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Why don't we consider quantum effects when thinking about reactions and their mechanisms?
When we're reasoning about chemical reactions and their mechanisms (in organic chemistry in particular), the way we model the behaviour of molecules is almost in a “common sense” way. In terms of tracing their trajectories, steric effects, positive and negative charges behaving how point charges would. The treatment to me feels very different to how we think about electrons, for example.
Is this because reactions occur in the bulk, and even though we are thinking about what is happening in terms of the behaviour of individual molecules, that molecule actually represents some sort of average behaviour of the bulk, and so quantum effects/weirdness are being averaged out and we can think of them acting in a more “common sense” way?
Of course I understand that the actual bonding and our understanding of chemical structures and things like that comes from QM based models.

Molecules are a good deal heavier than electrons.

Quantum and relativistic effects fortunately tend to be pretty subtle in case of most molecules' reactions. Still there's QM/MM used sometimes. In general, one models stuff as crudely as it's still viable.

Electron transfer involves tunnelling so that could count as your quantum weirdness. Molecules also diffract in a double slit type of experiment but thats not a reaction but is still 'weird'.

@IvanNeretin Hm. Molecules aren't small enough to where quantum effects are significant?

Pretty much so. One who thinks of electron as a point makes a grave mistake. One who thinks of nucleus as a point has a decent approximation for most cases.

I think you need to define "quantum effects" more precisely. For example, organic mechanism are often explained conceptually in terms of orbital interactions, and orbitals can only exist because of QM, so is that a "quantum effect" or not?

@IvanNeretin - I largely agree, but protons are also small enough that proton tunneling effects can also be relevant to chemical reactions, for example giving classically impossible isotope effects

Without QM, bonding as it is now understood would not be understood! Neither would atomic structure period. What is taught is an abstraction of that, though, simplified models, enough to be useful. So you may want to reconsider the premise of the question and add a clarification about what you mean wrt "quantum weirdness".

"Prague school" of quantum chemistry uses quantum chemistry for computational  screening of potential drug candidates and eliminates up to 99% even before anything is synthesized. But lately, there is heavy development of AI systems for that, that is probably much less computationally demanding.

The most common (and fairly useful) treatment of reactions is in the Born-Oppenheimer framework, ie assuming nuclei are slow and follow classical movement while electrons are treated with quantum mechanics. When you write reactions on paper the bond breaking-bond forming process is a quantum process, but the regular electron counting methods give generally a good guess about them. The BO approximation goes to the toilet eg when light atoms tunnel (PCET and similar cases) or in case of many photochemical reactions. The diagrammatic versions are sill a good first guess, though.

@Poutnik What is the "Prague school"? Also, prescreening of drug candidates with non-QChem tools (QChem are freaking expensive and give little insight to drug behavior) is pretty routine for decades and decades.

"Prague school" is informal name used by quantum chemist community for the group around recently passed Rudolf Zahradník,  former head of Czech Academy of Science. They realized the computarional screening is faster and cheaper than synthesis and primary testing.

@Poutnik No offense, but it is pretty pretentious to call this a "school". The idea of computational pre-screening is several decades old, especially for small molecules (like drugs). It is literally the whole basis of most chemoinformatics. Also, there are tons of properties that are not really efficient to screen based on QChem, so Qchem prescreening alone has its limitations.

@Greg Quantum chemists coined that informal term, not me nor prof. Zahradnik himself. I have never been personally involved in the field, so I cannot provide more info. But as far as I know, as Prague schoold is meant rather the scientist group and their general activities, rather than particular Qchem application. // In 1961, Zahradník became the head of the “Applied Quantum Chemistry” group, so decades.... //  More here

The answer in short is Born Oppenheimer Approximation (I'm a physics grad student, by the way), and the fact that electrons are much lighter, or "more quantum", than atom nuclei.
In more details:
Given nuclei are much heavier than electrons, and given typical temperature and pressure in chemistry. It turns out one can use Born Oppenheimer Approximation to treat electron quantum mechanically (using orbitals, for instance) while treating nuclei classically, and still get good enough results (for reaction barrier, for instance). Under Born Oppenheimer Approximation, the energy of a system, given a specific nuclei configuration, is given by electrons energy and electron-nuclei-interaction-energy calculated by quantum mechanics (assuming classical nuclei position) + nuclei energy calculated classically (assuming classical nuclei position).
There are a few cases where Born Oppenheimer Approximation fails in chemistry. For instance, proton transfer (movement of hydrogen) often need quantum mechanical treatment (like tunneling rate) if one wants to get good result (for reaction barrier or reaction rate, for instance) since hydrogen is the lightest of nuclei. Chemistry in astronomy and laboratory with extreme pressure and temperature might require quantum mechanical treatment of nuclei (for instance, superliquidity of Helium 3 requires quantum mechanical treatment of nuclei). Zero-point energy of nuclei vibration also require quantum mechanical treatment of nuclei.

When you mean proton transfer what exactly are you talking about? Would acid/base reactions count here?

@xasthor I'm not totally sure. I do know that {low barrier, long distance} reaction doesn't necessarily require quantum tunneling of H because thermal energy is enough to go over the barrier. {high barrier, short distance} reactions are strongly affected by tunneling because tunneling rate decay over distance quickly... continuing

@xasthor I'm more of a physicist than chemist, but I feel acid base reaction tend to have {low barrier, long distance} (low enough for thermal energy to go over the barrier), especially when facilitated by coordination of water around ions. I have only read about how quantum tunneling of H matters in the context of tautomerization-like reactions where H is moving within the same molecule while breaking covalance bond {high barrier, short distance}

If I understand you correctly, you are implying that in acid-base reactions, there is necessarily an intermediate phase where the H+ ion is solvated in water and this is transported over a distance, which isn't the case. Consider the acid-base reaction between HCL and NH3 for example. I don't think this would qualify as a "long distance" reaction

@xasthor I didn't say it well. I probably should leave the {especially when facilitated by coordination of water around ions} out of my explanation since it is not always the case and it fundamentally doesn't solve the problem. I should have said that typical acid base reaction tend to have low energy barrier (you can google "acid base reaction, typical energy barrier"), so it is not necessary to use tunneling to judge whether the reaction will happen or not. This is what I know for sure.

@xasthor Now my speculation: Furthermore, short distance indeed mean tunneling will happen way faster than long distance, but tunneling is still a slow process compared to  reacting by overcoming the barrier thermally. If so, then tunneling with {low barrier, short distance} might be relevant, in terms of getting the absolute value of reaction rate. However, percentage-wise tunneling is still not as relevant since reacting by overcoming thermal barrier can explain, say, 90% of the reaction rate. This is my guess.

@xasthor "facilitated by coordination of water around ions" is an attempt to explain why low barrier. HCL and NH3 reacting in air, has high barrier than its water facilitated reaction: HCL first becoming water coordinated F- and H3O+ which then react with NH3. I should not have said "I feel acid base reaction tend to have {low barrier, long distance},.. water coordination..". I should have said "acid base reaction tend to have {low barrier} and my guess is thermally getting over the barrier is still the most dominant process since tunneling tend to be a slow process overall"

@xasthor I guess the intuition should be: thermally overcoming barrier depends on temperature. Tunneling pretty much doesn't dependent on temperature at all. When barrier is very large compared to thermal energy, tunneling dominate reaction rate percentage wise (given mass is small, like H). When barrier is very small compared to thermal energy, thermal dominates reaction rate percentage wise because tunneling tend to be a slow process overall. However, when distance is very long, tunneling practically can't happen, so {high barrier, long distance} reaction just doesn't happen at all.

Okay, understood

Quantum effects most definitely are important in chemistry,and can easily be observed even in bulk samples, such as the photoelectric effect. Put $\ce{H2}$ and $\ce{Cl2}$ in a flask (in a dark room), and nothing happens immediately because the activation energy to dissociate $\ce{Cl2}$ has not been reached. Hit the flask with a bright red light, and still nothing. Use even a weak UV source, though, just a single near-UV LED, and BANG. Not rocket science, but certainly quantum chemistry, worth a Nobel prize for Einstein in '21.
See a demo of the reaction. If you want to try it, enough UV can go through an ordinary borosilicate test tube to work. Caveat: Both $\ce{Cl2}$ and the resultant $\ce{HCl}$ are somewhat poisonous, and, with the bang, the glass container can shatter!

This reaction has radical mechanism, no kicking out electrons, like in photoelectric effect.

@Mithoron, it is the same phenomena: wavelength longer than needed for activation has insufficient energy, no matter how many photons. Whether ionized, or moving to a higher energy level, e = hν is a quantum effect. Each photon is a quanta of energy.

Note for kiddies : "in '21" refers to 1921, not 2021. :-)

I agree that molecules absorbing specific frequencies is not the photoelectric effect and calling it so might be misleading to student (though both share the same underlying quantum mechanism).

When we're reasoning about chemical reactions and their mechanisms (in organic chemistry in particular), the way we model the behaviour of molecules is almost in a “common sense” way.

When the reasoning happens in our heads and on paper, it is easier to use heuristics than quantum mechanics. The chemical reactions are governed by quantum mechanics, of course, but you can predict quite a bit through analogies to known reactions in your head without doing a quantum chemical calculation.
This works particularly well in organic chemistry because many reactions involved functional groups of a limited set of atoms (C, H, O, N, ...). Of course, organic chemists use almost the entire periodic table when it comes to reagents, but those are a given for a type of reaction, and you ask whether you can apply them to a new organic molecule, having tried them for a lot of cases already.

Is this because reactions occur in the bulk, and even though we are thinking about what is happening in terms of the behavior of individual molecules, that molecule actually represents some sort of average behavior of the bulk, and so quantum effects/weirdness are being averaged out and we can think of them acting in a more “common sense” way?

No, each single molecule has to react independently, so it does not help that the molecules are present in bulk (like it does for equilibrium thermodynamics). In fact, the "cheaper" computational methods such as molecular dynamics are not so helpful when trying to predict chemical reactions.

Of course I understand that the actual bonding and our understanding of chemical structures and things like that comes from QM based models.

Traditionally, QM was used to study very simple systems (dihydrogen reacting with proton, for example), and you had to extrapolate to more complicated systems. This is changing as methods and hardware become more powerful. Once you have a handheld device that can do a full QM calculation on the reaction you are attempting in the lab, I'm sure chemists in the lab will do a quick QM calculation before starting the wet chemistry, but we are not quite there yet.

I think you say why don't we use quantum mechanics to calculate e.g., the trajectories of the molecules of a gas.
We could write a single wavefunction for all the molecules of the gas but remember since the Schrödinger equation is a partial differential equation, it gets very complicated.
So we use some approximation, which is not 'correct' but it is still a  good approximation for our experiment e.g., when the # of molecules gets so big in thermal equilibrium we can say that the direction of the motion of the particles is random.
That's why statistical mechanics was created in the first place.
Conversely some processes cannot be explained using only classical  mechanics and these processes were unexplainable before QM. So yes, we look for a model that is close to the outcomes of the experiments and we can use it to make predictions for what will happen if we change the setup of the experiments a little bit. And if we don't push these models to the limit of their range then we can use them to understand better chemistry.


