text:
Are polar bond stronger than non-polar bonds?
I came across Pauling's electronegativity scale in which he noticed that the experimental bond enthalpy of X-Y is always greater than the average of X-X & Y-Y and proposed that the difference, D, between the experimental bond enthalpy and the average of X-X & Y-Y gave a measure of the ionic character/polarity of the bond/ electronegativity difference
This seems to imply that the greater the electronegativity difference, the higher the bond dissociation enthalpy (as compared to the average of X-X & Y-Y).
I can't think of any explanation for this though. I came across one explanation stating that the partial charges result in an extra electrostatic attraction, but I can't find any reliable source for verification

https://chemistry.stackexchange.com/questions/129328/why-is-the-carbon-monoxide-triple-bond-stronger-than-the-nitrogen-nitrogen-tripl

This also indicates that ionic bonds are equally or more strong that a number of covalents one. Although the opposite is often claimed by books and teachers.

The argument is circular: electronegativity is a measure of the tendency of atoms to attract electrons. Therefore electrons are attracted more strongly to electronegative atoms. More attractive has no more meaning than "bound more strongly" and a measure of binding strength can be regarded as a mixture of ionization energy and electron affinity, that is, an electronegativity parameter as defined by Mulliken.
Pauling's original definition of electronegativity relied on the idea that the bond dissociation energy contains contributions from a QM interaction plus an additional purely electrostatic interaction (a classical Coulombic interaction not involving a QM description of chemical bonds). In homonuclear diatomic molecules the electrostatic term is not important because the atoms "tug" at the electrons with equal strength, and so only a QM term contributes to bonding. For heteronuclear diatomics one can estimate the magnitude of the contribution of the QM term as an average of the terms in the two homonuclear molecules. Subtracting this average from the dissociation energy for the heteronuclear diatomic then results in an estimate of an electrostatic contribution to bonding. The electrostatic contribution is argued to be due to the different electron attraction in the two atoms, which is encoded in Pauling's electronegativity parameter. This is summarized by the equation for the heat of formation postulated by Pauling (in $\pu{kJ/mol}$):
$$Q_f = 100 (\chi_A-\chi_B)^2$$
The additional electrostatic term enhances the binding energy:

The greater the separation of two elements on the electronegativity scale, the greater is the strength of the bond between them. The extra stability is the resonance energy between the normal covalent structure and the ionic structure.

The excerpt comes "from the horses mouth" (Pauling's General Chemistry textbook). If the explanation is not satisfying I recommend looking into more advanced theories.

the bond enthalpy seems to be effected not by the electronegativity of the atoms involved but their electronegativity differe. Fluorine has the highest electronegativity or tendency to atttract the bonding electrons, yet the F-F bond is much weaker than the H-F bond. There appears to be some correltion between the bond enthalpy and polarity


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Why is the standard enthalpy of formation of aluminium oxide higher than magnesium oxide?
$$\begin{align}
&\text{Aluminium oxide:} &\ce{4Al + 3O_2 &-> 2Al2O3} &(\Delta H_\mathrm f &= \pu{-1675 kJ/mol})\\[1.5em]
&\text{Magnesium oxide:} &\ce{2Mg + O_2 &-> 2MgO} &(\Delta H_\mathrm f &= \pu{-602 kJ/mol})\\[1.5em]
&\text{Iron (III) oxide:} &\ce{4Fe + 3O_2 &-> 2Fe2O3} &(\Delta H_\mathrm f &= \pu{-824 kJ/mol})\end{align}$$
Question:

Why is the enthalpy of formation for aluminium oxide higher than that of magnesium oxide, even though magnesium is higher in the reactivity series?
Why is enthalpy of formation for magnesium oxide lower than iron oxide even though its more reactive? Is this due to the number of bonds formed? Thus, magnesium is less exothermic as less bonds are formed?
What determines reactivity? Why is magnesium more reactive than aluminium and why is aluminium more reactive than iron? Does it have anything to do with valency of the electrons?

Because you are comparing apples with donuts.  If I eat enough apples they will add more calories than that donut I gobbled for breakfast.
To compare food calories properly you need to match serving sizes.  Similarly, to compare reaction energies fairly you need to use a common basis for the extent of reaction.  For oxidation of metals the reactivity of the metals is usually compared by using one mole of oxygen as the common basis, as in Ellingham diagrams.  The physical situation is if there is only a limited amount of oxygen available, which metal would that oxygen prefer to react with according to thermodynamics?  (Not necessarily kinetics.)
So for the aluminum oxidation you should divide all the coefficients by three to get a reaction with one mole of oxygen, and then the enthalpy of formation is similarly divided by three.  You would then get $-558\text{ kJ/mol }\ce{O2}$ versus $-602\text{ kJ/mol }\ce{O2}$ for magnesium oxidation.

source: Stack Exchange

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

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EMF - Intensive or Extensive Property
In electrochemistry, I know that standard emf of a cell (also called standard electrode potential (E°)) is an intensive property.
But, I wanted to know whether non-standard emf will also be intensive or extensive?
(Also, in general, is electromotive force intensive or extensive?) (I mean emf used in other topics like electromagnetism)

@Poutnik Most websites say that it will be intensive but when we change concentration of electrolyte then cell's potential also changes, why is that?

You pay twice as much, you get twice as much - that's what "extensive" means.

Being intensive or extensive property is not related to changing of system properties like concentrations. It is related to scalability of systems.
If you change the system scale, like doubling or halving it, and if the property remains the same, like temperature, density, composition, equilibrium EMF, it is the intensive property.
If you change the system scale and the property changes proportionally to the scaling, it is the extensive property, like mass, volume, energy.
If you change the system scale and the property changes, but not proportionally  to the scaling, it is neither extensive neither intensive property.  Like the object surface, galvanic resistance or the non-standard non-equilibrium EMF, where level of the electrode polarization can be scale dependent.

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


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CO2 Catalytic Converter
We all know how CO2, as a greenhouse gas affects the climate and causes global warming. I was thinking of a small device, which could be attached to the exhaust pipe of an automobile. It could convert the CO2 into something having negligible greenhouse effect, such as water. Is this feasible? Thanks!

Water doesn't contain carbon; where will the carbon go? More importantly, though, the question we face here is not from the realm of chemistry. The real question is "where is the money".

I found out that CO2 reacts rapidly with monoisopropanolamine to give amine carbamate. Could this be used?

The reaction surely does happen, as do many, many others. But what will you do with all that amine carbamate? Where will you get all that monoisopropanolamine? See, I was not kidding. It is all about money.

I see. Would there be any other way to convert CO2 into something harmless?

How about use of microorganisms like  Microalga Scenedesmus dimorphus?

There are many other ways, and they have one thing in common: they all cost money.

This question is presently studied by huge numbers of scientists, all over the world, in huge numbers of labs and factories. A valid solution has not yet been found. But they keep trying !

Many organisms use photochemistry to convert CO2 to useful molecules. They are called plants. But planting many trees isn't practical in a car.

@Maurice This question has not been studied much by any scientists because the thermodynamics and economics are ludicrous. What scientists have studied is whether industrial processes for converting CO2 into useful products or capturing it on a large scale are feasible at a sensible cost. That is irrelevant to what could work in a vehicle.

Unfortunately this is not feasible
Current catalytic converters are designed to remove noxious pollutants from exhausts. Things like carbon monoxide (poisonous) and nitrogen oxides (noxious and irritating).
The trouble with doing this for carbon dioxide is fundamental. Carbon dioxide and water are the ultimate products of combustion of hydrocarbons because they are the thermodynamically deepest products of combustion (ie forming them releases the most energy from combustion). We are not worried about water. But the trouble with being at the bottom of the thermodynamic well is that, to turn the compound into something else, you need to add energy.
There are known chemical reactions that convert carbon dioxide into other, more useful and less greenhouse-contributing products. We can go the whole hog back to methane or, more commonly, methanol. But these reactions are very energy intensive and require fairly special catalysts and conditions. To do them as a way of reducing carbon dioxide emissions would end up using most of the energy from hydrocarbon burning to run the reaction not to move the vehicle. And there are always energy losses in such reactions so we might end up greatly increasing fuel consumption overall to get a small reduction in emission. It is probably far better to make engines and vehicles more efficient.
On an industrial scale the preferred alternative to carbon dioxide conversion is carbon capture and storage (CCS). Separating carbon dioxide is much cheaper than converting it to something else. Even so, CCS plants are in their infancy and none are remotely economical even when subsidised (or at any feasible carbon price). and these reactions require large chemical plants and could never be viable in a car.

@AJKOER Your answer would be a good answer if the question was "are there potential future technologies for turning CO2 into useful products". But that isn't the question. The question is "can a small device attached to exhaust pipe of a vehicle reduce the emitted CO2". And your answer is irrelevant to that for simple thermodynamic reasons. Feel free to post the other question and your answer should get a positive score.

Actually, one should ask Math-Whiz of his conception of "small" and whether a battery cell (as alluded to in the breakthrough) would constitute an acceptable answer.  Even if he concurs, why do I suspect that it will have no impact on grading here even though it would clearly refute your speculated affirmation of not possible now (ostensibly wrong) or sometime in the future (more likely wrong). The word of the day is integrity.

I respectfully disagree. Vehicles come in different sizes and claiming some artificial 'size' argument is a bit weak. Catalytic converters are 'large' in my relative view so, a battery cell of the equivalent weight/volume that may be perhaps a bit larger is not, in my opinion, disqualitive. The truth is that your  definitive answer is (and remains) weak, and as such, should be subject to suspicion/attack, especially in the presence of already realized technology.

Further, as the alluded breakthrough technology generates electricity, which can be used to power/propel vehicles, why is 'size'  even considered as a driving parameter here? Clearly, it is not,  the paucity of your argument just becomes increasingly apparent and desperate.

@AJKOER Typical catalytic converters for a car have a volume of a few litres. Do some calculations–if you want to prove me wrong–and tell me the electrical storage capacity of any possible battery that size and the potential volume of CO2 it could absorb. Then convert to the distance such a battery could drive a car and the amount of CO2 consumed and compare to the distance a car could drive on a single tank of gasoline.

@AJKOER A vehicle could be as massive as a plane, or as small as a bicycle. It's all relative.

@matt_black You're right.

Forget battery size, as to answer the question: "on the distance a car could drive on a single tank of gasoline"? Well, if the car has a large electric engine receiving some power also from the battery that also consumes CO2 from the small gasoline engine also present, I suspect somewhat more than the exceptional gas mileage of current electric/gas vechicles.

@AJKOER Really? If your hypothetical battery is a primary cell it consumes Al prodigiously to convert CO2 to something like oxalic acid; if a secondary cell, it has to be recharged as the conversion process consumes energy even if it produces some voltage while doing so (check the thermodynamics of the proposed reaction).

Not talking about it as being the primary cell, clearly secondary. Interestingly, as the 2nd battery  (turning CO2 into store electric power in the main battery) can be of any shape, it could fix in existing unused space between the car's interior chassis and exterior frame. So, it actually may not even consume any currently used volume, but it adds weight. At some point, there is a 'benefit' to weight tradeoff. Benefit is defined to include pollution reduction.

