text:
Carbon monoxide formation in broken boilers
It's common knowledge, that if you're boiler is broken, it might produce carbon monoxide ($\ce{CO}$) as a byproduct, which is very toxic, even in small quantities. My question is, why do broken boilers do this?

The combustion of hydrocarbons works as follows:
$\ce{hydrocarbon + O2-> CO + H2O}$
The reaction can then continue to the complete combustion:
$\ce{CO + 1/2 O2-> CO2}$
An example of balanced equation is the combustion of methane:

$\ce{CH4 +  3/2O2 -> CO + 2H2O}$
$\ce{CO +  1/2O2 -> CO2}$

So, carbon monoxide is a product of the incomplete combustion of a hydrocarbon.
It is produced when there is too much fuel (eg: methane), but not enough oxidant (eg: oxygen from air).
I guess that a boiler which cannot mix properly air and fuel produces carbon monoxide.

And when/why does the reaction stop at the first stage?

It does not really stop at the first step. Both steps will occur but overall there is not enough oxygen to convert everything to carbon dioxide.


text:
Is there 'no actual equilibrium'?
If you look at the concentration verses time graph of typical reactions, you can see that the concentration of reactants and products, when time goes towards infinity, becomes the equilibrium concentration. But! It never seems to reach that concentration (it converges towards that , but never reaches it). Which means that there is no 'real equilibrium??
I think this is true in terms of theory, where the number of molecules is considered to be infinite, while in reality, the number of molecule is finite. Then, in reality, it reaches equilibrium??

Equilibrium is based on the speed of the forward and backward reactions. When the backwards and forwards are equal,there is basically no product being made (Net products that is). If you theoretically had infinite reactants, that means the forward reaction would always be at the highest is could possibly be, and it would never decreased, so it would never reach equilibrium.

The thing is, yes THEOCRATICALLY it can never reach equilibrium, but, in the graph of concentration verses time graph, when t is very very large, it is ALMOST close to the equilibrium state. But if that much difference in concentration is smaller than the concentration difference that can be achieved by a single molecule, then could we say that it reached equilibrium?

Yes there is actual equilibrium. The Achilles and the turtle kind of argument fails for three reasons, one of which you mentioned yourself:

Like you said, the number of molecules is finite, so the matter is not infinitely divisible; there must be a limit.
More importantly, nobody really cares about one molecule. Anyway, it's not like the system is going to hit just the right number with one-molecule precision. No, there are going to be mighty fluctuations involving millions of molecules, and they never stop (that is, unless you freeze the system down to absolute zero, in which case reaching the equilibrium may indeed become problematic). Like anything with molecules, equilibrium is a statistical thing, and once we're within certain error margin from the theoretical figure, we might just as well consider the goal achieved.
We may circumvent the argument altogether by playing with the conditions. Say, we want to get a saturated solution of some salt in water, and it just seems to take forever. But wait, what if we just heat the thing up a bit (which would make the solubility go up), dissolve the exact amount we need, and then cool it down again? Here is your saturated solution in a finite time. You may even make it supersaturated if you'd like.

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text:
What is the formula for evaporation rate of water?
I did my own research, and according to The Engineering Toolbox,the formula for the evaporation rate of water is as follows:
$g_\mathrm h = \Theta A (x_\mathrm s - x)$
$g_\mathrm h$ = amount of evaporated water per hour ($\mathrm{kg/h}$)
$\Theta = (25 + 19 v)$ = evaporation coefficient ($\mathrm{kg/(m^2\ h)}$)
$v$ = velocity of air above the water surface ($\mathrm{m/s}$)
$A$ = water surface area ($\mathrm{m^2}$)
$x_\mathrm s$ = humidity ratio in saturated air at the same temperature as the water surface ($\mathrm{kg/kg}$)  ($\mathrm{kg}$ $\ce{H2O}$ in $\mathrm{kg}$ dry air)
$x$ = humidity ratio in the air ($\mathrm{kg/kg}$) ($\mathrm{kg}$ $\ce{H2O}$ in $\mathrm{kg}$ dry air)
However, later I realized that something is wrong with the formula. There is no variable for water temperature.
The formula for humidity ratio in the air is:
$$x = 0.62198 p_\mathrm w/(p_\mathrm a - p_\mathrm w)$$
$p_\mathrm w$ = partial pressure of water vapor in moist air (Pa, psi)

$p_\mathrm a$ = atmospheric pressure of moist air (Pa, psi)

The formula for humidity ratio in saturated air is the same, except partial pressure of water vapor in moist air is replaced with saturation pressure of water vapor in moist air.
And the formula for water vapor saturation/partial pressure is:
$$p_\mathrm{ws} = \mathrm e^{\frac{77.3450 + 0.0057 T - 7235 / T}{T\cdot8.2}}$$
$$p_\mathrm w = p_\mathrm {ws} \cdot HU$$
$p_\mathrm{ws}$ = water vapor saturation pressure (Pa)
$T$ = dry bulb temperature of the moist air (K)
$HU$ = humidity ratio (%)
So, according to this website, temperature-wise the formula is determined solely by the air temperature without considering the water temperature. Is it because the formula assumes the equivalence of the two temperatures for simplification or because I'm missing a variable? I searched for other methods, but so far no website is as comprehensive as this one, and I don't want to re-program everything again.
This is a different question from the existing ones. I've read many of them and the answers are not specific enough for my situation. Here I need to know how the temperature of the water plays a role in the formula for the rate of evaporation of water.

http://chemistry.stackexchange.com/questions/55244/evaporation-rate-of-water

Thank you for your attention. But I saw that. This is a very different question, and the answers there don't satisfy my question.

there was another similar question, but generally there's no way to have exact formula.

I'm no expert in this field, but for all I know the formula I've found is empirical formula. I thought if chemistry experts saw the formula, even if they do not know such made-up formula, they might recognize either 1. the simplification of the formula to remove the water temperature variable or 2. my stupidity for missing the water temperature variable.

I have a feeling it's probably going to be assuming the equivalence of the ambient temperature and the water surface temperature.

I agree with Eashaan. The definition states: "xs = humidity ratio in saturated air at the same temperature as the water surface" . When you compare saturated air "xs" and air above the surface "x" they should have the same temperature. This would then be the same temperature as the surface of the water.

I am currently tussling with this empirical formula. I concluded, rightly or wrongly, that the air temperature must be the same as the water temperature as the phase change takes place at constant temperature. However in practice there will be a difference as nature always hunts around an equilibrium position due to leads and lags in the processes imvolved.
Also the moment the vapor moves away from the liquid the situation is likely to rapidly change.
Note how the invisible vapor rapidly condenses to visible water droplets a few inches from the spout.

Hi, I'm back! Also, haven't you answered the temperature dependence question already? As $T$ varies, so do $p_w$ and $p_{a}$. Considering the system, I doubt $\Theta$ would be as sensitive to temperature though. And, interfaces are assumed to be at equilibrium -- this might clarify things some more.

The temperature of water is implicitly covered by actual and saturated vapour content, or equivalent vapour pressures.
There is deterministic relation between relative humidity, air temperature and temperature of water, supposed to be in dynamic equilibrium.
In such a state, the cooling effect of water evaporation is balanced by the thermal transfer from warmer air.
This relation is used in meteorology/climatology as the standard method to determine humidity via the psychrometric tables of temperature of "dry" and "wet" precise thermometers.
Effectively, the wet bulb temperature is  somewhere between air temperature and the air dew point.
They are standard items in the white meteorological wooden box 2m above the lawn.
See psychrometrics particularly the sections dry and wet bulb temperature.
The main article Wet bulb temperature.

@EdV Wind speed, $v$, has been factored into the mass transfer coefficient, $\Theta$.

Air temperature is a factor in the vapor pressure value. Lower temps will have lower vapor coefficients.

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Does negative sign of s-orbital exist?
I know that the signs '+' and '-' are written in case of lobes of p and d atomic orbitals, which denote the sign of the wave function of that orbital. But s-orbital is also a wave. So, it should also have negative regions. However, most books show the s-orbital with a '+' sign only. What is the significance of this?
Again, the 2s, 3s, 4s, ... etc orbitals have some nodes. So does these nodes indicate the seperation of '+' phase and '-' phase? The 1s orbital does not have any node.So where is the '-' part of the 1s orbital ?

That's right, 1s orbital has no nodes and no negative part. Higher s-orbitals do have both.

I take a much simpler approach. Since s orbitals can interact in or out of phase (in bonding or anti-bonding interactions), then they could be represented by opposite signs (phases).

Although the answers are correct, I think that they do not help you because your question shows a misunderstanding. Those are not the signs of the orbitals. The signs often used represent the coefficients of a linear combination of orbitals (look for LCAO method). They have not sense in isolated atoms.
1s orbitals are not  special in this sense. Although the orbitals commonly used in practice are real valued functions,  they do not need to be at all. Even more, excluding  $n$s orbitals they are (before any transformation) complex valued functions, so the meaning of the signs would be very diffuse. Also, orbitals are defined up to a phase factor. If you multiply an orbital by a complex number (eg. -1+0i), it is as valid as before (it is not exactly true (only valid for the wave function), they loose some important property, but the orbitals you use are already transformed).

But then, when combining 1s and 1s orbital by LCAO, we have both sigma and sigma* molecular orbitals(as in hydrogen molecule).

@ShoubhikRajMaiti yes, so what? Excuse me, I do not get what you are trying to point out

@user1420303 A sigma orbital forms when orbitals of same sign overlap and sigma* when orbitals of opposite sign overlap. If s-orbital has only positive sign then there would be no antibonding orbital. Shouldn't it?

@ShoubhikRajMaiti , In my answer I tried to clarify that it is not true that A sigma orbital forms when orbitals of same sign overlap and sigma when orbitals of opposite sign overlap*, because there it is meaning less to speak about "sign of an orbital". We can say that: $\sigma = c_A \cdot f_{1s_A}+  c_B \cdot f_{1s_B}$ , where $f_{1s_A}$ and $f_{1s_B}$ differ from each other only by the coordinate origin (nuclei A or B). In such case '+' and '-' are related to $C$s coefficients and are not related to the orbitals $f$s. The equality in $f$s is arbitary. You can choose other options, but

@ShoubhikRajMaiti ... in such case you have to modify $C$s coefficients according.  Please note, if you take the more primitive and general form for $f$s, you'll get complex valued functions. In such case you can't speak about "positive" or "-" at all.
Also, note that if $f$ is a valid wave function, $z\cdot f$ (with $z$ being a complex number) is as valid as $f$.  So, signs in $f$ is meaning less. Even more: Consider the a real valued wavefunction of a particle in a box: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html
Is it negative or positive?

