text:
What is the accuracy of medical syringes?
I am interested in both low and high volume syringes.

Note that the accuracy is also limited by your ability to manipulate and read the plunger. (Digresseion:I once spent half a day moving single grains of a cmpound on and off a balance, trying to get another digit of precision, when I should instead have taken the mass I had and adjusted the amount of solvent... which would still have been overprecise for the intended use.)

The requirements for ordinary sterile single-use syringes made of plastic materials are standardized in the international standard ISO 7886-1:1993 Sterile hypodermic syringes for single use – Part 1: Syringes for manual use.
According to this standard, the tolerance on the graduated capacity depends on the nominal capacity of the syringe $V$ and the expelled volume $V_\text{ex}$.
For syringes with a nominal capacity of $V\lt5\ \mathrm{ml}$, the tolerance amounts to

$\pm5\ \%\ \text{of}\ V_\text{ex}$ for volumes equal to or greater than half nominal capacity $V$
$\pm{\left(1.5\ \%\ \text{of}\ V +2\ \%\ \text{of}\ V_\text{ex}\right)}$ for volumes less than half nominal capacity $V$.

For syringes with a nominal capacity of $V\geqslant5\ \mathrm{ml}$, the tolerance amounts to

$\pm4\ \%\ \text{of}\ V_\text{ex}$ for volumes equal to or greater than half nominal capacity $V$
$\pm{\left(1.5\ \%\ \text{of}\ V +1\ \%\ \text{of}\ V_\text{ex}\right)}$ for volumes less than half nominal capacity $V$.

The values apply to the volume of water at a temperature of $T=\left(20\pm5\right)\ \mathrm{^\circ C}$ expelled from the syringe when the line circumscribing the end of the piston traverses a given scale interval.

The accuracy would depend on the brand and type of syringe in question.
If you need highly accurate amounts of a liquid, you'd be best to look at reusable syringes (Hamilton make some excellent ones) which are calibrated individually.
For most needs, the disposable medical syringes you mention are adequate, but not accurate for analytical purposes.
As an example, a 1 mL syringe often has markings which seem to indicate the ability to measure liquid to the nearest 0.01 mL, but in practice they are only accurate to the nearest 0.05 mL.
Generally speaking, the larger the syringe, the more 'accurate' you can be, as the volume by which the syringe is off by will be smaller as a percentage of the total volume.
You can test the accuracy (precision may also play a part) by taring an empty syringe on the balance, then syringing up a volume of liquid, before placing the filled syringe back onto the balance, as water has a density of roughly 1 g/mL, you can see for yourself how accurate the syringe is (assuming you have an analytical balance capable of reading to 3/4 decimal places).


text:
Is menthol an especially volatile molecule?
I have been studying a bit about odors and plant essential oils. Several sources claim that the fragrant terpenes found in these oils, such as menthol, eugenol, limonene, linalool, etc. are volatile compounds.
I understand that all molecules are technically volatile, but since these molecules all have boiling points of over $170~\mathrm{^\circ C}$, and their vapor pressures are not particularly high, I don't see why they are considered 'volatile compounds'.
I have two guesses:
a) There is some fundamental aspect of chemistry that I am ignoring, and these compounds do have a set of properties that makes them especially volatile. This to me sounds very interesting, and I would really like to learn what is that makes these compounds especially well-made for being vaporized.
b) The compounds are not especially volatile. They are synthesized in large amounts, concentrated to high concentrations in the plant, and our nose is very sensitive to them. They are called 'volatile' compounds because we happen to notice their smell when they are in the air.
For my question I've chosen to focus on menthol. Is there anything about menthol that makes it especially volatile?

It's especially curious to see menthol described as being 'volatile' as its crystalline in its pure state, but I think in the context of flavour/fragrance 'science' the word volatile has simply been (mis) appropriated.   Many of these terpenes have incredibly low odour-thresholds so we can smell them even at concentrations lower than parts per billion (the human nose is pretty incredible, lookup the threshold for dimethyl sulfide), so its entirely possibly that its vapour pressure is enough that even a tiny tiny amount of menthol given off is sufficient.

For a solid, it is relatively volatile. If you keep a small phial in your pocket, for example, it is sufficiently volatile that it will sublime and recrystallise in the cooler parts of the vessel.

The term "volatile oils" is a traditional term.
From the 1916 book A TREATISE ON PHARMACY FOR STUDENTS AND PHARMACISTS 5th edition at page 217:

Although the boiling points of volatile oils are considerably above that of water, the oils pass over rapidly with the vapor of boiling water, and in the leading establishments in this country [USA] and Europe volatile oils are now distilled by passing steam under pressure into stills which contain the material on a series of perforated trays...

So it means volatile to the degree that it could be steam distilled.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Mechanism of dehydration of alcohols by iodine/triphenylphosphine
This paper describes the dehydration of an alcohol to form an alkene. What is the mechanism of alcohol dehydration with iodine? Why can only iodine of all the halogens be used?

Can you post an example of the kind of system you mean? In the broadest possible sense iodine can act as a Lewis acid to help the OH leave, but it would depend on the structure of the SM as to how the whole thing proceeded.

It is generally used for dehydration of alcohols.

My point is that asking for a mechanism without any indication of the structure is unlikely to yield a correct answer, as the mechanism for any process depends upon the steric and electronic demands of the molecule.  I'd also question whether iodine is considered a 'general' method for these reactions, quoting from March's Advanced Organic Chemistry: "Many other dehydrating agents159 have been used on occasion: P2O5, I2..." .. the examples I have seen of iodine used tend to be very specific examples where other more common reagents were tried and failed

I would like to see an example with mechanism where it is used. I am unable to provide one,

I'm pretty sure that iodine is effective in this way where other halogems fail because it acts as a good leaving group for the OH as @NotNicolau said. Also the iodide molecule is larger than the other halogens presenting more surface area which makes it leave easier, I think.

Marko, if you would like to see such an example, could you re-word your question to be "What is an example of a reaction and mechanism where iodine dehydrates an alcohol?"?

Do you have an example of this type reaction?

May this question be opened now or is any further correction  needed?

The mechanism possibly resembles that of the Appel reaction (https://en.wikipedia.org/wiki/Appel_reaction). It is mentioned in the article that $\ce{I2}$ can be used instead of $\ce{CCl4}$. To arrive at an alkene as final product, the last step would have to be elimination instead of substitution by the halide.

Seems reasonable.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Using Gelatin as Rocket Fuel?
I'm doing a DIY project on if it's possible for gelatin to be used as rocket fuel. This is just standard gelatin in a standard rocket. I know that gelatin is made from animal fat and that it has lots of sugar in it. So is it possible?

No. Jello-type gelatin, made per recipe, has way too much water to burn effectively. Even without the water the dry gelatin wouldn't burn with enough thrust to be an effective rocket propellant.

But like what if you make it super dry

Like the powder version

Powdered sugar is often used in chemistry demonstrations, in the presence of oxygen it is highly combustable (there have been significant explosions and fires at sugar processing plants in the past due to this).
The sugar is high in energy, and the fine powdered variety has a greater surface area with which to interact with oxygen, producing the burst of energy needed. The sugar by itself however, is probably not enough to get any propellant effect, you'd need to add some kind of oxygen source. A quick google suggests mixtures of powdered sugar and potassium nitrate (link).
Dry gelatine (the powdered version) is essentially just sugar plus, as you noted, animal extracts or vegetarian friendly alternatives, so the same should apply, though using actual sugar is probably a lot cheaper.
Edit:  As pointed out by @MaxW in the comments below, and as eluded to in my opening comment, igniting a fuel in the presence of an oxygen source can be highly explosive. Be sure to take necessary safety precautions (work on a small scale, away from buildings. Ignite from a distance, and be sure any spectators are also out of harms way etc.) and also ensure that you're not breaking any laws in your area.

So gelatin in powder form plus potassium nitrate can yield thrust?

I don't specifically know about gelatine, but based on your question, this is what your project aims to find out. If your gelatine contains enough sugar, and you can get enough oxygen around it when it ignites, then I see no reason why you shouldn't get some generation of thrust from the combustion of the sugar present.

There is a fine line between a bomb and rocket booster. So willy nilly mixing of sugar and nitrates to make a rocket booster doesn't seem like a good idea.

Ok...sure...well I guess that's what an experiment is, amirite?


text:
Comparing acid strength of oxoacids derived from oxides
Acidic strength has been compared before here and here, but the first question is about organic chemistry (and I am talking about inorganic chemistry), and the second does not discuss molecules such as $\ce{Cl2O7}$, $\ce{P4O10}$, $\ce{CO2}$, $\ce{N2O5}$, $\ce{Al2O3}$ etc. These molecules do not contain hydrogen. So what are the factors which affect their acid or base strength?
The following question brings me this doubt:

Which of the following is the correct order of acidic strength?
(A) $\ce{Cl2O7}$ > $\ce{SO2}$ > $\ce{P4O10}$
(B) $\ce{CO2}$ > $\ce{N2O5}$ > $\ce{SO3}$
(C) $\ce{Na2O}$ > $\ce{MgO}$ > $\ce{Al2O3}$
(D) $\ce{K2O}$ > $\ce{CaO}$ > $\ce{MgO}$

Orthocresol is correct, $\mathrm{p}K_\mathrm{a}$ is only used for Bronsted acids. But you can use a $\mathrm{p}K_\mathrm{a}$ argument if you assume these compounds are allowed to react with water. Lets start with the covalent oxides in (A). Allowing these compounds to react with $\ce{H2O}$ produces $\ce{HClO4}$, $\ce{H2SO3}$, and $\ce{H3PO4}$. When comparing the $\mathrm{p}K_\mathrm{a}$s, remember lower numbers imply greater acidity. A qualitative explanation could be based on electronegativity (which is affected by oxidation state), and resonance stabilization of the conjugate base. The more resonance structures the conjugate base has, the more stable it is. A more stable conjugate base implies greater acidity. The same logic can be used to compounds in (B).
Mixing metal oxides with water produce metal hydroxides and aqua acids (depending on the pH of the water). The acidity of the $\ce{OH}$ fragment is dependent on the metal's ability to withdraw electron density. Typically, complexes with more electronegative metals are more acidic.

The question quoted in the OP is designed to test a periodic table trend concept.
Particularly, within a given row of the periodic table, excluding noble gases, the element oxides form increasingly acidic aqueous solutions from left to right across the row.
See the lecture by Dr. Pinkington The Acid-Base Character of Oxides and
Hydroxides in Aqueous Solution for additional information.
Only answer (A) is increasingly acidic left to right across the periodic table.

in option B and D, all the elements are not in same period...however your answer is correct

@manshu two in the same row, but in the wrong order, is enough to exclude a given answer.  For answer B, carbon is to the left of nitrogen.  B excluded.  For D, potassium is to the left of calcium. D excluded.