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Is it possible to (naturally) change alkalinity without changing pH?
The Bjerrum plot shows how the different species of dissolved inorganic carbon (DIC) are distributed in sea water as a function of pH. The higher the pH, the more carbonate ion, less dissolved $\ce{CO2}$ (i.e. $\ce{H2CO3}$), and vice versa. And according to Takahashi et al. (1993), this is relevant for the $\ce{pCO2}$. That is, the higher the pH, the lower the $\ce{pCO2}$.
Are there processes in nature that can change alkalinity, as defined by Dickson et al. (2007) (i.e. sum of negative charges of non-conservative ions), without changing pH?
Here is the definition:

4.2 Total alkalinity

The total alkalinity of a sample of sea water is a form of mass-conservation
relationship for hydrogen ion. It is rigorously defined (Dickson, 1981) as “. . . the number of moles of hydrogen ion equivalent to the excess of proton acceptors (bases formed from weak acids with a dissociation constant K ≤ 10 –4.5 at 25°C and zero ionic strength) over proton donors (acids with K > 10 –4.5 ) in 1 kilogram of sample.” Thus

$A_T = [\ce{HCO3-}] + 2[\ce{CO3^2-}] + [\ce{B(OH)4^-}] + [\ce{OH-}] + [\ce{HPO4^2-}] + 2[\ce{PO4^3-}] + [\ce{SiO(OH)3^-}] + [\ce{NH3}] + [\ce{HS-}] + ...
- [\ce{H+}]_F - [\ce{HSO4-}] - [\ce{HF}] - [\ce{H3PO4}] ...$ (13)

where the ellipses stand for additional minor acid or base species that are either unidentified or present in such small amounts that they can be safely neglected.
$[\ce{H+}]_F$ is the free concentration of hydrogen ion

This last term is defined later in the text according to
$[\ce{H+}] = [\ce{H+}]_F\left(1+\frac{S_T}{K_S}\right)$ where $S_T = [\ce{SO4^2-}] + [\ce{HSO4-}]$ and $K_S$ is the $K_a$ for $\ce{HSO4-}$.
Zeebe & Wolf-Gradow (2001) show in their figure 1.1.3 that the air-sea exchange of $\ce{CO2}$ does not change alkalinity (the horizontal line). Although it does change pH (diagonal background lines). This is in fact a problem related to antrhopogenic climate change, as $\ce{CO2}$ is taken up by the ocean, causing acidification, and posing threats to marine ecosystems. I was wondering if there is a process that draws a line that is parallel to the pH isolines in the figure below.

What are non-conservative ions?

Note that this is a follow up question from Earth Science SE - Why does decreasing surface alkalinity increase surface pCO2?

Some participants have there an interesting idea that changing [OH-] does not change [H+] and therefore nor the pH.

I think conservative ions are those whose concentrations won't change due to changes in the environment (temperature, pH, salinity), such as Na+ or Mg++. Non-conservative ions, such as carbonate ions, do. Does it make sense? https://doi.org/10.1016/j.marchem.2007.01.006

It does make sense, but I would rather expected conserved as a passive form. Conservative sounds to me like such ions are actively conserving, protecting something. Or at least proponents of conservation.  But I am not an EN native. Perhaps conservative is closer to conserved than to conserving.

If there are 2 controlled processes, one increasing alkalinity, e.g. by dissolving $\ce{MgO(s)}$ or alkaline earth carbonates, and the other dissolving $\ce{CO2(g)}$, so the concentration ratios of $\ce{CO2(aq)}$, $\ce{HCO3-(aq)}$ and $\ce{CO3^2-(aq)}$ are constant, then yes.
It is theoretically possible, as the above constant ratios would form pH buffers with constant [$\ce{H+}$], [$\ce{OH-}$], but with variable concentrations and therefore alkalinity.
But in geological context, it is highly inprobable both processes occur in such a balanced way, so generally no, it is not possible and can occur just accidentally.

I just learned about the example: $\ce{CO2}$ absorption in sea water changes pH, but doesn't change alkalinity, since it adds a carbonate ion at the same ratio it adds a H+ (Figure 1.1.3 in Zeebe & Wolf-Gradow: http://geosci.uchicago.edu/~kite/doc/Zeebe_CO2_In_Seawater_Ch_1.pdf ). Since it is possible in one way, I was wondering if something similar could exist in the other direction too.

@ouranos Well, as the procedure of alkalinity determination involves titration with methyl orange pH indicator and heating the sample to get rid of CO2, its dissolution alone is not really relevant, if not followed by dissolving of alkaline matter.

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Concentration of products in tank reactor
I want to calculate the concentrations of all components in the outlet stream of the reaction: C$_2$H$_6$ -> C$_2$H$_4$ + H$_2$  (A -> B + C)
I have a tank reactor (isothermic and isobar): V= $305$ dm$^3$, C$_{A0}$ = $53.46$ mol/m$^3$, P = $2$ bar, E$_a$ = $130$ kJ/mol, A = $10^{13}$ s$^{-1}$. The in flow is $10$ Nm$^3$/h and has the same temperature as the reactor.
First of all, what is $10$ Nm$^3$/h (some places it says that it is normal cubic meter)?
I began by using the gas law to calculate the temperature, which I got to be $450$ K. Then by the use of the Arrhenius equation I got that the rate constant $k$ = $0.0081$ s$^{-1}$.
I then calculated the outlet concentration of A by using the material balance:
C$_{A0}$ $v$ - $k$ C$_{A1}$ V = C$_{A1}$ $v$
C$_{A1}$ = (C$_{A0}$ $v$) / ($v$ + $k$ V) = C$_{A0}$ / ($1$ + $k$ τ)
where τ = $0.305$ m$^3$ / $10$ Nm$^3$/h = $109.8$ s (assuming it is normal cubic meter)
So C$_{A1}$ = $28.3$ mol/m$^3$
But what I am having trouble with is how to calculate the concentration of B and C. How can I calculate B and C without having their start concentrations?
All help is appreciated!

You are interested in steady state conditions, that don't depend on start conditions.

@Poutnik I don't really understand what that means. How could I calculate B and C then? Also is the calculations I've done correct or do I have to convert the inlet flow by using STP? I am also uncertain now if I should calculate the temperature or assume that it is STP?

It was meant as a hint, was not going to write a full answer.  If you open a water tap above an unplugged bath tube, there will get established such a water level, where outlet flow equals the inlet flow. This level ( a steady state) does not depend on what was the water level at the beginning. That applies to all A, B and C. Rate of in-flow + creation of X = rate of out-flow + consumption of X. (I.e. all A, B, C have steady concentrations)

@Poutnik I'm not expecting a full answer. I just don't really understand how to use the fact that A, B, C have steady state concentrations. I thought about making a material balance like this: C${A0}$ $v$ + k (C${B1}$+C${C1}$) V = (C${B1}$+C${C1}$)V, but then I would have two unknowns, unless because of the stoichometry I can say that C${B1}$=C$_{C1}$?

There are 3 steady concentrations and 3 equations for them balancing their positive and negative rates.

@Poutnik From what I know the material balances are:
A: C$A0$$v$ - $k$C$A$V = C${A1}$$v$
B and C:  $k$C$A$(1-X)V = C${B1}$$v$ =C${C1}$$v$
But this would give me that the concentration of B and C is $13.33$ mol/m$^3$, which is more than the amount of A that reacted

Since you have a flow reactor with constant $P$ and $T$, you have to consider the generation of volume.
Stoichiometry
The reaction stoichiometry gives
$$ C_A = \frac{C_{A0} (1-X)}{1+\varepsilon X} $$
$$ C_B = \frac{C_{A0} (\Theta_B+X)}{1+\varepsilon X} $$
$$ C_C = \frac{C_{A0} (\Theta_C+X)}{1+\varepsilon X} $$
Where $\Theta_B = \dfrac{C_{B0}}{C_{A0}}$ and $\Theta_C = \dfrac{C_{C0}}{C_{A0}}$ are the ratios of initial concentration of species. $X$ is the conversion and $\varepsilon$ is the fractional change in volume flowrate:
$$ v = v_0 (1+\varepsilon X) $$
To calculate $\varepsilon$, we start with
$$ \varepsilon = y_{A0} \delta$$
Where $$y_{A0} = \frac{C_{A0}}{C_{A0}+C_{B0}+C_{C0}}$$
and $\delta$ is the change in stoichiometric coefficients,
$$ \delta = \frac{\gamma_B + \gamma_C - \gamma_A}{\gamma_A} = \frac{1 + 1 - 1}{1} = 1$$
Where $\gamma$ is the stoichiometric coefficeint of each species in the reaction considered.
Mole balance of $A$
$$\text{(In) - (Out) - (Consumed) = (Accumulation)}$$
At steady-state, accumulation is zero.
$$ F_{A0} - F_{A} - r_A V = 0 $$
Dividing by $F_{A0}$:
$$ \frac{F_{A0} - F_{A}}{F_{A0}} - \frac{r_A V}{F_{A0}} = 0 $$
Note $\dfrac{F_{A0} - F_{A}}{F_{A0}}$ is the conversion $X$,
$$ X - \frac{r_A V}{F_{A0}} = 0 $$
Considering the rate law $r_A = k C_A$,
$$ V = \frac{F_{A0} X}{k C_A} $$
Using the stoichiometry,
$$ V = \frac{F_{A0} X (1+\varepsilon X)}{k C_{A0} (1-X) } $$
Note that $F_{A0}/C_{A0} = v_0$,
$$ V = \frac{v_0 X (1+\varepsilon X)}{k(1-X) } $$
You can solve for $X$ using this last equation. After that, you can calculate whichever concentration - $C_A, C_B, C_C$ - using the stoichiometric relations.

How do you get those equations from the reaction stoichiometry?

Start from the definiton of conversion (of A): $F_A=F_{A0}(1−X)$ (it's important to know that conversion defined for absolute molar quantities, not concentration). When you divide by $v_0$ to get concentrations, the RHS $F_{A0}$ becomes $C_{A0}$, but the LHS will have a $v/v_0$ factor, which leads to the $1+\varepsilon X$ factor of volume generation.

@katara For the other species $B$ and $C$, you can derive the equations by considering the meaning of stoichiometry, that is $dF_A = -dF_B$. Integration, using the initial conditions and incorporating the volume factor to convert to concentrations will get you those equations.

@katara I'm trying to be brief and thus not very rigorous as to keep a short comment but please check these infos on your preferred Reaction Engineering textbook.