While not likely easily created, other products with air, $\ce{CO2}$ and water, in the presence of an electrical source, are conceivable in my opinion.
For example, there is some discussion around NAIs per this government source to quote:

One of the examples is the negative ion generator using an ultraviolet source to irradiate electrically conductive material, which was patented as early as 1964 (Patent No. US 3128378 A). In this patent, an ultraviolet lamp was used to irradiate metal materials, which photo-electrically eject electrons. The electrons then collide with air molecules and generate NAIs.

Assuming the chemistry involves superoxide radical anions created as follows:
$\ce{O2 + e- -> .O2-}$
And, in a medium of moist air, the superoxide is present as the hydroperoxide anion (see Wikipedia discussion here):
$\ce{H2O <=> H+ + OH-}$
$\ce{H+ + .O2- -> .HO2}$
In an albeit slow reaction, this is a precursor to hydrogen peroxide:
$\ce{.HO2 + .HO2 -> O2 + H2O2 }$
although this source notes an acceleration of the self-reaction in the presence of $\ce{CO2}$. It is also known that:
$\ce{H2O2 + CO2 -> H2CO4}$
as per a source: The Mechanism of Carbon Dioxide Catalysis in the Hydrogen Peroxide N-Oxidation of Amines, to quote:

"The reactivity of the peroxymonocarbonate ion, HCO4- (an active oxidant derived from the equilibrium reaction of hydrogen peroxide and bicarbonate), has been investigated in the oxidation of aliphatic amines"

Also, see Figure 2 at this source. Possibly, a more direct (and rapid) postulated radical reactions (which is in accord with the above):
$\ce{.HO2 + HCO3- -> .HCO4 + OH-}$
$\ce{.HCO4 + e- -> HCO4-}$
Here is an interesting supporting reference: Chemical effects of large impacts on the Earth's primitive atmosphere providing support for a prior significant presence of H2CO4 in the primitive earth's atmosphere, to quote:

Intense bombardment of the moon and terrestrial planets approximately 3.9-4.0 x 10(9) years ago could have caused the chemical reprocessing of the Earth's primitive atmosphere. In particular, the shock heating and rapid quenching caused by the impact of large bodies into the atmosphere could produce molecules such as HCN and H2CO4 which are important precursors for the abiotic synthesis of complex organic molecules.

Bottom line, some more research, and work may provide a path in the future.
[EDIT] The future is near. See The O2-assisted Al/CO2 electrochemical cell: A system for CO2 capture/conversion and electric power generation. To quote from the abstract:

Economical and efficient carbon capture, utilization, and sequestration technologies are a requirement for successful implementation of global action plans to reduce carbon emissions and to mitigate climate change. These technologies are also essential for longer-term use of fossil fuels while reducing the associated carbon footprint. We demonstrate an O2-assisted Al/CO2 electrochemical cell as a new approach to sequester CO2 emissions and, at the same time, to generate substantial amounts of electrical energy... On this basis, we demonstrate that an electrochemical cell that uses metallic aluminum as anode and a carbon dioxide/oxygen gas mixture as the active material in the cathode provides a path toward electrochemical generation of a valuable (C2) species and electrical energy. Specifically, we show that the cell first reduces O2 at the cathode to form superoxide intermediates. Chemical reaction of the superoxide with CO2 sequesters the CO2 in the form of aluminum oxalate, Al2(C2O4)3, as the dominant product. On the basis of an analysis of the overall CO2 footprint, which considers emissions associated with the production of the aluminum anode and the CO2 captured/abated by the Al/CO2-O2 electrochemical cell, we conclude that the proposed process offers an important strategy for net reduction of CO2 emissions.

As such, my opening sentence is completely accurate. Further, if one re-reads my proposed answer, there are some uncanny similarities.
What is quite amazing is that the cited reaction path also involves the presence of electrons, oxygen, the superoxide radical anion, and CO2 leading to a composite compound.
Assertions of a definitive negative answer here, is patently incorrect!

How can this be a solution to CO2 emission from engines? Will cars need to have huge electrical generators or UV lamps in future?

Yes, UV emission processing perhaps leading to less likely than complete CO2 conversion. However, there is already an apparent battery cell CO2 capture system. See my EDIT.


text:
Confusion in Raoults Law
In Raoults law for a solution of two volatile liquids do we measure the mole fraction of components at the time when we just add the two liquids and there is no vapour pressure or we measure their mole fractions at equilibrium condition that too in the liquid which is left after evaporation.

The latter.$;$

Im currently in class 12 and in my textbook in all questions they have taken the mole fraction of components when they are mixed initially when there is no vapour pressure. So is this an assumption? Maybe taking the mole fraction of the components at equilibrium condition in liquid state is out of the scope of my syllabus.

Well, if the air space above the liquid is not very large, the composition change will be negligible. That's the common assumption.

So basically if there are no assumptions involved then the mole fraction Xa or Xb of the components will be calculated in the liquid phase which is left in the vessel after evaporation and reaching equilibrium.And we will not consider the moles that are present in gaseous phase. Am i right

Basically yes, but that's pretty much never done. The assumption is good enough for everyone.

Thanks a lot Sir for your valuable time

Total mole fractions of a system are known, constant and can be measured or calculated at/for any time ( no reaction is assumed ).
Mole fractions within a gaseous or liquid phase are measured or calculated in context of Raoult law for the state of equilibrium, unless they are defined as an initial state for a particular scenario.

Ok so in raoults law when we write partial pressure of any component as suppose Pa Xa then Pa represents the pure vapour pressure of the component and Xa represents the mole fraction of component in the liquid which is left after evaporation at equilibrium. It might sound like im asking the same thing again but i just need assurity because my textbook is taking the mole fraction of components at initial condition when we just mix them and there is no vapour pressure. So can you please tell what assumption are they using

Yes, it is like you say  pa.xa is partial vapour p of a, which if pure has p.p. pa, , if xa is its molar fraction in liquid at equilibrium conditions. Said by other words, equilibrium saturated partial vapour pressure  is proportional to molar fraction of volatile liquid in mixture.


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Is tautomerism in ketimines possible?
In a reaction between an aliphatic primary amine and a ketone, I could expect a ketimine as a product or intermediate. My question regards tautomerization of this intermediate: has tautomerization between the two possible imine structures ever been observed, or is even possible? I tried to search imine tautomerism and transamination, but I can't find what I'm looking for. I imagine this would be a simple experiment to conduct, but I can't find anything in the literature on this.

Yes, this is common. One example is enzyme catalyzed transamination reactions that utilize pyridoxal phosphate as a cofactor.

I found something very similar to what I sought in Dose, Nature 179, 734 (1957) though it involves aldimines and a decarboxylation step. The author gives a very well explained mechanism for the process and reports varying transamination ratios of valine based on the substitution of the aldehyde. An example:


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What hydrocarbons burn in air without a change in volume?
Find all hydrocarbons(alkanes, alkynes, alkenes), which answers given conditions:
the sum of volumes of combustion air products equals to the initial sum of volumes of airs. ( volumes are measured in the same conditions 101.3 kPa pressure and 150 deg C)
So this is my solution:  let's write ideal air equation --> we get $PV=nRT$  therefore for the same conditions volumes are dependent on the moles.
So next I wrote general equations for combustion of every type of hydrocarbons.
Next I summed coefficients of products and equated to the sum of the coefficients of reagents.
For example for alkanes, we get  $(3n+1)/2 + 1 =n$  this doesn't have a solution in positive numbers so these kinds of hydrocarbons don't exist.

The reasoning is right, the math isn't.

Write the chemical formula for the complete combustion of an alkane then show the math formula which proves that after burning that alkane using an isothermal process that the volume must increase.

repeat steps for alkane using simplest alkene and alkyne. What happens?

Ok, the compound has to be a gas at 150 C, so the boiling point has to be lower than that. Write down the reaction for complete combustion of n-alkanes (methane, ethane, propane, butane, pentane) can you see a pattern in how many molecules on left and how many on right? Mathematical formula for pattern? // Remember isomers. So if butane would work so would isobutane. However the reverse is also true. If butane doesn't work then isobutane can't work either.

Earn some cred on the site. Write the chemicals reactions and the math formula in answer. -- You can answer your own questions here...

If we consider a possibly branched, but not cyclical  hydrocarbon ( unless we count each cycle as an extra double bond ):
$$\ce{C_nH_{2n+2-2d-4t}},$$
where

d is number of double bonds
t is number of triple bonds

then the combustion reaction is
$$\ce{C_nH_{2n+2-2d-4t}} + (n + 0.5(n + 1 - d - 2t))\ce{ O2 -> n CO2 } + (n+1-d-2t)\ce{H2O}$$
Comparing volumes, we get:
$$1 + 1.5n + 0.5(1 - d - 2t)=n + (n + 1-d-2t)$$
$$0.5(1 - d - 2t)=0.5n - d -2t$$
$$n=1 + d + 2t $$
With the restriction $$n \gt d + t$$
For alkanes (d=0, t=0): n=1, so $\ce{CH4}$.
For simple alkenes (d=1, t=0): n=2, so $\ce{CH2=CH2}$. For longer chain, the would have to be cumulated double bonds: $\ce{CH2=Cn=CH2}$
For simple alkynes (d=0, t=1): n=3, so $\ce{CH3-C#CH}$, $\ce{HC#C-CH2-C#CH}$ etc.

We can more simply render the hydrocarbon as $\ce{C_xH_y}$.  Render the combustion reaction
$\ce{C_xH_y + O2 -> CO2 + H2O}$
and balance the equation:
$\ce{C_xH_y + (x +\frac{y}{4}) O2 -> x CO2 + \frac{y}{2} H2O}$
Combustion takes place in the vapor phase and, from the Ideal Gas Law, the volume of the reactants or products will be proportional to the total number of moles of gas once the combustion products are brought to the same temperature and pressure as the reactants.  So equate the number of moles on both sides:
$1+x +\frac{y}{4}=x+\frac{y}{2}$
from which $x$ cancels out and we can solve for $y=4$.
Thus any hydrocarbon gas with four hydrogen atoms per molecule burns at constant number of moles of gas, which leads to constant volume once the combustion products are brought to the same temperature and pressure as the reactants.  Methane and ethene are the most common such hydrocarbons, but some others that remain gaseous with more carbon atoms are also possible.

Good answer  :-)  I have not realized that count 4.


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Why is the bond energy for a C=O bond higher in CO2?
I was looking at the table of bond energies, and what caught my eye was how the bond energy for a $\ce{C=O}$ bond is $\pu{736 kJ/mol}$ with the exception of the $\ce{C=O}$ bonds in $\ce{CO2}$ being $\pu{799 kJ/mol}$.
What causes $\ce{CO2}$ to make such a  big difference in a $\ce{C=O}$ bond?

That's not a big difference at all. Bond energies only give you a very, very rough estimate.

@IvanNeretin What a lack of curiosity!! You may be right, but you can probably come up with a better and more interesting answer than that. The presence of noise does not imply the absence of real differences in the signals.

Can anything be concluded from the resonant structures of CO2? they don't seem to contribute much to the actual structure...


