text:
Chemical species equations in a kinetic mechanism of combustion
In combustion, for a kinetic mechanism involving $m$ reactions and $n$ chemical species, of the form
\begin{align}
    \begin{cases}
        \nu_{11}' \mathcal{S}_1 + \nu_{12}' \mathcal{S}_2 + \dots + \nu_{1n}' \mathcal{S}_{n} &\ce{->[k_1]} \hspace{0.75cm} \nu_{11}'' \mathcal{S}_1 + \nu_{12}'' \mathcal{S}_2 + \dots + \nu_{1n}'' \mathcal{S}_{n} \\
        \nu_{21}' \mathcal{S}_1 + \nu_{22}' \mathcal{S}_2 + \dots + \nu_{2n}' \mathcal{S}_{n} &\ce{->[k_2]}  \hspace{0.75cm}  \nu_{21}'' \mathcal{S}_1 + \nu_{22}'' \mathcal{S}_2 + \dots + \nu_{2n}'' \mathcal{S}_{n} \\
        \hspace{2.63cm} \vdots & \hspace{0.45cm} \vdots \hspace{3.85cm} \vdots \\
        \nu_{m1}' \mathcal{S}_1 + \nu_{m2}' \mathcal{S}_2 + \dots + \nu_{mn}' \mathcal{S}_{n} &\ce{->[k_{m}]}  \hspace{0.75cm}  \nu_{m1}'' \mathcal{S}_1 + \nu_{m2}'' \mathcal{S}_2 + \dots + \nu_{mn}'' \mathcal{S}_{n}
    \end{cases},
\end{align}
we have the rate of the $i$-th reaction given by
\begin{equation}\label{eq1}
    q_i = k_i[\mathcal{S}_1]^{\nu_{i1}'}[\mathcal{S}_2]^{\nu_{i2}'}\dots [\mathcal{S}_n]^{\nu_{in}'},
\end{equation}
where $[\mathcal{S}_j]$ corresponds to the molar concentration of the specie $\mathcal{S}_j$, and $k_i$ corresponds to the Arrhenius equation
\begin{equation*} %ARRHENIUS MODIFICADA
    k_i = A_i T^{\beta} e^{-E_a / (RT)}.
\end{equation*}
The change in the concentrations of all species with time $t$ is, then, given by the
system of differential equations
\begin{equation}\label{eq2}
    \frac{\mathrm{d}[\mathcal{S}_j]}{\mathrm{d}t} = \sum_{i=1}^{m} (\nu_{ij}''-\nu_{ij}') q_{i}, \hspace{3cm} j=1,\dots,n.
\end{equation}
I received a computational code from a combustion researcher, in which the molar concentrations of the species, $[\mathcal{S}_j]$, in the equations above are replaced by the mass fraction of the species, $Y_j$. Thus, the code solves the system
\begin{equation}
    \frac{\mathrm{d} Y_j}{\mathrm{d}t} = \sum_{i=1}^{m} (\nu_{ij}''-\nu_{ij}') q_{i}, \hspace{3cm} j=1,\dots,n;
\end{equation}
with
\begin{equation}
    q_i = k_iY_1^{\nu_{i1}'}Y_2^{\nu_{i2}'}\dots Y_n^{\nu_{in}'}.
\end{equation}
Can this substitution be made so that the equations still make sense?

With mass fractions, calculated rates for particular reactions would be strongly affected by presence of components that do not take part in the reaction at all..