I am don't quite understand the $dF_A$=−$dF_B$ part.
There isn't anything like this in my textbook hence why I am asking on here. I have never come across a question like this where the products are calculated without some sort of information about them.

I am struggling with how the material balance should be for B and C because I somehow have to distinguish between them. Because can't I simply use that r$_B$ =r$C$ = kC${A0}$ instead of r$A$ = -kC${A0}$?

@katara Consider the mole balance of A: $F_{A0} - F_A - k C_{A} = 0$. What is the mole balance of B? $F_{B0} - F_B + k C_{A} = 0$. Notice that although the 2 first terms change, the reacion term is the same up to sign (positive for products and negative for reactants) and stoichiometric coefficient (it's just 1 for this reaction). Of course that comes from the fact that each mole of A reacted corresponds to 1 mole of B produced. Now consider adding these two equations, you'll get $F_{A0} - F_A = F_B - F_{B0}$.

@katara Now using the definition of conversion, $X = \frac{F_{A0} - F_{A}}{F_{A0}}$, you will get a relation between $F_B$ and $X$. Then you'll need to consider dividing by $v$ to get a concentration. However $v$ changes with the conversion, so you'll need to use $v = v_0 (1 + \varepsilon X)$.

Let us continue this discussion in chat.

Okay, so then I get that C$_B$ = X * C$_A0$ = $25.13$ mol/m$^3$ and then C$_A$ + C$B$ = $28.3$ + $25.13$ = $53.43$ and we had that C${A0}$ = $53.46$ mol/m$^3$. So then C$C$ =  C${A0}$ - C$_A$ - C$_B$ = $0.03$ mol/m$^3$. It is not how you calculated it but would this be correct?

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If O2 is heavier than air, why does it float in air?
If the molecular weight of air is 28.96 G/mol and O2 is 32 G/mol,
does O2 float in air? If yes, then why?

O2 and N2 are essentially idea gas in everyday condition. Suppose there is only O2 ideal gas, then it will distribute throughout the whole atmosphere. Now lets add N2 to the atmosphere as well. The mixture is still essentially ideal gas (ie, the atoms barely interact between different species as well as barely interact within the same species). Therefore, the spacial distribution of the two ideal gas is independent with each other and O2 is still distributed throughout the whole atmosphere.
The situation is different with two types of liquid, since liquid are not ideal gas (not close-to-none interaction). Their interactions within (or across) species can potentially segregate the two liquid.


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Can Hydrogen Peroxide solution bleach paper without touching the paper surface?
As many webpages suggest, 3% hydrogen peroxide solution can be used to bleach paper by soaking the piece of paper into the solution. I tried and it is really quite effective.
However, if the piece of paper cannot be soaked in water (as it may have a high chance to get damaged), is it possible to put it a few millimeters above the hydrogen peroxide solution, and let the hydrogen peroxide vaporize and perform the bleaching? Does this way of bleaching work as good, not at all, or with a lower efficiency? Would I need a higher concentration of hydrogen peroxide?

Not at all. Boiling point of hydrogen peroxide is about 150 deg C.

Not 3% hydrogen peroxide, but if you can get the 50% solution it might work. However, it is dangerous to work with.
As described in the comments, hydrogen peroxide is less volatile than water, so when a hydrogen peroxide solution is vaporized the vapor contains a smaller percentage of $\ce{H2O2}$ than the liquid. But using a more highly concentrated solution can compensate for this, so the partial and total vapor pressure data given here indicate at at 20°C, a liquid phase that's 50% hydrogen peroxide gives a vapor that is about 3% hydrogen peroxide. So you can hold the paper close to the concentrated liquid and have 3% hydrogen peroxide condense on it. The danger is that if you touch the paper into the concentrated solution, you are combining a flammable material with a powerful oxidizer, so handle with care and away from other flammable materials.


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Is there an intuitive answer why burning is very exothermic?
Water and carbon dioxide are much lower in energy than hydrocarbon and oxygen. Is there an intuitive answer for this? Is this because atom arrangements that have more polar bonds tend to be lower in energy? Or there are some other reasons? Or there is no meaningful rationalization here and it is just the way it is?

Is a fire hot? Giving off heat?

There is an article in JChemEd that blames it on the weak bonding of dioxygen.

From the time one sees a fire one intuitively realizes that wood , paper, gasoline etc. burn and when as a boy or girl scout asked to build a fire one gathered sticks to burn and stones to control the fire. Then one studies chemistry and learns about bond energies, Hess's Law, poorly shielded orbitals, and forgets that observation and perhaps even measurement is the most important aspect of chemistry. Observation, intuition, logic and measurement can be a powerful combination, HMMM wonder why rocks do not burn?

@jimchmst - coal burns just fine...

@Jon Custer my question is about why water and Carbon dioxide are much lower in energy in terms of their atomic position. My question is "Is there a reason X that leads to B, while B is equivalent to C". An answer "C (Burning is hot) leads to B (water and CO2 have lower energy)" is not what I'm asking.

@Karsten
I see that (on the first look) the paper didn't consider the polarity of a bond as a factor. For instance, it says "The bond energies in the fuel play only a minor role; for example, the total bond energy of CH4 is nearly the same as that of CO2". This is to be expected, given CO2 is more polar but also more "strained" in terms of orientation of its orbitals.

Also, some combustion reactions are less exothermic than others: https://en.wikipedia.org/wiki/Cool_flame

If we're going for pure intuition, perhaps the answer with the least amount of required underlying knowledge/assumptions (though not immediately obvious) is that combustion is kind of reverse photosynthesis, and photosynthesis consumes light. Burning organic matter is just opening a bag of stored sunshine.

If you add up the enthalpies of formation for a bunch of hydrocarbons and oxygen and compare it the the products (carbon dioxide and water) of rearranging the original atoms into those products, you will find that the products contain much less energy than the starting materials. That energy difference is the energy released by burning the hydrocarbons in oxygen.
This is one way of restating that the reaction is exothermic.
But, if the question is why do carbon dioxide and water have such low energy compared to the hydrocarbons plus oxygen that is more complicated.
Carbon dioxide has two strong bonds between carbon and oxygen. You could argue that this is because two polarised bonds (oxygen is keener to attract electrons than carbon) is a big part of that but this is probably an oversimplification. More complex arguments get into molecular orbital theory.
In truth there isn't a simple theory that explains why water plus CO2 have stronger bonds (lower energy) than the hydrocarbon-oxygen  mixture. But, if you do know their enthalpies of formation, it is easy to see why they often are the end-product of combustion. You could probably, with a lot of effort, calculate the thermodynamic parameters from some ab initio quantum calculations but that wouldn't give a simple satisfactory answer either.

I agree sometime there is no simple rationalization for how things work the way they do. Now "You could argue... polarised bonds... is a big part of that but this is probably an oversimplification. More complex arguments get into molecular orbital theory" is there a specific reason why you think this is probably an oversimplification? If anything, I feel =C= is strained and would tend to increase energy of CO2.

The C=O bonds is carbon dioxide are not at all "strained" but this probably needs molecular orbital theory for a good explanation. Simple explanations involving electronegativity or the ionic character of bonds rarely capture important issues like the geometry of the bond (CO2 has pi-bonds but the C-O pi-bonds are orthogonal). O2 is a diradical: how to explain that without molecular orbital theory?

Oxygen loves electrons, and tries to gain as much partial negative charge as it can. This leads to a huge release of energy when it forms bonds with carbon or hydrogen, both of which are less electronegative than it.
O2 -> Two bonds, both to another oxygen atom (well, technically 1, but MO theory isn't relevant here)
H2O -> Two bonds, both to a less electronegative hydrogen atom.
CO2 -> Two bonds again, though there's definitely diminishing returns with repeated bond formation with the same carbon atom.


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Why are crystals of carboxylic acid fluffy
I'm an undergrad student and we recently performed saponification of esters which resulted in the formation of carboxylic acid which we had isolated out of the reaction mixture.
After recrystallisation of the crude sample , the crystals came out to be super fluffy unlike any other crystals of previous compounds we've made yet
Is there a reason to this fluffiness?

This is know as the "crystal habit": https://en.wikipedia.org/wiki/Crystal_habit

What you observe is the effect of highly different growth rates of the crystallites along the different axes. There are some rules of thumb, but for most molecules, they are very hard to predict.


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Can halide be a sigma donor?
When I took inorganic classes, I was told that halide is a pi donor. Assume I am using metal hexahalide why I can't put the $p_z$ orbital of the halides directing to the metal centre? With that halide will just act as a sigma donor.

The two are not exclusive. Just because halides are pi donors does not mean they aren't sigma donors. If you orient the pz to be a sigma donor, px and py can also be pi donors.

Most ligands have both sigma and pi actions, involving different orbitals on both the ligand and the metal. Hydride is really the only exception, being sigma-only with the hydrogen 1s orbital.


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Using a Chi-Square distribution table to calculate fraction of gas molecules with activation energy
I couldn't decide whether to ask this to Chemistry, Physics or Statistics stack exchange. Hopefully I made the right choice.
According to this Wikipedia page, while the speed of particles in an ideal gas follows a Maxwell-Boltzmann distribution, the energy distribution in a collection of ideal gas molecules follows a Chi-Square distribution with however many degrees of freedom as the molecule. i.e. 3 for monatomic, 5 for diatomic (at moderate temperatures).
As an example computation to try this out, I wanted to find the activation energy of a reaction which would have 10% of the monatomic gas molecules eligible for reaction at temperature $ 300 \text{ K} $.
From the Chi-Square distribution, we should have $ \frac{E}{RT} \sim \chi^2_3 $. From the inverse Chi-Square table, at a significance level (i.e. right-tail proportion of the distribution) of 0.1, the critical value is 6.251, which implies that we need $ \frac{E_a}{RT} = 6.251 $ so the activation energy should be no more than $ E_a = 8.314 \times 300 \times 6.251 = 15.5912 \text{ kJ mol}^{-1} $.
This seems like a reasonable number but then I tried checking the answer by working it out directly from the distribution of energies (also given on the Wikipedia page). I worked out that
$$ P(E > E_a) =  \frac{2}{\sqrt{\pi } (R T)^{3/2}}  \int_{E_a}^{\infty } \sqrt{E} \exp \left ( -\frac{E}{R T} \right ) \ \text{d}E $$
Evaluating this numerically for $ E_a = 15591.2 $ and $ T = 300 $ gave $ P(E > E_a) = 0.005847 $, or about 0.5% of the particles, which is very different to the 10% I specified when solving it the first way, showing that something went wrong when I was working it out.
A third method is to use the commonly-cited (but it seems nowhere derived) result from the Arrhenius equation that $ \exp(\frac{-E_a}{RT}) $ is the fraction of particles with $ E > E_a $. Using this gives a proportion of 0.00193 = 0.2%, which differs yet again from both previous answers. I am very skeptical of this result though because neither that integral nor the Chi-Square inverse pdf have a nice closed form, and this is a suspiciously clean simple formula to say it's supposed to do the same thing.
Does anyone know the correct way to go about finding particle energy proportions? It would be nice to be able use Chi-Square correctly here as I find it interesting how different areas of science and statistics fit together.

