for sulfur shows the relative stability (in terms of cell potentials) of the possible aqueous oxidation states. What is the reason for the instability of higher oxidation states in the acidic regime and their stability in the basic regime ? Something I have always noticed but never been able to apply chemical intuition to.

Write out the half-reactions and you will see that oxidation of S almost always creates lots of protons.  So you can think of Le Chatelier's principle: if there are already lots of protons around, it will be harder to create more of them.  Here's the example for complete oxidation of sulfide to sulfate at pH ~7:

$$\ce{HS- + 4 H2O -> 8e- + 9H+ + SO4^{-2} }$$

Conversely, under alkaline conditions, the equilibrium favors the products even more, because of the "removal" of the protons by the alkaline conditions.


Which hydrogen does the nucleophile attack in E2 reaction? This reaction will be both an E2 and SN2, where the E2 will be major and the SN2 minor (If I have done it correctly). I was drawing out the mechanism as I noticed that I was making my nucleophile attack the wrong hydrogen, the one attached to the alpha-carbon. I know that the nucleophile is supposed to attack the hydrogen on the beta-carbons however how do I determine if the $MeO^-$ attacks the hydrogen on the second carbon to the left or to the right of the alpha-carbon? Or does it attack both and since the sturcture of the compound formed is the same either way, it doesn´t matter which hydrogen it removes?

Well done to narrow the options down to E2 and SN2 and recognize that E2 is likely favored over SN2.

As the comments suggest, in this case, it doesn't matter which one is deprotonated. The reason is that deprotonation at either beta-position gives the same product. If you are unsure, draw out both options.

In future similar examples, you may come across reactants with multiple possible reactive positions that are not equivalent. For example, consider 3-chlorohexane (instead of 3-chloropentane, as in the original problem). In this case, again there are two possible sites to be deprotonated (Ha and Hb). Looking at the two possible products, both are disubstituted alkenes, so they will be close in stability. There isn't much difference in the groups nearby Ha and Hb, so the kinetics of the deprotonation are probably similar. So which is the "correct" answer? In reality, both are correct. They may not be formed in exactly equal amounts, but they will probably both be formed in substantial amounts.

There are a couple of key lessons here for those studying organic chemistry:

1) Draw out all of the possible products that come from mechanisms that you think make sense. This might be lots of drawing! Nevertheless, it is much easier to realize the differences if everything is drawn out in front of you.

2) Organic chemistry is messy. Often times multiple mechanistic pathways will be viable and multiple products will form. Much of the creativity in organic chemistry is finding ways to ensure that only one mechanism can occur or one (desired) compound is formed as the major product.


What considerations are made in determining whether to use a sand bath or oil bath? As suggested by the title, I am curious as to what would make, for instance, a sand bath better suited to a specific heating application/context as opposed to an oil bath (and vice versa). Equivalently, I suppose I am asking what the "pro and cons" of both of these methods would be.

The following is the comparison list from the top of my head:

Oil for the oil bath is often available commercially, every batch has the same properties and is arguably easier to store and transport; sand bath requires sifted and refined sand (quartz sand, ideally). It all depends on what's easier to find a good supplier and storage for.
Oil bath heats up relatively quickly, especially if the oils isn't very viscous; sand bath heats up very slowly due to slow heat transfer in powdered/granulated material.
As a consequence of the former statement, oil bath is uniformly heated, whereas sand bath always has a temperature gradient with the maximum temperature at the bottom and minimum on top layer.
This requires that the tip of the thermal sensor is located at the same depth as chemicals in the reactor.
Oil bath can usually only be heated up to approx. 250 °C max depending on the oil used; sand bath can be heated up to 400 °C (provided a ceramic pod is used).
When operated at high temperatures, mineral oil in the oil bath may start to decompose producing volatile health hazardous organics; sand bath, if the sand was prepared correctly, doesn't deteriorate even at high temperatures.
Silicone oil is much safer in this regard, but is also considerably more expensive than mineral oil.
In case of temperature control fail, oil bath may even ignite (again, depends on oil used); whereas sand bath will just overheat – however, the aftermath may be comparable.
Oil bath can be used with rotavap; sand bath can't.


What solute should I use with dialysis tubing if I want a 100% retention rate? I purchased dialysis tubing from the Carolina company.

The molecular cut off is 12,000-14,000 Daltons and I've been using table sugar for my experiments with it to raise columns of water. While the sugar does work it slowly leaks out of the membrane over time if it's left to sit for an hour or so. What should I use if I want near 100% of the solute to stay within the dialysis tubing? Preferably something cheap and non-toxic that will also dissolve in water. Or maybe a different type of membrane I could purchase that would retain the sugar better?