This is the correct answer and it can be backed up easily by quoting pKa values. At first I thought the question might have been something interesting, but sadly it's just an extremely poorly phrased question. As you said, it's a question which is actually asking about the Bronsted acidity of the oxoacids derived from the oxides given, and not about the Lewis acidity of the oxides themselves. But it gives no hint whatsoever that that is what it's asking about! Sheesh.

Think about the products of each oxides reaction with water. The more electronegative elements will have a stronger attraction to the hydrogen atom's electron, making it easier to ionize. This means that the compounds with a more electronegative central atom will have stronger acidic character.

(A) $\ce{Cl2O7} (3.16) > \ce{SO2}(2.58) > \ce{P4O10}(2.19)$
(B) $\ce{CO2}(2.55) > \ce{N2O5}(3.04) > \ce{SO3}(2.58)$
(C)$\ce{Na2O}(0.93) > \ce{MgO}(1.31) > \ce{Al2O3}(1.61)$
(D) $\ce{K2O}(0.82) > \ce{CaO}(1.0) > \ce{MgO}(1.31)$

I've included the electronegativities of the central atom of each compound. The only answer that shows the compounds with decreasing central atom electronegativity is A, so it is therefore the answer. This also explains why C and D are incorrect.
Also I should note that the amount of oxygens on an oxyacid does affect its acidity. As oxygen is a very electronegative atom, the presence of additional oxygens will increase the acidity of the hydrogens in oxyacids. As a general rule though, the strongest weak oxyacid of an electronegative atom will be stronger than the strongest weak oxyacid of a less electronegative atom of the same period. This explains why B is incorrect (no oxyacid of carbon will be more acidic than an oxyacid of the more electronegative nitrogen).

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why is the Gibbs free energy for phase changes zero at constant temperature and pressure?
I get that $\Delta G$ measures the spontaneity/capacity of a system to do non-mechanical work, and that if:
$\Delta G > 0$, the reaction is not spontaneous
$\Delta G < 0$, the reaction is spontaneous
$\Delta G = 0$, the reaction is at equilibrium
So why is Gibbs free energy zero for phase changes at constant temperature and pressure?
If it is negative or positive why we say that freezing water releases heat to the surrounding and it is equal to $\Delta H$? If at equilibrium  $\Delta H$ is used to reform liquid how we say it is released to the surrounding. I know I miss something, anyone can tell me?

In the inequalities you wrote for chemical reaction, the $\Delta G$ represents a change in free energy from an initial thermodynamic equilibrium state of a system to its final thermodynamic equilibrium state.  Please precisely define the "system" they refer to and the initial and final equilibrium states (so that we can be sure you understand exactly what they are saying).  Also, are you aware that these relations are just a rule of thumb for beginners?  After we address these reaction issues, we can look at phase change.

I think the system here is the liquid/solid mixture. What I am not understanding in the phase change ok we have $\Delta H$ but this heat is going where at equilibrium? to the mixture or to the surrounding? Even when we dont have equilibrium $\Delta G = \Delta H - T \Delta S$ Here T$\Delta S$ represents $\Delta H$ at equilibrium right?

The initial and final thermodynamic equilibrium states of your system are as follows:
State 1:  $\pu{1kg}$ liquid water at $\pu{0 ^\circ C}$ and $\pu{1 atm}$
State 2:  $\pu{1 kg}$ water ice at $\pu{0 ^\circ C}$ and $\pu{1 atm}$
You want to find the change in enthalpy, entropy, and Gibbs free energy between these two states.
Now, we know that, to get the change in entropy between two thermodynamic equilibrium states of a closed system, we need to identify a reversible path between these states and then calculate the integral of $\mathrm dq/T$ for that path.  For the present situation, this can be accomplished by putting the system into contact with a constant temperature reservoir at a temperature only slightly lower than $\pu{0 ^\circ C}$.  As the liquid water freezes, its molecules lock into place (losing potential energy) and this results in a release of heat to the reservoir.  So the temperature of the system never deviates significantly from $\pu{0 ^\circ C}$. Since the process is at constant pressure, the change in enthalpy is equal to the heat transferred from the surroundings to the system:$$\Delta H = q$$where both $q$ and $\Delta H$ are negative for this process.  Since the process is also at constant temperature, the change in entropy, which is given by the integral of $\mathrm dq/T$, is:$$\Delta S=\frac{q}{T}=\frac{\Delta H}{T}$$Thus, for this change from State 1 to State 2, $\Delta S$ is also negative.  If we now use the change in enthalpy and the change in entropy to calculate the change in free energy between the two thermodynamic equilibrium states, we obtain:
$$\Delta G = \Delta H-T\Delta S=\Delta H-T\frac{\Delta H}{T}=0$$

Great, thanks. You said that the temperature of the reservoir is slightly lower than t= 0c thus it is a reversible process right?. But some of the sources on the internet says that $\Delta G = 0$ because we have an equilibrium between the 2 phases. In this case, how the liquid is reforming?  Is it due to &\Delta H& of freezing? If yes, so the system is unchanged, how there's a change in the entropy of the system?

We showed that $\Delta G=0$ whenever we have phase equilibrium.  The liquid is not reforming.  We are removing heat until all the liquid has changed to ice.  Look back at the initial and final states that we identified.  Whenever you are trying to determine changes in thermodynamic functions, it is important to precisely identify the initial and final states that you are considering.  The system is not unchanged.  Only the free energy is unchanged.  The entropy and enthalpy both change between the initial and final equilibrium states.

If ΔG = 0, why does melting take place at all?

Why would it not.  delta G = 0 just means that the two phases are in equilibrium.

Could you explain what changes when one considers @DavePhD's answer ?

So why is Gibbs free energy zero for phase changes at constant temperature and pressure?

It isn't necessarily.
For example if you supercool liquid water to -10 degrees C, let the water freeze and do this in a way that the final state is solid water at -10 degrees C, the change in Gibbs free energy will be negative. This is a spontaneous process.
See W. R. Salzman's Gibbs Free Energy and Temperature: The Gibbs-Helmholtz Equation for further explanation.

Here is a simpler answer.  $\Delta G$ is zero for a phase change at the normal phase change temperature.  For instance, for the melting of ice,

$\Delta G<0$ if at $T>\pu{0 ^\circ C}$ (spontaneous)
$\Delta G=0$ if at $T=\pu{0^\circ C}$
$\Delta G>0$ if at $T<\pu{0^\circ C}$ (not spontaneous)

At $\pu{0^\circ C}$ there is no preference for $\ce{H2O}$ being ice vs. water, so $\Delta G=0$.
For melting ice, $\Delta H=\pu{+6 kJ/mol}$ and $\Delta S=\pu{+0.022 kJ/(mol K)}$.  Using $\Delta G=\Delta H-T\Delta S$, you can plug in temperatures in Kelvin (eg. $263, 273, \pu{283 K}$) and see the $\Delta G$ variations (eg. $+0.2, 0.0, \pu{-0.2 kJ/mol}$).
Freezing always releases heat ($\Delta H = \pu{-6kJ/mol}$ for freezing water); it just sometimes isn't spontaneous.  If spontaneous, this lost $\pu{6 kJ/mol}$ of heat is being sucked out of the system by the sub-zero surroundings.


text:
Does the addition of chlorine to carbon-carbon double bonds proceed through a halonium ion?
I need some serious help with the mechanism of ionic addition, both my textbook and my teacher are at odds with internet videos (khan academy) which both seem to be at odds with quote, "recent evidence" on the topic.
I don't need to know much of this for my exam, but basically I want to understand it properly because I hate the idea that my foundation is built up badly from the start.
Here's the deal:
During the ionic addition of say, Bromine-Bromine to ethene, my textbook says that 'recent evidence shows' that there is a 'bridged structure' formed when the diatomic bromine breaks up and attacks the double bond.
I get this, the bromine becomes polarised when approaching the ethene at a right angle, leading to one bromine atom losing an electron to the other, more negative bromine.
This positively charged bromine which is two electrons short of a full shell now needs two bonds, right? So it attacks the ethene double bond, forming the 'bridged structure' my book mentions by bonding with each Carbon on the ethene molecule. Fine. Cool. Then eventually the other negatively charged bromine atom can form a bond on the opposite side of the molecule from the first bromine, leading to a lovely looking (1,2)-dibromoethane:

Except now, my book goes on to completely ignore what it just did during the exact same process with Chlorine gas, which makes no sense. No bridged structure, the pictures even say that the Chlorines end up on the same side of the molecule as seen on wikipedia and on any quick google image search of 'addition reaction':

I've seen this in different places on the web too... Bromine somehow undergoes a different ionic addition than Chlorine which makes no sense, it basically bonds to one carbon even though it needs two (like the bromine did), so no bridged structure.
I don't get it folks. Bromine and Chlorine have the same valency, and they're more electronegative than Carbon or Hydrogen. Why are these two processes different, and why do they end up with significantly different products?

The product is the same in both cases, so you are only really asking about the intermediate. Do not read too much into the chlorine atoms appearing on the same side. It does not mean anything because in this case there is free rotation about the C-C bond. Anyway, I was taught that both bromine and chlorine form a halonium ion (i.e. the three-membered ring), and that fluorine doesn't because it's too small. As such, addition of either Br2 or Cl2 proceeds with anti stereospecificity (which means they are added to opposite faces of the double bond).

...which also means that, in a sense, the diagrams that show the chlorine atoms on the same side are a little misleading - not wrong, because free rotation about C-C, but a little misleading. If the rotation was not allowed, for example in the addition to cyclohexene, you would exclusively get the trans diastereomer (halogens are on different sides of the ring). By the way, what textbook are you using?

That's extremely helpful. I understand it a lot better now, thank you. Out of interest, what effect does the size of the atom have on the ability to form a halonium ion? That's interesting, and it's mentioned nowhere in my notes/textbook.

Sort of a handwavy argument - the electron cloud of fluorine isn't large enough to reach over to both carbons at once. There is a bit in Carey and Sundberg that mentions that it is unstable relative to the corresponding carbocation - I don't have the book with me now but maybe someone else can find it.

Ok, thanks. I'll do some research on that one. Super helpful regardless.

@orthocresol looks like an answer to me


text:
Can PBE (and LDA) actually be a better choice sometimes?
Part of my work as an inorganic chemist is to investigate the magnetic coupling between metallic centers in coordination compounds. After some time, I've noticed that the classic PBE functional is the one giving me the best results (better than B3LYP, PBE0 or even M06) and, at first, I was convinced that this was a mere coincidence.
The cases in which PBE keeps giving me better results are adding up, so, I've decided to perform a quick test using the well known $\ce{[Mn(III)2O2(NH3)8]^{2+}}$ that have a reported experimental $J_{AB} = -186.5~cm^{-1}$.
My benchmark showed this:

To me, the results are that the Local and Gradient Corrected Functionals like PBE and LDA are, actually, better for this kind of properties in which orbital superposition and spin delocalization are important.
My point here, I think, is that I struggle to believe that such simple and "old" functionals like PBE and LDA are actually better describing a complicated phenomenon like magnetic coupling.
I'd love to hear what you, guys, think.
P. S. 1: I'm using ORCA, that gives me J calculated using 3 different formulas, that's why I have 3 J values.
P. S. 2: Sometimes I'm using 2 basis sets, a bigger one for Mn and a smaller for O, N and H.