I believe I figured it out - I was off by a factor of 2 in my scaling.
The variable $ \frac{2E}{RT} $ has a Chi-Square distribution with 3 degrees of freedom (on further research this appears to be unrelated to the physical degrees of freedom of the molecule and is always just 3 - could be wrong here though).
So once we have our critical value from the Chi-square table, rearrange to get
$ E_a = \frac{\chi^2_c RT}{2} $ as the activation energy for which (significance level)% of the molecules have above this energy.
Then, I got the distribution formula wrong, the correct equation should be
$$ P(E > E_a) = \int_{E_a}^{\infty} f_E(E) \ \text{d}E = 1 - \frac{2}{\sqrt{\pi}(RT)^{3/2}}\int_{0}^{E_a} \sqrt{E} \exp \left ( -\frac{E}{RT} \right ) \ \text{d}E. $$
In the example problem I gave, the corresponding activation energy is 7795.8 kJ / mol and this does produce the correct answer of 10% of the molecules.
The $ e^\frac{-E_a}{RT} $ factor representing this fraction of molecules with the activation energy does not work. I can only assume this is some commonly-taught thing in chemistry that nobody bothers to check the maths for because to me, it looks like it is completely wrong!


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How do we separate H2O from CO2?
Let's say I have a mixture of $\ce{CO2}$ and $\ce{H2O}$ gases. I've read that $\ce{CO2}$ is heavier (molar mass is $\pu{44.01 g/mol}$) than $\ce{H2O}$ (molar mass is $\pu{18.02 g/mol}$). So, if the mixture is supplied to a cylinder from the bottom, will only the water vapour reach the top? If so, is this a good way to separate the $\ce{CO2}$ and water vapour? Or is there a more efficient way to separate them?

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

What you propose does not work at all. Never.

@Karl Then, could you please write an answer for a better solution to this problem?

It is hard to see any source containing a mixture of CO2 and water as gasses. Most likely under normal conditions you would have water liquid under an atmosphere of carbon dioxide (with the water saturated with CO2). Or a stream of hot gases containing both. What circumstances are you considering?

It depends on what equipment you have available.  A mix of compression and cooling will work much better than relying on their molecular weights.
Water boils at 100 degC, while CO2 sublimates at -78 degC.  Simple cooling will ensure that water condenses long before the CO2, while compression will drive the formation formation of water-water bonds. The Clapeyron equation should let you calculate the needed values.

Separation of $\ce{H2O}$ from $\ce{CO2}$ has been done by all analytical chemists in the $19$th century, when they had to determine the formula of a new organic compound X.  The compound X was mixed with some $\ce{CuO}$ powder, introduced in a horizontal glass tube and heated by a burner in an oxygen current. X was burned int $\ce{CO2}$ and $\ce{H2O}$.  Then the oxygen gas current containing some $\ce{H2O}$ and $\ce{CO2}$ had to cross another glass tube filled with pieces of anhydrous calcium chloride $\ce{CaCl2}$ which fixed the water vapor, producing $\ce{CaCl2·6H2O}$. The $\ce{CaCl2}$ tube was weighed before and after the operation. The increase in weight gives the total amount of $\ce{H2O}$ in the mixture of gases.
Then the gas was sent to bubble through a concentrated $\ce{NaOH}$ solution, where the following reaction happened $$\ce{CO2 + 2 NaOH -> Na2CO3 + H2O}$$ The $\ce{NaOH}$ solution was weighed before and after the operation. The increase in weight gives the total amount of $\ce{CO2}$ in the gas.


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Constants in Enzyme Kinetics under the Rapid Equilibrium Assumptions
I am new to enzyme kinetics, and I am trying to write an rate equation for a rapid equilibrium random terreactant system. I have been consulting the books Enzyme Kinetics and Mechanism by Cook and Cleland and Enzyme Kinetics by Segel. One thing that have been confusing to me is that, sometimes they refer Ka (Michaelis constant for reactant A) and Kia (inhibition constant for reactant A) as dissociation constant in their books, and I am not sure which one to use.
From what I understand, Ki is dissociation constant except two systems, namely Ordered Bi Bi and Ordered Uni Bi (under Haldane Relationships In Cook and Cleland's book, as well as illustrated clearly through a lot of other cases in Segel's book). I feel like dissociation constant and inhibition constant should be two different concepts but it seems like they use both interchangeably.
For Ka, Ka is the dissociation constant when the system is rapid equilibrium Uni-Uni (specifically, when they discuss the Michaelis–Menten/Briggs–Haldane equation, as MM equation is derived with assumption of rapid equilibrium). In the case of random Bi-Bi, Cook and Cleland defined Ka as the dissociation of A from the ternary complex, and Kia as the dissociation of A from the binary complex; whereas Segel used dissociation constant KA for $\ce{E + A <=> EA}$ and $\alpha$KA for $\ce{EB + A <=> EAB}$ (i.e. the dissociation constant is modified by another substrate that is already bound to the enzyme with a factor of $\alpha$. What is the difference between them?
I guess what ultimately confused me is in the case of terreactant: Segel used the same method for defining the constants for each reaction step; Cook and Cleland just used "const" "coef A" etc. to describe the rate equations without further details. The paper they referenced, Methods in Enzymology 87, 353, only provides the deduction of possible mechanism based on the missing terms in the denominator without further defining them since they depend on the mechanism. The paper Rapid-Equilibrium Enzyme Kinetics by Alberty (2008) provided an explicit expression for random terreactant system, following Cleland's nomenclature and used a mixture of Michaelis and inhibition constants in the equation, but I do not understand how six constants (KA, KB, KC, KIA, KIB, KIC) cover all 12 edges in the system. I do not need the exact expression of the constants in terms of the rate constants (the small letter ks), but I would very much like to know what are the terms (Michaelis, inhibition or dissociation) should be used when describing the rate equation of a rapid equilibrium random terreactant system.

I do not have access to either books of Enzyme Kinetics and Mechanism by Cook and Cleland and Enzyme Kinetics by Segel. However, I read some of Cleland and coworkers' papers (e.g., Ref.1 & 2). In those and other peer-viewed papers by Alberty, my understanding of the nomenclature of rapid equilibrium of enzyme kinetics as follows (I avoid use of $K_\mathrm{a}$ and use $K_\ce{A}$ instead to avoid the confusion with acid dissociation constants):

As you have mentioned, $K_\ce{A}$ is the dissociation of $\ce{A}$ from the ternary complex such as $\ce{EAB}$ such that $\ce{EAB <=> EB + A}$. Thus, $K_\ce{A} = \dfrac{[\ce{EB}][\ce{A}]}{[\ce{EAB}]}$ (Ref.3).
You have mentioned $K_\ce{IA}$ is the inhibition constant. But I don't think it has anything to do with inhibition here. As Ref.2 and other paper mentioned, it is the dissociation of $\ce{A}$ from the binary complex such as $\ce{EA}$ such that $\ce{EA <=> E + A}$. Thus, $K_\ce{IA} = \dfrac{[\ce{E}][\ce{A}]}{[\ce{EA}]}$ (Ref.3).

As Ref.1 mentioned, before you are trying to write an rate equation for a rapid equilibrium random terreactant system or any other system, you may need to have an idea of the reaction mechanism (e.g., like Ping Pong type of mechanism). Accordingly you have to do a lot of assumptions (e.g., steady state steps) to get calculated some of those unknown contributions such as $coef\mathrm{A}$ and $coef\mathrm{B}$ in final equation.
Note: A good reference to read is Ref.4 and references therein to get some help for derive the equation with appropriate assumptions.

References:

Ronald E. Viola and W. Wallace Cleland, "[19] Initial velocity analysis for terreactant mechanisms," Methods in Enzymology 1982, 87, 353-366 (DOI: https://doi.org/10.1016/S0076-6879(82)87021-3).
Gerald Litwack and W. W. Cleland, "Studies on the Tyrosine Aminotransferase Mechanism," Biochemistry 1968, 7(6), 2072–2079 (DOI: https://doi.org/10.1021/bi00846a008).
Robert A. Alberty, "Determination of Rapid-Equilibrium Kinetic Parameters of Ordered and Random Enzyme-Catalyzed Reaction A + B = P + Q," J. Phys. Chem. B 2009, 113(29), 10043–10048 (DOI: https://doi.org/10.1021/jp9021097).
S. Cha, "A Simple Method for Derivation of Rate Equations for Enzyme-catalyzed Reactions under the Rapid Equilibrium Assumption or Combined Assumptions of Equilibrium and Steady State," J. Biol. Chem. 1968, 243(4), 820-825 (DOI: https://doi.org/10.1016/S0021-9258(19)81739-8).

There is another paper by Cleland in 1963 that has explanation similar to the Cook and Cleland's book. I am not sure why they used the term "inhibition constant" as well. The rate equation derived will be used for modelling, and I found two possible mechanisms for the enzyme, rapid equilibrium random terreactant is one of them, and I will be comparing both, with steady-state assumption included. I am not sure what should the constant for the dissociation EABC ⇌ EBC + A be called and what is its relation with K_A and K_IA.

Good point. I'll look at it when I got a chance today, which is kind of busy.


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What's the purpose of this vintage glassware?
I found this glassware in a phytochemistry lab abandoned in the 1990ies. The piece itself was shelved in the storage room, thus probably already obsolete back then.
The device, entirely glass, is a compact sequence of six bulbs (relative sizes: L, S, S, S, M, XS), 155 mm wide and 105 mm high. As a layman I'd guess it was used to separate, maybe distill liquid fractions.
Thank you in advance for identification or ideas.

Edit: flipped picture horizontally and marked inlet.

In the good old days Amoco R&D had its own glass shop and people would bring in their own designs for unique equipment. So there can be some pretty unusual stuff.

This is Geissler potash bulbs (German “Geißler’schen Kaliapparat”/“Kaliapparat nach Geißler”) designed and produced by Heinrich Geissler in 1860s and primarily used in the second half of 19th century to quantify carbon dioxide content in the air or among the products of combustion analysis of organic matter.
The apparatus has several advantages over the original version of potash bulb devised by Liebig in 1831 (which is depicted on the ACS logo since 1908), such as improved stability (convenient for placement on a table), but is was also prone to clogging by building up potassium carbonate and was more expensive and complex to build.
Several versions of the apparatus existed and they still were sold in 1910s (items 6011 to 6014, image courtesy of Science History Institute — Illustrated Catalogue of Scientific Apparatus, Chemical Glassware, and Laboratory Supplies, ca. 1910, p. 201):

Some times, I still wonder how these vintage people come-up with such creative ideas.