@user1420303 Thank you for your explanation. It is a bit difficult to understand for me but I get your point.

The sign of the wave function does not have any physical meaning since for every solution $\Psi$ of the Schrödinger equation $-\Psi$ is also a solution. In fact there are infinitely many equivalent solutions that only differ in their overall phase.
So the (meaningless) answer to your question is yes.
It should be noted that $\Psi \Psi^*$, that can be physically interpreted as a probability distribution for finding a particle at a certain point, is always a non-negative real number.

As noted in the comments, the 1s orbital does not have any nodes. In general an orbital with quantum numbers $n$ and $l$ will have $n-1$ nodes, $l$ of which are angular nodes, and $n-l-1$ of which are radial nodes.
The wavefunction for the 1s orbital (in a hydrogenic atom with nuclear charge $Z$) is commonly written as:
$$\psi_{\mathrm{1s}} = \sqrt{\frac{1}{4\pi}}\cdot 2Z^{3/2}e^{-Zr}$$
Since $Z$ and $e$ are both positive numbers, $\psi_{\mathrm{1s}}$ is positive for all values of $(r,\theta,\phi)$; indeed, the wavefunction has no dependence on the two angles. Thus, the most conventional representation of the 1s orbital depicts it as being purely positive.

However, note that if $\psi_{\mathrm{1s}}$ satisfies the Schrödinger equation for the hydrogen atom, so must $-\psi_{\mathrm{1s}}$ (and also any other phase-shifted form $\mathrm{e}^{\mathrm i\alpha}\psi_{\mathrm{1s}}$ for a generic angle $\alpha$; for $\alpha = \pi$, this prefactor evaluates to $-1$). It turns out that the absolute phase of a wavefunction, on its own, has no physical meaning. We cannot measure the wavefunction directly, only indirectly by way of measuring observables, and the values of all observables are independent of the phase of the wavefunction:
$$\langle O \rangle = \langle -\psi | \hat{O} | {-\psi} \rangle = \langle \psi | \hat{O} | \psi \rangle$$
Thus, the choice to display the 1s orbital as being purely positive is not mandated by a physical requirement, but is rather a conventional choice. The difference between the 1s orbital and a 2p orbital, for example, is that the 2p orbital contains regions that have opposite signs to one another: thus, no matter how you choose the phase of a 2p orbital, there will always be a positive and a negative region.
Note that the absolute phase of an orbital does not have the same physical meaning as the relative phases of two orbitals when constructing linear combinations (e.g. in molecular orbitals). For example, the bonding MO of dihydrogen is made up of an in-phase combination of hydrogen 1s orbitals. Here it is physically important, and mandatory, that the two contributing 1s orbitals have the same phase. However, exactly what phase this is does not matter: they can both be positive, or they can both be negative.

So you want to say that we tend to write 1s orbital with positive sign conventionally however it can be negative as well as that also satisfies the wave equation?

@Lllt Yes, and it goes beyond orbitals, it is true of any quantum state. $\psi$ is exactly the same state as $-\psi$.

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Do you get a new chemical compound if you mix sodium hydroxide and water?
Me and dad started fighting over this a couple of days ago and wagered on this.
He said mixing these two gives a chemical compound, I said that it doesn't.

Is what a chemical compound? The mixture?

Is it a chem compound the mixture of water and sodium hydroxyde.

Then you are right. NaOH is a compound, so is H2O, but their mixture isn't.

We need the exact sentence for your bet, because it is more tricky than you (and your father) think.

My dad said:
"In my work we combine NaOH and a chemical compound occurs"
Me:
"It is not a chemical fusion/compound, because there is no new compound being created, it's just a mix"

How much was the wager?

Sodium  hydroxide dissociates into sodium cation($\ce{Na+}$) and hydroxide anion ($\ce{OH-}$) in aqueous medium.

We cannot say that it is forming a compound or a mixture clearly. Actually what happens is that NaOH just breaks down into Na and OH when it added to water and the mixture posses the properties of Na, OH and water, so u may say that if NaOH is added to water then a mixture of Na, OH and water is formed.

According to Oxford's dictionary of chemistry "A compound is a mixture of two or more elements in a fixed ratio".
So if $\ce{NaOH}$ (in solid state) is added to water, it just dissociates into its constituent ions that are $\ce{Na+}$ and $\ce{OH-}$. You can add any ratios of two until you cannot add any more.

He doesn't want to admit I am right. When I showed him what you said he just clicked the power button on my laptop.

did he prove to you that it is a "compound"

Another take could be this: Assuming it is a chemical compound, it must mixture possess the characteristics a chemical compound must have, then show that that mixture of water does not have those properties, one of them being the fixed ratio thing, and if this is the case, our assumption is wrong that is mixture of water and sodium hydroxide is not a chemical compound.

@Alek It may be futile to argue with a brick wall...

What exactly counts as a compound is not exactly clear cut, and probably not a really useful distinction a lot of the time. It's fair to say that a solution of NaOH in water is not a new compound, though.
If you start with pure crystals of NaOH however, adding water can create new compounds: NaOH.1H2O and NaOH.4H2O (the tetrahydrate appears to be polymorphic, so there's another can of worms there).


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Can the molecule CH2(ONa)2 be stable?

While studying about Cannizzaro reaction, it came to my mind as a wild thought what if this HCH(ONa)(ONa) molecule could pop out in some step of this reaction (not giving much details about how I thought it as it would be irrelevant). Then I thought this molecule might have some problem with stability. The $\mathrm{sp^3}$ hybridized carbon might have some have problem with spacing two -ONa groups close together.
So, is this molecule stable or not? If not why?

I guess it's sort of follow-up of http://chemistry.stackexchange.com/questions/55692/is-the-compound-ch2ohona-possible and answer is similar I guess - it's highly basic intermediate.

Oxygen and sodium will hardly form a covalent bond.

The species you are proposing would be a methanediol dianion (CH2O22–) with two sodium cations “attached” to it in some way, depending on the nature of the solvent.
While methanediol can indeed readily form an anion, with a pKa of 13.3, I find no mention of the dianion or any measure (or prediction) of the second pKa of methanediol.


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Why is the solubility of sodium acetate trihydrate so much lower than anhydrous sodium acetate?
According to Wikipedia:
Solubility of anhydrous sodium acetate in water at 0 $^{\circ}$C is 119 g/100 mL (AKA 1.495 mol/100mL)
Solubility of trihydrate sodium acetate in water at 0 $^{\circ}$C is 36.2 g/100 mL.
This makes no sense to me because it should take MORE of the trihydrate sodium acetate (about 1.495 * 136.08 g) to dissolve the same amount of sodium acetate.
So this begs two questions:
1.) Is my understanding of hydrates incorrect? Are their bonds somehow harder to solvate than the anhydrous version? I think this can't be correct because the melting point of sodium acetate trihydrate is 58 $^{\circ}$C compared to the anhydrous melting point of 324 $^{\circ}$C.
2.) Is my understanding of solubility incorrect? I thought solubility is a measure of the quantity of solute that could be dissolved in a solvent. In other words, it seems that MORE sodium cations and acetate anions can be dissolved in 100 mL of water if they came from an anhydrous crystal rather than a trihydrate crystal...
How can that be true?

See https://fr.wikipedia.org/wiki/Ac%C3%A9tate_de_sodium there's 362 not 36.2 - someone put decimal point in wrong place.

The figures in the English Wikipedia for the solubility on anhydrous sodium acetate in water are most probably wrong.

@Mithoron That still doesn't make sense because that is literally three times the amount than anhydrous can be dissolved...@aventurin I can't seem to find anywhere another source of the solubility of anhydrous sodium acetate

The trihydrate already has it's water of crystallisation in the crystal lattice so it would take longer to dissociate in water , it would be less hygroscopic .

@Mithoron The values in French Wikipedia are given per liter, so they're the same.

@vapid Oh, indeed :(

The values of solubility for these salts are $100\%$ correct. Your understanding of solubility is also correct. What you didn't know (probably) is that solutions can exist in a non-equilibrium state. In case of anhydrous sodium acetate, it dissolves in water forming so-called supersaturated solution, which is metastable. Eventually, crystals of sodium acetate trihydrate will precipitate out of the solution, forming a system which is in the thermodynamic equilibrium.

Crystals of sodium acetate trihydrate would precipitate out of solution (of anhydrous sodium acetate) until it reaches the solubility of the normal sodium acetate trihydrate solution?? That is incredible. Then what is the point of naming both solubilities when one will eventually equilibrate to the other?

@Nova Why name it otherwise? It is solubility. All substances dissolve in water to the point where they reach the equilibrium of dissolution and crystallization processes. Most substances dissolve starting from zero concentration and reach the saturation. There are some substances, however, that can exceed the saturation concentration, then reach the equilibrium from the other site - by crystallizing out of the solution. Sodium acetate is a classical example. There is nothing incredible, really.

@Nova You can picture it that way (it is a bit of simplification): when you dissolve the anhydrous sodium acetate you get a solution which is saturated in respect to the anhydrous salt and supersaturated in respect to the trihydrate. You will eventually get a saturated solution of trihydrate and excess of crystalline trihydrate.


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Is there a way to fine tune a laser to excite the electrons of a particular molecular bond into dissociating?
I don't know if this makes any sense, but I came up with this in my head: if each molecular bond in any molecule had a specific *signature wavelength of dissociation", this would be the exact wavelength of the laser beam necessary to excite the electrons involved in that one particular bond and no other bonds in the molecule into dissociating homolytically. Is this theoretically sound in any way?