I know this question is about density functionals, but I have to suggest that you should also test "conventional" quantum chemical methods like HF and MP2. These should be definitely within reach in terms of CPU time.

@uLoop, actually, I did. HF gave me something like -36 cm**-1 and MP2 was comparatively bad.

I have never worked with transitional metal complexes, but AFAIK they have a tendecy to be nasty multireference problems. That would be consistent with HF and MP2 giving bad results. Unfortunately, multireference methods tend to be quite CPU intensive.

Can PBE (and LDA) actually be a better choice sometimes?

Of course, they can. This is in fact one of the major problems with DFT: there is no systematic way of improving a functional, so we never know a priori which will work better for some particular problem. The only thing we know is that some functionals perform systematically better that the others for some cases (see a bit more on that at the very end).
The choice of a functional (and even more generally the level of theory) is dictated by which property has to be calculated with which accuracy and is usually based on extensive benchmark studies. A rather quick search by "magnetic coupling dft benchmark" terms found few papers claiming that hybrids indeed work worse than pure functionals for that particular property. For instance, here, it is explicitly claimed in the abstract that

The inclusion of exact exchange does not lead to improvements but
  worsens results drastically.

Exactly the same thing happens in your case, and this is not a big surprise for a number of reasons.

First, hybrid functionals are primarily better for energetics, and not just for any property. Many of hybrid functionals were even crafted with that particular aim in mind, i.e. to improve the description of energetics. But for other properties hybrid functionals might provide almost no improvement comparing to pure functionals and the results can indeed be even worse.
Secondly, the might be some cancelation of errors out there. In some cases you can get lucky and have a better results using a lower level of theory with a smaller basis set due to cancelation of errors.
Finally, benchmark has to be extensive. When it is said that some functionals perform better, the meaning is that they perform better systematically, i.e. for a wide variety of cases at average they spit out "better" numbers. There is absolutely no guarantee that these functionals will perform better for each and every case.

So, I recommend to:

Always start from the literature search for benchmark studies on the property of interest which were already done and reported. They very likely exist and can be of a great help.
On the basis of extensive benchmarks choose just a few combinations of methods and basis sets which are known to perform systematically well for the property of interest and benchmark them on your system, or, even better, on a number of similar systems which you might want (or even planned already) to study in the future.
Pick one or two methods and stick to them.

Thank you for your answer. You're right: I'm searching for other benchmarks and as soon as I have something consistent I'll be sharing it here.


text:
Characterization of products of non-stoichiometric combustion
To give some background, I'm looking at this from an aerospace standpoint as to the calculation of the efficiency of a propellant reaction at a given oxidizer-fuel ratio. That efficiency is a prerequisite to obtaining other results such as the adiabatic flame temperature, the specific impulse ($I_{sp}$) and other such data.
How is it that the relative abundances of products can be obtained from a non-stoichiometric combustion in an analytic manner (i.e. without needing to obtain empirical data)? This is particularly important because rocket engines rarely run stoichiometric, due to exhaust considerastions.
From the Wikipedia article on combustion, it shows that incomplete combustion results in primarily carbon dioxide, carbon monoxide, water and elemental hydrogen gas. However, it also states that:

For stoichiometric (complete) combustion, $z = x + ¼y$. When $z$ falls below roughly 50% of the stoichiometric value, $\ce{CH4}$ can become an important combustion product; when $z$ falls below roughly 35% of the stoichiometric value, elemental carbon may become stable.

This makes sense, as anyone who has left the barrel of a lit Bunsen burner shut has seen the sooty smoke it produces; but is there a manner in which these products may be analytically identified (and quantified)?
To cover all bases, I've had some experience with the use of NASA's CEA program, and it appears to have some ability to calculate thermodynamic parameters of propellants that aren't given as choices (i.e. RP-1, $\ce{H2}$, etc.). I've read the manual on its theory of operation, but I'd like a more gentle introduction to the underlying behavior, if at all possible.


text:
Why did the red colour in a flame test for strontium disappear and appear again on longer heating?
We did a basic flame test, and I understand that different salts produce different spectra because of atoms present in the salts.
When I was doing the test, I noticed that when I dipped the nichrome wire into the solution of $\ce{SrCl2}$ and placed it on the tip of the flame, I got the expected red color. However, what would happen is I would get the red color, and then it would disappear, going back to the typical bluish flame, the nichrome would glow orange hot, and the red colored flame would return.
Essentially, the red color appears, disappears, and then reappears. I thought that it might be "wicking" the strontium chloride from another part of the dipstick, but I was careful to only submerge a small part in the solution.
What's happening here?

Good, careful observation! Here's my assumption of what happened:

The $\ce{SrCl2}$ solution on the wire boiled violently, releasing a large amount of the salt into the flame.
Some remaining anhydrous $\ce{SrCl2}$ stuck to the nichrome after all water boiled off, and melted when the wire reached ~$\pu{874^\circ C}$, coating the loop with a thin layer of the salt.
Additional heating, after melting the salt, went into raising its temperature until it approached the boiling point, ~$\pu{1250 ^\circ C}$. Nichrome can withstand ~$\pu{1400 ^\circ C}$ before melting, but you don't even need to reach the b.p of $\ce{SrCl2}$ to have a bit evaporate and color the flame. Ergo, the flame turned red again.

By the way, this implies you could melt $\ce{SrCl2}$ in a ceramic crucible, insert two nichrome wires connected to a battery, and electrolyze the molten salt to see a bit of strontium (which will catch fire as soon as it rises to the surface, with its characteristic flame color). Do this only under a teacher's supervision.


text:
How does baking soda works as an desiccant?
The thing is that I did a project from a website which is about the science of mummification and I can't find a way to explain the process of desiccation of the hot dog with the baking soda.
By this I mean to answer questions like: Does the baking soda absorb or adsorb water? If it adsorbs then which type of adsorption is it? What's happening in details between the hot dog and the surrounding baking soda?
Here is the link to the project
My results did show a decrease in mass and size which is explained in the lost of water from the hot dog but how?

Frankly speaking, dehydration occurs because of osmosis. In fact, cellophane is used for "peel-before eating" hot dogs and also for osmosis experiments. Natural sausage casing (from intestinal submucosa) are also semipermeable membranes.
You might try an experiment demonstrating dialysis using sausage casing: in a cellophane tube, put plain tap (or distilled) water and tie the ends of the tubing. Weigh the watery "weiner". Immerse the tube in a concentrated solution of baking soda for a day or so, then weigh the sausage again -- some of the water should have leaked out into the outer solution, faster than larger sodium and carbonate ions can infiltrate, so the sausage should shrink.

Well for this project I removed the casing of the sausage. Actually that brings me to another question... Is the baking soda absorbing the water directly from the sausage or just the humidity inside the sealed box?

Even without an artificial membrane, each cell of meat has a cell membrane, which is semipermeable. To check if desiccation was due to lowering the humidity in the box, you could try the experiment again without covering the frank. BTW, CaCl2 would be more effective desiccant than NaHCO3; it is available as a driveway deicer.

authors: هاجر سلمان قاسم


text:
Does Gibbs free energy of formation always equal zero for elements in standard state?
I was looking through the appendix in my chemistry textbook when I noticed that every compound that had a $\Delta H_f$ of zero also had a $\Delta G_f$ of zero. So of course the compounds that have a $\Delta H_f$ of zero are the elements in their standard states but why does that make their $\Delta G_f$ zero? Does every compound that has a $\Delta H_f$ of zero also have a $\Delta G_f$ of zero? What exactly is $\Delta G_f$?

Do you know about relationship between $\Delta G$ and $\Delta H$?

Yeah $\Delta G$ = $\Delta H$ - T $\Delta S$ where T is the temperature in Kelvins and S is entropy.

so from that relation, $\Delta H = 0$ does not always mean $\Delta G$ is not equal to 0.

Well isn't that a little different from what I am describing? That is just $\Delta H$ for a reaction or physical process not the $\Delta H_f$ of a compound right?

Is the formation of a compound from its constituent elements not a reaction or a physical process?

If you look more closely through your textbook, you will find it pointed out that the heats of formation and free energies of formation of all elements in their natural states at 298 and 1 bar are taken to be zero by convention.  This in no way limits your ability to use these tables to determine the changes in enthalpy and free energy for arbitrary chemical reactions involving compounds comprised of these elements.

@ChesterMiller is correct, and should make that the answer. While various groups do try to establish absolute Gibbs free energy numbers, nothing prevents you from using values relative to a standard state.

Does Gibbs free energy of formation always equal zero for elements in standard state?

No.  Elements occur in different allotropes.
$\Delta H_f^o$ and $\Delta G_f^o$ are defined to be zero at 298K, 1 bar for the lowest energy allotrope, with the exception that the values for white phosphorous are defined to be zero even though it is not the lowest energy allotrope.
The other allotropes do not have zero $\Delta H_f^o$ and $\Delta G_f^o$ even though they are still elements.

So of course the compounds that have a $\Delta H_f$ of zero are the elements in their standard states but why does that make their $\Delta G_f$ zero?

$\Delta H_f = 0$ does not imply $\Delta G_f = 0$
The two values are independently defined to be zero for a particular allotrope as explained above.

Does every compound that has a $\Delta H_f$ of zero also have a $\Delta G_f$ of zero?

If a compound (other than by definition) happens to have $\Delta H_f = 0$, then $\Delta G_f$ could be any value, it is not necessarily zero.

What exactly is $\Delta G_f$

It is the change in Gibbs energy upon forming a compound in its standard state from its constituent elements in their standard states at 298K.

For allotropes of elements which have both $\Delta G_f$ and $\Delta H_f$=0, is $\Delta S_f$ also zero? I have never seen this hold.