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Is it possible to add R chain that contains a carbonyl (ketone) to the alpha carbon?
I'm trying to synthesize 2‐(3‐oxobutyl)cyclohexan‐1‐one from cyclohexanone:

I thought about LDA, then adding an R group that has a ketone. Is that possible? As far as I know, we can add halogen or alkyl chain only. If it's not possible how can I get this compound from cyclohexanone?

Consider a Michael reaction with methyl vinyl ketone

Yes, i figured it sould be conjugate addition! but @Waylander is it possible to add R group that has a ketone as an " alpha carbon alkylation reaction " ?

Yes it is  https://pubs.acs.org/doi/pdf/10.1021/jo01031a058. Consider forming the silyl enol ether of the cyclohexanone before introducing the methyl vinyl ketone

Here is a possible approach  https://pubs.acs.org/doi/abs/10.1021/ja00319a040


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Why does vapor pressure of ice decrease faster than water when temperature decreases?
In melting point depression, I have learnt the explanation from the textbook that (e.g. aq. solution):
Definition of melting point: ice and water exist in equilibrium (have the same vapor pressure).
(1) aqueous solution generally has a lower vapor pressure than the pure solvent (water)
(2) Vapor pressure of ice drops faster than water when temperature drops
Therefore, at 0°C, the vapor pressure of water in aq. solution is lower than ice. When the temperature drops below 0°C, the drop in vapor pressure of ice can 'catch up' with water, hence an equilibrium can be reached i.e. melting point depression.
Question: I would like to know what is the rationale behind (2) since I have no idea on how to explain its validity.
Thank you.

It is not limited to ice+water, but is valid for solid+liquid phase of all compounds. The reason is that enthalpy of sublimation is higher than for evaporation, so the vapour pressure raises with temperature faster. See Clausius-Clapeyron equation. Related is slower decrease of vapour pressure with T for most organic solvents compared to water, as they have lower evap. enthalpy than water.

Got it. Thank you very much for letting me know the equation!


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How to draw xenon hexafluoride in rdkit from SMILES?
In SMILES I would think that $\ce{XeF6}$ would be F[Xe](F)(F)(F)(F)F or F[Xe](F)(F)(F)(F)(F). A quick Wikipedia search showed that the first result is the correct SMILES string. However, rdkit isn't able to convert either SMILES strings into a structure.
Here is my code:
from IPython.display import SVG
from rdkit import Chem
from rdkit.Chem import rdDepictor
from rdkit.Chem.Draw import rdMolDraw2D
# Create mol object from smiles string
# mol = Chem.MolFromSmiles('[B](C7OC7)(CCC3CCCC31N2CN([P]5CCCC8C5)C2(CO8)CC1)(C6CCC6)')
# The above molecule was me learning SMILES, below is the Xenon Hexafluoride string
mol = Chem.MolFromSmiles('F[Xe](F)(F)(F)(F)')
molSize=(900,300)
mc = Chem.Mol(mol.ToBinary())
if not mc.GetNumConformers():
    #Compute 2D coordinates
    rdDepictor.Compute2DCoords(mc)
# init the drawer with the size
drawer = rdMolDraw2D.MolDraw2DSVG(molSize[0],molSize[1])
#draw the molcule
drawer.DrawMolecule(mc)
drawer.FinishDrawing()
# get the SVG string
svg = drawer.GetDrawingText()
# fix the svg string and display it
display(SVG(svg.replace('svg:','')))

You may want to post this question in Matter Modeling SE instead which is well suited to answer such questions. If you do decide to migrate the question please delete the current one. Good luck!

Missing the sixth F?

Oh yeah Andrew you're right (doesn't fix issue though) and will do Buck Thorn.

With the last version of RDKit (2020.09) it works for me.

@SohilKollipara If you are still interested in this or found an answer since then, you can copy the question to Matter Modeling and remove the version here.

I agree, this is better suited at [mattermodeling.se] and there are way more experts who can help there, but it is too late for migration. In the meantime I have adjusted the tags and used the recommended way of forcing code highlighting.

If you start with RDKit, I suggest to work through the examples of RDKit's cookbook.  With minor adjustments, paste the following into a cell of a Jupyter notebook as a minimal, self sufficient working example (MWE)
from rdkit import Chem
from rdkit.Chem.Draw import IPythonConsole
from rdkit.Chem import Draw
IPythonConsole.ipython_useSVG=True

mol = Chem.MolFromSmiles("F[Xe](F)(F)(F)(F)F")
mol

to obtain the following illustration:

The setup used: Jupyter 6.4.5, Python 3.9.9, RDKit 2021.09.2 as currently repackaged e.g., by Linux Debian 12/bookworm, branch testing (package tracker). Stay tuned to the project's GitHub page to know if there is a new release (currently 2021.09.4, just two days old) more frequently packaged for conda.

There is something odd with the code provided by you.  Copy-pasted into the same Jupyter notebook, the molecule plot contains a dot about an unpaired electron:


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Reversible equation in Chemical Kinetics — Should I use definite integral when solving the differential equation?
I found this problem:
We have the following reversible reaction: $A \ce{<=>[\ce{k_1}][\ce{k_2}]} B $, described by the kinetics equation:
$$\frac{d[X]}{dt} = k_1([A]_0-[X])-k_2([B]_0+[X])$$
Where
$[X]$ is the increase of molar concentration;
$[A]_0$ and $[B]_0$ are the initial concentrations of $A$ and $B$.
Find $[X]$ as a time dependent funcion.

What I've tried:
The differential equation above is the same as:
$$\frac{d[X]}{k_1[A]_0-k_2[B]_0 - (k_1+k_2)[X]} = dt$$
Now, we should integrate both parts.

My questions are:

Should I use the definite integral? Why? Am I right when I state that the equation becomes this?
$$\int_{[X]_0}^{[X]} \frac{d[X]}{k_1[A]_0-k_2[B]_0 - (k_1+k_2)[X]} = \int_{t_0}^{t} dt$$

If $[X]$ is the "increase" of molar concentration, then is $[X]_0 = 0$?

Is $t_0 = 0$?

Thank you!
If I am wrong, please tell me the corect way of solving this. Thank you so much!

(1) Yes, why wouldn't you? You do this for a regular integrated rate law as well.

(2) Probably, but that depends on the statement of the problem. My reading is yes. (3) That also depends on the statement of the problem. My reading is yes.

Also, do not integrate to a bound on your integral that is the same as the variable of integration. That is, do not write $\int_{a}^{x}dx$. Use a dummy variable: $\int_{a}^{x}dx'$


text:
How are there two C3 rotation axes in ammonia?
If we look at the character table for the $C_\mathrm{3v}$ point group (which $\ce{NH3}$ is an example of), we see that these are the symmetry elements ${E, 2 C_3, 3 \sigma_v}$ present in the molecule. My question is — why are there $2 C_3$ rotation axes? Shouldn't there be only one, which is perpendicular to the plane containing the three hydrogens?
The $C_\mathrm{3v}$ character table:
$$\begin{array}{c|ccc|cc} \hline
C_\mathrm{3v} & E & 2C_3 & 3\sigma_\mathrm{v} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & z & x^2+y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & -1 & R_z &  \\
\mathrm{E} & 2 & -1 & 0 & (x,y), (R_x,R_y) & (x^2-y^2,xy), (xz,yz) \\ \hline
\end{array}$$

There is indeed only one rotation axis of order 3, and it stands for two symmetry elements: rotations by $120^\circ$ and by $240^\circ$.

In a character table for $C_{3v}$, $ 2C_3$, means that the element is actually 2 $C_3$ elements in the same conjugacy class.  That means the two $C_3$ operations are related in some way.
In fact, it means they share the same single axis.  One $C_3$ operation is a $120^\circ$ turn around the axis, and the other $C_3$ operation is a $240^\circ$ turn around the same axis in the same direction - or equivalently a $120^\circ$ turn around the same axis in the opposite direction.
In the image below the two elements $C_3$ and $C_3^{-1}$ are the two symmetry elements of the $2C_3$ class, which are rotations about the same axis out of the plain of the image.

Symmetry elements of the $C_{3v}$ point group
Image taken from Fig S5 in supplementary information of T. Karin, et al, Phys. Rev. B 94, 041201

(+1) I guess I would still be careful to make the distinction between a symm. element and a symm. operation — in my experience, at least, the two rotations are the (distinct) operations, and the element is the shared axis — although actually, I'm not sure if this is formalised anywhere better than my Wikipedia link..? (I could look in a book later, though I haven't the time right now.)

The top row of the character table does not refer to symmetry elements, but rather symmetry operations. (For the difference between the two, see Wikipedia:Molecular symmetry.)
In this particular case, the difference is that there is only one $C_3$ symmetry element in ammonia (i.e. the rotation axis), but there are two operations: one corresponds to the clockwise rotation, and the other to the anticlockwise rotation (which is equivalent to two clockwise rotations, should you prefer that).
It is the symmetry operations, not the elements, which form the so-called "group" in group theory (see Wikipedia:Group (mathematics)).

In mathematics, a group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that three conditions called group axioms are satisfied, namely associativity, identity and invertibility.

(Emphasis mine.) The "binary operation" for symmetry operations is to simply perform one operation after another, which is somewhat akin to composition. Note that the word "element" in the quoted paragraph refers to an abstract member of the group: it is mildly unfortunate that the elements are symmetry operations, not symmetry elements.

What if I prefer two anti-clockwise rotations though? ;)


text:
Which acetate salt is blue and turns yellow when dissolved in water?
***Update
Some questions I am considering to try to answer this question;

When dissolving the powder in acetic acid, which ions are present?
From what I've read, I haven't found any acetate salts that are not soluble in water. What other metal salts could be forming that aren't soluble?
Which ions might be present in the rock that would not readily react with the acetic acid?
What would be a good compound to replace the acetate ion in a double replacement reaction such that the resulting metal salt would precipitate out of solution?
Are metal acetate salts prone to reaction with oxygen? Are the metal ions in the acetate salt more attracted to the oxygen or the acetate ion? (i.e. will my sample 'rust'?)

I am leaving the main question as-is, posting this update because asking questions is part of my problem solving process.

I was experimenting with the (air) dried residual powder residue left over from cutting this rockin my wet saw in an attempt to determine the composition of the metallic components of the rock I was working with.
The rock appears to be a crystalline silicate rock of some variety. It has minimal to no reactivity with concentrated HCl in its raw form. The geology of the area indicates it was once shallow ocean, with lots of limestone and cryptocrystalline quartz and crustaceous fossils in the surrounding area.
The compound in question was created by soaking the residual powder in 7% white vinegar (acetic acid) for several days with constant stirring. The remaining solution was separated under vacuum filtration and residual solid was again left to air dry for several days. After several days I noticed that the top layer of the filtrate had turned brown with blue splotches. The bottom layer was white with bright blue speckles mixed in.
% isopropyl alcohol I scraped off the top layer and set aside for solubility tests. It was nearly insoluble in  and soluble in water. When the mixture dissolved in water, it turned yellow.
My hypothesis was that the blue compound was copper acetate, the dark brown solution is iron(II) acetate and the yellow in solution is an iron/copper coordination complex. I also suspect there is aluminum in the rock as a USGS survey indicated the presence of muscovite in the region and calcium due to the abundance of fossilized sea shells and limestone. Though I may be completely wrong. I would love if anyone could shed some light on this for me or point me in a different direction for determining the trace elements in the rock.