Slightly off-topic, but can you explain why the numbering of the figures is out of the expected order?

@LokiRagnarok - I would presume poor communication between the artist and the person numbering the items.

@MathewMahindaratne They didn't have TV.

Could something like I_O's piece be (safely) repurposed into a Geissler tube lamp? Very cool find and amazing it's survived!


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Comparison of two different coloured dyes using UV-vis spectra
I know that Both solutions are a shade of purple due to the lack of absorbances in the red and blue regions, however am I correct in thinking that the peak slightly to the right of the 2 in this spectrum (showing both spectra on one report) would be more blue in colour since it absorbs more at longer wavelengths?

Since you apparently collected the absorption spectra, do you have pictures of the solutions?

@Aurora. One of the dye absorbs the yellow strongly and more selectively. It must appear purple. The other dye does not absorb the yellow so strongly, but it absorbs more strongly the red and the blue. The optical effect of absorbing more red + blue must be similar to just and only absorbing yellow.

Yes, however I very foolishly didn’t label the report properly and am now trying to rationalise which solution corresponds to which spectrum


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Would a hypothetical Og2 +235 form a chemical bond?
In a hypothetical (?) Og2+235 we would have a simple sigma bonding orbital occupied by one electron (leading to a bond order of 1/2). But how to take into account the giant repulsion of the two nuclei? Generally how are nucleis taken into account for energy calculations from molecular orbitals in MO theory? So far I've always read about the electrons but very few about the nuclei in the system.
Same in He23+, half bond order but very large repulsion. My intuition says that He21+ should be more stable, even if both have the same bond order.
Where in the framework of MO theory is exactly that taken into account? How does nuclei repulsion change the MO energy scheme?

Oh, no, not again ;> For diatomics you get +2, tops. Some absurd charge is out of question.

Apply the common sense of classical electrostatics and you would see that the monotonous steep slope of potential energy of nuclei cannot be disturbed by very tiny bump caused by interaction of electrons.

https://chemistry.stackexchange.com/questions/106826/does-the-hydrohelium-dication-exist

@Poutnik And why is it then so often taught that bonds are formed when one simply takes mo theory into consideration and only looks at electrons? Nuclei repulsion is hardly mentioned in mo theory

Well, gosh, it's not like you need to consider dications all that often. Even multiatomic break up pretty fast.

@FlawC https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation - calculating MOs first involves the assumption that that the nuclei are stationary (have fixed positions) and then optimising the electron wavefunction. The nuclear repulsion term is not ignored in the overall energy, but it is just a constant (for a given set of nuclear positions).

https://chemistry.stackexchange.com/questions/146481/why-is-oxygen-more-stable-than-oxygen-dication https://chemistry.stackexchange.com/questions/112087/why-doesnt-h%e2%82%84o%c2%b2%e2%81%ba-exist

@orthocresol ok so considering a system just with the "Og2 +235", at what distance apart would they be stationary and form a bond? I mean what would be the hypothetical bond length of such a compound?

@FlawC Short answer: never. Long answer: what you need to do is to plot the energy of such a compound as a function of bond length. (You can google 'hydrogen bond energy curve' for an example of what this looks like for H2.) Now, your curve for $\ce{Og2^{235+}}$ would be so biased towards large distances, and the 'dip' in the curve would be so small, that it can hardly be considered a molecule. To be technical, it wouldn't admit a vibrational state which is required for something to be considered a molecule. https://goldbook.iupac.org/terms/view/M04002 So there is no bond at any length.

That said, the original question isn't silly at all (although I guess you did choose a rather extreme example). I would love to write a proper answer, but it's very late where I am now. Feel free to ping me again tomorrow and maybe I can get round to it.

@orthocresol that would be great :-) I've played a bit with a electrostatic system, two positive charges (p) x apart and one negative charge (e) x/2 apart, basically I've found out that the term: -pe/(x/2)^2 + p^2/x^2 describing the electrostatic force on one (and therefore the other) nuclei becomes zero at a "e to p ratio" of 1:4, so relative any more proton in the nucleis than 4*N(e) leads to a net repulsion at any distance. But of course we deal with quantum mechanics and I can't assume a electron being in the middle. I don't know enough about the quantum mechanical "forces" :-(

@orthocresol But I assume at a far enough distance apart (still repulsive, but it approaches the x axis asymptotic) the repulsion get overcompensatet by a type of quantum mechanical force which comes from the fact that the system would probably have an even lower energy than if the nuclei went to infinity apart and no longer felt any repulsion at all? (So can those electron interactions at some very very large distance overcome electrostatic forces in my example?)

@orthocresol OK so basically the only force here is actually electrostatic, then the only way I can explain it is that the atoms stay together when they are far apart and the fact that the electron interacts by staying in the area between the two nuclei, the moment where the electron would be @ x/2 doesn't contribute to the cohesion, only those where it's much closer to one or the other nuclei. Like a child that tries to hold two trucks together that rolling apart, but does not have the strength to hold them together in the middle, it always runs from left to right and pulls one back again?

I'm afraid that here you've gone off on a tangent which I don't really understand any more. There are no special quantum forces, it's all electrostatics. The problem is that the electrons are not point charges. They're charge distributions. So it's rather more involved than just doing Coulomb's law.

@orthocresol That's what I thought while writing, but unfortunately I know absolutely nothing about the nature of charge distributions, hence the analogy with the child. But if you have time to write an answer that would make me happy :)

I would not bet on it. We apparently can't even get $\ce{He2^{3+}}$, let alone higher single-electron diatomic ions. From Wikipedia:

$\ce{He2^+}$ was predicted to exist by Linus Pauling in 1933. It was discovered when doing mass spectroscopy on ionised helium. The dihelium cation is formed by an ionised helium atom combining with a helium atom: $\ce{He^+ + He -> He2^+}$.[1]
The diionised dihelium $\ce{He2^{2+}} (1Σ_g^+)$ is in a singlet state. It breaks up $\ce{He2^{2+}->2 He^+}$ releasing 200 kcal/mol of energy. It has a barrier to decomposition of 35 kcal/mol and a bond length of 0.70 Å.[1]

Thus even with only two positive charges the helium dimer is prone to breaking up due to electrostatic repulsion. Although the dication above is metastable, we should expect the stability to only become worse with an additional positive charge and only one bonding electron instead of two.
A simple classical electrostatic calculation is instructive. Suppose you have a negative charge and two positive charges, the latter equidistant from the negative charge on opposite sides. The net force is attractive if all charges are one unit, but this net attraction is lost if we give two or more (and certainly 118!) units to both positive charges. The true quantum mechanical calculation is of course much more complicated, but can give only less favorable results because as the positive nuclear charges approach the electron cloud cannot stay fully between them. Thereby $\ce{H2^+}$ is predicted to be the only single-electron homonuclear diatomic ion that remains bound.
Cited Reference

Grandinetti, Felice (October 2004). "Helium chemistry: a survey of the role of the ionic species". International Journal of Mass Spectrometry. 237 (2–3): 243–267. Bibcode:2004IJMSp.237..243G. https://doi.org/10.1016/j.ijms.2004.07.012.


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How can I achieve the following reaction where the iodide gets replaced by an aldehyde?

The question is what other chemicals need to be added so that the reaction is possible.
Although the solution shouldnt be that hard, I couldnt find the answer to this one.

Welcome to ChemSE! Through the good graces of @Waylander, you received a correct answer. Your question will likely be considered as homework and be voted down. You need to make some effort to answer the question yourself. Take a look here about homework policy. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange.

Note that the reaction adds a carbon to the struture.
This can be done in two steps:

React the iodide with NaCN or KCN in DMSO to form the nitrile
Reduce the nitrile to the aldehyde with di-isobutylaluminium hydride Dibal-H


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Hydroboration–oxidation reaction without THF
The first step of hydroboration–oxidation usually involves $\ce{BH3}$ accompanied by THF. Is it possible for hydroboration to occur without the presence of THF? Borane is incredibly reactive. Wouldn't it just react by itself without being stabilized by THF?

Borane dimerises to diborane which does all the reactions of borane, including hydroborating alkenes. Diborane gas is commercially available if you wish to use it rather than stabilised borane, and have the equipment and skills to handle it. A number of other stabilisers are used for borane and are commercially available e.g. dimethyl sulfide, ammonia, triethylamine, pyridine.

But practically speaking, you need a stabilizer without which the reaction will not take place?

@user28684  Not so. If you use gaseous diborane the reaction will take place.


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Difficult in grasping "the extent/limit of conformational space of a compound until it change its stereochemistry"
I have heard that a conformational landscape encompasses all conformers that a compound has for a specific stereoisomer. I think it makes sense verbally, because if all conformers of a stereoisomer include all conformers for all stereoisomers and one (mistakenly) designates those all stereoisomer conformations to be conformers for a specific stereoisomer, it will require an interconversion (or change in permutational position) of one or multiple bonds to be broken to make a different stereoisomer.
While it seems to make verbal sense to me, my visual thinking is really against this. Searching on Google, "stereoisomer conformation" and "conformer" are always treated as two different realms having no connection in between.
I feel there should be an image (or someone has done a research article about this) showing a conformational landscape with a functional group bonded to a particular atom, showing differences in spatial 3D position. It suffices to show that "this coordinate position is still the same stereoisomer" but that if the functional group is extended, whether freely translated or rotated in drawing to somewhere, it will make a different stereoisomer and it is convincing enough that its stereochemistry interconversion will have to break some bonds first.
Any example would be fine, but I prefer the case of cis-trans conversion of a cyclohexane in its chair conformation with few functional groups, because my visualization is that its functional group stereochemistry interconversion still does not need bond breaking (but why it would have a different stereoisomer though?).
Thus, here is the general question. What is the extent or limit of conformational space of a molecule (or specifically, a particular stereoisomer) that is spatially unique to it, but if further extended to all random positional permutations will change its stereochemistry so that it will have to break some bonds first to let that interconversion happen?

How about being less verbose and maybe write something that is clear and specific?

Also it's a bad example. Butene is good, even blatant, one.

I want to know more about this though, this is important in stereochemistry education. +1 for you.

It would definitely help if you were more concise and provided the simplest example of the problem you are facing. Try to use more punctuation (shorter and clearer sentences). It appears that this is about nomenclature, and specifically the meaning of the word "conformation". If so, a brief answer would be that it depends on context.

@Mithoron What makes you think a butene is a better example than the author given even it is just a sitting planar that doesn't make any prominent conformational space.

@làntèrn Exactly because of that.