If you look more closely through your textbook, you will find it pointed out that the heats of formation and free energies of formation of all elements in their natural states at 298 and 1 bar are taken to be zero by convention. This in no way limits your ability to use these tables to determine the changes in enthalpy and free energy for arbitrary chemical reactions involving compounds comprised of these elements.


text:
Oxidizer/reducer in decomposition of potassium chlorate
I'm in the process of tutoring a close friend of mine in basic redox chemistry, and this problem shows a little bit of my chemistry atrophy (It's been two years since I've been active in chemistry work).
The problem stated is the decomposition of potassium chlorate into potassium chloride and oxygen gas, so
$$\ce{2KClO3 -> 2KCl + 3O2 ^}$$
Obviously, the half-reactions of this process can be written as the following:
$$\ce{2Cl^5+ + 12e^- -> 2Cl^-}$$
$$\ce{6O^2- -> 3O2 + 12e^-}$$
The issue I'm encountering is in successfully demonstrating the identity of the reducing agent and oxidizing agent. It's clear from the half-reactions which entity is oxidized versus reduced (and the fact that $\ce{K^+}$ is a spectator ion), but my friend has prior worked examples that appear to show that the identities of the reducing/oxidizing agent are that of the reactants. If so, can $\ce{KClO3}$ be both the reducing and oxidizing agent? If not, would this information be stated as the individual ionic entities listed above?

Yes, it is both the reducing and oxidising agent.

Indeed, there is nothing exceptional in having both the reductant and oxidant within one compound.

If I remember my introductory lab course correctly, there are two reactions going on, one after the other. First, $\ce{ClO3-}$ disproportions according to the following equation:
$$\ce{4 ClO3- -> Cl- + 3 ClO4-}$$
Which has these two half-equations:
$$\ce{ClO3- + 6 e- + 6 H+ -> Cl- + 3 H2O}\\
3(\ce{ClO3- + H2O -> ClO4- + 2 e- + 2 H+})$$
Where the formally required water is merely spectator.
Only upon further heating does the perchlorate anion decompose according to:
$$\ce{ClO4- -> Cl- + 2 O2}$$
Which again can be broken down to:
$$\ce{ClO4- + 8 e- + 8 H+ -> Cl- + 4 H2O}\\
2(\ce{2H2O -> O2 + 4 e- + 4 H+})$$
And again formally requires water as a spectator.
Note that it is not at all unusual for one entity to be both; the simplest example is probably the disproportion of hydrogen peroxide according to:
$$\ce{2 H2O2 -> O2 + 2 H2O}$$
Where one molecule is reduced by another of the same type.

Great answer - was not aware of the perchlorate steps involved. So in the circumstance that one entity is both reductant and oxidant; it is that of cooperation - that is, one serves temporarily as the oxidant and vice-versa?

Sorry, not sure what you mean there @ecfedele =C

I'm trying to grasp how one molecule can reduce/oxidize itself in a reaction such as this. To give an example, above you give the decomposition of hydrogen peroxide - what makes one of those hydrogen peroxides become a reducing agent toward itself?

That’s an interesting question that I can’t answer out of the blue. From a thermodynamic, macroscopic point of view it is certainly that the combined stabilities of $\ce{O2}$ and $\ce{H2O}$ are better than that of 2 $\ce{H2O2}$. So maybe it’s just random deviation; by chance one oxygen is almost losing its electron to give to another or something.

If you consider the oxidation states of chlorine and oxygen, oxygen is REDUCING the chlorine as it goes from +5 to -1 (oxygen in turn is OXIDISED from -2 to 0). This makes no sense as O is more electronegative than Cl and I can only assume that since the reaction requires heat, this energy is driving a non-spontaneous reaction.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How to Dissolve AgNO3 in ZnAc Solution?
In the preparation of $\ce{Ag}$ doped $\ce{ZnO}$ thin films by sol-gel method, I dissolved zinc acetate in 2-propanol while adding DEA and the solution is ready.
After that I added different concentrations of $\ce{AgNO3}$ to the solution, stirring on a stirrer, starting with $0.005$ concentration, but I found that it didn't completely dissolved even after $20$ minutes, and it gets harder for higher concentrations. Can I get any help for this?

We all learn right away (Chem I) that most silver salts are insoluble - except nitrate - in water, but how soluble are they in isopropanol?  That lonely alcohol functional group may not give the same "umph" that water would for the dissolution of the Mighty Silver Cation.
Normally, you'd start with a ZnOAc dihydrate salt, so there's gonna be some water in your mix to dissolve the nitrate, but if you're going for an {aq.} soluble salt in ethanol, did you consider just adding DI to the mix?
What about changing your starting components to include the silver and zinc salts together in dissolution before adding the alcohol?  Add them both gravimetrically ratioed as salt to your alcohol solvent and use the DEA to assist in solvation.  There's docs on this, let me know if you can't find them.

Update:(moved from a second answer, ed.)
So we actually just tried using water to help dissolve the ZnOAc and AgNO3 in the ethanol solution and we found that adding DI water dropwise did help to dissolve the ZnOAc, but after aging overnight a cloudy precipitate had developed, after googling around a bit, I saw that water in ethanol could react with ZnOAc to form ZnOH. In order to test this, I added ~3M Nitric Acid dropwise, and it cleared up the solution. Unfortunately, the solution then failed to stick to our slides when spin-coating.
The good news is, the solution with the cloudy ZnOH actually produced similar XRD results to the solution without water added. We have been using monoethanolamine as a stabilizer to help dissolve the ZnOAc. MEA acts as a bidentate ligand to hold the Zn+ ions in solution. I should have a parametric analysis on a ZnO sol-gel process published by the end of semester - will post it when available.

Additionally, note from Ahmed's comment below.  Silver salts tend to be extremely photosensitive, so the smallest amount of exposure of their solutions may cause discoloration.  We have had improved success by performing as many steps as possible in darkness, and/or by covering glassware with aluminum foil.

I tried adding  silver and zinc salts together before adding the alcohol, increased the spinning period, and it worked and the solution dissolved, giving gradient brown colors for different concentrations of Ag salt from 0.005 to 0.5.


text:
Calculating equilibrium constants: adding two or more equations
If two or more reactions are added to give another, the equilibrium constant for the reaction is the product of the equilibrium constants of the equations added.
How is this true? What is the reason?

This is a direct result from the law of mass action.
When you have an equilibrium
$$\ce{A <=> B}$$
with the equilibrium constant
$$K_1=\frac {[\ce{B}]}{[\ce{A}]}\tag1$$
and a second equilibrium
$$\ce{B <=> C}$$
with the equilibrium constant
$$K_2=\frac {[\ce{C}]}{[\ce{B}]}\tag2$$
you can combine both chemical equations:
$$\begin{align}
\ce{A \;&<=> B}\\
\ce{B \;&<=> C}\\
\hline
\ce{A \;&<=> C}
\end{align}$$
Then you can rearrange Equation $\text(1)$
$$\begin{align}
K_1&=\frac {[\ce{B}]}{[\ce{A}]}\tag1\\[6pt]
\Leftrightarrow [\ce{B}]&=K_1\cdot[\ce{A}]
\end{align}$$
and substitute $[\ce{B}]$ in Equation $\text(2)$
$$\begin{align}
K_2&=\frac {[\ce{C}]}{[\ce{B}]}\tag2\\[6pt]
K_2&=\frac {[\ce{C}]}{K_1\cdot[\ce{A}]}\\[6pt]
\Leftrightarrow K_1\cdot K_2&=\frac {[\ce{C}]}{[\ce{A}]}
\end{align}$$
Thus the equilibrium constant for the combined equilibrium is the product $K_2\cdot K_1$.

Because reactant $\ce{B}$ doesn't show up in the net reaction this doesn't mean that its concentration is zero right?

The equilibrium constant is related to the Gibbs free energy change: $K \propto \exp(-\Delta_0 G / RT)$. When adding equations, you can add the free energies:
$\Delta_0 G = \sum_i \Delta_0 G_i$,
so the equilibrium constant becomes the product:
$K\propto \exp(-\Delta_0 G / RT) = \exp\left(-\sum_i\Delta_0 G_i / RT \right) = \prod_i \exp\left(-\Delta_0 G_i / RT \right)$.

You're right, my $\Delta G$ was not per mole but per particle which is not very usual. I corrected it, thanks.


text:
Is this a possible structural isomer for Butene(C4H8)
Today in school, my teacher was explaining the different types of alkenes. Examples included ethene, propene, and butene.  She also showed us their structural formulas.
I tried to draw butene's formula and came up with this:

Is it correct?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Correct; and the more common example is 1-butene: H2C=CH-CH2-CH3

And there are also the cis- and trans- forms of 2-butene. And if one just went by the molecular formula, there's cyclobutane and methylcyclopropane as well.

Yes, the formula you drew is an example of isobutylene.  Isobutylene has the "official" IUPAC name of 2-methylpropene, and is an important industrial chemical.  Because it:

has a formula of $\ce{C4H8}$, and thus contains four carbon atoms;
contains a carbon-carbon double bond;
and doesn't contain any other functional groups

... it is indeed an isomer of butene (sometimes known as butylene).  You got this one right!$%edit$


text:
Strip away a small amount of aluminium with household chemicals
I'm trying to turn a 6″ diameter mirror (I have spares) into a two-way mirror. It shouldn't be too hard to remove the backing, but then I need to remove most, but not all of the Aluminium plating.
My current plan is to remove the backing with paint-stripper, then soak the mirror in citric acid. However, I'm not sure how long it needs to be left in the citric acid, or even if that will work at all.
Are there any household, or easily acquired, chemicals I can use that will react with Aluminium in a controlled fashion?
Any help or pointers in the right direction would be appreciated.

Citric acid will not etch pure aluminum on any reasonable timescale, at room temperature, though commonly-used aluminum alloys, high temperatures, or the presence of salt can change that. However, it may be better to try some other chemicals: here is a decent resource for finding metal etchant solutions. The two that are probably accessible with household chemicals are hydrochloric acid (can be found at hardware stores with pool supplies) and sodium hydroxide (often found at grocery stores as lye). Neither is super dangerous, but you'll want to take precautions to avoid getting them in your eyes, on your skin, or on your clothes and HCl produces hydrogen gas as it dissolves aluminum, so don't put big pieces of aluminum in concentrated HCl unless you have adequate ventilation.
It will probably take some experimentation to get the concentration right: you want an etch rate fast enough that you're not there for hours etching, but slow enough that you can control it. The big problem with wet etching aluminum is that most etchants take a long time to get through the surface layer of aluminum oxide (and the time it takes is not really reproduceable due to differences in thickness and such) so when you put a piece in the etchant, nothing may happen for several minutes, then you look away for a few seconds and turn back to find all the aluminum gone. This can be avoided by adding HF, which happily removes the oxide, but is very nasty stuff that is rather dangerous to work with.
It might be possible to do this by picking a low concentration etchant and checking the piece frequently, but you may run into issues if the surface is dirty (I recommend washing the surface with acetone, isopropanol, and deionized water in sequence) or depending on how the aluminum was deposited. Some deposition techniques produce large variations in film thickness across the surface and you might run into the situation where the aluminum will be completely etched away in one place but is still totally reflective in another.

Thanks a lot! That's really helpful. So there's no way to slow down the reaction with $\ce{Al}$ without affecting the reaction with $\ce{Al_2O_3}$?