Obviously if you have a diatomic then yes, you can put energy in and the bond will break. There are three cases; (a) as long as the photon has enough energy it can excite a dissociative excited state, (grey in sketch) that is a state that is only repulsive. The absorption is continuous here so its energy has to be above a certain value. (b) The second case has a curve crossing in the excited state. This means that two excited states cross , a bound state and a dissociative one. Then if you hit the energy at the crossing point, dissociation will occur, but not at higher or lower energy. This is called pre-dissociation. (The read & grey line crossing show pre-dissociation). It occurs in many molecules not just diatomics, e.g. benzene. (c) The third case is when a bound state is excited above its dissociation limit, e.g. above approx 10000 in the blue curve, 5500 on the black curve . However, reaching these energies may be hard because the Franck-Condon factors may be vanishingly small. (roughly speaking the FC factors ensure vertical transitions)

Now the case with polyatomic molecules is tricky. Many experiments have been tried with varying success, the area is sometimes called 'mode selective chemistry' and/or 'coherent control'.
In a molecules there are many vibrations, called normal modes. Although we think generally of one bond vibrating, in actuality many bonds vibrate in unison (i.e. in a fixed phase relationship) in a normal mode. These modes are governed by the symmetry species of the molecule. Each normal mode is supposed to be orthogonal, so that vibration of one does not effect vibration of another. But in fact vibrational anharmonicities allow vibrational quanta to flow from one mode to another. Thus exciting one normal mode soon involves energy spreading out into all other modes and thus even if there is enough energy to break a bond it soon becomes diluted.
In coherent control experiments one uses one or more very short light pulse ( < 100 femtoseconds) to excite the molecules but control the phase of the light (hence pulse shape) to control how long the molecule takes absorb the photon. This is done in the hope that once the energy starts to spread out (as a vibrational wavepacket) it can be controlled to end up in just the bond you want to break, i.e. to try and use the vibrational spreading out tendency to advantage. The pulse shaping is done using an electro-optic or acousto-optic device in a feedback loop using say, a genetic algorithm linked to a detector that observed product formation.

The first paragraph concerns electronic excitations (excited states), whereas the rest vibrationally excited states.
The dissociation energy of a given bond is the minimum amount of energy you need to put into a molecule to break that bond. There are several ways, including heating, etc... You could even imagine two-photon mechanisms, e.g. first ionizing the system and then breaking the bond, if the ionized state is easier to dissociate.

@ ssavec yes heating  would work but the question asks about using a specific laser wavelength. Also two photon transitions would work if you wanted to get to a state that selection rules forbid for one photon  transitions.  Once in that state its the same thing as a one photon transition. Ionisation is not necessary. Electronically excited state also have vibrational normal modes provided that they are not entirely repulsive in nature, i.e have a potential well as in the case of the red and blue curves in the figure.

So, Is there a way to imagine a simplified version of reality in which this would be possible. What I mean is, take of all the practical things, and leave just theory, would it then be possible to (no matter how big the molecule is) fine tune a frequency that would disrupt that one particular bond of that one particular molecule if you fired the laser at a bunch of different molecules (for example, you're firing the laser at a solution of several complex molecules in gaseous state). What I mean, take off the technicalities of the real world and leave a theoretical shell, like Bohrs model

Just by enthalpy reasoning, I'd guess it is very difficult for larger (more than 4 atoms) molecules. You have to add enough energy to break the bond, approx 100 kcal/mol ~ 4.336 eV ~ 285.88 nm ~ UVB radiation. So if it should be one photon, it has to be UV and therefore electronic excitation, as shown on the plot. Nevertheless, see
https://en.wikipedia.org/wiki/Photodissociation#Multiple_photon_dissociation

The important point about a polyatomic molecule is that energy flows among vibrational normal modes very quickly indeed so its generally not possible to get it into the one bond you want to dissociate, and then to  keep it there  long enough to break.


text:
Is there any formula that can be used to find loss of mass due to binding forces in atomic and sub atomic particles?
Atomic weight of Br-79 is 79.641 if you add the masses of protons and neutrons. However, in periodic table, it is less than the value given here. How is the difference arrived at for all the elements listed in the periodic table?

This is really a question about nuclear physics, not chemistry. See http://physics.stackexchange.com/q/60802/.

What do you mean by "arrived for"? Mass of nuclei can be measured but it's not chemistry.

@Mithoron It means atomic weights mentioned in periodic table are borrowed from the works of nuclear physics.

What do you mean by the atomic weight of Br-79 being 79.641? It's clearly not an atomic mass but a molar mass

Yes, there is the Semi-empirical mass formula which will give you the binding energy/mass of the nucleus.
Once you subtract the binding energy/mass from that of the nucleons, you have the mass of the nucleus.  Then you should add the mass of the electrons.  Finally, in principle at least, you would subtract the binding energy/mass of the electrons (in other words the ionization energies of the electrons), but this is a very minor correction.

Thanks. However, if you could provide the calculations for Br-79 it will be useful to understand how the average atomic mass is arrived.

Average atomic mass would mean the weighted average  of the masses of br-79 and br-81 https://en.m.wikipedia.org/wiki/Isotopes_of_bromine

You are right and I was working on that.

Yes - they call it the Theory of Relativity. E=mc^2
Take all the mass of the protons/neutrons lost in the nuclear reaction (you need to know your isotopes here). That mass is a tiny one, but then multiple that by the speed of light squared, and you have the energy created from the nuclear reaction.
Bang! Thats it! [A very big bang in fact...]


text:
Is a gas leak considered incomplete combustion?
This may be a dumb question, but if I leave the stove-top valve open (or I have a gas leak due broken tubing), would that be considered incomplete combustion?
From my knowledge, I figured a reaction is necessary for combustion to occur and that just the leaking of a gas would not qualify.

Your right, why doubt?

It is incomplete only of you think“no combustion” is part of the scale of incomplete combustion. Most people wouldn’t agree.

In climate science we call these leaks fugitive emissions. The chemical composition of the emissions are different. If you just have a leak, the emissions will consist of what ever your fuel is made out of, probably methane. If you burn your fuel and the combustion is not perfect you might sill have a littel bit of the original fuel left but most will have been used up and formed CO (incomplete combustion), $\ce{CO_2}$ (complete combustion) and $\ce{H_2O}$ .


text:
CuSO4.5H2O >> CuSO4 + 5H2O : chemical or physical reaction?
In a textbook it says it's a chemical reaction which involves breaking bonds.
But I don't think so since I think water molecules are just trapped in the crystal. What's the right answer and why ?

Copper II forms many types of complexes with water. The aquo ion is $\ce{[Cu(H2O)6]^{2+}}$ where the oxygen ligates to the metal atom. The structure is a distorted octahedral one, two bonds are longer than the other four. (The Jahn-Teller effect$^*$ is behind the fact that two ligands are not bound as strongly as the other four). It is possible with ammonia to form various hydro complexes; $\ce{[Cu(NH3)(H2O)5]^{2+}}$... $\ce{[Cu(NH3)5(H2O)]^{2+}}$. Similar complexes form with chloride ions. Copper sulphate is coloured in solution (due to $\ce{[Cu(H2O)6]^{2+}}$ ion) and in the crystal; the anhydrous crystal is colourless and this seems to indicate that the waters are bound, due to metal to ligand transitions, as in other transition metals. (The reason that sulphate is not significantly bound to the copper in aqueous solution, as it is in the crystal, is due to the vast excess of water (water is 55.5 M) which will displace the sulphate.)
In the crystal of $\ce{CuSO4.5H2O}$ the metal is, as usual, 6 coordinated, with four water oxygens in the plane and one oxygen from a sulphate occupying each axial position. The extra ($5^{\text{th}}$)  water molecule is hydrogen bonded between the second oxygen of the sulphate and a bound water molecule in the plane. This same water's oxygen is also $\ce{H}$-bonded to a sulphate oxygen bound to the next $\ce{Cu}$ atom. The $\ce{H}$ bonds are not quite of equal length  0.287 and 0.294 nm respectively.

(Fifth water oxygen thick black circle, sulphate hashed, $\ce{Cu}$ small circle, other four oxygens, large circles. Source: Wells 'Structural Inorganic Chemistry')
So bonds are broken, $\ce{Cu}$ to oxygen, also oxygen to oxygen and oxygen to sulphur hydrogen bonds as water is removed.
$^*$ The Jahn Teller theorem shows that any non-linear molecule in a degenerate state will undergo a distortion which will lower its symmetry and split the degenerate state. The theorem does not indicate what the distortion will be, or how big it will be. Nevertheless it is extremely important in describing the geometry of metal complexes.

Thanks. But I have two problems now: you said "the ion [CuSO4 (5H2O)5]2+..." why 2+ ?  Also i didn't somehow understand the second hydrogen bound "to a sulphat oxygen bound to the next Cu atom" would you explain it a bit more please? Thanks again!

I will rewrite this bit and will try and find a picture to make structure clear.

in your structure you have shown 2 sulphates bound to one copper, the formula indicates there should be only one ... see wikipedia https://en.wikipedia.org/wiki/Copper(II)_sulfate#cite_note-12 .... each sulphate is bound to 2 coppers and each copper to 2 sulphates

Yes, this is the experimentally measure crystal structure.

If water was acting as ligands and went from a complex to individual molecules, then it would be a chemical reaction.
There is no way to tell the difference without experience if that was the only expression explaining the compound.
There are more exact methods of expressing complex ions.

The nomenclature infers this is water of crystallization. Thats what the "dot" means.
This water is en-trained in the crystal structure.
The process of sublimation drives of the water (with heat for instance) and the crystal degrades. This change of state is a physical one, however there can be complex things happening at the molecular level in some systems such as organics that.
What you have here is Chem 101 - no chemical change for the context of the question.
If this was not water of crystalization, then the O would all be noted together in the formula, something like CuH10SO9. The other contributors are assuming there is a chemical change, but have missed the other information inferred by the notation.

I've thought about this same question quite a bit in the past and have come to the conclusion that there isn't really a single concrete answer.  If you consider the hydrate and the anhydrous to be distinct compounds, then yes, this would by definition have to be considered a chemical reaction.  The hydrate and the anhydrous compound have different physical properties, with color being the most obvious difference.  However, I generally view this process as a physical change.  The hydrate is, in essence, an ionic compound with water molecules contained within.  On strong heating, you ultimately end up with exactly the same compounds which you began with; the only thing that has changed conceptually is location.
It seems to me that if you are going to consider this process a chemical change, then you must also consider dissociation of an ionic compound in solution as a chemical change.  I don't think I've ever heard anyone try and make that argument before.  While dissociation clearly involves interactions between the solvent and ions, the overall process is widely considered to be a physical change.