Watch your terminology:
Conformers are also stereoisomers. What you probably mean is how to partition conformational space based on configurational isomerism.

https://chemistry.stackexchange.com/questions/169059/is-there-a-formal-definition-of-identical-molecules

@BuckThorn I think the vote system is a bit disgrace to me as there are many questions have bad forms with short lines of "can you read my mind" and I am just telling my process of reading and infering a bad school textbook because as with the site's policy, I have to demonstrate how I can conclude to that question though.

@làntèrn Thank you for the vote anyway. The vote does enhance a reward for people who are curious and seems the site's policy doesn't adhere to this a bit.

@Mithoron I don't know why butene is a good example. You might want to flex your answer with stereochemistry of mathematical knots and braids, perhaps?

I just re-read the first paragraph and I think I understand the question a little better, maybe because of the edits. My first comment was not to give you a hard time or as an explanation of up or downvoting (by myself or anyone else). I just felt confused by what you had written. It could be because it is a difficult question, simply!

You have the right idea that conformational space can be partitioned by configurational isomers. The clearest cases are where you have to break bonds to end up with another configurational isomer, but that is actually not always the case. Atropisomerism also gives rise to different configurational isomers, even though they are just separated by a bond rotation with a very high barrier.
Your example with cyclohexane is actually a clear-cut case, since you really need to break bonds to switch from cis to trans. I visualized all possible chair conformations and the partitioning into configurational subspaces. There are more conformers that I left out for simplification.

If you still don't see why you need bond breaking, I recommend playing around with a molecular model kit.
In summary, the partitioning of conformers into distinct configurational isomers happens when they are interconverted by a process with a high barrier (i.e. bond breaking, hindered rotations, polytopal rearrangements, etc.).

I will edit the comment a bit while writing.

I get it now. Thank you very much for clearing up and sharing this. Seems like I got the wrong terminology classification then (as you noted in the question's comment). Let me conclude this. So, a molecule has a global stereoisomer space (not a global conformational space) that can be partitioned into whether the interconversion of this space permutation will make a different configuration (configurational space, not stereochemical space) or not (conformational space).

The configurational space can be generated by some processes that the bond breaking process is not always a sole requirement to make a different configuration because it can be interconverted by another process requiring high barrier i.e. a hindered rotation to make atropisomers. I believe enzymatic catalysis performs this more commonly than the frequency it mentioned in the textbooks though.

Thank you very much for sharing this. The chair conformations in each configurational subspaces of cis and trans configurations are aesthetically pleasing to me as I can imagine with the given equlibrium arrows. I will affirm it later with my slo-mo molecular dynamics simulation to see each process. Again, thank you very much.

By the way, where do you obtain the picture? The link you gave is quite a bit different source one

@Leticia I drew it myself while focusing on double Newman projection. Btw. the intermediate conformers you can expect during a ring flip process are well known: link.

Fun fact: In theory you could also argue that the two configurations are separated by a polytopal rearrangement with a very high barrier instead of bond breaking. Such processes are in fact known for certain amines.

Ah I see, the diagram looks improved and surely it looks like a diagram that is ready to be published in a journal. Thank you for the detailed diagram anyway.

It seems like the research progress in this field is blooming very well. From the link you have given to me, I found a lot of new concepts from there. Ring puckering of various heterocycles and macrocycles, pyramidal inversion, polytopal rearrangement like a Rubik cube, molecular machine, and so on. Seems like I have already jumped into a rabbit hole deeper.

Welcome to the world of applied computational chemistry ;)


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I tested the effect of temperature on the pH of carbonated water but the results seem to go opposite to what was expected. Could anyone explain?
We carried out an experiment to investigate the effect of temperature on the acid carbonate equilibrium in carbonated water. We tested 5 different temperatures (room temp, 2 above and 2 below) by heating carbonated water in a closed test tube. After leaving the test tube for 10 minutes in the water bath, we opened it and quickly recorded the temperature and pH of the solution using probes.
We got the results that as temperature increases the pH decreased (so the solution got more acidic) and vice versa, however published results show that the pH should have increased for an increase in temperature. I understand this is because the increase in temperature decreases the solubility of carbon dioxide, decreasing the amount of carbonic acid and hydronium ions in solution and therefore increasing the pH.
The entire system should have shifted as per the equation:$2CO_{2(g)}+3H_2O_{(l)}\rightleftharpoons CO_{2(aq)}+2H_3O^{+}_{(aq)}+CO^{2-} _{3(aq)}$
I originally thought that the opposite trend in my data was due to the endothermic nature of the ionisation of carbonic acid making the system overall endothermic, but published data indicates that the system is overall exothermic.
It would be great if anyone had any ideas about how to explain the opposite shift in equilibrium that our results show.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

In carbonated water, content of $\ce{CO3^2-}$ is negligible, as it occurs in very alkaline solutions(50 % of it at pH near 10.3), while the water is acidic.

Did you account for the temperature dependence of $K_w$?

The experimental design is unfortunately not robust. Certainly, the pH of pure water decreases with temperature (slowly) but the biggest source of error is most like the pH probe. If you are using a pH-meter that does not have a built in temperature compensation, the pH readings are not reliable at any temperature except the one where it was calibrated. Secondly, pH meters operate in a limited temperature range and the pH calibration buffers are not meant for different temperature.
Secondly, pH probes have a response time, i.e., it take a certain amount of time to produce a stable reading, usually a minute or less. Since you mentioned "quickly recorded", this is another source of error.

Would the issues with the pH probes have made the results skewed enough to go completely in reverse from what was expected? Because we tested three trials for each temperatures and they all matched up, and the overall data made what would have been the correct trend except going in the opposite direction

High temperature does reduce pH because Kw of water is a function of temperature. You have to do controls as well. Measure pH of pure water exactly the same way.

If the data are reproducible, you did measure something; the question is What? The easy possibilities are that the system is not equilibrated, reversing the order might detect that, the closed tube is bothersome.  Try to find a way to work at constant pressure. The prepared 4 and 7 buffers have the temperature pH on the bottle; calibrate at the temperature and do not use the ATC probe. Record the millivolt values to ensure the probe is responding properly. Use a combo electrode with a flowing junction not a gel filled.  you disagreed with the lit. find out why!


text:
Effect of coal and natural gas burning on particulate matter pollution
I sometimes hear people talking about how we should replace coal burning plants with natural gas ones, to alleviate the case of particulate matter pollution. What exactly is the difference between coal fuel and natural gas that makes the latter seem "cleaner"?

At the same energy outcome, natural gas produces less carbon dioxide than coal. In a way, natural gas is half way between coal and hydrogen.
Coal produces smelly smoke, solid particles, sulfur dioxide and minor or trace heavy metal pollutants.
It is less known to common people, but power plants burning coal are more significant source of radioactive pollution than nuclear plants. This pollution is very diluted, but rather significant in absolute amount. Coal ash, used in past as a filler for some construction materials, has lead in some cases to significantly increased content of radium-226 in building walls. This radium is a product of long term decay of natural uranium. It further decays while producing radioactive gaseous radon-222, which is dangerous in long term inhalation because of lung cancer. As it stays in lungs as polonium-218 and its decay products.

See e.g. Uranium produced from coal ash

... the uranium concentration in the ash pile is about 150-180 parts per million, about 1/4th of the concentration often thought of as commercially viable for ISL[In Situ Leaching] mining. However, coal ash piles have some physical characteristics that might help overcome that disadvantage since they may be easier to drill and it might be easier to protect the local groundwater from contamination. ...

See Radon in building materials by Czech government agency for radiation protection.

I think the second point answers my question about PM pollution. Also, is there a source for the third claim? Sounds interesting to me.

@jonahuang Ahmed et al.. 2020; IEA 2020; Duke Surgery 2018 and of course Kurzgesagt: Why Don't We Shoot Nuclear Waste Into Space? ;)

Dealing specifically with particulates would depend on the coal and burning conditions. You could look up a specific coal you are interested in; However generally it will contain a few % of each sulfur, iron, silica, alkali earths, and smaller amounts of things like nickel. The burning conditions make a great difference whether the contaminant become fly ash or slag and clinker in the furnace. As a boy in Chicago in the 50s, our apartment was heated with coal as were the great majority. That meant someone had to dig about a cubic foot of ash and clinker from under the furnace firebox each month. Something missed with gas furnaces.  I am tying to say there is hardly a comparison of relative particulates produced by gas versus coal.

While the details depend strongly on the specific burning conditions, gas is almost always cleaner than coal and will produce far fewer particulates.
There are two main reasons for this. One is related to the level of non-carbon contaminants in the two fuels. Gas contains far fewer contaminants and most of them are gases many of which are not sources of pollution of any kind (gas is mostly methane but some gas has CO2, some has small levels of higher hydrocarbons like ethane). There is usually very little sulfur (if the original source has a lot it is usually stripped before distribution unless you count the very low levels of sulfur odorants added for safety reasons). Coal, even the best anthracite, can be thought of as mostly carbon but often contains several percent of a variety of contaminants. Some will be non-volatile minerals; some will be sulfur or nitrogen compounds and a variety of other non-carbon materials. These often burn to create particulate-forming compounds (sulfates often react to create particulates in the atmosphere).
But the second reason is the way even the pure carbon in the fuels burns. No burning reaction is "perfect" and side reactions leave stuff other than carbon dioxide (and water). Since burning the polymeric solid carbon in coal involves breaking up large chains or networks of carbon atoms, one of the common side reactions leaves incompletely burnt up lumps of carbon or partially reacted carbon. These form very small dust particles (in other words particulate pollution) and there is no perfect way to scrub them from the emissions. The combustion of methane is far cleaner as there are no long molecules of carbon that need to be broken up (though a very poorly controlled burning process could encourage some to form). So the expected level of imperfect reactions leading to particulate pollution will usually be far, far smaller when gas is the fuel.
In short, coal is a much dirtier fuel, even if all you are worried about is particulates. Add to this that, for a given amount of energy output, coal will produce a lot more CO2, and gas is always a better fuel for the environment.


text:
When refining copper through electrolysis, do the copper ions from the electrolyte copper sulfate get reduced at the anode?
Copper sulfate is the electrolyte that is brought up most commonly when we talk about the refining of copper. Do any of the copper ions from the copper sulfate solution reduce at the anode like the copper ions that were oxidized from the cathode?

Oxidation is at the anode and reduction takes place at the cathode. Copper ions get reduced to copper metal at the cathode.

At the cathode/anode is ongoing reduction/oxidation, what means the cathode/anode is kept at more negative/positive external potential in the cell electrolytic mode. While in the galvanic(active) mode, the cathode/anode has the  more positive/negative potential (i.e. marking of electrodes switches places).