The relation between molar concentration and mass fraction is
\begin{align}
  [S_j] = \frac{N_j}{V} = \frac{m_j}{M_jV} = \frac{Y_jm}{M_jV} \rightarrow
  [S_j] = \left(\frac{\rho}{M_j}\right)Y_j \tag{1}
\end{align}
where $M_j$ is the molar mass of chemical species $ j $ and $\rho$ is the mass density.
It seems that your hope of writing the rate law by analogy, i.e. changing $[S_j]$ for $Y_j$, is true under very restrictive conditions: (1) irreversible elementary reaction $A \rightarrow B $, and (2) constant volume process. We prove this by using Eq. (1) in a batch reactor, like your equations show, where there is no inlet or outlet of any chemical species
\begin{align}
  \require{cancel}
  \text{[Rate of acumulation of A within the system]}
   =& \text{[Rate of flow of A into the system = 0]} \\
   -& \text{[Rate of flow of A out of the system = 0]} \\
   +& \text{[Rate of generation of A within the system]} \\
  \frac{dN_A}{dt} &= \nu_A rV \\
  \frac{d([S_A]V)}{dt} &= -k[S_A]V \\
  \bcancel{V}\cancel{\left(\frac{\rho}{M_A}\right)}\frac{dY_A}{dt} &=
  -k\cancel{\left(\frac{\rho}{M_A}\right)}Y_A \bcancel{V} \\
  \frac{dY_A}{dt} &= -kY_A \therefore r' = k Y_A
\end{align}
The condition of constant volume, in your system, is satisfied by the way you have written the differential equations. The only way you can arrive to that form, is by taking out $V$ from the time derivative, and cancelling it on both sides. I will continue with this assumption, and since the mass is constant, then the mass density $\rho$ also is.
It is unusual to carry reactions, specially gas-phase reactions like combustions, in this manner. I will leave at the end the mathematical restriction that it imposes.
Nevertheless, lets try to obtain an expression according to your case.
1. Rate law in terms of $Y_j$ Combined with Eq. (1) it gives
\begin{align}
  q_i &= k_i \prod_{k = 1}^n [S_k]^{\nu_{ik}'} \\
  q_i &= k_i \prod_{k = 1}^n \left(\frac{\rho Y_k}{M_k}\right)^{\nu_{ik}'} \\
  q_i &= k_i \rho^{\sum_{k = 1}^n \nu_{ik}'} \prod_{k = 1}^n
         \left(\frac{Y_k}{M_k}\right)^{\nu_{ik}'} \\
  q_i &= k_i \rho^{\nu_{i}'} \prod_{k = 1}^n
         \left(\frac{Y_k}{M_k}\right)^{\nu_{ik}'} \tag{3} \\
\end{align}
where I have defined $\nu_i' := \sum_{k = 1}^n \nu_{ik}'$. If we go back to the the reaction scheme, and stare at row $i$, this guy is the sum of all the stoichiometric coefficients of the reactants for that row.
2. Mole balance in terms of $Y_j$ The rate of change of the concentration of species $j$, using Eqs. (1) and (3), yields
\begin{align}
  \dfrac{d[S_j]}{dt} &= \sum_{i = 1}^m (\nu_{ij}'' - \nu_{ij}')q_i \\
  \left(\frac{\rho}{M_j}\right)\frac{dY_j}{dt}
  &= \sum_{i = 1}^m (\nu_{ij}'' - \nu_{ij}') k_i \rho^{\nu_{i}'}
  \prod_{k = 1}^n \left(\frac{Y_k}{M_k}\right)^{\nu_{ik}'} \\
\end{align}
$$ \boxed{\frac{dY_j}{dt} = M_j
   \sum_{i = 1}^m (\nu_{ij}'' - \nu_{ij}') k_i \rho^{\nu_{i}' - 1}
   \prod_{k = 1}^n \left(\frac{Y_k}{M_k}\right)^{\nu_{ik}'}} \tag{4}  $$
Eq. (4) is as far as we can get, but is the desired expression, as we have the ODE in terms of the mass fraction of species $j$.
3. Constant volume process In the simplest of cases, the mixture of gases will obey the ideal gas law
\begin{align}
  \rho &= \frac{P}{MRT} \\
  \ln\rho &= \ln(P) - \ln(MRT) \\
  \frac{d\ln\rho}{dt} &= \left(\frac{1}{P}\right)\frac{dP}{dt} - \left(\frac{1}{\cancel{MR}T}\right)
  (\cancel{MR})\frac{dT}{dt} - \frac{1}{M\cancel{RT}}(\cancel{RT})
  \frac{dM}{dt} \\
  \frac{d\ln\rho}{dt} &= \left(\frac{1}{P}\right)\frac{dP}{dt} - \left(\frac{1}{T}\right)\frac{dT}{dt}
  - \left(\frac{1}{M}\right)\frac{dM}{dt} \tag{5} \\
\end{align}
Where $M$ is the molar mass of the mixture. At any instant in time, it has a value of $ M = \sum_{k = 1}^n M_ky_k $, and continuing with Eq. (5)
\begin{align}
  \frac{d\ln\rho}{dt} &= \left(\frac{1}{P}\right)\frac{dP}{dt} - \left(\frac{1}{T}\right)\frac{dT}{dt}
  - \frac{1}{\sum_{k = 1}^n M_ky_k}
  \frac{d}{dt}\left(\sum_{k = 1}^n M_ky_k\right) \\
  \frac{d\ln\rho}{dt} &= \left(\frac{1}{P}\right)\frac{dP}{dt} - \left(\frac{1}{T}\right)\frac{dT}{dt} - \frac{1}{\sum_{k = 1}^n M_ky_k}
  \sum_{k = 1}^n M_k\frac{dy_k}{dt} \tag{6} \\
\end{align}
In consequence, if the mass density is to remain constant, by Eq. (6)
$$ \boxed{\left(\frac{1}{P}\right)\frac{dP}{dt} = \left(\frac{1}{T}\right)\frac{dT}{dt}
   + \frac{1}{\sum_{k = 1}^n M_ky_k}
   \sum_{k = 1}^n M_k\frac{dy_k}{dt}} \tag{7} $$
Eq. (7) must be satisfied at any instant of time while solving the system of differential equations. It requires the need of the energy balance, in order to obtain an expression of $dT/dt$. Of course, you can always sweep it under the carpet, and the pressure will "adjust" itself to meet the demands of the process. Pretty much like a normal force in a body diagram. However, if you can couple Eq. (7) to the system and solve it, then perhaps high values of pressure will be obtained (e.g. if the system of reactions is excessively exothermic). This is an indication that the isochoric operation may be unfeasable in reality.