This discusses crystalline solids. The difference between bound and unbound is not a dichotomous yes/no issue. Interatomic distance between two atoms may vary continuously from clearly bound to dispersively bound to unbound. If the interatomic distance between two atoms is significantly less than the sum of their van der waal radii, then it is assumed they are chemically bound. In answering the question about CuSO4.5H2O, all you need to do is identify each of the 15 atoms of the water molecules and determine (ignoring intramolecular distances) each of their closest neighbors and compare those minimum interatomic distances to the Van der Waals distances. I'm pretty sure you'll find distances substantially less than VdW for most, if not all, of the 5. [An alternate technique is to map the 3D electron density in the crystal, but I view distances as being 'cleaner' and hence more definitive.] To repeat one of my main points, it is often the case that interatomic distances are intermediate between a "full" covalent bond and just dispersive interaction. You may have two atoms 22.3% bonded, for instance. Hypothetically, since I've not studied the pentahydrate you ask about, what if the 5 waters are 0%, 20%, 40%, 60%, and 80% bonded, respectively? Perhaps a more useful answer for a novice is to ask whether the water molecules occupy definite positions in the crystal structure. If they are distributed randomly, then describing them as (physically) trapped might be justified. The implication is that the stoichiometry can range from 0 up to a maximum occupation (say 5) and this is NOT what we have for the pentahydrate; the water occupy specific locations in the crystal structure and their loss destroys the crystal structure. Making them much more than physically absorbed.


text:
What is ideal water hardness for brewing coffee?
While procrastinatingworking, I am wondering whether it is optimal to use softened (or deionized) water for brewing coffee, rather than tap water which is in my area rather hard.
My gut feeling is that extraction of the coffee beans' aromatic molecules will be made more efficient if there are no ions on the water, but is this backed up by science?

I realize there is already an unanswered question about tea brewing, but there are large differences between tea brewing and coffee percolation I believe…

Related: Does the hardness of water matter when making coffee?

There can be to much of a good thing. Where I live the water has too many ions in it making for terrible water. It is technically brackish. You can easily test this for a rough idea of where your water is by doing a total dissolved solids (TDS) test. I required a reverse osmosis system, after the water softener to get my tap water to a good baseline quality, Brita filters and the like do not lower the TDS. The RO system was money well spent for my water quality and taste, but test first. You can also call the water company to ask what additives they add so you know how to deal with them.

According to a discussion I had on this matter with a friend in the coffee processing industry, you should definitely consider yourself lucky for your naturally hard water.  We live in a region with very soft water and he pre-treats all of his coffee-making water with a bed of ground granite (as a physical substrate) and powdered gypsum $\ce{(CaSO4 *2H2O)}$.  For individual use he suggests mineral water, defined in Wikipedia as:

"The U.S. Food and Drug Administration classifies mineral water as
  water containing at least 250 parts per million total dissolved solids
  (TDS), originating from a geologically and physically protected
  underground water source."

The "taste testing" conferences attended by my friend have rigorously determined that DI water gives the "blandest" taste and water high in calcium and sulfur containing compounds to be the best.
Unfortunately, the chemical basis for this is poorly understood.  It could certainly be pH related, and that my friend's treatment bed just raises the pH which in turn better extracts the desirable components of coffee relative to the undesirable.  The same could be true of the mineral water and hard water in general should have a higher pH than soft water.


text:
Nucleophilic addition reaction of ketone to form alcohol - no acid treatment?
I'm preparing for a chem prac on reduction of a ketone group (in particular, the C=O in benzoin) using a nucleophilic addition reaction.
In the experimental method, all we do is react benzoin with $\ce{NaBH4}$, the nucleophile. The solvent we're using is a mix of ethanol/water. From what I've learned, the reaction forms a $\ce{C-O^-}$ alkoxide group which then needs to grab a proton from something to form the alcohol.
In the procedure, we mix ethanol/water, and that's it - no addition of dilute acid to then protonate the alkoxide. We're told to add hot water. Am I right in saying that addition of hot water provides a source of protons to protonate the alkoxide?
Thanks :)

The nucleophile is the H-, a hydride. In your SP3 ketone, the tiny Hydride will have no issue producing a high yield of the alkoxide.
The heat will simply make the reaction faster and drive the equilibrium further.
Ethanol + water is a mild acid, so no issue with protons to complete your alcohol, and the heat makes that mix slightly more active as a proton donor.
You end up with a new equilibrium of your 2 alcohols competing for the proton, however with plenty of methanol and some careful washing with hexane or toluene you should be able to get your alcohol out of that mixture...


text:
Why does zinc react while its outermost suborbital is full?
Why does zinc lose electrons while its outermost configuration is 4$\rm{s}^2$ 3$\rm{d}^{10}$ both outermost suborbital are full.
Why does it react or lose electrons?

You could ask the same question about the alkali earth metals with their $\rm{s}^2$ valence shell. While there is an energy cost to removing electrons, in a spontaneous reaction (e.g. $\ce{Cu^2+_{(aq)} + Zn_{(s)} -> Cu_{(s)} + Zn^2+_{(aq)}}$) there will be a compensatory transfer of energy elsewhere (like the reduction and deposition of $\ce{Cu^2+}$) which leads to a more stable overall system.

The further out the electron orbital the less energy is associated with a move. The issues is not so much about the stability of the zinc, but the total system. If a lower energy state can be reached by the electron changing orbitals then that is what will happen.
ie it's energetically preferable for 2Zn & O2 to exist as 2ZnO than as the original materials.

Zinc has quite a stable configuration, but that does not mean it does not want to attain noble gas configuration. You should also observe that $Zn^{++}$ is more stable that ground state of Zn. This is due to the fact that both $e^-$ from $4s^2$ orbital are removed and the Zn ion has a fully-filled d-orbital, which is much more stable than the ground state configuration.  Thus Zn reacts to get more stability provided by the fully-filled d-orbital.

FYI: Zn has positive ionization energy, therefore $\ce{Zn^{2+}}$ is less stable than $\ce{Zn^0}$

You got it the other way round. If the ionization energy is higher (it's always positive, FYI), it indicates that the configuration is more stable.

@downvoter, want to disclose the mistake?

Ionization is a reversible process.  If it takes X kJ/mol to go from M+ to M2+ then it releases that energy going from M2+ to M+.  A large portion of this answer is just balatently wrong.


text:
Find the maximum temperature attained by the gas during the change
The main question is:

2 moles of an ideal gas is changed from its initial state $(16\ \mathrm{atm},\ 6\ \mathrm L)$ to final state $(4\ \mathrm{atm},\ 15\ \mathrm L)$ in such a way that this change can be represented by a straight line in $P{-}V$ curve. Find the maximum temperature atained by the gas during the above change. Take $R=\frac{1}{12}\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$.

My approach :
I plotted the initial and final conditions on a $P{-}V$ curve using a Cartesian coordinate system. Joining them gives me the temperature transition.
I also made two isotherms (hyperbolic curves) on the same graph, passing through the points a previously plotted. The values of initial and final temperatures I calculated are
$$\begin{align}
T_\mathrm f&=360\ \mathrm K\\
T_\mathrm i&=576\ \mathrm K
\end{align}$$
I tried finding a function in $T$ which I could maximise, but in vain.

T is maximum when PV is maximum.  So get the equation for the straight line through the two end points, expressed in the form $P=aV+b$.  Then write $PV=aV^2+bV$.  Find the maximum of the function $aV^2+bV$.  It is at $V=-b/(2a)$. So the maximum of PV is $$(PV)_{max}=-\frac{b^2}{4a}$$ This is equal to $nRT_{max}$
From fitting the straight line to the end points, the slope obtained is a = -4/3, and the intercept is b = 24.  So, from the equation above, the maximum value of PV is $$(PV)_{max}=108\ liter-atm$$

The ideal gas law $pV = nRT$ allows us to relate one of $p$, $V$, and $T$ to the two other variables, and is the main equation to work with here.
Rephrasing the original question to make it more suggestive, we are interested in maximizing $T(p,V)$ subject to the constraint $p(V)$, a linear function in $V$ readily found from the conditions specified in the problem. Substituting $p(V)$ for $p$ to get $T = T(p(V),V) = T(V)$, we have reduced the problem to one amenable to simple calculus techniques, and the answer follows from taking a derivative.

Remark. In general, extremization problems are not so easy, and we will need to use the more powerful tool of Lagrange multipliers.


text:
Reactivity change of periodic table
How does reactivity change as you move from one side of the periodic table to the other or if you move from the top to the bottom?

please define reactivity more thoroughly

I'm assuming you're asking about how the CHARACTER changes as we move all about the periodic table, and based on that how does the reactivity change.
As we move from left to right, the electropositive character decreases, and the electronegative character increases. For example, take the 3rd period. Sodium's reactivity is pretty high, because it readily gives away an electron to get a stable configuration. As we move towards the right, the reactivity decreases, but again it starts rising and reaches a high vue when we get to chlorine, which readily accepts an electron to get a stable configuration.
When we go down a group, electropositive character increases, but electronegative character decreases. Thus, if we may see the Li-family, reactivity increases down the group, but if we look at the halogen family, i.e. the F-family, reactivity decreases.


text:
Phase transition for an ideal fluid
Does a phase transition occur also for an ideal fluid (from ideal gas to ideal uncompressible liquid)?
Or ideal fluids can be modelled only as ideal gas or uncompressible liquid at all, without the possibility for a phase change?
I hope the question is well posed, thanks

Ideal gas and ideal fluid are not the same. Ideal fluid has molecular interaction that is even throughout different entities, and has no viscocity, and no compression.
Yes, ideal fluid is modeled so that it will go through phase change, but you would have to consider the gas from ideal fluid as a pseudo- real gas, since it has molecular interaction.
An ideal gas has no volume, has no interaction, will not go through any sort of phase change.
Hope this helps.

For sure any answer is helpful! So, correct if I'm wrong, if I assume that water vapour is an ideal gas, it will be always in the gas phase and hence the saturation dome will not exist. In the same way if I assume water as an (ideal) incompressible liquid?