Polyethylene oxide, AKA polyethylene glycol or PEG has MW > 100,000. You can find higher-MW PEG in disposable diapers and similar absorbent products.

You could also use agar, pectin or gelatin, available in groceries and from lab supply companies.

One of my favorite demos was to make an agar solution with a bit of phenolphthalein added and put it in sausage casing (AKA dialysis tubing). Holding the hotdog over an open bottle of ammonium hydroxide turned it red, and over hydrochloric acid, back to a clear frank.


How do 1s and 2p orbitals overlap? In the following figure

we can see that the p-orbitals overlap 1s orbital (though relatively very little). How can an electron in p-orbital, be simultaneously in the 1s orbital at any given point in space and time?

The question is interesting, but slightly ill-defined.
If you take s and p orbitals as a solution of Schrodinger equation of hydrogen-like atom, then they are eigenfunctions of Hamiltonian, therefore are orthogonal. That means - no overlap. Do not be misled by the figure, if you do the math properly, $\langle1s|2p\rangle = 0$.

If you speak of just one electron, the solution is reasonably simple as stated in the article and the electron does not simultaneously reside in both orbitals. Even if you would think of interaction with excited states, it would not help, as single excitations do not interact with ground state. You can of course excite it, but than it is not "simultaneous".

In many-electron system, the picture gets more complicated, but this was not a question.


Thermochemistry problems? Question 1:

A compound A reacts according to the following hypothetical equation and has a molecular weight of 48.36 g/mol.

3 A (s) +B (aq) ––> 2 C (aq) ∆H° = ?

A sample of A, weighing 0.152 g reacts in a flask containing 250.00 g of water and the water temperature increases from 24.85 °C to a temperature of 26.26 °C. Calculate ∆H° for the reaction as written in the equation.

Answer: -1408 kJ

Question 2:

For which of the following reactions is ∆H° = ∆H°f, the heat of formation?

i. C (s) + 2 F2(g) ––> CF4 (g) ∆H° = – 221.0 kJ
ii. H(g) + Br (g) ––> HBr (g) ∆H° = –366.2 kJ
iii. 2 C(s) + H2(g) + 3 Cl2(g) ––> 2 CHCl3 (g) ∆H° = –268.2

Answer: i only

Also, could someone explain exactly what delta H means in terms of Thermochemistry. I would like to have a better understanding of it.

Thank you very much!  :)

For the first question I attempted to use the q=m x s x (delta) T for the water and I attempted to use stochiometry for substance A but I always receive the wrong answer. Could someone help me on where to start with these types of problems?

Water calculation:

(250) x (4.184) x (26.26-24.85)

= 1474.86 kJ

H is enthalpy.

By definition H = U + PV (where U is internal energy, P is pressure and V is volume).

For question 1, your work says:

  (250) x (4.184) x (26.26-24.85)

  = 1474.86 kJ

Because you failed to show units in your work, you are off by a factor of 1000.

But the main reason your answer is different from the desired answer is you have calculated the amount of heat released by 0.152 g reacting, whereas the desired answer comes from calculating the heat released by 3 moles of A reacting, considering change in enthalpy of the system to be negative when heat is released to the surroundings.

For question 2:

Answer ii is wrong because the monoatomic forms of hydrogen and bromine are not the standard states of the elements.  Answer iii is wrong because the equation is written for 2 moles being formed rather than 1 mole.


Which molecule is more acidic: CHF3 or CHBr3? I know that fluorine is more electronegative than bromine. However, because of the size of bromine, it is more stable with a negative charge. In the case of $\ce{HF}$ vs. $\ce{HBr}$, to me, $\ce{HBr}$ is without a doubt the stronger acid. When it comes to comparing $\ce{CHF3}$ with $\ce{CHBr3}$, we are supposed to compare the stability of their conjugate bases.

In $\ce{CF3^-}$, the fluorine can hold the electronegative charge quite well. It also exerts a strong inductive effect.

In $\ce{CBr3^-}$, the bromine is less electronegative than the fluorine but is more polarizable, so I think it can hold the negative charge more efficiently.

With this understanding, I am concluding that $\ce{CBr3^-}$ is more stable which would mean $\ce{CHBr3}$ is the stronger acid. However, I've been told otherwise.