@Zaz I'm not sure on this point. I've never bothered timing it with commercial etchants, but it makes sense that they would be related. Waiting a few minutes isn't really a big deal if it means you get a controllable Al etch rate. Increasing the temperature can also be used to increase the etch rate, so it might be possible to start hot for a bit, then cool it down, but it's probably pretty difficult to control. Your best bet is just to watch carefully.

Thanks. Am I right in thinking paint stripper isn't going to do much to the aluminium?

An ordinary solvent-based stripper should be fine for aluminum.


text:
How to isolate and determine an unknown molecule in a known heterogeneous mixture using HNMR
I am currently working on an assignment as part of my second year lab course. Essentially I have to determine the unknown molecule which is present at 5-15 wt% in 1 gram of Aspartame.
I am planning to do the HNMR of a pure aspartame sample then the HNMR of the heterogenous mixture. Hopefully there will be some distinctive peaks but most likely there will be many overlapping peaks between the Aspartame and the unknown.
My question: Is there any way I can subtract the HNMR of the pure sample from the heterogenous mixture. Therefore I could be left with the HNMR of just the unknown compound.

Technically, this is a software recommendation, isn’t it?

The software that comes with the instrument (Topspin for a Bruker machine,but Varian has a similar product made by them) will certainly be able to do this, as can third party software such as MestreNova or the free ACD/NMR.  You should check with the NMR service in your department as to what software they have licenses for.
The method you describe of 'subtracting' two NMR spectra, whilst possible, is not however very good in practice. For two spectra (more accurately, for two FIDs) to be subtracted, they must be IDENTICAL. That is, the peaks for aspirin in both the pure sample, and the contaminated one, must be perfectly aligned (same peak shapes, intensity, shift etc), which is practically different due to slight differences during each acquisition.
The easiest thing for you to do is probably to purify the sample, for instance by running some chromatography, or perhaps re-crystallising the aspirin, which would hopefully leave your impurity behind, which you could then analyse.
Failing this, and if you have access/inclination, some 2D NMR experiments would be useful to you. A simple COSY would allow you to see which peaks correspond to which peaks (thus allowing you to see which peaks aren't aspirin). More complex methods (HSQC, HMBC, DOSY) would also be of great help, but are probably beyond your needs for a second year lab course.


text:
Why is the acylium ion stable?
Why is the acylium ion stable? Doesn't the inductive effect of oxygen destabilize it? Compared to an alkyl cation, is an acylium ion more stable or less stable?

The acylium ion is stabilized by the positive mesomeric effect of the lone pair of electrons on the oxygen atom onto the empty p-orbital of the $\ce{C+}$. This helps achieve octet completion of the carbon atom, and thus imparts stability to the carbocation.

(Image by Jü CC BY-SA 4.0, from Wikimedia Commons)
Resonance effects mostly dominate the weak inductive effect, so here the destabilizing negative inductive effect of oxygen atom has only a negligible role.
An alkyl carbocation has no such positive stabilizing mesomeric effect, rather only hyperconjugative effects, which are not able to provide as much stability as is possible in the acylium ion.

In addition to what Gaurang already wrote, I would like to add another, not quite as popular, interpretation.
A case can be made to treat the $\ce{CO}$ moiety as a ligand to the (then positively charged) alkyl group $\ce{R}.$ The set of (major) resonance structures gains another member:
$$\ce{
R-C+=O <-> R-C#O+ <-> R+\bond{<-}^-C#O+
}$$
This should of course highlight the linear coordination of the carbon, and the importance of the two π bonds. Furthermore depending on $\ce{R}$ there will be additional stabilisation via hyperconjugation, and/ or conjugation.
A bent structure (as it is often depicted in the literature) of acylium ions is not stable, and does only exist as a transition or excited state.

authors: Сергей Егоров


text:
How many basis functions used in STO-3G and 6-31+G** for the water molecule?
I am a little confused about how to decide how many basis functions that are used in a particular basis set for a given molecule.
STO-3G is a so-called "minimal basis set", meaning that only one basis function is used for each atomic orbital in the atoms of which the molecule is made from.
The water molecule has two H atoms and one O atom, with 1s1 and 1s2 2s2 2p6 electron configurations, respectively. When STO-3G is used, would that then mean that three 1s type basis functions, one 2s type basis function, and 3 p-type basis functions are used, totalling at 7 basis functions that are used, each being made up from a linear combination (LC) of three simple Gaussian functions?
The 6-31+G** is a split-valence basis set, which includes both diffuse and polarisation functions. The two asterisks tell that polarization functions are used for both the H atom and the O atom, p and d polarization functions respectively. I am not sure "how many" polarization functions are used per atom - just 1, or one for each of the p-orbitals? (Some basis sets are of the form 6-31G(3pd); does that mean that 3 p and 1 d polarization functions are used, or 9 p functions?) The single "+" sign means that just one diffuse function is used for the O atom.
So, I think that the 6-31+G** for the water molecule equates to a total of 16 basis functions:

1s in oxygen: one LC of 6 GTOs
2s in oxygen: one LC of 3 GTOs + 1 GTO
2p in oxygen: one LC of 3 GTOs + 1 GTO (for each of px, py, and pz)
1s in hydrogen 1: one LC of 3 GTOs + 1 GTO
1s in hydrogen 2: one LC of 3 GTOs + 1 GTO
Diffuse: one diffuse basis function for the oxygen atom
Polarization: one polarised basis function for the oxygen atom
Polarization: one polarised basis function for the hydrogen atom
Total: 16

Is this correct? I do not feel confident carrying out this analysis, especially the point concerning how many polarization functions for the p/d-orbitals that should be used.

6-31+G** is a synonym for 6-31+G(d,p), so one polarization function of d-type is added for O atom and one polarization function of p-type for H atoms.

You could just run a simple HF calculation and print the basis to check the correctness of you thoughts.

6-31G(3pd) is essentially  6-31G(3p1d), i.e. it is  6-31G plus 3 p-type functions and 1 f-type function on non-hydrogen atoms.

Do not really have easy access to Gaussian. So after doing some searching, a d-type polarization function consists of six (not five) functions, and a p-type polarizastion function is three functions. Hence the correct total should be 23?

Ouch. I was not very careful with words. Say, for 6-31+G(d,p) one set of polarization function of d-type is added for O atom, not just one d-type function, of course. There are always just 3 p-type functions, while how many d- and f-type functions are there depends on do you use spherical or Cartesian harmonics.

The question does not specify anything concerning the coordinates. Is it usual to assume spherical or Cartesian coordinates?

You don't need Gaussian, any program would do it. A set of d-functions indeed contains 6 Cartesian Gaussians, but if one restricts basis functions to spherical harmonics then out of these 6 functions of d-type 5 spherical Gaussians of d-type + 1 of s-type will be built and the later will be excluded.

Spherical vs. Cartesian Gaussians has nothing to do with coordinates.

Ahh, that should be "coordinates" and not "gaussians". Concerning the diffuse basis function, is that only one, regardless of whether it is used for oxygen or hydrogen? What do you mean "any program will do"?

You can use any program to check the number of basis functions. But be aware of spherical vs. Cartesian harmonics issue.

Okay, you are a couple of steps ahead of me. I only know of Gaussian. Is there a free option that you would recommend? OS X compatible would be awesome, but Windows works too.

@Yoda I know you have likely figured something out for this, but for future people viewing this question, I can recommend Psi4 and I have heard good things about ORCA, both of which are free to use.

@Tyberius Yes, if possible I now use ORCA over Gaussian. Some awesome features in the most recent version :) Psi4 I know of, but have never used.

Notation

** is just a synonym for (d,p), so that 6-31+G** basis is just a different name of 6-31+G(d,p) one.
The meaning of (d,p) is that one set of d-type polarisation functions is added for heavy atoms and one set of p-type polarisation functions is added for hydrogens.
6-31G(3pd) is essentially 6-31G(3p1d), i.e. it is 6-31G plus 3 sets of p-type polarisation functions and 1 set of f-type polarisation  functions on heavy atoms.

How many functions are there in a set?

There are always 3 functions of p-type, but the situation with d- and f-type functions is a bit more involved.
There are either 5 or 6 functions of d-type and either 7 or 10 functions of f-type depending on is a calculation restricted to use only spherical harmonics or not.

Cartesian vs. spherical harmonics
Basis functions can be written in two forms: as spherical harmonics
$$
  N(n, \alpha) Y_m^l r^n e^{-\alpha r^2}
$$
or Cartesian ones
$$
  N(l_x, l_y, l_z, \alpha) x^{l_x} y^{l_y} z^{l_z} e^{-\alpha r^2} \, .
$$
And for $l \geq 2$ the number of Cartesian harmonics is greater than spherical harmonics. For instance, a set of d-type function can have five spherical harmonics ($d_{3z^2-r^2}$, $d_{xz}$, $d_{yz}$, $d_{xy}$, $d_{x^2-y^2}$) or six Cartesian harmonics ($d_{zz}$, $d_{xx}$, $d_{yy}$, $d_{xz}$, $d_{yz}$, $d_{xy}$). Note that these six Cartesian harmonics can be transformed into five above mentioned spherical harmonics of d-type plus one function of s-type that can be neglected. Similarly, there are 10 Cartesian f-type functions and 7 spherical ones, and the later  could be constructed from the former with 3 p-type functions as a by-product.
The choice which harmonics to use can be controlled through the corresponding option in an input file and default choice varies from program to program. If you're specifically interested in Gaussian, then, if I remember correctly, by default Cartesian d-type functions and spherical f-type functions are used for the Pople 6-31G basis set. Thus, there will be 6 functions in a set of d-type functions and 7 functions in a set of f-type functions. This might be the case that Pople basis sets were even originally defined to use Cartesian d-type functions and spherical f-type ones, but I'm not sure.

Yes, the Pople-style sets use Cartesian d-type functions, and "pure" f-type functions: "All of the built-in basis sets use pure f functions. Most also use pure d functions; the exceptions are 3-21G, 6-21G, ... 6-31G, ..."

@GeoffHutchison, true for Gaussian, but not for, say, Molpro.

Indeed, I was simply commenting on your statement "if I remember correctly" - this is true for Gaussian. As you say, some programs convert to only-Cartesian or only-spherical.

Yeah, true. That is why it is usually impossible to reproduce Gaussian results obtained with Pople bases in other programs: you just can't have Cartesian d functions and simultaneously spherical f ones. Harmonics are either all Cartesian, or all spherical in most of the programs.


text:
Clash between thermodynamic and kinetic processes
I have often come across reactions where kinetic and thermodynamic processes give completely different outputs. I was told that in such cases check temperature, if temperature controlled, then take thermodynamic product otherwise if reaction is very fast (not temperature controlled) then consider the kinetic product.
I was also told that if there is a case where you can't decide whether it is temperature controlled or speed controlled, then always take the thermodynamic product.
My question is: Why will thermodynamic processes have an upper hand in the latter case?
CH2CHC(CH2)CH3 what MAJOR product does it produce on giving 1 equivalent HBr?