Yes. If you assumed water vapor as an ideal gas, no saturation dome would exist. Ideal incompressible liquids will  have vapor pressure, and will saturate in a finite container given that there is sufficient amount of liquid.


text:
Geometry optimization: dependence on initial geometry
One would think that initial configuration should not significantly impact the final optimized geometry. Starting with two moieties, the final "optimized" geometry is not unique and depends highly on how they are placed initially. Is there a way to avoid this? Otherwise one could never be sure that they have ended up with the global minimum.

Starting with two moieties, the final "optimized" geometry is not
  unique and depends highly on how they are placed initially. Is there a
  way to avoid this?

Yes and no. No, because, as OP anticipated, starting from an input geometry the optimisation algorithm will indeed find the closest (from the algorihtm point of view) local minimum and not necessarily the global one. And yes, because by performing an extensive conformational analysis, one in principle can find structures that correspond to all possible local minima and then identify the global one.

Operative words on that last sentence being in principle

@F'x, sad but true. In practice, even for relatively small molecules, an extensive conformational analysis requres too much (CPU) time, so that often one has to resort to the help of "chemical intuition" to reduce the cost or even avoid conformational analysis completely.

@Wildcat - I'm not sure I'd say that "extensive conformational analysis" requires too much time. Exhaustive (systematic) searches are exponential in the number of rotatable bonds, but there are many conformer search strategies that are pragmatic.

@GeoffHutchison, but such pragmatic non-systematic strategies (that use e.g. random sampling or "chemical intuition") could not provide 100% guarantee of finding the global minimum.

No, but "I'm not going to do any conformational sampling because I don't have a 100% guarantee of finding the global minima" seems a bit extreme. Trying some strategy is better than nothing and can at least indicate if the property of interest has much dependence on the conformer geometry.

Your first comment implies it's OK to avoid conformational analysis. I'm saying that I personally don't recommend avoiding conformer sampling, and there are pragmatic solutions.

Trying some strategy is definitely better. And I think it can also be mentioned in the answer that some properties aren't so sensitive to small changes in geometry, so that finding the global minimum might not be so important in the first place.

And no, I didn't mean that conformational analysis has to be avoided. On the contrary, I personally always do it (and I do it as systematic as possible). In my first comment I described a scenario (replacing conformational search by "chemical intuition" ) that I don't like and don't support.

One would think that after years of development, computational packages like Gaussian must have incorporated ways to escape local minima automatically during optimization. The necessity of manual conformational analysis or chemical intuition or any user intervention could be minimized by strong optimization algorithms. My question was about specific keywords or techniques in Gaussian package that incorporate stronger optimization.

@ChemInformationist - Gaussian certainly does not. To my knowledge, no standard computational package includes any type of global optimization strategies built-in. In many cases people want a local minima because they have some chemical intuition or the molecule is rigid (e.g., benzene, water). Moreover, there are many such techniques in chemistry and it's hard to have a one-size-fits-all approach because of the diversity of molecular systems and the large # of degrees of freedom.

@ChemInformationist There's no general algorithmic way to know if a given local minimum is global or not; the first two sentences of Deathbreath's answer are inescapably correct. All you can do is come up with methods that give you a certain confidence that you've explored "enough" of the conformation space. One interesting system studied by such a method is periodane.

It is not possible to derive an algorithm that finds the global optimum independent of the initial structure. This is a general fact of non-convex optimization.
The potential energy surfaces of stable compounds are not globally convex because the "dissociation" energy of any two components is finite at infinite separation.
Although there are so-called "global" optimization methods, no method that does not search every point of the search space can in general guarantee to have found the global optimum (i.e., there is always a function for which the algorithm fails to find the global optimum or visits every point in the domain).
That said, global optimization methods putatively reduce the sensitivity of the result to the initial point. There are a large number of these, some of which are deterministic (lists definitely not exhaustive since it's missing my favorite level-filling method :-).
Finally I would like to point out that the specific formulation of your problem influences how well your algorithm works, e.g., the chosen coordinate system or how any constraints are enforced. The lack of global optimization is actually useful if a certain bonding topology is supposed to be maintained; otherwise, t-butane would always optimize to n-butane, for instance.


text:
Which of these are more reactive: the lanthanides or aluminium?
I am not sure whether to consider the standard reduction potential or ionisation enthalpy. Which of these are more reactive: the lanthanides or aluminium?

Reactive towards what?

That has not been mentioned.The statement says "Lanthanides are more reactive than aluminium,is this statement correct or incorrect?" and I believe it's pretty ambiguous to ask such a thing.

Since the lanthanides are neither in the same period nor in the same group as aluminium, comparision is actually not feasible. Factors like atomic size and effective nuclear charge tend to predominate such comparisions, rather than ionization enthalpy or reduction potential.
Bear in mind that due to presence of electrons in f-orbital for lanthanides provide poor shielding of electrons farther away from the nucleus. This poor shielding is due to the fact that f-orbitals are quite diffused in nature. Yet, we cannot say anything about reactivity of lanthanides when compared to aluminium.
If you are still a high-school student, I suggest you do not ask such hypothetical questions.

Well this was asked in our country's entrance examination and I had the same views as you. I felt it was quite ambiguous to compare them and probably there was some more in depth explanation which I did not know of.


text:
how do carriers actually stabilize nitroglycerine i.e. Dynamite? Do they disrupt the shockwave or free radical chain?
To reiterate-does it prevent a shockwave from propagating, or does it disrupt the light, heat or free radical chain necessary for Detonation?? On a related side-question, does the Blasting Cap then work by breaking up the individual "cells" or by providing free radicals to initiate the explosion?

I believe the answer is that by binding the explosive chemicals they are as not free to interact with each other.

It might, of course, be that the key thing that is altered is the activation energy of the initiation step. This is pretty low for liquid nitroglycerine but much higher for dynamite.

Nitroglycerine is the only chemical involved (actually, with Dynamite, they add sodium nitrate and other additives.) The nitrogens combine and the oxygen burns up the carbon.


text:
Graphitic acid-Its structure and properties?
What exactly is graphitic acid? I know that it's obtained when you react graphite with concentrated $\ce{HNO3}$ and that its structural formula is $\ce{C11H4O5}$. But I can't find or deduce its structure. And also I am unable to find any of its properties, except that it's a yellow-green liquid insoluble in water.

$\ce{C4H11O5}$ is not a valid formula for any uncharged-non radical compound.  Perhaps this is the empirical formula?

Sorry it's C11H4O5

https://en.wikipedia.org/wiki/Graphite_oxide

Graphitic acid is a compound derived from graphite.
You may find it named "Graphite oxide" as well: as you see, it is obtained by reacting graphite with an oxidizing agent ($\ce{HNO3}$, for instance).
The structure resembles that of graphite, but it carries several oxygen-containing substituents, such as hydroxyl and carbonyl functions, thus having a less regular geometry, compared to that of graphite.
The exact features of the structure depend a lot on the method used for its preparation, and the layers may show some "defects", not described by this ideally planar model, but the key concept is this.
Here is a simple proposed structure for a single layer (source: Wikipedia)

Note that, just like graphite is made of several layers of graphene, graphite oxide is made of several layers that, upon reaction with a base, become graphene oxide.
Here you can find further information, and a nice image of a single graphene oxide film


text:
Confusion about the choice of axis for Molecular Orbital Theory
This video from MIT OpenCourseWare says that pi bonds and anti-bonds are for 2px and 2py, i.e., $\pi_{2px}$, $\pi_{2py}$ and $\pi_{2px}^*$, $\pi_{2py}^*$ and sigma bonds and anti-bonds are for 2pz i.e., $\sigma_{2pz}$ and $\sigma_{2pz}^*$.
But in my reference books it says pi bonds and anti-bonds are formed for 2py and 2pz i.e.,  $\pi_{2py}$, $\pi_{2pz}$ and $\pi_{2py}^*$, $\pi_{2pz}^*$ and sigma bonds and anti-bonds are for 2pz i.e., $\sigma_{2px}$ and $\sigma_{2px}^*$.
Which is actually the right one? Please explain.

It depends on the choice of the bond axis: I guess it was chosen to be $z$-axis in the MIT lecture, but $x$ in your book. Hence the difference.

Representing it in either way is correct? If so, can we choose $y$-axis as the bond axis? Add your answer below so that I can mark it as the correct answer.

Right. In principle, you're free to choose any direction as the bond axis. But the convention is to choose $z$-axis for this purpose.

Add your answer below so that I can mark it as the correct answer.

Although we could not see what is written in OP's "reference books", the mentioned difference most likely is the result of different choices for the bond axis.
Conventionally, and as it is (implicitly?) done in the MIT video lecture, it is chosen to be the $z$-axis, in which case $\pi$ bonds are indeed formed from $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals while $\mathrm{p}_z$ can take part in $\sigma$-bonding. If, however, the bond axis is chosen to be the $x$-axis, then $\pi$ bonds are formed from $\mathrm{p}_y$ and $\mathrm{p}_z$ orbitals while $\mathrm{p}_x$ takes part in $\sigma$-bonding.

As other people says x y and z directions are arbitrary, However, the tradition that open courseware uses comes from the diatomic molecules. There the atom-atom axis is assumed to be the z direction and then the y and z, being perpendicular, correspond to the direction of pi bonds. A similar use is to design sigma to the atom-atom direction and pi to the perpendicular ones. This tradition of diatomic molecules is then exported to polliatomic examples. See any good book on molecular spectrocopy (Jeanne L McHale, Molecular Spectroscopy) for instance.


text:
aluminum as cathode electrode
Aluminum really wants to and will form aluminum oxide. Upon research, aluminum oxide is fairly unreactive. I wanted to ask if I had aluminum anode which would quickly form aluminum oxide (which is unreactive), will  that inhibit the aluminum from being oxidized? Basically, will aluminum metal loose electrons and become an aluminum ion?

Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions.
As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere.
In an aqueous solution, if the potential is high enough, $\ce{OH-}$ and $\ce{O^2-}$ are generated at the anode, which can then react with the aluminum to produce aluminum oxide.
$\ce{Al^3+}$ can also be generated directly.
The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the $\ce{Al^3+}$ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode.
In this way, coatings thicker than the normal passivation in air can be produced.
However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made).
This is the normal behaviour in aqueous solutions at near-neutral pH (5–7). [1]
However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface.
One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed $\ce{OH-}$, leaving pores in the oxide layer through which the ions can travel and continue to react.
These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.

ok so is it fair to say that aluminum would not be a good anode material in a cell due to the incapability of not being able to transfer/carry electrons due to the oxide layer

It depends on the specific application and if the potential is low enough, it is possible to avoid significant oxide formation, but you are right that they do not make good anodes in a lot of cases. However, there have been some interesting developments in things like aluminum-air batteries to overcome issues with aluminum anodes.