Many university geology depts have instrumentation to do elemental analysis

I would like to find an answer that uses chemistry more so than postal mail :) With that in mind, I was thinking about dissolving the solution again and separating with column chromatography, then recrystallizing and trying to use some algebra (partial fraction decomposition) to determine the composition.

If you have a relatively pure sample and access to reagents, there are many flowcharts available online for "qualitative analysis of cations" that provide systematic methods for distinguishing common cations from one another. That's how they did it in the days before mass spec.

@Andrew Can you provide any such flow charts? I've looked in a couple of texts but came up short. The closest I've come was my old chm lab manual for determining an unknown salt, but the info there was specific to the salt in question and isn't comprehensive.

Here’s one example: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.9%3A_Qualitative_Cation_Analysis. A basic internet search will turn up other variations


text:
Why do we need weak or strong acids to displace the anions of other weak acids from salts?
I have already looked at Why does a strong acid displace salt of weak acid? but I couldn't understand the answer and the analogy there. And I mean the above question in the following context :
$\ce{CO_3^{-2}}$ ion (from a salt) when treated with $\ce{H^+}$ from a dilute strong acid or a relatively stronger acid (i.e. relative to the acid from which the anion originated for example, for carbonate anion that would be $\ce{H_2CO_3})$ releases carbon dioxide gas. For example :
$$\ce{Na_2CO_3 + dil.H_2SO_4 \rightarrow Na_2SO_4 + CO_2 + H_2O}$$
Another one would be that when thiosulfate ion (from a salt) is treated with $\ce{H^+}$ ion from a dilute strong acid or a relatively stronger acid then two of the products will be $\ce{S}$ and $\ce{SO_2}$ or for example when sulphite ion (from a salt) is treated similarly one of the products will be sulfur dioxide gas.

What particularly do you not understand ?

Why do we need a strong acid to displace the anion of a weak acid ?Even my organic chemistry teacher told me that for an acid base reaction the conjugate acid formed must be weaker than the reactant acid, but why ?

@Poutnik my best guess i.e. why this reaction takes place is that sodium carbonate is a basic compound and hence an acid base sorta reaction takes place

To be said at first: There is no real displacement. There is an equilibrium reaction of a weak acid/base and their salts, that form mutually a conjugate pair acid/base ( like acetic acid/acetate or ammonium/ammonia ).
All is then matter of chemical equilibrium. The stronger acid is, the more is it's dissociation equilibrium in favour of its anion and vice versa. So stronger acid releases hydronium ions and anions of weaker acids captures them, because their dissociation constants tell them so.
Sodium carbonate dissociates when dissolved:
$$\ce{Na2CO3(s) -> 2 Na+(aq) + CO3^2-(aq)}$$
Strong acids dissociate while forming hydronium ion:
$$\ce{H2SO4(l) + 2 H2O -> 2 H3O+(aq) + SO4^2-(aq)}$$
Meeting each other, sodium and sulphate ions are  just "spectator ions", while an acido-basic reaction happens:
$$\ce{2 H3O+(aq) + CO3^2-(aq) <=>> 2 H2O + H2CO3(aq)}$$
That happens because hydronium is much stronger acid than carbonic acid, while carbonate is much stronger base than water.
Stronger acid A  + stronger base B <=>> weaker base A + weaker acid B.

Carbonic acid is unstable and releases gaseous oxid:
$$\ce{H2CO3(aq) <=>> H2O + CO2(g)}$$
What remains in water are sodium and sulphate ions that eventually precipitates as sodium sulphate decahydrate:
$$\ce{2 Na+(aq) + SO4^2-(aq) + 10 H2O <=> Na2SO4.10 H2O(s)}$$
The similar things happen with thiosulphate.


text:
If bonds are not physical things, what is meant by bond rotation?
I’m not sure if this is a stupid question or not, but since bonds are not physically there as such, what does it mean when it can rotate? And how do pi bonds prevent rotation if they are not physical?

Bond rotation implies that there is a change in the relative positions of the nuclei with respect to each other- which looks like as if we have rotated the "bond".

Atoms rotate around the bond. Come to think of it, you may rotate any everyday object, like a book or a cat. Is the axis of rotation physically there? No, it is a mathematical abstraction. Yet we use it all the time, and not without merit.

If you are using your hand to screw or unscrew a screw or a wing nut, your hand makes a rotation movement around the axis of your forearm. The forearm is like the bond between hand and elbow. It can rotate..

Also, covalent bonds are not kind of rod glued to different atoms. But I wouldn't call them not physical. They are physical indeed, else there won't be the molecule. That their description is rather complicated is something else.

Chemical bonds are physical things.  They are manifestations of  electron wavefunctions.


text:
Do Lewis acids and bases turn change colour of litmus
Well, the question is short. Can lewis acids and bases change the colour of a litmus paper? I can't seem to find the answer online (at least according to my high school level chemistry search history).
The intuitive answer I feel is that they shouldn't be able to, as lewis acids and bases are defined only by their ability to donate or accept electrons. On the other hand, Arrenhius and Broensted-Lowry acids seem to be able to change the colour of litmus, possibly due to the presence of H+ or OH-
As a bonus question, is there any indicators that can be used to test the presence of lewis acids that change their colour after donating a lone pair of electrons?

Let's see. Will litmus react to AlCl3, a Lewis acid? Without water, probably no, but then again, without water many ordinary Broensted-Lowry acids and bases won't react either. With water, it will, because AlCl3 will partially turn into a BL acid due to hydrolysis.

@IvanNeretin, yes you are right to point that out. I think dry HCl is a good example. But when it dissolves in water, i learnt it forms H3O+ and OH- which is are required for the litmus paper to change colour, without which  HCl wouldnt change colour. Would AlCl3 also form a similiar ions?

Yes, Anhydrous $\ce{AlCl3}$ reacts violently with water, producing insoluble $\ce{Al(OH)3}$ and gaseous $\ce{HCl}$. If enough water is present $\ce{HCl}$ will soon get dissolved producing $\ce{H3O+}$ which will dissolve $\ce{Al(OH)3}$.

You should keep in mind that the concept of acid-bases is not restricted to water. Lewis acid definition is applicable to gaseous, aqueous and non-aqueous systems. Consequently, all Bronsted-Lowry acids are Lewis acids. The idea applies to bases. Note that the converse is not true just like an apple is fruit but not all fruits are apples. The idea of litmus test is rather restricted to aqueous systems. So in short, the all relatively strong Lewis acids and bases in water can affect the litmus.
If you put a blue litmus is very dry HCl (gas), it would still turn it red because paper is never dry. It always has moisture absorbed in it.


text:
New relationship between exponential factors and entropy of a reaction?
Consider the reaction $\ce{A <=> B}$. Let $k_f$, $E_A$, and $A$ denote the forwards rate constant, activation energy, and pre exponential factor, respectively. Let $k_r$, $E_A'$, and $A'$ denote the backwards rate constant, activation energy, and pre exponential factor, respectively. I seem to have determined the following relationship:

$\ln{\frac{A}{A'}} = \frac{\Delta S}{R}$

First, is this a correct equation? Second, is this a known equation? I have not seen it in any of the textbooks that I have read. Third, is this a useful equation? I imagine it could be used to determine the ratio between the pre exponential factors from the entropy.

My derivation is below:
The equilibrium constant $K$ is equal the ratio of the forward and backwards rate constants $\frac{k_f}{k_r}$, so $\ln{(K)} = \ln{\left(\frac{k_f}{k_r}\right)}$. Furthermore, the Arrhenius equation expresses forward and backwards rate constants as $k_f = Ae^{-E_a/RT}$ and $k_r = A'e^{-E_a'/RT}$, where $A$ is the pre-exponential factor of the forward reaction and $A'$ is the pre-exponential factor of the reverse reaction. Substituting these expressions yields $\ln{(K)} = \ln{\left(\frac{Ae^{-E_a/RT}}{A'e^{-E_a'/RT}}\right)} = \ln{\frac{A}{A'}}-\frac{(E_a-E_a')}{RT}$.
In Chapter 5 of Atkins' Chemical Principles 7th edition, the difference in activation energies of the forwards and backwards reactions $E_a-E_a'$ is equated to the reaction enthalpy $\Delta H$. Applying this relationship, we obtain: $\ln{(K)} = \ln{\frac{A}{A'}}-\frac{\Delta H}{RT}$.
It is also known, that since $\Delta G = -RT\ln{K} = \Delta H - T\Delta S$, $\ln{(K)} = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$.
Combining these two equations yields: $\frac{\Delta S}{R} = \ln{\frac{A}{A'}}$.

Your expression seems to be correct. I have never seen it before.

Yes, your expression is correct. It's nothing new though. Entropy of activation can be interpreted as part of the pre-exponential factor in the Eyring equation:
$$k=\bigg(\frac{k_B T}{h}e^{\frac{\Delta^\ddagger S^\circ}{R}}\bigg)\cdot e^{-\frac{\Delta^\ddagger H^\circ}{RT}}$$

In case it is not obvious to OP - note that $\Delta S_{rxn}$ is the difference between the two entropies of activation, so if you use this equation to rewrite the $\frac{A}{A'}$ term in the original question, the right side reduces to the left side. The Eyring form is much more informative and useful than the derived equation.


text:
Does the enthalpy of formation include translational and rotational energy?
I can look up the enthalpy of formation of water vapor and get -241,826 KJ/mol. I want to use this to compute the total energy of water vapor at STP from the energies of O2 and H2 gas. What I don't know is whether I should use the electronic energies of H2 gas and O2 gas or the total energies of H2 gas and O2 gas, which include the contributions from the translational and rotational partition functions.
Basically, which of these two is the correct usage of Enthalpy of Formation?

ETotal(H2O) = Eelectronic(O2) + Eelectronic(H2) -241,826 KJ/mol + ETranslational(H2O) + ERotational(H2O)
ETotal(H2O) = ETotal(O2) + ETotal(H2) -241,826 KJ/mol

What about vibrational energy and zero point energy correction?
Thermal corrections include all degrees of freedom that are excited at the specified temperature. No idea why you get the idea to dump some degrees of freedom in the first place...

Btw. Gaussian has a full reference for how thermal corrections are done: https://gaussian.com/thermo/

What about translation and rotation of H2 and O2 ?

@Kexanone at room temperature, the vibrational modes are all frozen out. We can ignore them comfortably in the scenario I proposed due to the exponential dependence of the partition function on the eigenstate energy, but you are correct for sufficiently hot gases they are relevant. Most textbooks cover this "dumping" of certain degrees of freedom under the section on "Equipartition of energy"

@Poutnik I am using total energy to refer to the translation and rotation added to the electronic (DFT) energy. So when I say the total energy of O2 or H2 above, that is what I am referring to.

@colinb I mean, it is rather inconsistent to explicitly involve them in the equation for H2O, but not for O2 and H2. Regarding vibration energy, it would be quite complicated to involve different degree of influence of quantum vibration quantisation for all involved vibration modes.

@ColinB Yes, you can dump degrees of freedom, if the temperature is too low to excite it. Maybe your textbook dumps it, but actual computational software like Gaussian doesn't. It's up to you how accurate you want to be. Btw. under ideal gas conditions $H = E_{total} + RT$.