Can you give a specific example?

An example from organic chemistry.

I edited the question.

I'm having trouble understanding what you are asking.  Is it something like:  what is the difference between solving a problem using the equilibrium constant and using the reaction kinetics equations?

Not quite. Look at the example that I gave. If we assume CH2CHC(CH2)CH3 to follow temperature controlled reaction when provided with 1 equivalent of HBr, the product it gives (I think) will be different from the one if we assume the reaction to be very fast and not quite affected by temperature.

I think what you are saying is that there is a sequence of reactions occurring, and the rate constants for these reactions vary differently as functions of temperature.  Is that a correct interpretation?

Quite true. I will explain. We know that the rate constant of a reaction is a function of temperature . So naturally If temperature changes then the reaction path might also change(considering various factors like activation energy etc..).  i.e. Of heat is supplied then it may take a part with a higher activation energy. However if the reaction is super fast, i.e. much much less compared to the minimum time required to reach a path with a higher activation energy, then the reaction might not get the time to reach another higher activation energy path, even when supplied with energy.

So now as my question stands, while writing reactions, we always do not know the time in which the reaction occurs. So I was told that for almost all cases, if such controversial situations rise, take the thermodynamic product to be the major one. Why?

My understanding is that you have several reactions occurring simultaneously, and they all have different activation energies.  Maybe the activation energies for undesirable by-products are high, so you want to keep the temperature as low as possible.  But there are also time constraints related to the economics of the process.  So you might want higher temperature to have higher reaction rates and lower the size of the equipment.  Is that correct?  Also, please define the thermodynamic product.  It this what you end up with at equilibrium?

Sir, I am a Student, who is being given Equations to write down in an exam I am not Experimenting things in industry. The equation I gave as an example was an organic reaction given to me to write down the major product. What I had thought was that in the equation above, if I considered it to be slow enough for temperature to affect it's path to equilibrium, the product it is likely to produce as major would be different from the major product of the reaction if I considered the reaction to be too fast for any temperature to affect it's rate or path to equilibrium.

Remember, her Bromine atom can add to either the tertiary carbon or the secondary carbon atom .

I'm sorry.  I'm not able to follow what you are asking.  I think I need to drop out of this discussion.


text:
Converting an Alcohol to Alkyl Chloride (Ibuprofen)
In the synthesis of Ibuprofen you can shake 1-(4-Isobutylphenyl)ethanol to produce 1-Chloro-1-(4-isobutylphenyl)ethane
$\hspace{7 mm}$
In a separatory funnel at room temperature.
Usually its only tertiary alcohols that can proceed this way so fast, under mild but in this reaction its a secondary alcohol. Is it to do with the alkyl group with the hydroxyl activating the aromatic ring or maybe the para isopropyl group?
Any advice on this would be greatly appreciated :)

The presence of the aromatic ring next to the position where the reaction is happening it what would allow the reaction to take place under the mild conditions described.
The HCl first protonates the alcohol, causing it to leave (i.e. the first step of an SN1 reaction). This is made 'favourable' by the fact that a cationic charge can be stabilised at the benzylic position (you can push the arrows so that the benzenes pi-electrons move, delocalising the positive charge around the ring). The chlorine anion (produced when the alcohol is protonated) may then attack the cation at the benzylic position, giving the product.

Naively I would expect that if the resonance structures of the carbocation that you drew are indeed the correct explanation, then chlorination could also happen at other positions as well.  However, I also think I must be wrong otherwise the indicated procedure would not be widely used for synthesizing 1-Chloro-1-(4-isobutylphenyl)ethane.  What's wrong with this line of thinking?

That had crossed my mind, I think the resonance argument works for explaining the stability, but not selectivity. An MO approach would (hopefully) show that the positive charge has a greatest density at the benzylic position.


text:
Why does (2z²-x²-y², x²-y²) have -1 trace in the character table of Td for C3 rotation?
I'm trying to understand the irreducible rep for $(2z^2-x^2-y^2, x^2-y^2)$ in the $T_\mathrm{d}$ point group. Specifically, for $C_3$ rotations, a trace of $-1$ is listed in the character table for this group. For any matrix, one can rotate the coordinates (basis for the matrix) - i.e. an orthogonal transformation, yielding a matrix with different entries, but the trace will remain the same. So in the case of $120^\circ$ rotations around an axis through a vertex and opposite face of the tetrahedron, one can rotate the coordinates so that the $z$-axis aligns with rotation axis making the calculation of the trace easier.
Any rotation in this basis will leave $2z^2-x^2-y^2$ unchanged, ie. a $1$ on the diagonal. However, the projection of a rotated $x^2-y^2$ back to its own basis will yield $(\cos 2\theta)(x^2-y^2)$. The trace will then be $1 + \cos 2\theta$.
Can anyone help me where I went wrong?

How much group theory do you know? I ask because depending on your level of knowledge your question might require a very lengthy answer.

I think I know now.
Choose rotation axis that makes equal angles with the $x$-, $y$-, and $z$-axes so that the $C_3$ rotation essentially permutes the axes. Without loss of generality, I could choose a $C_3$ rotation that permutes $x$ to $z$, $z$ to $y$, and $y$ to $x$. Rotation in either direction will give the same trace with this $C_3$ rotation. With this rotation (permutation), $(2z^2-x^2-y^2,x^2-y^2)$ becomes $(2y^2-z^2-x^2,z^2-x^2)$. Multiplication of basis function vector a by $2\times 2$ matrix in this representation yields
$$\begin{pmatrix}
  -1/2 & -3/2\\
  1/2 & -1/2
\end{pmatrix}
\begin{pmatrix}
  2z^2-x^2-y^2 \\
  x^2-y^2
\end{pmatrix}
=
\begin{pmatrix}
  2y^2-z^2-x^2\\
  z^2-x^2
\end{pmatrix}
$$
The trace of this matrix is $-1$ as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.
The tricky thing about this example was that I needed to use a rotation in a three-dimensional vector space and project it into a rotation in a two-dimensional vector space.

This is answer number #2:
Choosing the $z$-axis as the rotation axis just made things more complicated in the end, that is, in trying to simplify the calculation of the trace in this example.
Other examples of pairs of quadratic basis functions commonly used in point groups such as $D_{n\mathrm{h}}$ includes: $(x^2-y^2,xy)$.
For this, one can replace $x$ and $y$ with $(x\cos\theta - y\sin\theta)$ and $(x\sin\theta + y\cos\theta)$ respectively, which represents a rotation around the $z$-axis by an angle $\theta$. After using some trigonometric identities and algebra, one finds that the matrix for this representation is
$$\begin{pmatrix}
  \cos 2\theta & -2 \sin 2\theta\\
  (\sin 2\theta)/2 & \cos2\theta
\end{pmatrix}$$
yielding a trace of $2 \cos2\theta$ for $C_n$ rotations about the $z$-axis. So, in this example of finding the character for representations associated with $(x^2-y^2,xy)$ for $C_n$ rotations, using the $z$-axis turns out to be useful because the the result of the operation could be expressed as a linear combination of your basis functions which you can not always do.

The following is a solution alternative to OP's own. As OP correctly observed, any $C_3$ rotation belonging to the group $T_d$ is similar to the $C_3$ rotation about $\hat{\textbf{n}} = (\hat{\textbf{x}}+\hat{\textbf{y}}+\hat{\textbf{z}})/\sqrt{3}$, the axis making equal angles with all three coordinate axes (of course, with appropriate choice of the coordinate system). This rotation switches the coordinate variables such that $x\to y$, $y\to z$, and $z\to x$.
Notice that $C_3$ rotations are also elements of $\mathrm{SO}(3)$, the group of three-dimensional rotations. The five $d$ orbitals furnish an irreducible representations of $\mathrm{SO}(3)$ corresponding to $l=2$ (angular momentum quantum number). If we take the five $d$ orbitals to be the ones with definite magnetic quantum number $m$ ($-2\le m\le 2$) and the axis of rotation to be $\hat{\textbf{z}}$, the rotation by $2\pi/3$ radians (the angle of rotation for $C_{3}$) is equivalent to multiplying the phase factor $e^{-2\pi i m/3}$ to each $d$ orbital. Therefore, the character of the corresponding matrix is given by
\begin{equation}
\chi(C_{3}(\hat{\textbf{z}});l=2) = \sum_{m=-2}^{2}e^{-2\pi i m/3} = -1.
\end{equation}
Because rotation about any axis is similar to the rotation about the $z$ axis with the same rotation angle, we have
\begin{equation}
\chi(C_{3}(\hat{\textbf{n}});l=2) = -1.
\end{equation}
Next, we know that in the group $T_{d}$, the $l=2$ representation formed by the five $d$ orbitals are no longer irreducible: it further splits into the $e_{g}$ and $t_{2g}$ representations. Hence,
\begin{equation}
\chi(C_{3}(\hat{\textbf{n}});l=2) = \chi(C_{3}(\hat{\textbf{n}});e_{g}) + \chi(C_{3}(\hat{\textbf{n}});t_{2g}).
\end{equation}
Observe that $C_{3}(\hat{\textbf{n}})$ simply switches between $t_{2g}$ orbitals such that $xy \to yz$, $yz \to zx$, and $zx \to xy$. Therefore, the diagonal entries of the corresponding matrix are zero, meaning that $\chi(C_{3}(\hat{\textbf{n}});t_{2g}) =0$. It then follows that
\begin{equation}
\chi(C_{3}(\hat{\textbf{n}});e_{g}) = \chi(C_{3}(\hat{\textbf{n}});l=2) - \chi(C_{3}(\hat{\textbf{n}});t_{2g}) = -1.
\end{equation}


text:
Mechanism of the Synthesis of Cyclooctatetraene from Pseudopelletierine
We've been looking at total synthesis in class and the uses of the products. An example was Willstätter's use of pseudopelletierine to synthesise cyclooctatetraene.

I've just been wondering how the mechanism would proceed and if someone could show me the entire reaction scheme then that would be great.
I think the first step to the mechanism involves a sodium radical attacking the ketone. I'm just a bit intrigued how it would all work.

Are you looking for the mechanism of the step involving sodium? or of the entire sequence?

The entire sequence would be really helofull, thanks :)

I also think there will be intermediate radical in first reaction. BTW it's old and rather primitive synthesis

http://www.organic-chemistry.org/namedreactions/bouveault-blanc-reduction.shtm


text:
Chromatography types and techniques
Are HPLC and GC comparable in term to size exclusion, ion exchange, sequential injection chromatography? Or they are two different terms, HPLC and GC refers to the equipment and phase of the analyte, while IEC, HIC, SEC are the principles that would refer more to the column itself?

These are very "googlable" questions. Maybe a good starting point for you would be to figure out what these acrynoms stand for. Eg. GC means gas chromatography.