Very much so. All Aluminum you have ever seen has had a thin layer of aluminum oxide coating it.  The outer most layer of aluminum metal instantly oxidizes in air and forms aluminum oxide.  This aluminum oxide layer protects the aluminum from further oxidation because the oxygen ions formed are too tightly bound in the matrix to diffuse through to the inner aluminum atoms.
If this layer was permeable to oxygen on a reasonable time scale than the aluminum would degrade rapidly.  For example, if mercury is added to aluminum the oxide layer becomes permeable to oxygen and you can watch aluminum break down rapidly.
Youtube vid: https://www.youtube.com/watch?v=Z7Ilxsu-JlY, thanks to Mr. Gray for his video.


text:
The effects of H2o2 (hydrogen peroxide) with oils
I have some 35% $\ce{H2O2}$ that I've been cutting down to 3% to create a product. Cheaper than buying pre cut solution from store.
However, I've been thinking about trying to try it out with producing a new product which would induce mixing of the $\ce{H2O2}$ with pure oils and essential oils. I am not sure if the $\ce{H2O2}$ would break down the oils and render them useless in the product or not.
I select $\ce{H2O2}$ cause I want it to be able to kill germs to help kill a particular smell. But I want the oils to maintain their properties in which they keep there defined smells. So pretty much the $\ce{H2O2}$ is a catalyst to make to oils more liquidity able to spray and also helping hide original scent.

Oils will oxidize (and burn, too). With dilute H2O2, it's not a rapid reaction (in laymans terms)... some oils (like linseed) will oxidize more rapidly than others (like mineral oil).
Essential oils are not really oils... they are more like phenolic and terpenic extracts.

And hydrogen peroxide decomposes much more rapidly when contaminated (except by phosphates). You'd wind up with partially oxidized "oils" and water.

Would yall have a suggestion of a better substance to use then? I am pretty much trying to make something that will mask normal smell and replace it with the smell of an oil:ex ( acorn, pecan etc..). To make a smell mask i mix H2o2 with baking soda. { H2o2 to kill germs, baking soda to deodorise a neutral smell} . Hence the reason i suggested using H2o2

@JonathanSzychulski How about febreze? The active masking-ingredient in febreze is hydroxy-propyl-beta-cyclodextrin (plus various scents).

yes i was just reading about what was inside febreze. Thing is i cant really have a "fragrance" smell. It is for hunting, in which i own a outdoor brand. so therefore i need to hide to human scent . (which i can already reproduce) but have been working on a way. To mask the human scent but add a natural smell from the woods. I can accomplish this by just boiling up a tea type substance outta items found in the woods but after a week- a month it breaks down. so trying to find away to make it last longer which id assume finding the right chemicals would do the trick

@JonathanSzychulski hydroxypropyl beta-cyclodextrin is the thing that would mask your scent. I can't recommend a good preservative for dirt tea (for use as a cologne). I might recommend geosmin (aka petrichor oil) as a scent.


text:
Do modern dispersion-corrected DFT methods give more accurate molecular geometries?
I was discussing with a colleague the use of modern dispersion-corrected density functionals. I take it almost as a given that the methods generally produce "better" (for some definition) geometries, in addition to more accurate predictions of thermodynamic and other properties.
Naturally, since the dispersion correction involves intermolecular interactions (e.g., van der Waals complexes), they should do better than "regular" functionals. I'm curious about intramolecular geometries.
Are there any benchmark comparisons on bond lengths, angles, etc. comparing different computational methods (DFT and otherwise) to crystal structure or other geometries?
In other words, how accurate are different computational methods at getting geometric properties of individual molecules?

I remember that I did my own very small benchmark a couple of years ago and found almost no difference for geometries of few small organic molecules from my studies. But in my systems there were only (intramolecular) H-bonds, there was no any dispersion-bonded complexes. At that time the only more-or-less extensive study I found was this one. See specifically the "Geometry Optimization with DFT-D" section at pp. 564-567.

I quote from the above mentioned article: "The empirical dispersion correction [...] remarkably improves the intermolecular geometries if at least a TZVP basis set is used. This applies especially to the dispersion-bonded complexes, where the RMS error is reduced up to 25 times [...]"

@Wildcat - I appreciate the comments about intermolecular complexes and geometries. I should have been more specific that I was curious about intramolecular parameters and geometries. I have the sense that should be a bit better, since there are intramolecular hydrogen bonding and dispersion effects too, but haven't found much about it.

A surface science paper by Dr. Greely discussed the effects of vdw correction in furfural adsorption and hydrogenation.
http://ac.els-cdn.com/S0039602813003518/1-s2.0-S0039602813003518-main.pdf?_tid=3f046148-5e4f-11e6-958b-00000aab0f27&acdnat=1470760693_12023ff08a1227d76768632b8300805b

@GeoffHutchison I'm not aware of an extensive review, but there are several papers looking at the influence of vdW corrections in peptides. Also don't Truhlar's fitting papers on "weakly bound" complexes include dispersion corrected functionals.

Although most papers look at intermolecular effects, these should include the information necessary to decide on intramolecular structural effects. This paper is a nice summary of effects. Erin Johnson is very active in dispersion as is Alexander Tkachenko. One of their papers might have what you seek.

I did a benchmark study on a few simple dimer systems (about 8 atoms in total) with little to significant dispersion components depending on the conformation.  DFT-D did better in general for these types of systems (full optimizations) than more straightforward DFT approximations.  (EDIT: the geometries gave rise to better binding energies.  I didn't compare the actual bond lengths, angles, etc. or anything).

Another example where dispersion is critical is: Angew. Chem. Int. Ed. 2010, 49 (41), 7440–7443

Disclaimer: These articles are from my old group and myself. But they may serve as a starting point for someone's own investigation, even if one disagrees with our conclusions.
The Grimme group, me included, looked at experimental, back-corrected rotational constants of smallish to medium-sized organic molecules a while back. Our conclusion was that a modern dispersion-correction, i.e. D3(BJ) is beneficial for geometry optimization at virtually non-existent computational cost. The references are: T. Risthaus, M. Steinmetz, S. Grimme, J. Comput Chem., 35, 1509 (2014), M. Steinmetz, S. Grimme, Phys. Chem. Chem. Phys., 15, 16031 (2013).

With that first link, your anonymity is now toast. ;-)

I am afraid I do not know of a paper specifically discussing the effects of DFT dispersion corrections on intramolecular properties. However, it should be noted that for short-range descriptors (such as bond lengths and bending angles) the dispersion corrections will not influence those as they correspond to distances below the typical distance at which the corrections take place. The various schemes differ in how they work, but typically the corrections (in Grimme's schemes) are damped below a distance of ~6 Å. It is also my experience that such local structural characteristics are not affected by dispersion corrections.
However, you are right that dispersion corrections will influence intramolecular structure at larger scale, and in particular for “soft” or “flexible” long molecules with many conformations. There, it is worth noting that the so-called “dispersion” corrections actually correspond to the lack of accounting in local DFT functionals of middle-to-long range correlation effects. Thus, it is actually expected that the corrections — or nonlocal functionals including dispersion terms — actually perform better for those too.
Unless, like Grimme's 2006 two-body dispersion correction scheme (often called “D2”), the dispersion corrections happen to be often too strong and make intramolecular distance somewhat too small (instead of much too large)!

Edit: Now that I come to think of it, one extreme example of intramolecular system that has been studied in detail w.r.t. dispersion corrections is that of helicenes. It was shown (for example there and there) that dispersion corrections improve the accuracy of computational structures and energies compared to experimental (or high-level computational) data.

We're finishing up a benchmark of conformer scoring for exactly this reason. Our working assumption is exactly that middle-to-long-range effects (e.g., between conjugated segments or intramolecular hydrogen bonds) are not well treated by existing force fields used heavily in conformer generation, nor previous quantum methods. Thus, using at least dispersion-corrected semiempirical methods is needed.

@hBy2Py Just remove the last forward-slash and it works. (DOI: 10.1021/jp404252v)

Yes, we've done some helicene work so I'm aware of those papers.

Our group just published a benchmark, considering ~6500 conformer geometries across ~650 molecules. Using high-level DLPNO-CCSD(T) / def2-TZVP energies, we found a huge increase in accuracy (from several metrics) when using dispersion correction at minuscule cost in time.
'Assessing conformer energies using electronic structure and machine learning methods' Int. J. Quantum Chem. 2020
For example, the mean $R^2$ correlation between B3LYP and DLPNO-CCSD(T) energies was 0.706 without dispersion and 0.920 with -D3BJ dispersion correction. Similar effects were seen with other functionals.
Our conclusion is that intramolecular non-bonded interactions are handled better, and as mentioned by F'x reflect medium-to-long range behavior of standard chemical functionals.
tl;dr - use dispersion correction for any density functional calculation. Even on molecular geometries it's generally better.


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difference between hexane and hexane fraction
i'm doing a chemistry assignment and cant find the difference between hexane and hexane fraction. i think it means there are impurities but not sure so it would be nice if someone could help. for the the fuels i am doing hexane fraction and 1-hexanol. if it means there are impurities that will impact the experiment.
when i search google "hexane fraction" i get stuff on hexane not hexane fraction.

Hi ethan, welcome to Chemistry StackExchange! Your question is not very clear as it is, but perhaps this will help: fraction has a very specific meaning in chemistry, in separation processes.

The difference is simple. Hexane, if that is what the label says it is, should mostly be n-hexane. But hexane fraction is any hydrocarbon that distills at close to the same temperature as n-hexane. This mostly means a mixture of other six-carbon hydrocarbons (e.g. 2-methylpentane) most of which boil between 55 and 70 °C. You might find that the mixture contains some C-7 hydrocarbons as well.
For many reactions what you need is some volatile hydrocarbon and it isn't that important that it is a mixture. Occasionally, though, this will affect a crystallisation (especially if the hydrocarbon is bound in the crystal lattice). Pure hexane will work better in that case.
The reason why they are sold as different things is that pure hexane is a lot more expensive to produce as it will require careful distillation to separate it from the other, similar boiling point, hydrocarbons. So you don't want to use it if the cheaper mixture works. The cheaper mixture will just be a fraction obtained from crudely distilling petroleum at around 60–70 °C.