@Kexanone ah perfect, that's the relationship I was looking for! Yes already the rotational only contributes about 4 millihartree for this system, and DFT is at absolute best accurate to 1 millihartree, and since the vibrational is going to be at least an order of magnitude less, it's not gonna matter.

To add to @Kexanone's answer above, the enthalpy of formation refers to how much heat is evolved or adsorbed from the environment in a reaction. So it's the total energy, and in an ideal gas it has the form: dft + translational + rotational + vibrational + PdV, where the PdV refers to the energy needed to make space for the gas (grand canonical ensemble; in the canonical ensemble (fixed volume) this does not show up). At room temperature the vibrational and PdV terms are typically small compared to DFT energy errors (order 1 millihartree), with the PdV term only coming into play at pressures within an order of magnitude or two of those found in the detonation of thermonuclear bombs.

Nope, there is no work. SCF energy + thermal corrections for all relevant degrees of freedom is all there is for the internal energy. Again, I strongly advice to use https://gaussian.com/thermo/ as a reference.

At T = 298 K, and p = 1 atm, the $p \Delta V$ term for the formation of $\ce{H2O(g)}$ is –12.3 L atm/mol = –1.3 kJ/mol.  That's comparable  to 1 millihartree (=> 2.6 kJ/mol).  At 10 atm, the   $p \Delta V$ term becomes 13 kJ/mol, significantly exceeding your stated DFT energy error. Saying this term only becomes significant at pressures within an order of magnitude of those in a thermonuclear explosion is silly.
Ideal gases have no internal structure, and thus no vibrational energy.

Is fine. We are working here with a hybrid model. We still neglect the eigenvolume and interaction between the particles, but allow an internal structure, and thus a more complex heat capacity.

@Kexanone I don't do DFT, so maybe there is some looseness of the term there but, substantively, it seems incorrect to refer to such a hybrid as an ideal gas (one of whose essential characteristics is a lack of vibrational DOF's) rather than as, say, an ideal gas hybrid.

I'm not sure either. It still follow the ideal gas law $pV=nRT$ though.


text:
Why is Chlormephos highly toxic despite requiring metabolic activation?
According to the book "The Chemistry of Organophosphorus Pesticides", the insecticide Chlormephos (S-(chloromethyl) O,O-diethyl phosphorodithioate) has an oral LD50 in rats of 7 mg/kg. For reference, the oral LD50 in rats for diisopropyl fluorophosphate is around 5 mg/kg. Why is chlormephos only slightly less toxic than DFP despite the fact that it would presumably require metabolic activation before being able to inhibit acetylcholinesterase due to the presence of a P=S bond? I would presume that the presence of a chloromethyl instead of methyl group contributes to the toxicity of chlormephos in some way, as the methyl analog is several times less toxic than chlormephos, but how? And why is chlormephos only slightly less toxic than DFP despite the fact that chlormephos would require metabolic activation before being able to inhibit AChE, while DFP would not?

If metabolic activation creates a really nasty toxin when an otherwise harmless compound is ingested, the initial harmlessness doesn't matter.

@matt_black according to that same book, parathion has an ld50 in rats of 6.4 mg/kg. So I suppose a better question would be "Why is the toxicity of chlormephos on par with that of parathion?"


text:
Calculating Enthalpy
I’m typing this from my phone...sorry if the quality is sub par.
I have a very involved spreadsheet from a colleague of mine that was put together in 1994. The spreadsheet calculates the excess air involved in combusting landfill gas in an enclosed flare.  I’m trying to step my way through it little by little.
At one point in the spreadsheet, there is a calculation for enthalpy of the inlet gas stream. For the methane portion of the inlet stream, assumed to be 50% of the LFG stream, the formula being done is very clearly:
$$MC_p\Delta{T}$$
The inlet gas stream is assumed to be $\pu{60 ^\circ F}$. The $C_p$ value for methane is assumed to be at $\pu{77 ^\circ F}$.
My question is: Why is $\Delta{T} = \pu{60-77 ^\circ F}$ and not $\pu{77-60 ^\circ F}$?  My understanding is that it doesn’t matter as the change in temperature is an absolute value.
EDIT 1
The following image is the particular place in the workbook where I am having difficulty understanding what is happening.  How does the inlet stream have any enthalpy at all?  Nothing has happened to it yet, it hasn't undergone any reaction.

Adding a negative value is equal to subtracting a positive value.

I guess I’m asking why have it 60-77 as appears to 77-60. It matters because in one case you get an exothermic reaction and in the other way it’s endothermic.

If you want help with this, you're going to need to provide a much  more complete and  precise description of your system.  For instance is your "inlet stream" a mixture of 50% air and 50% methane from a landfill that you are combusting?  And what is the entire system whose temperature is being increased by 17F?

50% methane, 50% CO2 for simplicity. Combustion air is obviously oxygen and nitrogen. Don’t think I need to provide a balanced combustion reaction for everyone.  The temperature increases in the entire system. System consists of CO2, CH4, O2, N2. Basic combustion reaction.

Stilll doesn't make sense.   If you combust a 50% air/methane mixture adiabatically, the temp increase is going to be far more than 17F.   So you're not just heating up the air/methane mixture, you're heating up a bunch of stuff around it as well.  That's the system to which I'm referring.

Maybe I just don’t understand what the person who put the spreadsheet together is trying to do. I’d post it but honestly I don’t know if I’m allowed to.

Sounds like the best approach for you is to at least get to the point where you understand the system well enough to be able to pose clear questions about it; maybe you could have someone you know look over the sheet.

If I had to guess, I think your colleague was trying to set $\pu{77 ^\circ F}$, the value $C_p$ was for, as the "reference state." The full equation for enthalpy at a point is:
$$H - H_{ref} = mC_p(T - T_{ref})$$
At your reference point, the enthalpy is $0$, so in that stream, you get:
$$H - 0 = mC_p(T - 77^\text{o} F)$$
This is okay to do because enthalpy is a state function, meaning that you're always focused on the change between states (by increasing or decreasing temperature/pressure/volume), not necessarily how it was done. For instance, if you wanted to get from point $\ce{A -> B}$ and you went $\ce{A -> B or A -> C -> D -> B}$, the change in enthalpy would be the same in both scenarios.
The value that you get in that cell is the enthalpy of methane relative to the reference state.
In your energy balance calculations, the reference states typically cancel out. Here's a quick example for flow in and out of a "box" (note that since this would be a steady flow through the box, the mass flow in would equal the mass flow out, so $m_{in} = m_{out} = m$):
$$(H_{in} - H_{ref}) - (H_{out} - H_{ref}) = mC_p(T_{in} - T_{ref}) - mC_p(T_{out} - T_{ref})$$
$$H_{in} - H_{ref} - H_{ref} + H_{ref} = mC_p(T_{in} - T_{out} - T_{ref} + T_{ref})$$
$$H_{in} - H_{out} = mC_p(T_{in} - T_{out})$$
You can find more information for this by googling something like "enthalpy reference state", but I hope that clears some of it up.
Update 1 based on Edit 1
Enthalpy tells you how much energy your stream has at a specific state compared to the reference. You don't need to have a reaction or something happening in that stream to get the enthalpy. The only way it would be $0$ is if it was pure methane at the same reference state, $\pu{77 ^\circ F}$ and I'm assuming $\pu{1 atm}$.

I'm going to take a guess at an answer based on the OP's limited description of the set-up:
I'm going to switch terminology from my comments above.  Let's let the system be the reacting mixture, and the surroundings be everything that's in thermal contact with the reacting mixture.
The way calorimetry usually works is you burn (or otherwise react) a small quantity of a substance (the system) in a large heat bath (the surroundings).  You ignore the heat capacity of the system, and calculate the thermal energy released by the reaction from $q_{p, surr} = m_{surr} C_{p, surr} \Delta T_{surr}$.  [I've added the "$p$" subscript to indicate you have a constant-pressure system, which is implied by the fact that you are calculating $\Delta H$, rather than $\Delta U$, from the heat flow.]
In an exothermic reaction, $\Delta T_{surr} >0 $, and thus $q_{p, surr} = \Delta H_{surr} > 0 $.  But whatever heat flowed into the surroundings must have flowed out of the reacting mixture (the system), and thus $q_{p, sys} = \Delta H_{sys} = -q_{p, surr}$.
It sounds like your colleague just reversed the sign of $\Delta T$, changing it from positive to negative, to get the correct sign for the heat flow relative to the system (negative) from the direction of heat flow relative to the surroundings (positive).

I just added a screenshot to my original post in an attempt to get into more of the specifics.


text:
How to find the equilibrium constant under non-standard conditions?

Given the following thermodynamic data at $\pu{298 K}:$
$$
\begin{array}{lrrr}
\hline
\text{Compound} & \Delta_\mathrm{f}H^\circ/\pu{kJ mol^-1} & S^\circ/\pu{J K^-1 mol^-1} & C_p^\circ/\pu{J K^-1 mol^-1}\\
\hline
\ce{HI(g)} & 26.5 & 206.6 & 29.2 \\
\ce{I2(g)} & 62.4 & 260.7 & 36.9 \\
\ce{H2(g)} & 0.0 & 130.7 & 28.8 \\
\hline
\end{array}
$$
determine the equilibrium constant $K$ for the following reaction at $\pu{475 K}$ and $\pu{1 bar}:$
$$\ce{I2(g) + H2(g) <=> 2 HI(g)}$$
NB: Do not assume that the enthalpy and entropy of reaction are independent of temperature. What assumption(s) do you need to make?

At first I thought I could work out the enthalpy change using Kirchhoff’s law of thermochemistry
$$\Delta H(T_2) = \Delta H(T_1) + \int_{T_1}^{T_2}\Delta C_p\,\mathrm dT$$
and then find the entropy change of reaction similarly. Then apply
$$\Delta G = \Delta H - T\,\Delta S$$
and use the equation relating $\Delta G$ and $K,$ but I know that is only for the Gibbs free energy change under standard conditions. The question tells me to assume the enthalpy and entropy depend on temperature, so I’m not sure what my approach should be for this.

The first statement about $\Delta_rG$ is not correct. It should be correct if $K$ is replaced by $Q$

Wouldn’t it be -RTlnK+RTlnQ?

I don't think so. Look ! If $Q$ is approaching $K$, your formula would get $\Delta G = 0$, and $K = 1$.  This is not correct.

Do you know how to quanify the effect of temperature on the equilibrium constant?

The Van‘t Hoff equation? Not sure how I should use the Cp

That's correct.  But first you need to use the Cp's to express the $\Delta H^0$ as a function of temperature.  Then you have to integrate the Van't Hoff equation taking into account the temperature dependent of the $\Delta H^0$.

That makes sense. So I don’t use the deltaG= deltaH-TdeltaS at all? I don’t need to find the entropy of reaction?

You need that to get K at 298K.

Ok. So is the assumption that Cp is independent on temperature? Or is it that deltaH and deltaS only vary slightly with T?

And is it ok to just substitute the modified deltaH into the already integrated regular form of the Van’t equation?

They expect you to assume that the heat capacity is independent of temperature, so that $$H^0(T)=H^0(298)+C_p(T-298)$$It isn't OK to use the integrated regular form.  You need to integrate the new version.