text:
Effect of changing of the concentration of reactants when concentration of the initiator and the concentration of the initial reactant are very small
A + I---> p
What happens to the rate of the reaction when the concentration of the reactant is changed in a system that its concentration of its initiator and the concentration of the initial reactant is very small? Generally, we add a small amount of initiators with respect to the concentration of reactance.eg: I= 1*10^-4M, A= 1M
Can we take the order of A zero under these conditions?
For example, styrene polymerization is initiated by benzoyl peroxide.

styrene used (mmol)
benzoyl peroxide (mmol)

0.45
1.5

0.93
2.48

2.27
1.5

Why do you think you could?

It is because the reaction is likely limited by the concentration of the initiator. Generally we add small amount of initiators respect to the concentration of reactance.

Try to reformulate the question as the scenario is not very clear. Be aware that posts containing just questions are not good posts. // Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, misunderstanding, objections, question down-voting or even question closure.

Is it a kind of polymerization, like e.g. styrene polymerization initiated by benzoyl peroxide?

Yes. exactly styrene polymerization initiated by benzoyl peroxide

Sorry my bad, I reformulated the question

Let us continue this discussion in chat.

As you written the reaction in an irreversible fashion, that will be my starting point. You cannot state that the partial order of $A$ is zero, still the rate law has the form
$$ r = kC_A^\alpha C_I^\beta \tag{1} $$
However, if you put an excess of $A$, then its concentration as the reaction takes places, will not deviate much from the initial concentration $C_{A0}$. Thus, Eq. (1) now is
$$ r = \underbrace{kC_{A0}^\alpha}_{k'} C_I^\beta = k'C_I^\beta \tag{2} $$
Eq. (2) is helpful because it allows to at least determine $\beta$ in this limiting case. For example, if you have data, you can couple Eq. (2) with a mole balance in a batch reactor at constant volume
\begin{align}
  \frac{dC_I}{dt} &= r_I \\
  \frac{dC_I}{dt} &= -\nu_Ir \\
  \frac{dC_I}{dt} &= -k'C_I^\beta \rightarrow
  \ln\left(-\dfrac{dC_I}{dt}\right) = \ln(k') + \beta\ln(C_I) \tag{3}
\end{align}
If you have $C_I$ as a function of time, you can plot the left-hand sight (by numerical differentiation of your data) vs $\ln(C_I)$. If it results in a straight line, then the slope and the y-intercept will inform you that
$$ k' = \exp(b) \hspace{1 cm} m = \beta \tag{4,5}$$
Eq. (4) won't give you anything valuable, since $ k' = kC_{A0}^\alpha $, and we don't know $\alpha$. However, at least you have $\beta$ for this set of experiments.
If the reaction mechanism is more complicated, as Poutnik pointed out, then the algebra changes. However, you can always approximate at every step that $C_A \approx C_{A0}$ to simplify matters.