I'm not sure this is the case, for the hydrocarbons distilling between various temperatures (in the UK at least) they'd be sold as 30-40/40-60/60-80 petroleum ether (Pet Ether).  With hexanes vs n-hexane denoting whether the bottle was pure n-hexane or  a mixture of C6 isomers.


text:
Heaviest atom with which chemistry has been performed?
The nucleosynthesis of superheavy atoms such as transactinides is an exciting area of nuclear physics, but not much of a playground for chemistry because
My question is: what is the heaviest atom ever used in chemistry? For this I define chemistry as “making a compound with the atom (molecule or material) and characterizing it to some extent”.

The one source I found on the topic is “Critical evaluation of the chemical properties of the transactinide elements (IUPAC Technical Report)”, dated 2003, but it seems quite old for such a rapidly developing field. Also, despite being title “chemical properties” it mostly reports on physical properties.

I'd say chemistry ends somewhere around Z=100. Sure, there were some examples of reactions involving heavier elements, but it's not like they were used in chemistry; it's more like an elaborate chemical setup built around a tiny bunch of atoms in the hope of catching a glimpse of their fleeting existence. You can't really characterize a lone molecule. Or maybe you can, but it's more about physics.

The Chemistry of the Actinide and Transactinide Elements, 2010 edition, reports compounds up to hassium (Z=108), namely $\ce{HsO4}$, and  as-of-then unsuccessful attempts at chemistry up to copernicium (Z=112), dated around 2002-2003. Not sure about anything more recent. Wikipedia agrees that meitnerium (Z=109) is currently the first element with no reported experimental chemistry.

Although it's completely hypothetical, Periodic Videos  postulates the heaviest possible tetrahedral molecule.

According to Adsorption behavior of super-heavy elements (Z
[greater than or equal to ] 112) on metal and inert surfaces :

Nowadays the research focusses on the chemical properties of
  the element 114. So far, two gas chromatography experiments to study the interaction strength of
  element 114 with a gold surface have been performed with conflicting results. One experiment [reference 1]
  reported a weak interaction of element 114 with a gold surface, leading to adsorption only at very
  low temperatures of approximately –90°C, while in the second experiment [reference 2] adsorption on gold
  has been observed at the room temperature, indicating a much stronger bond between element 114
  and gold.

See also the earlier: Chemical characterization of element 112 :

We therefore conclude that the stronger adsorption interaction of element 112 with gold involves formation of a metal bond, which is behaviour typical of group 12 elements

The heaviest element with a known compound, as of this writing, is copernicium, atomic number 112, which has been combined with selenium:

Because the lighter group 12 elements often occur as chalcogenide ores, experiments were conducted in 2015 to deposit copernicium atoms on a selenium surface to form copernicium selenide, CnSe. Reaction of copernicium atoms with trigonal selenium to form a selenide was observed, with $-ΔH_{ads}^{\ce{Cn}}(t-\ce{Se}) > 48$ kJ/mol, with the kinetic hindrance towards selenide formation being lower for copernicium than for mercury. This was unexpected as the stability of the group 12 selenides tends to decrease down the group from ZnSe to HgSe.[1]

Cited Reference

Paul Scherrer Institute (2015). "Annual Report 2015: Laboratory of Radiochemistry and Environmental Chemistry" (PDF). Paul Scherrer Institute. p. 3.


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Excited-states properties with traditional density functional theory
Recently I was shown the book "Density Functional Theory II, Relativistic and Time Dependent Extensions" by Nalewajski in which a chapter from E. K. U. Gross et al. is printed. [1] This chapter contains an interesting statement in it's introduction:

Excited-state properties, however, are notoriously difficult to calculate within the traditional density functional framework and time-dependent phenomena are not accessible at all.

What are those excited-state properties that are "notoriously difficult to calculate" and what makes it hard to calculate them without the TDDFT framework?

Literature:
[1] Gross, E. K. U.; Dobson, J. F.; Petersilka, M. Density Functional Theory of Time-Dependent Phenomena. In Density Functional Theory II; Nalewajski, R. F., Ed.; Topics in Current Chemistry 181; Springer Berlin Heidelberg, 1996; pp 81–172.

Conventional density functional theory (DFT) is strictly limited to describing the electronic density of ground electronic states. This is because the Hohenberg–Kohn theorems, on which it is based, are restricted to non-degenerate ground states (with no magnetic field). Also, because DFT aims at solving the time-independent Schrödinger equation (i.e. it explicitly finds stationary solutions), time-dependent phenomena are not accessible.
From the ground state density, any property of the system including excited state properties could in theory be derived given an appropriate functional[1] $X[\rho]$ that maps the ground state density $\rho$ onto the property $X$. However, in practice, no known functional exists that maps the ground state density to excitation energies or other excited-state properties. Thus, excited states and their properties are in practice unreachable by DFT: I suppose this is what is meant by Gross when stating them as “notoriously difficult to calculate”. I myself would have said impossible.
Finally, to give a short list of properties that can be described with TDDFT but not with DFT — are thus all excited state properties. In particular: bandplots/bandgaps, excitation energies, transition moments, photo absorption spectra, frequency-dependent response properties (optical & dielectric properties), etc.
Specifically, one has to be careful about band plots and band gaps, which are routinely reported from time-independent DFT calculations: they are a representation of the Kohn–Sham energies, i.e., the energies of the fictive non-interacting Kohn–Sham system, which has no physical interpretation at all.

[1] Note here that we are talking about a functional in the generic, mathematical meaning of the word. It is not the exchange–correlation functional.

There are quite a few papers out there highlighting some examples. I particularly like the following paper titled "Failures of TDDFT in describing the lowest intramolecular charge-transfer excitation in para-nitroaniline" because this is TDDFT getting it absolutely wrong for small molecule like para-nitroaniline.

Welcome to Chemistry.SE! This answer is basically a 'link-only' answer. If the content the link points to changes or is removed, this answer becomes worthless. It also points to an article that might be behind a paywall for some users. Feel free to improve your answer by summarizing some of the key points of the article you linked.


text:
Why elements with high atomic mass (like tungsten) are used in the production of X-rays, in the Cathode-Ray Tube?
I was studying about the production of X-rays, and was struck at how elements having high atomic mass (like tungsten) are used for the production of high energy X-rays. I am wondering why low atomic mass elements are not used for the production of high energy X-rays.
I have read that "tungsten has the highest melting point of any metal, and because over 99 percent of the energy put into the X-ray tube to produce the beam is turned into heat energy, the ability to dissapate the heat without loosing structural integrity is critical".

If a sample containing a particular element is irradiated (e.g. with photons or, in case of an X-ray tube, with electrons) with an energy high enough to excite inner electron orbitals, X-rays are emitted in the de-excitation. In contrast to the continuous bremsstrahlung energy spectrum, the energy liberated in the transition from the excited to the ground state takes the form of a characteristic X-ray photon whose energy is given by the energy difference between the initial and final states. If the temporary vacancy has been created in the K shell of an atom, then a characteristic K X-ray is liberated when that vacancy is subsequently filled. In particular, if the concerned electron comes from the L shell, than a Kα photon is generated whose energy is equal to the difference in binding energies between the K and L shells.
The K-series X-rays are generally of most practical importance because their energy is greatest. Their energy increases regularly with atomic number $Z$ of the element, for example, only about $1\ \mathrm{keV}$ for sodium ($Z=11$) and about $100\ \mathrm{keV}$ for uranium ($Z=92$). An empirical law concerning the characteristic X-rays that are emitted by atoms is expressed by Moseley’s law.
Hence, in order to produce characteristic X-ray photons with a high energy, K-series X-rays of a heavy element are preferred.
Since only a small fraction of about $1\ \%$ of the energy generated in an X-ray tube is actually emitted as X-rays and the rest of the energy is released as heat, the temperature of the anode target can be very high during the operation of the X-ray tube. Therefore, the anode has to be made of high-temperature materials. You already mentioned the high melting point of tungsten ($Z=74$); whereas for example the heavier elements lead and bismuth cannot be used because of their low melting points.

Since I just had to look this up myself: X-ray spectroscopy uses a set of historical names for the electron shells, X-ray notation.  "K" and "L" correspond to principal quantum numbers 1 and 2, respectively. The notation seems only to accommodate a subset of the orbitals, for instance there are three 2p orbitals but only two L-codes for them (this may be because some of the 2p orbitals produce indistinguishable emission lines).

So, practically is it possible to use heavy metals like Os, Ir, Pt etc. which have high melting points?

In places like X-ray crystallography systems and hospitals, copper (Z=29) is often used for the anode. I think this partly because of the good thermal conductivity but it may be that the lower energy K-shell X-rays are better suited for those applications.

Secondary issue: Yu don't want any ir that might get in to turn your metal into oxide liquids/vapors, so the oxides have to be refractory too. Osmium would be bad.


text:
Why Cu looses 2 electrons to become stable instead of gaining one electron to fill its 4s2 3d10 configuration?
Cu has 4s1 3d10 configuration so why it does not gain 1 electron and complete its suborbital. Why it looses its 2 electrons to become Cu++ which is more stable. Why?

http://chemistry.stackexchange.com/questions/11069/why-can-cu-have-an-oxidation-number-of-2

@Mithoron This question is about Cu2+ versus Cu-, while the question you linked is Cu2+ versus Cu+.

Copper can gain 1 electron to become Cu-.
Copper has an electron affinity of 1.236 eV, so Cu- is stable to electron loss by 1.236 eV.
See also "The Dissocationi Energy and Heat of Formation
of the Molecule NaCu" Z. Naturforsch. 28a, 316-317 which says:

But unlike as for NaAg a noticable ionic contribution
  to the bonding is observed for NaCu

Also there is Resonant two-photon ionization spectroscopy of LiCu :

The unusual shape of this curve is thought to derive from avoided
  crossings between the $\ce{Li^+Cu^-}$ ion pair state and covalent states, with the $\ce{Li^+Cu^-}$ ion pair state
  ultimately correlating to the ground electronic state of the molecule.