Ok, thank you so much for this. Has this been done before so I can check the final form of my integrated expression?

Have some confidence and integrate it yourself.

I have certainly given it a go, but I have no idea whether it is right or not

Using the Gibbs- Helmholtz equation with deltaG(standard)=-RlnK and integrating with the equation with the modified deltaH i end up with the same as I would have gotten had I just substituted it after. Is the T in the (T-298 ) not just a constant and not a variable? It would be the T we are wanting to find the new K at. Sorry I am taking up so much of your time. I really appreciate it

Ok with a fresh mind I realised what I said was just silly. I integrated properly (i hope) and got K=137. I would write out my full expression but I’m not sure how to do that on here. Does my final answer look right?

@MarinaCalder You can post an answer to your own question, it's a good practice on SE sites. If you are still having difficulties with formatting and typesetting math, post it the best way you can assure it's readable, and someone will edit it. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

There should be used the similar integrating formula for temperature dependence on reaction entropy, with the temperature in the integral denominator ( Cp . dT/T = dQ/T = dS ).


text:
Can vapour pressure be used to 'generate' work for free?
Given a bottle of water, a closed system, some of the water molecules in the liquid phase will have enough energy to escape that phase, forming water vapour, contributing to vapour pressure.
Now consider the simple setup below:

Water is used as an example, does not have to be water. The work for pumping water is converted to gravitational potential energy of water. But the system seems to be capable of doing extra work on the mass while retaining that gravitational potential energy gained. Where does this capability come from? Doesn't the existence of this capability already violate First Law?
P/S: I know the pressure increment due to formation of vapour will be small, but just because it is small and so negligible does not mean that the 'minor violation' of First Law can be overlooked, so I believe something must have been wrong with the argument above.

Anywhere you see a bias in movement you can attempt to exploit it to do work, even if it may not be worth your while.

The increase of pressure is of the order of 3% at room temperature. So this phenomena may produce a negligible amount of work

But at boiling water temperatures, this force is sufficient to drive locomotives and engines all over the world.

If you partially-filled a water bottle at 25 °C, and then replaced the remaining air with completely dry air at 1 atm and sealed it, the total pressure inside would increase to 1.03 atm, because the vapor pressure of water at 25 °C is 0.03 atm.  However, the air in water bottles is probably already partially saturated with water when the bottles are sealed.  If it's, say, 80% relative humidity, then the increase in pressure is only going to be 0.006 atm (20% of 0.03 atm).  That's why sealed water bottles don't seem to be pressurized (assuming you're at the same $T$ at which they were sealed).

Yeah, the pressure increment is small, and could produce small amount of work, doesn't that violate First Law? Or are you saying 'violation of First Law by a little' does not matter?

I believe the vapour produced in the step 3 of you experiment is due to the particles with high order speed in the Maxwell distribution of speeds curve that escape because of their higher kinetic energies. When escaping, they will reduce the average kinetic energy of particles in the water sample, which will then reduce the net temperature of sample.
This will then prompt the system (water sample) to take energy, in the form of heat, from the surroundings to reestablish the thermal equilibrium. Hence taking back the work done by system on surrounding via expansion of the bottle through heat to maintain its temperature. Although, as you can see in the Maxwell distribution of speed, the particles with enough kinetic energy to escape from the water surface are very less at low temperatures.
So, if you take a system with just enough temperature and manage to isolate it from the surroundings, I think you should see a visible drop in the temperature of water proportional to the work done in expanding the water bottle.
Mathematical description of the above Explanation:
Let's say the total energy change taking place after the particles escape water surface is $\Delta U$ out of which $W$ is used to do work on the surroundings. The remaining energy ($\Delta U - W$) is used to establish an equilibrium temperature inside the bottle and hence is given back to the air-water system inside the bottle.
This step may be happening simultaneously with the next steps I am writing but I am assuming that the air- water system inside the bottle will reach an equilibrium first before any work is done. This is purely to make the math easier and because thermodynamic is the study of equilibriums rather than processes so it won't affect the results as long as this equilibrium is established even when we are studying the final change taken place.
So the net change in temperature inside the bottle is $$\frac{-W}{C_{system}}$$ where $C_{system}$ is the heat capacity of the system.
Net temperature difference between system and surrounding $$\Delta T=\frac{W}{C_{system}}$$
So the system will absorb energy in the form of heat to bring the temperature back to equilibrium $$Q=C_{system}\Delta T=>Q=C_{system}\biggr[\frac{W}{C_{system}}\biggr]=>Q=W$$
So we can see mathematically that the system will take back all the energy it gave to the surroundings by doing work in the form of heat.

Using equations will make your answer clearer .(Equilibrium and Thermodynamic Expressions)


text:
ORCA: How to plot an adiabatic potential in dihydrogen H2 molecule?
I stared to study ORCA, and I try to obtain classical results for dihydrogen as for example. I need to get a starting point to understand what needs to be done.
So, How to plot an adiabatic potential in dihydrogen H2 molecule with ORCA calculation?
! RHF OPT def2-QZVPP
%geom Scan
B 0 1 = 1.0, 3.0, 12
end
end

* xyz 0 1
   H        0.000000       0.00000        0.00000
   H        0.800000       0.00000        0.00000
*

An optimisation only finds you the minimum of the curve, it doesn’t trace out the entire curve; for that you need a scan.

@orthocresol How and what output files should I use to build?

In addition to the answer, plus the ORCA manual and input library which I suggested on your last question, I can also recommend https://youtube.com/playlist?list=PLxm4o04Xz3Bb7q5EKZNHnA8peEuJQYk0L.

By adiabatic, I presume you mean the Born-Oppenheimer approximation (which is usually used). If you need a plot of potential against H-H distance, then you need to do a relaxed potential energy surface (PES) scan.
! UHF OPT def2-QZVPP
%geom Scan
B 0 1 = 0.3, 1.3, 30
end
end

* xyz 0 1
   H       -4.61685        1.79381        0.00000
   H       -4.14986        1.26166        0.00000
*

The Scan directive in %geom section will perform a relaxed geometry scan (although in this case you have only two atoms so there is no difference between relaxed and unrelaxed). The B 0 1 = 0.3, 1.3, 30 line will scan the bond (B) between atom 0 and atom 1 (in Orca, atom counting starts from 0). The scanning will start from the bond distance 0.31Å and will end at 1.3Å and will go through 30 steps (i.e. the PES will have 30 points in total).
Another thing is that you need to use the unrestricted hartree-fock (UHF) because when the H atom splits the two electrons go into different orbitals, so RHF will give the wrong behaviour at higher bond distances (it will force double occupancy in one orbital, resulting in $\ce{H+}$ and $\ce{H-}$ fragments). UHF gives the right behaviour but may not reproduce the correct energy. Multi-configurational methods might be required.
The input file that I have shown will show you the energy behaviour of the ground state (in your graph, $\mathrm{U_S}$). Plotting the first excited state (antibonding) i.e. the $\mathrm{U_A}$ graph would be more difficult. You would probably need to do some sort of orbital rotation to converge to the excited triplet state, and then use that as the guess for the geometry scan (or maybe do TDDFT). I don't know much about that.
After running the calculation, the energy values at each bond length will be printed out at the end of the .out file, and also as a .dat file. You can import the .dat file into Excel or other programs to get the actual visual plot.
Edit: If I am not mistaken, the $\mathrm{U_A}$ graph indicates the first excited triplet state, so converging to that UHF solution is easier, all you have to do is to set the multiplicity to 3 (*xyz 0 3). Using UHF/def2-SVP, the scan looks like this—

As you can see, the two curves are crossing whereas they shouldn't really cross. I am not entirely sure why this happens, but I suspect that using multiconfigurational methods will solve this. Also note that the energy plotted is the absolute SCF energy, so the graph's don't go to zero at high bond distances.

"absolute SCF energy, so the graph's don't go to zero at high bond distances." Of course, graph's go to -0.5 Eh.
"I suspect that using multiconfigurational methods will solve this" How can I do this in ORCA?

@Sergio These two guides https://sites.google.com/site/orcainputlibrary/cas-calculations/tutorial-active-space-selection , https://sites.google.com/site/orcainputlibrary/cas-calculations have a lot of information on setting up CASSCF on Orca.

If you need higher level methods (for very accurate energies) you can try CASPT2 (for that you have to look into the orca manual)

Where in that input file does it say that the atoms will be hydrogen atoms? I referenced your answer here: https://mattermodeling.stackexchange.com/a/4830/5.

@user1271772 I only posted the modified part of the input as the question had the full input. I have edited it to include the coordinates as well, so now it's a full input file.

You can feel free also to edit my answer or add one of your own!


text:
Can I use lipophilicity (partition coefficient) as a way to quantify solublity?
I am looking at investigating factors affecting the solubility of drugs using databases. So far, I am thinking of having my independent variables being the polar surface area, melting point, and size. My dependent variable would be lipophilicity. The reason why I am not directly measuring the solubility is that databases provide inconsistent solubility values.
Would quantifying solubility as lipophilicity be valid?

The same partition coefficient can be ratio of big numbers for one drug and small numbers for other drug.


text:
Can G, H, and S be negative?

Consider Gibb's free energy of a reaction in a closed system $G = H - TS$. Can $G$ ever be negative? (note, I am not talking about $\Delta G$)

I would suspect no because there is always some energy on every system.

What about $H$ and $S$? Are they nonnegative?

I would suspect yes because if there are bonds then they have internal PE and KE. I wouldn't know how to interpret negative disorder.

Is $H \ge TS$ always?

Variation on 1)

Are you asking about absolute values, values relative to some reference state, or changes in these functions?

Asking about absolute "measured" values and not changes. I am trying to understand them and interpret them.

G and H are like altitude.  They are not absolute; rather their values are always relative to whatever you set as a reference. So they can be either positive or negative.  There is no general constraint on the sign of their values.  For this reason, we usually talk about $\Delta G$ or $\Delta H$, since it's the changes that are meaningful; giving them arbitrary values isn't typically useful.
By contrast, there is a non-arbitrary zero for entropy.  In classical thermodynamics, it's the entropy of a pure substance at absolute zero. Since the temperature can never be lower than absolute zero in a system at thermodynamic equilibrium, $S$ (as opposed to $\Delta S$) can never be negative.
In statistical mechanics, there are various equivalent ways of defining the entropy, depending on the ensemble you are using.  The simplest definition is that for the microcanonical ensemble:
$$S = k_B \ln \Omega,$$
where $\Omega$ is the number of possible microstates of the system, and $k_B$ is the Boltzmann constant.  Since the minimum number of microstates is 1 (which happens at absolute zero), the minimum value for $S$ is zero.

If $H = U + pV$ where pressure $p \ge 0$ and $V \ge 0$, could internal energy $U$ be negative? Is there any way of computing it (i.e. wrt some reference) that allows it to be negative? I am curious if you can provide an example... I am especially curious about how to interpret it if it were negative.

@SeanRoody (1) Sure, $U$ can be negative. After all, you can define your reference for enthalpy such that $H<0$.  And if $H<0$, and $pV>0$, then it must be the case that $U<0$.