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text:
Is it necessary criteria that we draw Ellingham diagram for only one mole of oxygen consumed for a given reaction
In my textbook and many other reference text shows Ellingham diagram for only one mole oxygen consumed. Why is this done?

Ellingham diagram is not about the number of moles at all. Two moles of anything are just as stable (or unstable) as one mole.

A magazine article where the headline is longer than the article text would be rejected by magazine editors. Questions on CH SE site are not that much different.// Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, misunderstanding, objections, question down-voting or even question closure.

(Disclaimer: during composition of this answer I had many issues with typing and syntax, and I may have lost some battles. Improvements in these areas are much encouraged.)
Yes, a fair thermodynamic comparison of the oxidation processes requires using one mole of $\ce{O2}$ in the reaction.
For example, if we look up free energies of formation for magnesia and alumina at $25°\text{C}$, we get (from (https://en.wikipedia.org/wiki/Standard_Gibbs_free_energy_of_formation):
$\ce{MgO}: -569.3 \text{kJ/mol}$
$\ce{Al2O3}: −1582 \text{kJ/mol}$
It looks like the free energy release for alumina formation far exceeds that for magnesia formation. But now look at the balanced chemical reaction equations:
$\ce{Mg + (1/2)O2 -> MgO}$
$\ce{2Al + (3/2)O2 -> Al2O3}$
Thus the free energy data, each reported per mole of the compound, corresponds to one-half mole $\ce{O2}$ for magnesia versus 3/2 mole $\ce{O2}$ for alumina. So to match the oxygen consumption and get the fair thermodynamic comparison, we have to multiply  the quoted bagnesia value by $2$ and the quoted alumina value by $2/3$
$\ce{MgO}: -1139 \text{kJ/mol } \ce{O2}$
$\ce{Al2O3}: -1055 \text{kJ/mol } \ce{O2}$
So when we match oxygen bases, we find that magnesia has the lower free energy of formation.
When we reduce free energy of firmation to one mole of $\ce{O2}$, this free energy becomes equal to $RT\ln(p_{O_2})$ or $RT\ln(f_{O_2})$, where thecargument of the logarithm is the equilibrium ideal-gas pressure or fugacity of $\ce{O2}$. The significance of the magnesium/magnesia couple being lower than the aluminum/alumina couple on the Ellingham diagram is that the former drives oxygen pressure or fugacity down to a lower level and thus magnesium is a thermodynamicalky more powerful reactant for aluminum than oxygen at $25°\text{C}$.
You'll also see that the magnesium/magnesia couple is actually lower than the aluminum/alumina couple only up to a temperature of about $1400°\text{C}$. At hotter temperature aluminum becomes more powerful thermodynamically than magnesium. Thus occurs because magnesium is rather volatile and boils off into a gas. The large entropy increase in the metal when this happens makes reaction with oxygen to form the oxide (which is still solid) less favorable. Aluminum does not boil off at temperatures commonly used in industry and so retains its fill reducing power. In contrast, with iron it is the oxide that boils off, so that the increased entropy of the gaseous oxide makes iron eventually catch up with the normally more reactive zinc.

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text:
Soap equation using potash
I'm trying to understand the chemical formula of soap using potash ($\ce{K2CO3}$) as they did in antiquity instead of using sodium hydroxide ($\ce{NaOH}$) or potassium hydroxide ($\ce{KOH}$). I've found the formula for making soap with sodium hydroxide on the Khan Academy, but I didn't make much progress on my own.

Can somebody help me understand how the soap and the glycerin molecule would look like?
Or is the saponification not possible and ancient soap makers first needed to create potassium hydroxide from potassium carbonate?

Closely related, would be a duplicate if there was an answer. But it has useful comments.

I closed the related answer as duplicate now that we have multiple answers here.

Unrelated to the potassium carbonate question, but where does the oxidation (carbonyl) on the central carbon come from? Saponification should produce glycerol (and/or related alkoxides), not dihydroxyacetone.