So if Cu is to make a compound with an electronegative element, Cu should lose electrons, but with an electropositive element it can gain an electron.


text:
Best ingredients for a homemade desinfectant
I am trying to determine the best recipe/ingredients for a homemade desinfectant/germ killer. I've been researching these as of now:

Isopropanol (or even Ethanol)
White vinegar
Borax
H2O2
Specific Essential oils (that are supposed to kill germs)

What I want is something liquid, that you can either spray on your hands or on a toilet seat for example, and kills the most germs possible while being easy to make/mix. It should not be corrosive/toxic etc.
Which of these products can be mixed together? Is there any other products I didn't research? Will mixing any of these cancel the effects of the other?
Thank you!

This question probably belongs in the DIY or Biology SE.

I think it may belong at least a little bit here, as I'm looking to mix some "chemicals" to produce the best possible "homemade" sanitizer/desinfectant.
And for the DIY, I'm saying "homemade" because I want it to be easy to make, without requiring an actual lab. Maybe "homemade" is not the best term.

What is the metric for success of this experiment?

How would you recommend measuring the success of it? I bought a UV light, but that might not be enough (or maybe not effective at all).

The goal would be to kill enough bacteria/germ so that the immune system can take care of the rest. I guess.

Isopropanol or ethanol. Then set it on fire.

H2O2 is caustic and reactive, but of the list above, it would be the most complete germicide. H2O2 is an oxidizing agent. H2O2 compares well with bleach (bleach is arguably, the king of disinfectants). However, typically the usage level of H2O2 is higher (3%) than other disinfectants. You cannot mix bleach with H2O2 (H2O2 reacts rapidly with bleach). Also, you cannot mix bleach with acid, although many people will mix vinegar and bleach in tiny amounts to give it an extra kick (which it will)... this is DANGEROUS. H2O2 can generally be mixed with (most) acids- especially organic acids like citric acid or vinegar.
Essential oils (especially phenolics like clove, cinnamon, and/or thyme oil) are the most germicidal e-oils, but e-oils are kind of like gasoline (plastic dissolving-staining etc.). Other, terpenic e-oils (like peppermint, pine, citronella, lemongrass, etc.) are somewhat germicidal. Limonene and pine oil are often used as a germicide, although it is comparitively weak to most essential oils. However, they is not as caustic or toxic as clove (for example)- although they do have some toxicity. Pine oil, thyme oil, peppermint, and limonene are popular choices for germicidal use. I have seen thymol and limonene mixtures before. Peppermint is common for school mop-products (becuase in schools, its common to clean up vomit, and peppermint helps quell nausea). Cinnamon, thyme and peppermint are common in mouthwash. Usage levels for e-oils is usually much less than 1%... they are not as effective (but still helpful) as other disinfectants at such levels, but the lower usage is balanced between the noxious-toxicity of the oil and the efficacy of the germicidal effect. Since antibiotic e-oil products would need to contain a noxious and cost prohibitive amount of e-oil to be effective (as a stand-alone antimicrobial), most antibiotic e-oil products (pine-sol for example) contain ammonia (high pH), and simply list pine as a scent (it would help kill bacteria- but to claim antimicrobial on a bottle, the product must be EPA registered and there are lots of problems with e-oils... again, cost, toxicity, and efficacy to name a few). You can mix e-oils with soap and alcohol. Other chemicals (acids, bases, and oxidizers) can/will break e-oils down, but not very rapidly. And the break down/byproducts would probably still have some degree of antimicrobial activity.
Vinegar is used by many DIY people, but it is not a popular commercial product because it smells kind of bad, it's not very effective, and if it is mixed with bleach (WARNING) it can cause a serious health hazard. Also, 5% acetic acid is not extremely germicidal (it will work to kill bacteria in food after a few days); many spores will survive for hours in acetic acid. Vinegar is most effective against non-spore forming (gram negative) bacteria- which are often the more nasty kinds of bacteria (Salmonella, E. coli, Serratia, Pseudomona, etc.)... but it's less effective against spore forming (gram positive) bacteria (Candida, Bacillus, Streptococcus, Micrococcus, etc.). Again, it is an effective preservative in food... as is 5% ethanol (ie., beer). 50% acetic acid is much more effective and it works faster... as does 50% ethanol. pH is a factor, so vinegar is a bit more effective than ethanol, but in general, a high pH is much more effective than a low pH. So, ammonia or lye solutions are more effective than organic acids.
70% isopropanol and 60% ethanol are a common hand and wound disinfectants. They are fairly antimicrobial and antiviral, but again, spores (and other gram positive bacteria) can survive in dilute alcohols (for a little while).
Borax is not germicidal (I don't care what the popular opinion on the internet is). The insecticidal efficacy of borates is also relatively poor, but since it doesn't leave or break down, it is effective over time (weeks, years, etc.)... persistence is why it is used as an effecitive termiticide for wood-framed houses. And although borax is claimed to be effective for roaches... mixed with bait and such- I have tested it thouroughly and it's not very good. But more at the subject here- I would not consider borates to be antimicrobial... unless you consider pH to be the antimicrobial agent. In which case, ammonia or lye are extremely antimicrobial (they really are). But you can't mix lye and ammonia for cleaning (DANGER). Also, bleach has a relatively high pH (about 11).
Notably, but not mentioned, SOAP is the originial antimicrobial and insecticide. It works great; pound for pound- it works better than vinegar. It's been used for centuries, and it has very low toxicity. But, since it has very low usage levels (generally less than 1% in water) and it is more for removing bacteria- so it is not purely germicidal. Generally speaking, soap more strongly inhibits gram positive, spore forming bacteria (at say 0.1% concentration) and inhibits most gram negative bacteria at around 4% concentration. Also, soap manufacturers don't need to pay hefty tolls for registering all their products as antimicrobial because people buy soap anyway. But adding a strong antibiotic (usually triclosan, which tends to work better on gram negative bacteria) they can easily skate through EPA with an affordable antimicrobial product (affordable especially compared with the cost of the claim which costs MILLIONS of dollars).
To claim "antibiotic" with EPA, a product must kill most bacteria (I think it's something like 99.9%, minimum, within 5 minutes against several standard/representative microbial species). You really can't do that without something like 60% ethanol, or 0.15% bleach (ie., 3.33% of a 5% bleach solution), or 3% H2O2, or 2% clove oil (ooh that stings!) or 0.1% triclosan in soap (which gets diluted 100x to about 0.001% in common use) or some antibiotic molecule in high enough concentrations to kill.
Some antimicrobial combinations are synergistic, some are antagonistic, some are additive, and some are deleterious. Most of the reliable combinations of over-the-counter antibiotic materials have already been tried and the best products (soap, hydrogen peroxide, alcohol, bleach, ammonia, and yes, even fragrance) have stood the test of capitalism and time.
Here is some good reading about this: Guideline for Disinfection and Sterilization in Healthcare Facilities, 2008

So let's say I make a mix of:

49% of 3% H2O2 (I suppose the 97% remaining is water?)
49% of 70% Isopropanol (I suppose the 30% remaining is water?)
2% any phenolic oil

Would that make an effective germ killer/desinfectant (I do not really care about EPA standards at this point in my research)?

Could some lye also be added to the mix? Isopropanol / H2O2 / Lye / Phenolic oil.

@FredericHoule no. absolutely not. Lye and H2O2 will react. Lye will also decompose e-oils pretty quickly (by comparison with H2O2).

@FredericHoule H2O2 will mix with organic acids...

@Bel Welborn so just half/half H2O2 and Isopropanol, and a few e-oils. That would make it?
(Thanks for your answer by the way, very very interesting read. I tried to upvote but not enough reputation to do it)

@FredericHoule a solution of 0.04% lye in water should be pretty germicidal, and caustic. The pH will be around 12. ....... If you have 2% thyme oil in soap it will be germicidal... and it will probably burn your skin (not as badly as clove oil though- don't get any in your eye). ....... 1% H2O2 would be kind of a threshold... so yeah you could mix it with IPA, but the benefit of alchol would be very minor... you could mix H2O2 with water and have a similar effect. I might recommend mixing half H2O2 with say half 10% solution of citric acid (so that's 5% citric and 1.5% H2O2).

@FredericHoule ammonia and alcohol mixtures are common

@FredericHoule Essential oils and alcohol mixtures are common too. I sometimes use clove and isopropanol to remove permanent ink (sharpies).

Why trying to reinvent the wheel? 70% ethanol in water is more than enough considering the examples you gave. You can add some essential oil (mainly for the smell).
Note that, to my knowledge, pure alcohol is not a disinfectant. It needs a fair amount of water.

The problem is, if I want to commercialise that product, it can't have only ethanol or isopropanol, from a marketing point of view. It needs at least 2-3 ingredients, no matter if they are super effective or only minor enhancement. That's why I'm considering mixing 1-2-3 things together.


text:
Difference between Gibbs Free Energy and Exergy
I'm wondering if there is any difference between Gibbs Free Energy and Exergy because their definition seems to me the same $\Delta H - T_\text{env}\cdot\Delta S$.
Exergy represents the maximum amount of work that can be extracted from a system going in equilibrium with its environment, while the Gibbs Free Energy is evaluated at $T,P$ of the system in equilibrium with the environment.
Practically what is the difference, if any?

https://en.wikipedia.org/wiki/Exergy

Gibbs free energy is calculated without possible PV-work, ie. without volume changes.

Exergy is equal to the Gibbs free energy if you happen to be in the isothermal--isobaric ensemble $(N, P, T)$, i.e. in a closed system if the pressure $P$ and temperature $T$ are constant. This is indicated in Eq. 7 of the Wikipedia page for exergy.

I would add a personal note: I've done thermodynamics (and taught it) for many years without ever using the concept of exergy… Although it has some historical relevance, I am not sure how useful it is in a modern presentation of statistical thermodynamics.

Thanks for your precise and clear answer. So Exergy is a way to "call" the free energies depending on the process considered (constant volume-->Helmholtz free energy, constant pressure-->Gibbs free energy)? For example if I considered a stream of hot gas, the maximum work that can be extracted, i.e. its exergy, is the Gibbs free energy if I assume to be at constant pressure?