(2)There is no special significance to $U$ being negative; whether it's negative or positive depends on what you've set as your zero. E.g., suppose you want to measure all altitudes relative to Denver, CO (1600 m above sea level). Then the altitude at sea level becomes –1600 m. Is there any special signficance to the altitude at sea level now being –1600 m? Does it change anything about the physical properties of the atmosphere at sea level if you use this new system? Does it change how high you actually are if you use this new system? Answers: no and no. All you've done is to reset your zero.

As a specific example for OP: potential energy of electron/nucleus interactions is usually defined with respect to 0 being the state where electron and nucleus are infinitely far apart. Thus PE for an electron bound to a nucleus is always negative. The choice of 0 is arbitrary.

One could get somewhat philosophical and say that there is an absolute zero of energy, namely the absence thereof (say, at 0 K). Would that be an "arbitrary definition"?

@BuckThorn  It's an interesting thought but, at $0$ K, you would still have PE based on the pressure, so that would need to be specified also ($p=0$?).  Of course, even if you selected a pressure, you'd still have mass-energy which, IIUC, is based on the invariant mass (which used to be called the rest mass).  There's the KE of the sys. as a whole, but that's explicitly ignored in calculating int. energy. But there's also the energy from the expansion or  contraction of space-time (see http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/ ); is that part of $U$?


text:
Comparing bond strength by IR frequency: symmetrical modes
In literature the stretching mode of $\ce{CO}$ is often said to be the diagnostic criteria for strength of back-donation by virtue of weakening the $\ce{CO}$ bond. However, when speaking about homoleptic carbonyls, the vibrations of individual $CO$ ligands interact, forming symmetric and asymmetric modes with considerably different frequencies.
For example, let's say I want to compare strength of back-donation in $\ce{Cr(CO)6}$ and $\ce{Fe(CO)5}$ and compare it with a free molecule?

$\ce{Cr(CO)6}$ has modes $2119 (A_{1g}), 2027 (E_{g}), 2004(T_{1u})\ cm^{-1}$
$\ce{Fe(CO)5}$  has modes $2121 (A_{1}^{'}), 2041 (A_{1}^{'}), 2034 (A_{2}^{''}), 2013 (E_{1}^{'})\ cm^{-1}$
free $\ce{CO}$ vibrates at $2143 cm^{-1}$ )

What modes I should use for comparison?


text:
Use of powder of kernel of tamarind seeds for water purification
In the chapter Environmental Chemistry in our textbook it's given that Powder of Kernel of Tamarind seeds can be used for water treatment as a substitute for alum.
But if I understand correctly, alum ionizes in water and these ions neutralize the electrical double layer surrounding colloidal particles. How does a substance like 'Powder of Kernel of Tamarind seeds' produce the same effect?

The key is to know the composition and behaviour of this powder what is mostly unknown to chemists. My understanding is that alum hydrolyzes, forming colloidal hydroxide, which adsorbs these colloidal particles. The powder may form organic colloids of similar properties, settling later  down as aluminium hydroxide does.

mostly unknown to chemists – better expression may be "unknown to most chemists".

Alum is a flocculating agent which causes  suspended solids, like fine clay particles, to form aggregates and precipitate.

In my understanding,, Al^3+ in near neutral environment hydrolyzes faster than it gets chance to interact with colloids.


text:
Why does contaminated brake fluid conduct electricity? Using a voltmeter to test for water contamination in brake fluid
Contaminated brake fluid conducts electricity. In the automotive repair world, brake fluid is considered contaminated by two ways:

Brake fluid is hygroscopic. Higher water content in brake fluid means lower boiling point. Brake pads/rotors get hot which can boil the water near near the pads/rotors. Boiled water is steam which is a compressible gas. Having a gas in the brake lines leads to a spongy brake pedal. So, high water content absorbed by brake fluid is considered contaminated.
Brake lines contain copper. When microscopic bits copper get in the brake fluid, they can build up in the ABS unit causing failure of the ABS system. Thus, high amounts of copper mean contaminated brake fluid.

Since pure water and copper are both nonelectrolytes, how can brake fluid conduct electricity? Is there some other contaminant I am missing here that is an electrolyte? Is the water being absorbed, somehow, by the brake fluid not strictly H2O?
I am asking because I want to know if I can use a voltmeter to test if there is too much water contamination in brake fluid.

It is possible break fluid contains organic compounds that are weak electrolytes and partially dissociate in water.

Beside the point that Poutnik made, copper wouldn't be floating around in the brake fluid as metal flakes. The copper would be as ions so the copper ions would be conductive.

Even pure water is somewhat ionized. If water is more conductive than pure brake fluid, the test is valid. Why don't you try this as an experiment? To sample(s) of brake fluid fresh from a new, unopened, can, add distilled water, drop by drop, stirring each time and look for a change in conductivity. Measure the proportions, and you can graph the effect of moisture on conductivity.
N.B. There are two types of automotive hydraulic brake fluid in common use: glycol-based and silicone-based. You should test the conductivity of each, and the effect of moisture on each.
BTW, unless so much metal was suspended in the fluid to make a thick slurry, that would not affect conductivity. It is possible some metals, such as aluminum, would react with the fluid and ionize, but those metals should not be part of the brake system.

Yes. Experimenting is a good idea. I will be testing the conductivity of freshly opened DOT 3, 4, and 5.1 brake fluids in separate open containers (each container will be the same type). I’ll test each fluid’s conductivity at 15 minute intervals. If the conductivity rises for each fluid as time passes, then it is confirmed moisture absorbed from the hygroscopic brake fluids causes an increase in conductivity. This will be good as I know I can test brake fluid for water contamination with a voltmeter.

If the conductivity does not rise, well then I’ll have to use boiling point for testing water contamination. I will be conducting this experiment on Saturday, Feb 13 2021. If there are any ideas as to how this experiment can be improved, please comment below. Results will be posted the next day. Thanks!

@JoshM. How did the experiment turn out?


text:
What is Resonance, exactly?
I know the that it(Resonance) is the delocalisation of pi electrons but how and why are those electrons delocalised? What is the driving force that causes the effect?

Delocalisation as in resonance and indeterminacy have to do each other just because the latter is a general principle. It is at work even with an isolated hydrogen, or even with a single electron. No, they are not the same. Sorry @DrMoishe Pippik but this is a case when downvoting comments would be a needed function.

Resonance is a conceptual tool to help make sense of the real observed structures of molecules
Resonance is not a real physical phenomenon. That is the first important thing to realise. It is merely a tool to bridge the simple view where all bonds consist of integer numbers of electrons and the real structures of molecules.
The simple view of bonds often fails to represent the "real" observed structure of molecules. The archetypal example is benzene. A simple view of the bonding of benzene has the molecule consisting of alternating single and double bonds (which might be called cyclohexatriene). This theoretical molecule has the bonds neatly grouped into simple bonds that use up all the available electrons into either single bonds (one pairs of electrons) and double bonds (two pairs of electrons). But this would imply that the bond lengths alternate. But we know the structure of benzene and the bond lengths are all the same. We can explain this with molecular orbital theory when we recognise that some of the bonds are delocalised, but that gets fairly sophisticated fairly quickly and chemists prefer easier ways to keep track of the electron counts when writing structures on paper and explaining reaction mechanisms.
One halfway house that saves the ability to write structures where simple electron counts are possible is the idea of "resonance". Instead of needing to invoke molecular orbital theory all the time we can write the structure of benzene as a combination of two structures where the double and single bonds alternate. We pretend that the real structure is a "resonance hybrid" of the two hypothetical structures. This is like the mathematical average of the two competing structures.
More complex molecules can also be represented as combinations of (more complex) resonance structures. We often say that the structure consists of resonance hybrids. But resonance is not a real physical phenomenon. The different structures are not real independent structures that rapidly interchange: they are merely a convenient model so chemists can talk about electron counts and draw structures without having to invoke much more complex bonding theories.
So there is no such thing as "resonance" as a physical phenomenon. What drives the structures of molecules is the way orbitals in atoms combine together in molecules and that is more complicated than simple views of bonding can represent. The underlying driving force is quantum mechanics and energy minimisation of the possible ways atomic orbitals interact in molecules. And that is sometimes far more complicated than simple views of bonding can easily represent.

So, you are implying that resonance is not a real thing a nd is just a simple way of understanding the experimental and observed facts which otherwise would have been really complex to understand...right?

@DivyanshKailkhura exactly.

The canonical structures don't exist independently, but the electrons are actually free to zip through (or whatever the quantum mechanical equivalent for that) the conjugated centres, right? Just to clarify.

Resonance rather than a physical process is our description of a molecule.
Basically we mixes different descriptions (limiting structures or canonical formulas in the frame of valence bonds theory, which is the one more reflecting the way we draw chemical structures using Lewis based notation).
We need such a mixed description all the time that a single canonical form turns out inadequate.
I just recall here the the mix, so-called hybrid of resonance, is the only one physically describing the molecule and is, thus, real. The canonical forms are just hypothetical structures for which properties can only be wisely reconstructed or computed, depending on the level of treatment.
Shall we adopt different and more advanced ways to arrive at the molecule bonds description, we would see that the orbitals describing certain electrons would be spread over part of the molecule, or even its whole.
It is in such a sense that the electrons of a resonance hybrids are delocalised.
For what are the forces driving this delocalisation one can answer by big principle, that is the system has lower energy in the way it is. For what physically keep electrons around the nuclei fixing them in that molecular structure, well this is electromagnetic in nature, with the constraints of quantum mechanics. This is independent from actual delocalisation (resonance) taking place or not.
There is no special force for having resonance. It just happens that under the given potential, the electrons have a minimum in energy when spread rather than being under the influence of just one or a couple of nuclei.
If you really wish a kind of "meccano" analysis of the reason, let us take the archetypal benzene although it will be a kind of chicken / egg discussion.
A common perception at school is that is the sextet of pi electrons that drove benzene in its shape. However, a more natural picture with both energetic and symmetry grounds, is that the sigma-bond skeleton rather opposes its deformation (how it would be in the hypothetical form with localized double bonds wherein the bonds must shrink) and forces less bound pi-electrons to delocalise as a ring.
Finally I often heard that delocalisation is due to the Heisenberg principle. This is wrong. At least, it is true just in the sense that the above principle is general and forces us to treat electrons as spread particle for which only a wavefunction description is possible (the orbital, in atoms and molecules). In such a sense, of course, electrons are delocalised. But their are always so and is not the same (level of, or concept of) delocalisation which pairs with resonance.
In ethene, or in the hypothetical canonical form cyclohexatriene, the electrons are spread in orbitals. Still the pi-orbitals are localised in double bonds.
In benzene, the electrons, that still obey the quantum rules, are in spread orbitals. Indeed, as we know, there are no double bonds. But the principle of indeterminacy is alway there, and so is even for the only electron of one hydrogen atom.