Anode = Oxydation

Oh yeah sorry I got confused with the names of the electrodes,, Thanks for pointing it out

Like commenters pointed out the direction of the process is the other way round.
Apart from that the answer is yes, copper ions from the sulfate will be reduced at the cathode but at the same time copper ions from the anode replace the now "missing" ions from the sulfate.
So at the beginning of the process only ions from the original copper sulfate are reduced and soon after the amount of ions originating from the anode increases continuously until they all the ions in the solution have been replaced.

Such processes are usually stirred to retain current so it is diminishing returns on the original copper ions.

@jimchmst Could you explain a bit more please?

@chem_student The net movement speed of the ions in the electrolyte is limited. By stirring you level out concentration differences along their path. This helps to increase the current flow and thus the transport rate of ions through towards the cathode.

"Cathode is where the cations go" to be reduced.

In order to keep charge balance oxidation takes place at the anode.
With aqueous CuSO4 solution, if copper is not present in the anode, one can very conveniently use the reaction:
$$\ce{2 H2O -> 4 H+ + 4 e- + O2 gas - 1.23 V}$$
With an impure copper anode, copper and any metal with a less negative oxidation potential, such as iron, will go into solution preferentially:
$$\ce{Cu -> Cu++ + 2 e- - 0.34 V}$$
$$\ce{Fe -> Fe+++ + 3 e- + 0.04 V}$$
$$\ce{Al -> Al+++ + 3 e- + 1.66 V}$$
That goes with reduction at the cathode:
$$\ce{Cu++ + 2 e-  -> Cu metal + 0.34 V}$$
Metal drops out, O2 gas goes away (or other metals stay in solution), H+ replaces Cu ++, SO4 does not react.
Note that, at the cathode, copper will "go first" because it has a higher positive reduction potential than H+:
$$\ce{2H+ aqueous + 2 e- -> H2 gas  0.00 V}$$
as compared with + 0.34 V for Cu ++.
Note: reactions are written in the order they proceed at anode and cathode. More positive V values signify order of reaction.

What you say, Robert DiGiovanni, is correct if the anode is not made in copper. The trouble is that ChemStudent speaks of the refining of copper. This happens when the anode is made of impure copper. Here copper gets dissolved at the anode giving $\ce{Cu^{2+}}$ ions which go into solution. The impurities of the impure copper block are not dissolved and fall down the electrolysis container. So the solution is not going to be more and more acidic, as you state. The copper ion concentration does not change much during electrolysis, and pure copper is deposited at the cathode.

@Maurice yes, what an interesting way of refining copper.  Thank you. "The copper ion concentration does not change much during electrolysis".  You might lose your job if you threw any copper away.  All can be plated out (and re-refined).  I come from industry.  We have to turn a dime.


text:
QM9 atomization energies
I am doing experiments with the QM9 dataset from

http://quantum-machine.org/datasets/

and began with some simple checks on results from the literature. Consistent with what the authors of the paper

C.R. Collins, G.J. Gordon, O.A. Von Lilienfeld and D.J. Yaron, Constant size descriptors for accurate machine learning models of molecular properties. J. Chem. Phys. 148 (2018), 241718.

report, I find 3993 small QM9 molecules with 7 or fewer heavy atoms, which serves as their training set.
But I cannot reproduce the Null row of their Table II, probably because I don't understand well enough what they actually do there. In Section V.A they say:

A “null” model is used to provide a baseline measure of the difficulty of the prediction task. The null model always predicts the mean value of the training data.

They don't specify precisely which quantity they predict, but on p.6 they report values in kcal/mol, and in Table II they talk about atomization energies, so I suppose the target is one of the energies provided by QM9. But there energies are in Hartree and hence must be converted to kcal/mol.
Using the atomization energy (QM9 property 13), I get 319.48 Ha for the mean atomization energy on the training set. With this mean as constant null predictor I obtain an MAE of 30.84 Ha = 19,353 kcal/mol on the training set, and of 31,319, 48,632, and 58,000 kcal/mol, on the first 20,000, 50,000, and 133,000 QM9 molecules, respectively. (The data used are in https://arnold-neumaier.at/nullQM9.txt. Calculating the mean and the MAEs for the mean is a standard triviality.) But these numbers are over 100 times larger than the numbers stated in the Null row of their Table II. (The table is also on p.10 of the arXiv preprint https://arxiv.org/abs/1701.06649 - Section V.A is there called 5.1.)
Without conversion, the numbers would be far too small. The other properties from QM9 don't fare better.
Perhaps I misunderstood what they meant by the Null model, or there is an undocumented preprocessing step for getting the target energies for prediction?
Could anyone please resolve this discrepancy?

This question is missing the crucial part: What exactly are you doing to get to your numbers? Apart from this, this appears to be more a question about the computations or calculations and less about the chemistry or even the database. It's probably better suited at [mattermodeling.se]. I'll close the question to resolve these issues first.

I think the edit has not made the question any clearer. How do you calculate the energies? I unfortunately don't have access to the publication you cite, so I can't look at the context. The first reply from [mattermodeling.se] concurs with me, that data like an input file would be very helpful. I'm still hesitant to send this question over.

I used the energies tabulated in the QM9 dataset as Property 13. All I did was calculating the mean over the training set specified in the paper, and the mean absolute deviation of it (i) on the training set, and (ii) on the three initial subsets of the database for which values were given in Table II. The computation of these is trivial (and I had it checked by someone else), but I can add programs if really wanted.

@Martin: I also linked to the arXiv preprint of the paper.

I checked out the matter modeling site. There seems to be much less QM9 expertise than here, so that site seems less suitable for the question.

I agree with @Martin-マーチン that this is better suited at Matter Modeling. It's a newer site, so yes there are fewer "QM9" questions, but there are far more ML experts.

IIRC, their null model is not the mean of a dataset, but rather taking the mean energy from each individual atom. It's a given that $\ce{C6H6}$ will have a different atomization energy from $\ce{C5NH5}$ because they have different elemental composition. So the null model is a constant for each element, so $6 E_c + 6 E_h$ vs. $6 E_c + E_n + 5 E_h$, etc.

@GeoffHutchison: I don't think the problem is a machine learning problem since I obtained the discrepancy described without appyling any machine learning. I'll try your suggestion about what the null model is.

@GeoffHutchison: How does one get the $E_X$? Fitting them by least squares gives a much better MAE than reported in the table, so cannot have been their way of getting them.

In that case, you need to e-mail David Yaron. Despite him living and working nearby, I haven't seen him. (I certainly thought that was the null model for this paper.)

First off, let me stress that it's important to have a null model or baseline model for comparison in machine learning. In other words, does the ML model presented in the paper do better than a simple alternative.
Atomization energies are calculated from the electronic structure calculation of the molecule and the relevant atoms. In other words, how much energy does it take to dissociate the molecule into the constituent atoms (e.g., 6 carbon and 6 hydrogen atoms for benzene). Consequently, atomization energies depend on the number and type of atoms in the molecule. The atomization energy for benzene is higher than that of methane, because you're breaking more bonds.
My reading of this paper, was that the null model consisted of the average energies for each atom, e.g. $E_{C_6H_6} = 6E_C + 6E_H$ and $E_{C_5NH_5} = 5E_C + E_N + 5E_H$ with the various $E_C$ for each element fit from least squares. (Certainly that's what I'd consider as a null model for this, because standard atomization energies use the isolated atoms, so least-squares fit would at least adjust for atoms in a molecule.)
Your comment suggests that least-squares null model performs better than the reported null model in the paper.
In that case, I don't know, but if/when I see David Yaron next, I'll ask.

Incidentally, I'd strongly suggest avoiding the QM9. I know it has become a standard set, but the molecules are really strange compared to "typical molecules" from PubChem or similar sets.

The least squares appropach is not their null model but their model 1 - though they use ridge regression, hence get slightly different values. The null model must have gotten the $E_H$, etc from an unspecified, different source.


text:
How do we know the orbitals of multielectron atoms look like the orbitals of hydrogen?
How do we know that the solutions of the time-independent Schrodinger equation for multielectron atoms are the same with the solutions for the hydrogen atom if we add a effective potential due to the shielding of the charge of the nucleus from inner shell electrons?

We know the opposite: they are not the same. There is a vague resemblance, hardly more.

But if they are not why do we still use them to predict behaviour of chemical elements?

Because of the said resemblance, that's why.

@IvanNeretin, Would you mind expanding your comment into a useful answer? Who is "we" and what is the meaning of vague resemblance? I know at least one theoretical physical chemist who considers all this orbital teaching business for higher elements as complete non-sense.

Does they have a better approach that is somewhat understandable?

By the way, the question in your title doesn't quite match that in the body of the post. I answered the question in the post, which refers to isolated atoms (modeled using Slater type orbitals) not molecules.

The electronic Hamiltonian for a hydrogen atom can be written as
$$\hat{H}_e = -\frac{Ze^2}{r_{Ne}} - \frac{\hbar^2}{2m_e} \nabla^2$$
The relative nuclear charge for an H-atom is $Z=1$ (the atomic number). When approximating the single electron Hamiltonian using an effective nuclear potential and otherwise ignoring electron-electron repulsion, which is missing in the above Hamiltonian, $Z$ is set to $Z_{eff}$, the difference of the nuclear charge and the screening constant.
The effect of an increase in $Z$ is to contract the size of an atom. The Hamiltonian could be written as
$$\hat{H}_e = \frac{e^2}{r^*_{Ne}} - \frac{\hbar^2}{2m^*_e} \nabla^{*2}$$
where $r^*_{Ne} = r_{Ne}/Z$ is an effective distance from the nucleus, $m^*_e = m_e Z^2$ is an effective electron mass and $\nabla^* = \sum_i \vec{e}_i  \frac{\partial}{\partial x_i^*}$.  The change in the dimensions of $r$ therefore does not alter the mathematical form of the eigenfunctions of the Hamiltonian above, it only rescales the mass and distance dimensions. The solutions will be the single electron hydrogenic wavefunctions but with the effective radius $r^*_{Ne} = r_{Ne}/Z$ in place of $r$ and the momentum of the electrons altered by the change in effective mass. An effective Bohr radius can be computed under these circumstances as
(setting $\mu=Z^2 m_e$) $$a_\circ ^* = \frac{4 \pi \epsilon _\circ \hbar^2}{Z^2 m_e e^2} = \frac{a_\circ}{Z^2}$$
illustrating the contraction in the extent of the wavefunction due to the increase in $Z$.
Note that this is an oversimplification of the true situation, but the approximation suggests for instance that Slater-type orbitals might be useful as a first step to describing the properties of atoms. For molecules it is likewise inaccurate but can be helpful to grasp how behavior scales with nuclear charge.

Do we know that the additional contribution of electron-electron interactions in bigger atoms does not make a big contribution to the orbital shapes?

And not just the electrostatic repulsion which is stated in this answer maybe the Pauli repulsion plays a role as well?