You are mixing apples, oranges, airplanes and fatty acids


text:
Products of Electrolysis of NaCl
I have gone online to see many times that the electrolysis of $\ce{NaCl}$ dissolved in water produces Lye: $$\ce{2NaCl + 2H2O -> 2NaOH + H2 + Cl2}$$ Other times $\ce{NaClO3}$ or some other compound. I just want to know with one solid answer (anyone), what does this reaction produce?
Thank you.

Related: https://chemistry.stackexchange.com/questions/134780/ https://chemistry.stackexchange.com/questions/127701/

The one solid answer is that everything you've heard is right, given the appropriate conditions.
If the cathode and anode spaces are properly separated with a porous membrane or a salt bridge, then it is just like that:
$$\ce{2NaCl + 2H2O -> \underbrace{2NaOH + H2}_{cathode} + \underbrace{Cl2}_{anode}}$$
If they are not, then the products inevitably mix with each other and react further:
$$\ce{2NaOH + Cl2 -> NaCl + NaClO + H2O}$$
or, at elevated temperatures,
$$\ce{6NaOH + 3Cl2 -> 5NaCl + NaClO3 + 3H2O}$$
See, chemistry is diverse. Oh wait, and there are more options. Think of the mercury cell.


text:
Hypervalency of elements
I know that elements of the 3rd period and beyond can use their empty 3-d oribtals for bonding. But why is there a limit on their covalency?
eg. Al has 5 3d orbitals. And it has an empty 3p orbital. Which means that it's maximum covalency should be 9. But it is 6. Why is it so? Is it because of the small size of Al leading to increased repulsions? Also, why do the elements of the 3rd period use 3d before 4s for hypervalency? 4s has less energy so shouldn't it be used first?

Without using the 3d orbitals it is impossible to rationalise the geometries of hypervalent complexes as predicted by VSEPR. Nevertheless, the contribution of 3d orbitals in hypervalency has been shown to be minimal at best. Search this site or the Internet for more details. As for aluminium, its 3d orbitals are too high in energy for efficient overlap with ligand orbitals.

Those last two questions are the core ones. Basically what ortho said: $n$d orbitals are essentially not used by main group elements.

First off, as suggested in the comment above, the contribution of 3d orbitals in hypervalency is probably minimal.
It's clear from sampling the inorganic literature that "beyond the octet," the size of the atom matters, as well as the shape of the ligands, size of the surrounding coordination sphere, etc. To some degree, it's a matter of fitting X atoms around the central atom.
If you have a 5th or 6th row element (2nd or 3rd row transition metal), lanthanide, actinide, etc., higher coordination is relatively common. For example, for lanthanide ions, 8-coordinate is the most common coordination number. Still, 8-coordinate iron complexes have been found, e.g. "Combined Mössbauer Spectral and Density Functional Study of an Eight-Coordinate Iron(II) Complex" Inorg. Chem., 2015, 54 (17), pp 8415–8422 or "Square-Antiprismatic Eight-Coordinate Complexes of Divalent First-Row Transition Metal Cations: A Density Functional Theory Exploration of the Electronic–Structural Landscape" Inorg. Chem., 2015, 54 (4), pp 1375–1383.
I'm not aware of a general overview of hypervalency and "what coordination numbers can you get", but there's a nice overview on modern descriptions using multi-center bonding.

Your questions are some of those that bug me the most when chemists follow the rationalisation of Pauling and invoke d-orbitals. Most notably the last two:

Also, why do the elements of the 3rd period use 3d before 4s for hypervalency? 4s has less energy so shouldn't it be used first?

They are basically my core point for saying ‘it makes no sense to use 3d at all if 4s is also empty and available.’
All your remaining questions are basically follow-up ones of that central one. And there are more to add, e.g. why should phosphorus use only one d-orbital while sulphur uses two and chlorine may use three?
In general tetrahedral environments such as $\ce{PO4^3- , SO4^2-}$ and $\ce{ClO4-}$, the d-orbitals are represented by the irreducible representations $\mathrm{t_2 + e}$. That means, they should either appear in a group of three or in a group of two. This is inconsistent with the picture drawn for phosphate where we would need one — and for the similar $\ce{XeO4}$ we would need four, also inconsistent. (In case you’re immediately wondering: Systems such as ethene usually have a lower symmetry which does not contain degenerate irreducible representations; i.e. we would always get a single representation per orbital.)
And also, all elements should be able to have a maximum valency of nine if they used d-orbitals for additional bonding — however, the highest oxidation state of a main group element is $\mathrm{+VIII}$ to the best of my knowledge, only accessed for xenon.
All things put together, there is such a can of worms to explain that we should apply Occam’s razor and omit d-orbital discussions for main group elements.


text:
Estimation of the bond angle of water
We know from experimental data that $\ce{H-O-H}$ bond angle in water is approximately 104.5 degrees. If its two lone pairs were bonds (which is unfortunately impossible) also $\ce{O-H}$ bonds and a perfect tetrahedron resulted, then VSEPR theory would predict that the bond angle would be 109.5 degrees - this number can be easily derived using the geometry of a tetrahedron. However, how would people give estimates of the actual bonding angle of water, which is caused by a slightly greater repulsion by the lone pairs than there would be if they were bonds? What physics would be involved in the calculation? Thanks!

how would people give estimates of the actual bonding angle of water
What physics would be involved in the calculation

Background
That's a very good question. In many cases Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.
$$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

For example, in the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.
How can we use this approach to answer your question?
Unlike methane, water is not a perfectly tetrahedral molecule, so the oxygen bonding orbitals and the oxygen lone pair orbitals will not be exactly $\ce{sp^3}$ hybridized.  Since addition of s-character to an orbital stabilizes electrons in that orbital (because the s orbital is lower in energy than the p orbital) and since the electron density is higher in the lone pair orbital than the $\ce{O-H}$ orbital (because the electrons in the lone pair orbital are not being shared with another atom), we might expect that oxygen lone pair orbital will have more s-character and the oxygen $\ce{O-H}$ orbital will have less s-character.
If we examine the case where the $\ce{O-H}$ bond is $\ce{sp^4}$ hybridized, we find from Coulson's Theorem that the $\ce{H-O-H}$ angle is predicted to be around 104.5°. So indeed, in agreement with our prediction, removing s-character from the oxygen $\ce{O-H}$ orbital gives rise to the observed bond angle.
Note on reality
For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized.  The general description is that there are two equivalent O-H sigma bonds and two equivalent lone pairs orbitals.  The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the hybridization changes as described above and the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.
With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs).  Now, the hybridization of water is described as follows:

2 $\ce{sp^4}$ O-H sigma bonds
one lone pair in a p orbital
and the second lone pair in an $\ce{sp}$ orbital

As you said, lone pairs result in slightly more repulsion than bonds, because the lone pairs themselves expand a little around the atom. Common practice that I've learned is just to subtract about 2.5 degrees from the bond angle for every lone pair. For instance, a molecule such as NCl3 has 4 negative charge centers, giving it a tetrahedral group geometry, but it has one lone pair, giving it bond angles of about 107 degrees. Following this pattern, water would have bond angles of about 104.5 degrees.
Here's a picture that sort of shows how the lone pairs expand, reducing the bond angles slightly.


text:
How are reactants ad-/ -desorbed during Haber process?
Collecting information for a presentation [academic high-school] about catalysts, I stumbled upon the terms desorption and adsorption as important parts of the Haber process.
However, I could neither find a good explanation for these processes (why and how do they occur?) nor the difference between ad- / desorption and the process
of the reactants "diffusing through the phase boundary to the active site", as it's stated on German Wikipedia.

All chemical catalysis can be grouped into 1 of 2 categories: homogeneous or heterogeneous catalysis. Homogeneous catalysts are ones that are in the same phase (i.e. solid, liquid, gas BUT USUALLY IN SOLUTION) as the reactants they are helping to react. Heterogeneous catalysts are in a different phase than the reactants they are working on. Industrial processes (like the Haber-Bosch process for making ammonia) frequently use heterogeneous catalysts due to their comparatively inexpensive nature, and there ease of handling on a large scale. Doing it this way can cause some added problems, namely the reactants need to bind to the catalyst's surface (adsorption) for the catalyst to work. In the H-B process this is gaseous N2 and H2 binding to a mainly iron catalyst. To help this process along the iron is made with a high surface area with holes and such (think like the texture of a sponge except on a much smaller scale). After the reaction happens the products then need to release from the catalyst (desorption) so they can be separated from the catalyst and be collected.  The active site simply refers to the place on the catalyst where the reaction happens. Adsorption is helped by the high temperature and massive pressures in the H-B process, as the gasses are flowed through a chamber at high pressure containing the solid heterogeneous catalyst. Not all of the molecules stick and not all that do stick react, so the underacted gas is recycled and passed over the catalyst again. Long story long...the very question you asked won the Gerhard Ertl the 2007 Nobel Prize in Chemistry.

The first few slides seem to give a nice simple explanation. There is a Van der Waal force interaction between the solid and a gas. In your case, nitrogen, hydrogen and the catalyst. http://www.ece.eng.wayne.edu/~jchoi/Adsorption%20of%20gases%20on%20solids.pdf

This is intriguingly close to a link-only answer; it would be better if you could include more of the process in your answer since we do not know how long the link will be reachable.


text:
How can titanium burn in nitrogen?
I was going through the properties of titanium when a certain thing caught my eye: It was the reaction of burning of titanium in nitrogen. I was astonished to read it as I knew that neither is nitrogen a supporter of combustion nor does it burn itself. But then how can titanium burn in nitrogen? Moreover it is one of the few elements to do so.
It was written that titanium burns in pure nitrogen.

Titanium, and many (most!) other transition metals, form very stable nitrides. The trick is that the N2 molecule is actually pretty tough to break up, so only the nitrides with the most negative free energies of formation can manage to 'burn' in nitrogen. With ammonia it is a different story...

@Abhi Titanium can burn in nitrogen for the same reasons as anything burning in anything else: it is a fiercely exothermic reaction which is not prohibited by kinetics. Does not it surprise you that water can burn in fluorine?

From 2019, https://www.spacex.com/news/2019/07/15/update-flight-abort-static-fire-anomaly-investigation   SpaceX claims that a "reaction between titanium and nitrogen tetraoxide at high pressure was not expected."  Not pure Nitrogen, but relevant.

There are two questions one has to answer when asking ‘will $\ce{X}$ react with $\ce{Y}$?’

Thermodynamics
This basically asks ‘will the products of a hypothetical reaction be more stable than the reactants?’, i.e. whether the entire process will release energy. A quick web search tells me that $\Delta_\mathrm{f} H^0 (\ce{TiN}) = - 337.65~\mathrm{\frac{kJ}{mol}}$ and therefore the reaction is thermodynamically possible.
Kinetics
This is asking ‘can I overcome the activation energy required for said hypothetical reaction to occur?’ And this is where it gets complicated.