Why do electrons delocalise?
Delocalising is a stabilizing force in a way such that when the electron has a greater number of orbitals to move into the energy spreads over a larger area than in a single orbital as the distribution of energy in multiple orbitals brings extra stability in the system.
There is also a possibility that after energy gets distributed there is a lowering down of the total internal energy of the molecule had there been a single orbital holding the energy ie. if there were no delocalisation.
Have a look at this one.
Why does delocalization of π electrons bring stability?


text:
Why are the "extents of reaction" the same for all reactants and products?
Consider the follwoing reaction:
$$\ce{a A + b B -> c C + d D}\tag{1}$$
where $\ce A$ and $\ce B$ are the reactants, $\ce C$ and $\ce D$ are the products, and $a (<0)$, $b(<0)$, $c(>0)$, and $d(>0)$ their respective stoichiometric coefficients. The extents of reaction are respectively defined for these reactants and products as:
$$\begin{align}
\xi_A(t) &= \frac{n_A(t) - n_A(0)}{a}\tag{2.1}\\
\xi_B(t) &= \frac{n_B(t) - n_B(0)}{b}\tag{2.2}\\
\xi_C(t) &= \frac{n_C(t) - n_C(0)}{c}\tag{2.3}\\
\xi_D(t) &= \frac{n_D(t) - n_C(0)}{d}\tag{2.4}
\end{align}$$
However, I have found in many references (e.g. source) the extent of reaction defined for the whole reaction as:
$$\xi(t) = \frac{n_A(t) - n_A(0)}{a}
         = \frac{n_B(t) - n_B(0)}{b}
         = \frac{n_C(t) - n_C(0)}{c}
         = \frac{n_D(t) - n_D(0)}{d}\tag{3}$$
I do not understand how you can justify that the extent of reaction of the different reactants and products are equal.

It has to be, because of the stoichiometry of the reaction: if your reaction consumes $x_\ce{A}$ moles of $\ce{A}$, then it must also consume $x_\ce{B} = x_\ce{A} \cdot b / a$ moles of $\ce{B}$. Now note that (by definition) $x_\ce{A} = n_\ce{A}(t) - n_\ce{A}(0)$, and $x_\ce{B} = n_\ce{B}(t) - n_\ce{B}(0)$. (I might have been a bit careless with the signs, but it doesn't affect the argument.)

So, you seem to start your reasoning from $\left|a\right|x_A = \left|b\right|x_B$. But, form where this expression is coming from? You refer to the stoichiometry of the reaction, but I was thinking that the stoichiometry (only) says:
$$\begin{align}
ax_A + bx_B + cx_C + dx_D &= 0 \tag{1} \
\
-\left|a\right|x_A -\left|b\right|x_B + \left|c\right|x_C + \left|d\right|x_D &= 0 \tag{2}
\end{align}$$
I have the impression I am still missing something...
Did you actually mean that:
$$
\left|a\right|x_A = \left|b\right|x_B = \left|c\right|x_C = \left|d\right|x_D \tag{3}
$$
?

How should some constituents of one reaction have reacted to a different extend than the others? That makes no sense.

Re. eq. (3): yes, that is dictated, enforced, by the stoichiometry of the reaction. I am sorry, but I have no idea how to explain that any better. It is a very fundamental concept in chemistry, as you can judge from Karl's comment. You cannot consume B without also consuming a proportionate amount of A (the proportionality factor being $a/b$); you cannot produce C or D without also consuming a proportionate amount of A.

There is a slight error, though, in eq. (3) of your comment: the $x$'s should be divided by the stoichiometric coefficients, not multiplied. In eq. (3) of your question it is correct.

Oops, so my relation $(3)$ must be written as
$$
\frac{x_A}{\left| a \right|} =
\frac{x_B}{\left| b \right|} =
\frac{x_C}{\left| c \right|} =
\frac{x_D}{\left| d \right|}
\tag {3}
$$

Let's take a numerical example: $\ce{3 A + 1 B -> 8 C + 2 D}$
Here : $a = -3, b = -1, c = 8, d = 2$.
Let's start from $100$ $mol$ of $\ce{A}$, $120$ $mol$ of $\ce{B}$, $1$ $mol$ $\ce{C}$, and no $\ce{D}$. Suppose that $10$ $mol$ $\ce{B}$ have reacted with $30$ $mol$ $\ce{A}$, and producing $80$ $mol$ $\ce{C}$ and $20$ $mol$ $\ce{D}$. At the end of the reaction, the remaining amounts are :
For $\ce{A}$ : $100$ $mol$ - $30$ $mol$ = $70$ $mol$ $\ce{A}$.
For $\ce{B}$ :  $\ce{120 $mol$ - $10$ $mol$ = $110$ $mol$}$ $\ce{B}$
For $\ce{C}$ : $\ce{1 $mol$ + 80 $mol$ = 81 $mol$}$ $\ce{C}$
$$\begin{align}
\xi_A(t) &= \frac{n_A(t) - n_A(0)}{a} =\frac{70 - 100}{-3} = 10\tag{2.1}\\
\xi_B(t) &= \frac{n_B(t) - n_B(0)}{b} =\frac{110-120}{-1}  = 10\tag{2.2}\\
\xi_C(t) &= \frac{n_C(t) - n_C(0)}{c} =\frac{81 - 1}{8}    = 10\tag{2.3}\\
\xi_D(t) &= \frac{n_D(t) - n_D(0)}{d} =\frac{20 - 0}{2}    = 10\tag{2.4}
\end{align}$$ The extent of reaction are the same for all products.

Many thanks! When you wrote "suppose that $10$ $mol$ of $\ce{B}$ have reacted with $30$ $mol$ of $\ce{A}$, and producing $80$ $mol$ of $\ce{C}$ and $20$ $mol$ of $\ce{D}$", you are actually using the (formal) expression:
$$
\frac{\Delta N_A(t)}{a} =
\frac{\Delta N_B(t)}{b} =
\frac{\Delta N_C(t)}{c} =
\frac{\Delta N_D(t)}{d}
$$
where $a$, $b$, $c$ and $d$ are the stoichiometric coefficients/numbers (with the sign convention).
Is that correct?

I actually found it in "Physical Chemistry from a Different Angle", Eq. (1.15). Maybe I was just missing the "different angle" from my other sources ;-) | https://link.springer.com/book/10.1007%2F978-3-319-15666-8


text:
In electrolysis, why does each atom wait to turn into gas until they reach a particular electrode?
More specifically why don't bubbles of both oxygen and hydrogen appear everywhere?
If the electricity just breaks apart bonds wouldn’t the hydrogen and oxygen just immediately turn into gas and bubble up wherever they split apart? Why does each atom wait to turn into gas until they reach a particular electrode?

Electricity doesn't breaks apart bonds. Moreover, electricity isn't a thing at all.

You can always [edit] your question to make it correct.

Those who are happily downvoting this question and closing it, do they actually understand the mechanism of oxygen evolution reaction at the electrodes? If not, please don't expect a novice to understand that either.

@M.Farooq agreed and +1 and an adjustment to the title to better match the body of the question.

$\ce{H2O}$ is not simply split apart by electricity, as you say. No ! What happens on one electrode is not related to what happens to the second electrode. Let's start by discussing what is happening on the negative electrode, the cathode.
The negative electrode behaves as if it contains plenty of electrons ready to react with anything able to do it. It could be positive ions. But in neutral solutions it may be water that does react with these electrons, according to :
$$\ce{2 H2O  +  2 e^- -> H2 + 2 OH-}$$ So, some bubbles of gaseous Hydrogen $\ce{H2}$ are produced on the cathode and the solution become basic because of the appearance of $\ce{OH-}$ ions in solution.
On the other electrode, the anode, which is the positive electrode, something must occur to produce electrons. Once again, it is water that manages to be decomposed in order to produce electrons. It is done this way : $$\ce{ 2 H2O -> 4 H+ + O2 + 4 e-}$$ Here some oxygen bubbles are produced on the anode. And the solution becomes acidic, if it was neutral in the beginning.
A last consequence of these chemical transformations is the fact that the solution becomes acidic at the anode and basic at the cathode. Of course, by simple mixing the $\ce{H+}$ and the $\ce{OH-}$ ions neutralize one another, producing water. So the only visible effect of the electrolysis is the production of $\ce{H2}$ at the cathode and $\ce{O2}$ at the anode, as if $\ce{H2O}$ has been split. Only "as if".

So what happens to the OH- ions in solution afterwards? And if 4 hydrogen are produced on the positive electrode, why does only the oxygen become a gas?

The $\ce{OH-}$ ion gets into solution, as ions are usually stable in solution, and cannot pass in the vapor phase. Same thing for the $\ce{H+}$ ion

So they don’t combine and turn into water but just stay there doing nothing?

@user11937382 What they do depends on what they can do. If electrolysis happens in a single solution, they will eventually recombine (stirring will accelerate the process). If there is no stirring or if the solution is not truly continuous (e.g. two beakers with a salt bridge), they will hang around. And sometimes (e.g. the electrolysis of a sodium chloride solution where the second equation doesn’t happen but chlorine gets oxidised instead) other fancy reactions can occur.

In electrolysis, why does oxygen only appear on the anode?

Easy way to remember this is that oxidation occurs at the anode (both start with vowels), and oxidation implies a loss of hydrogen. If we lose hydrogen from water, we are left with oxygen.
If you wish to understand it crudely, the concept is that anode the surface in the solution where the electrons are being swept away by a "pump" (=read battery). Hence the anode needs electrons to balance the electrons, which were pump away by the battery towards the cathode. Recall a vacuum cleaner sucks air at one end and lets it out from another end. There is no creation or destruction of electrons.
Now depending on the voltage, the anode material can withdraw electrons from anything. Which is the most abundant material around the anode? The solvent. Hence, if we withdraw electrons from water, we get the overall reaction, which is written everywhere:
$\ce{ 2 H2O -> 4 H+ + O2 + 4 e-}$---- eq (1)
Now note that oxygen evolution reaction is still subject of interest to electrochemists. They wish to understand how does equation (1) proceeds at a molecular level. Lookup Wiki for Oxygen Evolution Reaction. The pictorial mechanism is

The ‘most abundant material’ doesn’t matter as much as the most easily oxidised material though …

Also, I like to remember it by saying Anodic Oxidation is the Alpha and Omega.

Electron transfer between the 'species in the solution' and the 'electrode' take place at the surface of the electrode. So, considering water with some acid
Oxidation at anode: $\ce{2 H2O(l) -> O2(g) + 4 H+(aq) + 4e−} \quad E^\circ = \pu{+1.23 V}$ (for the reduction half-equation)
This oxidation (electron transfer from water to the electrode) occur at the electrode surface. So, $\ce{O2}$ gas should evolve at the surface of the electrode. Similar is the case of hydrogen, $\ce{H+}$ in the solution gets reduced to $\ce{H2}$ gas at the surface of the cathode.
https://en.wikipedia.org/wiki/Electrolysis_of_water

Why does each atom wait to turn into gas until they reach a particular electrode?

There are no oxygen atoms or hydrogen atoms formed in the solution. There is also no "electricity" going through the solution. The processes happen at the electrodes because the cathode is able to provide electrons (to reduce hydrogen ions to dihydrogen) and the anode is able to take up electrons (to oxidize hydroxide ions to dioxygen, with the hydrogen turning into hydrogen ions).

If the electricity just breaks apart bonds wouldn’t the hydrogen and oxygen just immediately turn into gas and bubble up wherever they split apart?

In solution, water auto-dissociates into hydroxide and hydrogen ions. This process is not electrochemistry, and it is insufficient to form elemental hydrogen and oxygen.
The other answers nicely explain what actually happens.