I'm aware that the $\ce{C-Br}$ bond length is longer than the $\ce{C-F}$ bond length. I'm wondering if that short bond length causes $\ce{CF3^-}$ to be more stable than $\ce{CBr3^-}$. If this is true, is it okay for me to think about $\ce{CF3^-}$ dissociating vs. $\ce{CBr3^-}$ dissociating? In that scenario, I believe $\ce{CBr3^-}$ dissociates more easily than $\ce{CF3^-}$, meaning $\ce{CF3-}$ is more stable.

Note: This logic wouldn't apply to the instance of $\ce{HO-}$ vs. $\ce{HS-}$ because $\ce{HS-}$ is more stable than $\ce{HO-}$ (even though $\ce{H-O}$ bond length is shorter than $\ce{H-S}$ bond length).

The following table contains some relevant data, the $\mathrm{p}K_\mathrm{a}$'s of the various haloforms along with the Pauling electronegativity of the corresponding halogen.

\begin{array}{|c|c|c|} \hline
\text{Haloform} & \mathrm{p}K_\mathrm{a} & \text{Electronegativity}\\ \hline
\ce{CHF3} & 25\mathrm{-}28 & 3.98\\ \hline
\ce{CHCl3} & 15.7 & 3.16\\ \hline
\ce{CHBr3} & 13.7 & 2.96\\ \hline
\ce{CHI3} & \sim 25 & 2.66\\ \hline
\end{array}

Looking at the $\mathrm{p}K_\mathrm{a}$'s, the first thing we notice is that the two haloforms at the beginning and end of the series are weaker acids than the two haloforms in the middle of the series. This behavior is often seen when there are at least two factors affecting acidity and they are in opposition to each other.

Let's consider inductive, steric and resonance effects and examine their expected effect upon haloform acidity.

Inductive effects should correlate with electronegativity.  Based only on inductive effects we would expect fluoroform to ionize and produce the most inductively stabilized cabanion ($\ce{CF3^{-}}$), while iodoform would ionize to produce the least inductively stabilized carbanion ($\ce{CI3^{-}}$).
Steric effects are often discussed in terms of bulky groups coming too close to one another, but what we are really saying is that the electron clouds around the groups are coming too close to one another causing electron-electron repulsion. Iodoform is a very bulky molecule.  The iodine substituents are very large and there are many lone pair-lone pair repulsions.  Upon ionizing iodoform can relax from a crowded $\mathrm{sp^3}$-like geometry to a more open and spread out $\mathrm{sp^2}$-like geometry.  As the halogen gets smaller there is less of a steric \ electron repulsion driving force for haloform ionization.
Resonance effects are not a significant stabilizing factor in the haloform series. Significant low energy resonance structures cannot be drawn. Thanks to TanYongBoon for noting this.

Conclusion: Inductive effects and steric effects (steric meaning electron-electron repulsion) control acidity in the haloform series. The two effects are in opposition to one another.  The inductive effect strongly stabilizes the fluoroform anion and plays a lesser role as we move down the series.  Steric \ electron repulsion effects strongly stabilize the iodoform anion and plays a lesser role as we move up the series.  Said differently, only one effect stabilizes the fluoroform (inductive) and iododform (steric) anions, but both inductive and steric effects play a role in stabilizing the chloroform and bromoform anions making chloroform and bromoform the most acidic molecules in the series.


When is it okay to use $Q=mc\Delta T$, is this equation only for calorimetry questions? When is it okay to use $Q=mc\Delta T$, is the equation only for calorimetry questions? When a question talks about heat how do I know if i'm measuring $\Delta H$ or Q?

The equation $Q = mc \Delta T$ is appropriate for calorimetry at a constant volume. $Q$ here refers to the net heat transfer of the isolated system. Change in enthalpy, $\Delta H$, on the other hand, is really the total change in energy of the system, that is:

$$\Delta H = \Delta U + P\Delta V + V\Delta P$$

Where $P$ is pressure, $V$ is volume, and $U$ is the internal energy of the system, which is the sum of the system's net heat transfer ($Q$) and the work done on the system ($W$): $\Delta U = Q + W$. If we substitute the definition of $\Delta U$ into our equation for enthalpy, we get:

$$\Delta H = Q + W + P\Delta V + V\Delta P$$

Because any mechanical pressure-volume work performed on a closed system is equal to $-P\Delta V$, the equation further reduces to:

$$\Delta H = Q + V\Delta P$$

From this it becomes apparent that $Q$ and $\Delta H$ are only equal if pressure is constant (i.e., $\Delta P = 0$). The upshot of all this is that you should not assume $\Delta H$ and $Q$ are equal unless you're explicitly instructed to assume constant volume and pressure. When a problem refers to heat, it's almost certainly referring to heat transfer, $Q$, and not enthalpy change ($\Delta H$).