The process of titanium burning can be split into the following sub-equations:
$$\ce{2 Ti + N2 -> 2 TiN} \\
~\\
\ce{Ti(s) -> Ti(g)}\\
\ce{Ti(g) -> Ti^3+ (g) + 3 e-}\\
\ce{N2(g) -> 2 N(g)}\\
\ce{N(g) + 3 e- -> N^3- (g)}\\
\ce{Ti^3+ (g) + N^3- (g) -> TiN(g)}\\
\ce{TiN(g) -> TiN(s)}$$
Broken down in this way, we should immediately realise:

The first step is generally okay-ish but may require some activation. (We don’t need to make the entire material gaseous in one step; an atom a time is enough.)
The second step requires energy (no surprises there).
The third step is extremely hard to do, since the $\ce{N#N}$ is so strong.
The fourth step also requires energy since nitrogen has an endothermic electron affinity.
The fifth and sixth steps are the ones where huge amounts of energy are liberated; especially the sixth one where lattice enthalpy is liberated.

Now if we consider a block of titanium, as DavePhD mentioned, the the first step is already rather hard to do. The block’s crystal lattice is rather stable as is so we cannot easily extract atoms from there. However, the larger the surface to volume ratio gets the easier this can happen, especially since we have multiple ‘attacking’ points at the same time. Thus, solid metal blocks are usually hard to burn, while metal powder often burns just by putting it in pure oxygen. (Iron wool doesn’t even need to be a powder, the surface to volume ratio is small enough already to oxidise in pure oxygen at room temperature.)
As Jon Custer mentioned in a comment and as I mentioned earlier, the cleavage of dinitrogen molecules is also extremely difficult. It will take a lot of activation energy to do so. This means, that a lot of energy must be gained in the formation of the ion lattice so that at least the dinitrogen cleavage can be overcome and combustion can continue.
Thus, the kinetics part of the question can be answered with: ‘it can, but.’
The question remains how to supply activation energy. Since we are most likely dealing with powder, it could be too slow to ignite titanium in air and then immediately immerse in nitrogen. Since nitrogen cleaving is so hard, many other ignition ideas that rely on, say, a match’s flame won’t work (the match is too weak). Maybe the activation energy is ‘low’ enough that heating to a few hundred degres will suffice. Unfortunately, I do not know the experimental details.

A block of titanium won't burn even in air, let alone nitrogen.
However, if the surface area to volume ratio is high enough (for example titanium powder) it can burn in air and even nitrogen (forming titanium nitride).
See Principles of Fire Protection Chemistry and Physics, 3rd edition, by Raymond Friedman, at page 155.


text:
It is easier to iron cotton clothes when slightly wet, why?
As the title says, why is it easier to iron cotton clothes when slightly damp ? Has this got something to do with the hydrogen bonding in cotton ?

cotton is pure cellulose linked by hydrogen bonds. Water inserts into this bonds, allowing shifts between molecules. Water isn't the best grease here, it requires 200+ celsium to have significant and clearly visible effect. Liquid ammonia is much more efficient, allowing to bend thin timber planks.

It probably steams the clothes a bit more than just the iron by itself.

@permeakra Water is an excellent plasticisier for all natural fibers. Steaming at 100°C makes even thick wooden beams turn into a soft rubbery material, hardening into any new form during cooling. That's thousands of years of shipbuilding tradition.


text:
Can Carbon Form bonds without Hybridization?
Carbon has two electrons in its p orbital which should be able to form bonds, are there any examples in which this occurs instead of carbon hybridizing before bonding?

Carbon (much like other period 2 elements) wants pretty badly to have a full octet, which this arrangement would not provide.

The premise is incorrect. Carbon’s hybridisation is a mathematical model to help us understand its binding modes. All carbon compounds, when calculated ab initio start off from unhybridised carbon — and you can calculate a methane molecule without requiring pre-hybridisation. The nice thing is: If you then re-localise your bonds, you get something like an $\mathrm{sp^3}$ hybrid molecule on carbon.

Hybridisation is a mathematical concept. It is one way of deriving chemical bonds from elements but not the only way. A different one (that computational chemistry typically uses) would be to just plant atoms at certain positions in space, and then mix their unhybridised atomic orbitals with each other to see what comes out. One can then back-calculate this to localise the molecule-spanning orbitals to single bonds and thereby deduce a hybridisation that would have worked to build up the molecule. Take the following example of methane (taken from Professor Klüfers’ internet scriptum for basic and inorganic chemistry at the university of Munich):

On the left you have four hydrogen s-orbitals, on the right you have the orbitals of carbon. As you see, there is no hybridisation assumed at all and the ground state is taken to be a fully populated $\mathrm{2s}$ orbital and three $\mathrm{2p}$ orbitals populated with two electrons amoung them for a total of $\mathrm{2s^2~2p^2}$. The hybrid orbitals are gained from symmetry considerations (in a tetrahedron, the p-orbitals transform as $\mathrm{t_1}$) and hydrogen orbitals were selected as was seen fit. The electrons are added a posteriori. Spin conservation would have produced a twice-excited state with very high total energy, so immediate spin pairing is assumed.
To localise the bonds, linear combine all four orbitals. If you take every orbital with factor $+1$, you arrive at the $\ce{C-H}$ bond to the upper right hydrogen. By shifting around a factor of $-1$ among the three $\mathrm{t_1}$ molecular orbitals during linear combination, you can address every other $\ce{C-H}$ bond. The resulting bonding orbital looks like n $\mathrm{sp^3}$ orbital mixed with a hydrogen $\mathrm{s}$ orbital.

Hybridization isn't actually an operation carried out on the orbitals. It is an operation carried out on the mathematical functions that define the orbitals (refer chapter 1 of A Guidebook to Organic Reaction Mechanisms by Peter Sykes). As @Ivan Neretin said, Carbon, like all other period 2 elements, wants to attain octet badly. It does this by promoting two 2s electrons to 2p orbitals. The energy required for this is much less than that released during formation of two extra bonds (for eg.: formation of 2 C-H bonds in methane would release 97kcal/mol).
Now, carbon atom could form 3 directed bonds, which are mutually perpendicular, using three 2p orbitals and one non-directed bond using 2s orbital. Whereas in fact, it is known that the 4 C-H bonds are disposed at an angle of 109°28'. This can be accounted for if we use the concept of hybridization in functions defining the orbitals.

The first two sentences started out so well, and then you go on about a hypothetical excited state … sorry, $-1$ for proposing that that formal excitation happen in the real world.

No offence, but it is all in chapter 1 of peter sykes' book. I have just referred to it and answered the question. Please feel free to edit it.

@Jan Could you please tell me what's wrong with my answer?I am in high school and it would really help me.

It is wrong to assume excitation before bond formation if hybridisation is not done; see my answer. So sentences 4 and 5 of your first paragraph are wrong. I cannot edit without conflicting with the author’s (i.e. your) interest, though, since you clearly build up on that ‘lost energy’.

So, we need to use MOT to explain bonding instead of that excitation stuff I said earlier? Which book can I refer to understand this concept?

Terribly sorry but I cannot help you with books since I never really read any learner’s book =C But note that once you start arguing with orbitals you are already in MO territory. If you just say ‘I have a carbon atom with four electrons and want to draw Lewis structures’ you can draw them as four dots since the dots themselves do not indicate which orbitals the electrons populate until you add that information via MO.


text:
How polarization effects from cations affect H-bond strength
The average hydrogen-bond strength in pure water is higher than the average H-bond strength between the first and second hydration shells around a $\ce{Al^3+}$ ion (or any cation, for that matter). This leads to a shift in the OH stretch toward the lower frequencies.
This has something to do with the induced polarization resulting from the positive charge, no doubt, but I fail get why the H-bond becomes weaker. The H-bond would get stronger if the electron density around the H-atom participating in the H-bond gets smaller (larger coulombic attraction between H and O). When $\ce{Al^3+}$ approaches the O-atom of a water molecule, electron density will be forced toward the positive charge, ultimately affecting the H-bonded H-atom, which would give stronger H-bond, at least in my head.
So what's wrong with this picture?

Okay, I realized what was wrong with my picture above.
It is not the hydrogen bond that is weakened, but the OH bond within a water molecule that is weakened, and then stretches at a lower frequency.
When the cation is introduced, and is near the oxygen atom, the electron cloud in the water molecule is moved toward the cation (polarised), leading to electron density in the OH bond being reduced => weaker bond => lower frequency vibration.
The hydrogen bond is not really of interest here, although the same effect takes place when two water molecules are H-bonded. Hence, the OH stretch of an isolated water molecule will vibrate a bit faster than the OH stretch in a H-bonded water molecule.


text:
Propanoic acid reaction with potassium permanganate
I was conducting a lab where we have to identify/justify how functional groups affect the oxidation of a molecule.
I used propene, propanone, and propanoic acid for this reaction. From what I have learned in class, ketones and carboxylic acids cannot get oxidized. However, in my experiment the ketone somehow is not getting oxidized while the propanoic acid is. I tried using butanoic acid just to be sure but it turns out that it gets oxidized as well. I have also made sure that there are no contaminants. I used 4 ml of 100% propanoic acid with 0.003 mol of $\ce{KMnO4}$.

How do you know the carboxylic acids are being oxidized?

I know it because when KMnO4 is added it reacts immediately, ie the pink/purple colour goes away suddenly as it did with the propene. It stays pink with the ketone but not with the carboxylic acids. I asked the Chemistry teachers but they don't know any reason either. @jerepierre

The propanoic acid cation is not being oxidised here to form proponone . An acid/base reaction is occurring where the permanganate is acting as the base. An aldehyde may react with the KMnO4 to form the ketone .

Can you detail the concentrations of both the acid as well as $\ce{KMnO4}$ and also the time it took to de-colourise?
I tried an experiment however, I only have Glacial Acetic Acid, which didn't seem to react ( as the smell was still prominent).
I can suggest you a simple trick to analyse whether it is decomposing. Smell Propionic acid before adding $\ce{KMnO4}$. Then smell it after the colour of $\ce{KMnO4}$ is gone. Keep adding $\ce{KMnO4}$ till the pink colour stay(similar to titration). Then smell again. If the smell is gone then the carboxylic acid is oxidised.
There are 2 possible oxidation products of carboxylic acid:

Ketone
Carbon Dioxide

Both can be pretty obvious. Ketone will smell differently than the acid(more like nail polish remover) and Carbon Dioxide should be visible because of effervescence.

Could it be that there is an isohypsic (non redox) reaction that changes $\ce{KMnO4}$ into another $\ce{Mn(VII)}$ compound? According to wikipedia, it has been known to happen with other acids.
It would be nice to know if there were other methods available to analyze the reaction mixture (such as TLC).