@matt_black An effective nuclear charge attempts to account for a large chunk of the e-e repulsion, but does not account for correlation. Screening effects can predict electronic properties (although semi-empirically) with a fair degree of accuracy for moderately sized atoms.

So "How do we know that the solutions of the time-independent Schrodinger equation for multielectron atoms are ..." - the answer I give is that the math for the simplified version of the Hamiltonian for a single electron that sees an effective nuclear charge is the same. But that doesn't mean that the result is physically very accurate, especially for molecules. However you can look at Slater's papers and see what nice predictions (such as IEs) are possible for atoms using that model.

Also if you properly include the electron-electron interactions the whole orbital picture no longer applies - an orbital is a single electron wavefunction and in the strictest sense only applies to hydrogenic atoms. Their use in many electron systems is a (very useful) approximation.


text:
Where are the N and C-termini located in hemoglobin?
I am trying to locate the N and C terminus of hemoglobin and cannot tell if the C is at Val-1 and the N is Arg-141.

The N-terminus has the lowest residue number, and the C-terminus has the highest residue number. When showing a sequence, it is given from N-terminus to C-terminus when writing left to right. The specific sequence depends on whether you are looking at the alpha or beta chain, whether you are looking at fetal or adult hemoglobin, and which organism you are looking at.


text:
Why are certain hot polycarbonate films stickier than others? Molecular weight? Hydrolysis?
In van der waals dry transfer, we pick up flakes of materials such as graphene using polymer films. The material is exfoliated onto an SiO2 covered wafer and picked off using the polymer film. It's obviously pretty bad if the film sticks to the SiO2 itself, otherwise you wind up tearing and distorting your film trying to get it back off.
It's often necessary to try and pick up the flakes at elevated temperatures -- around 100-120 degrees C. It's at this point that some films get extremely tenacious and will not peel off the SiO2. The films that are the best seem to be the ones made on days with high humidity. Occasionally, water steam treatment seems to help. It would therefore seem to me that the PC is being hydrolyzed and broken into lower molecular weight pieces. Just a guess -- but if so, why would this make the film less sticky? If not, I would love to know why!


text:
Why don't we use sodium bicarbonate to neutralise base spills?
According to this document on lab safety (http://faculty.washington.edu/korshin/Class-486/AEESP-safety-notes.pdf), it says to not use $\ce{NaHCO3}$ to neutralise specifically base spills, stating "Do not use acetic acid or
sodium bicarbonate to clean a base spill. The sodium bicarbonate will not neutralize the spill, and acetic
acid could react strongly with the base."
Could someone explain why sodium bicarbonate won't neutralise a base spill? I have always learnt that it is amphiprotic and therefore can be used to neutralise BOTH acid AND base spills.

NaOH + NaHCO3 -> Na2CO3 + H2O, with sodium carbonate being still caustic, even if less than NaOH.  For neutralization of residues after water washing, it is adviced 2% NaHCO3 for acids and 2% acetic acid for bases, which are advised to kept handy as stock solutions.

I am going to argue that the currently-accepted answer is incorrect.
As the current accepted answer correctly pointed out, the bicarbonate ion dissociates into a proton and a carbonate ion, and that an equilibrium between the bicarbonate ion and the carbonate and hydrogen ion is established.
However, Sam202's calculation assumes that there is no driving force that would shift the equilibrium towards one side or the other. In the case of an acid/base reaction, this is simply not true. Following from Le Chatelier's principle, the equilibrium can be shifted to favor the products by removing products as they are formed. In the case of sodium bicarbonate reacting with sodium hydroxide, hydrogen ions formed by the dissociation of bicarbonate are removed by reaction with hydroxide ions to form water, causing the equilibrium to shift to favor greater dissociation of bicarbonate, driving the reaction to completion.
As a sidenote, acid/base titrations assume that the reaction between the acid and base is stoichiometric. If Sam202's answer were correct, titrating a weak acid with a strong base or vice-versa would be impractical.
As to the question at hand, the reaction of sodium bicarbonate with sodium hydroxide would produce water and sodium carbonate as pointed out by commenter Poutnik. While sodium carbonate is a weaker base compared to sodium hydroxide (pKa of sodium carbonate is 10.33 compared to 15.7 for sodium hydroxide, according to Wikipedia) it is still basic (For reference, pKa of ammonia is 9.25, pKa of sodium bicarbonate is 6.3). Thus, the reaction of sodium bicarbonate with sodium hydroxide would still leave a base spill, as Poutnik correctly deduced.

I would like to ask, for me it appears the current answer attempts to find how much bicarb is needed for the pH to be 7, whereas for the NaOH + NaHCO3 reaction, the salt will be basic, meaning, more bicarb will be needed for a neutral pH than just what is required to complete the reaction, hence the calculation. What are your thoughts on this, what calculations would you make for a neutral pH?

Upvoted. The pH of a typical baking soda solution is in the neighborhood of 8.3 or 8.4. Adding sodium hydroxide raises the pH: the pH has to be intermediate between that of the baking soda and NaOH. It cannot be 7 unless vast quantities of pH 7 water are used to dilute the baking soda and NaOH to ultra-trace levels.

@Quippy What Ed said. You won't ever get to pH 7 by adding bicarb to sodium hydroxide, because bicarbonate is itself slightly alkaline.

Suppose we have a strong basic aqueous solution $B$ that we attempt to neutralize with $\ce{NaHCO3}$ working as an acid $A$.
The source of excess $\ce{OH-}$ ions comes from the base, and since it's strong, we can consider:
$$n_{Bo}=n_{\ce{OH-}}\implies C_{Bo}{V_B}=C_{\ce{OH-}}V_B$$
The source of excess $\ce{H+}$ ions is bicarbonate, but since it's a weak acid, we have to consider the equilibrium:
$$\ce{HCO3-(aq)<=>H+(aq) + CO3^{2-}(aq)}$$
The equilibrium expression in terms of moles after mixing both solutions is:
$$K_n=K_a\;(V_A+V_B)=\frac{x^2}{C_{Ao}V_A-x}$$
For complete neutralization to take place, both excess $\ce{H+}$ and excess $\ce{OH-}$ ions need to completely react with each other, so:
$$x=C_{Bo}V_B$$
Substituting above:
$$K_a\;(V_A+V_B)=\frac{(C_{Bo}V_B)^2}{C_{Ao}V_A-C_{Bo}V_B}$$
For simplicity, let's consider equal molar concentrations of both solutions and a volume of $\pu{0.1L}$ for our basic solution that needs neutralizing:
$$C_{Ao}=C_{Bo}=\pu{1mol/L}$$
$$V_B=\pu{0.1L}$$
The acid dissociation constant of bicarbonate at 25°C is approximately:
$$K_a=5.012\times10^{-11}$$
Substituting all values:
$$5.012\times10^{-11}\;(V_A+0.1)=\frac{0.1^2}{V_A-0.1}$$
Solving for $V_A$:
$$V_A=\pu{14125.2L}$$
In other words, bicarbonate is considerably weaker as an acid (about 400 times) than it is as a base, so attempting to neutralize a strong basic solution with it would require an enormous amount.

Sorry once again but could you explain how you got $K_n=K_a(V_a+V_b)$ thanks :)

$$K_a=\frac{C_C^c;C_D^d}{C_A^a}=\frac{\left(\frac{n_C}{V}\right)^c;\left(\frac{n_D}{V}\right)^d}{\left(\frac{n_A}{V}\right)^a}=\frac{n_C^c;n_D^d}{n_A^a}\left(\frac{1}{V}\right)^{\Delta n}=;K_n\left(\frac{1}{V}\right)^{\Delta n}$$
$$\Delta n=1$$
$$V=V_A+V_B$$

Sorry I am really struggling with the notation, just note I'm still a highschool chemistry student. What is $\Delta n$ and what chemicals are subscript $C, D$ and $A$? Could you perhaps write working as $K_a=\frac{\ce{[H3O^+][NaCO3^-]}}{[\ce{NaHCO3}]}$ or something like that thank you!

I just represented the reaction: $\ce{HCO3- <=>H+ +CO3^{2-}}$ as:
$$\ce{A<=>C + D}$$
$\Delta n$ is the difference between the sum of product molar coefficients and reactant molar coefficients of the reaction:
$$\Delta n=c+d-a=1+1-1=1$$

Thanks got it and finally, where does $K_n=\frac{x^2}{C_aoV_a-x}$ come from? What does $x$ represent and how did you derive the expression? (Sorry for all the questions but I feel like it'll be so beneficial for me to understand)

If you're not familiar with ICE tables for equilibrium problems, you should look it up to fully understand these expressions.

I am, but I learnt to put it in using concentration values not no. of moles which I assume is what $C_aV_a$ gets you, in other words, why does $[\ce{HCO3^-}]=C_aV_a-x$ not $C_a-x$?

Let us continue this discussion in chat.

It seems to me that your calculation assumes that there is nothing that would cause the equilibrium to shift to one side or the other compared to the native dissociation rate. But wouldn't the reaction of H+ and OH- ions remove the product of the bicarbonate's dissociation and drive the equilibrium for the dissociation of bicarbonate further towards the products as per Le Chatelier's principle?

Also, acid/base titrations work on the assumption that the reaction between the acid and the base is stoichiometric. If your assertion were true, it would be impossible to do a titration without knowing the pKa value of the substance whose concentration one has to determine.

The calculation is intended to find what amount of $\ce{H+}$ ions from the source $\ce{HCO3-}$ is required to achieve complete neutralization. $\ce{HCO3-}$ in the presence of excess $\ce{OH-}$ will react as an acid, not a base, since the equilibrium: $\ce{HCO3- + H2O <=> H2CO3 + OH-}$ is heavily favored for reactants according to LC, as opposed to the equilibrium: $\ce{HCO3-  <=> H+ + CO3^{2-}}$, which is favored for products under said conditions.

As the bicarbonate solution is gradually added to the strong basic solution, the equilibrium shift you're describing is taking place, but the calculation accounts for the entire volume of bicarbonate already being added, since the total volume is $V_A+V_B$, and not $V_A$ plus some fraction of $V_B$.

Also, acid/base titrations necessarily being stoichiometric is only true for the case of strong acid+strong base. This case involves a very weak acid (bicarbonate), so its pKa needs to be considered.

"Also, acid/base titrations necessarily being stoichiometric is only true for the case of strong acid+strong base" Not true, titrations with various combinations of weak/strong acid/base are described, including weak acid/weak base (Such as ammonia and acetic acid, and the difference in pKa values between ammonia and acetic acid is less than between sodium hydroxide and sodium bicarbonate).

Such combinations exist for titrations but not all of them are stoichiometric. Having equimolar amounts of $\ce{NaOH}$ and $\ce{CH3COOH}$ does not result in equimolar amounts of $\ce{H+}$ and $\ce{OH-}$ in solution.