Ashes have been thoroughly studied more than $100$ years ago. See the Dictionnaire de chimie (in French) by A. Wurtz, Hachette ed., Paris, 1872, p. 787 - 789. Potassium hydroxyde $\ce{KOH}$ was never found in all ashes studied (Wood, fruits, vegetables) : in the presence of air, potassium hydroxide $\ce{KOH} $ is quickly transformed into potassium carbonate $\ce{K2CO3}$. The composition of wood ashes is $\ce{CaCO3}$ $5 - 50$%, $\ce{K2CO3}$ $12 - 32$%, $\ce{SiO2}$ $5 - 42$%, $\ce{Ca3(PO4)_2}$ $5 - 15$%, $\ce{MgCO3}$ $2 - 6$%, and $\ce{FePO4}$ $0 - 2$%. So, wood ashes do not contain any $\ce{KOH}$, as the $\ce{OH-}$ ion is entirely transformed into carbonate in the presence of air. But, as mentioned by Oscar Lanzi, this $\ce{CO3^{2-}}$ ion is hydrolyzed in water and produced about $1$% $\ce{OH^-}$ ion, which is able to saponify the vegetable oils. But the yield is poor.

Once you introduce water, potassium carbonate can generate potassium hydroxide through the hydrolysis reaction
$\ce{CO3^{2-} + H2O <=> HCO3^- +OH^-},$
and this reaction is fairly efficient because carbonate ion is a rather strong base ($\pu{pK_b} = 3.75$ from the conjugate acid $\pu{pK_a}$ reported by Wikipedia). That basicity combined with the good solubility of potassium carbonate in water allows a $\pu{pH}$ between $12$ and $13$ to be obtained.
In addition, the moderately strong-base carbonate ion can promote nucleophilic attack by the water itself, by acting as a base to remove a proton from the nucleophilic substitution product:
$\ce{RCOOR' + H2O -> RCOOH + R'OH}$
$\ce{RCOOH + CO3^{2-} -> RCO2^- + HCO3^-}$

With spending about 9% of carbonate, pH will drop to 11.3. With 50% to 10.3.

So I propose a mechanism that directly uses carbonate ion as the reactant.

Hmm, I am not sure about that, but I am not an organic chemist.

In response, I switch roles to carbonate acting as a proton remover, which seems better documented in the litersture.

We don't need to boil off CO2 or specifically create more KOH. But we may need hotter water to make the soap more effective.

The penultimate equation appears to defy the leveling effect in which the weaker base R′OH is protonated as opposed to carboxylate.

The proton stars off on the water molecule and gets kidnapped by the stronger carbonate ion.

Not the last equation, the one before it.

Can somebody help me understand how the soap and the glycerin molecule would look like?

Once you add water, you have potassium ions and hydroxide ions in solution, and the products look the same as with aqueous potassium hydroxide.
In contrast to the reaction with sodium hydroxide, which gives the sodium salts of the fatty acids, you would obtain the potassium salts of the fatty acids with potassium hydroxide or with potassium carbonate.
If you use potassium carbonate, there would be some form of carbonate as a side product. Depending on the final pH, it could be carbonate, hydrogen carbonate or carbon dioxide or some combination. According to this source, traditional recipes used an excess of wood ash, so the pH would be fairly alkaline.

I am afraid that with comparable ratio bicarbonate/carbonate, the saponification would be already effectively stopped due too low pH (pKa2 10.32) , not even speaking about CO2.

@Poutnik I added that you would usually work with an excess of wood ash. I am wondering whether it is possible to change the pH after the fact (this is soft soap, so semi-liquid).

Yeah, I have noticed that after posting my comment. From CO2 it would be just a small step to precipitation of free fatty acids. I think there would be needed much more potash than caustic potash, as with the former pH would quickly drop.


text:
Addition of cyclohexanone to indole with MeONa and MeOH
I need to find this reaction mechanism and i have run out of ideas. I have tried using -OMe on both indole and on cyclohexanone in order to create an enolate but i cant think of anything else. Thank you in advance

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Basically an enamine reaction

it seems like enamine reaction but the N doesnt get attached to any carbon. Am i mistaken?

The N remains unsubstituted.

As to your comment that " N doesn't get attached to any carbon", neither does nitrogen get attached to carbon in enamine reactions.

Is this a real reaction or an example from an exam paper? I can find no literature that reports such a reaction pathway under the reported conditions

it is an exam question. i couldn't find any citations either, that's why i made this post. I just made up a mechanism and i m still not sure if it is right

@user55119 but it does, doesnt it ? a similar reaction i found was stork enamine alkylation

With allylic halides initial attachment is to nitrogen, but then they rearrange to carbon. It is possible to write a base catalyzed mechanism for your reaction. If I were going to try to do the reaction, I would use an acid catalyst. I looked up this reaction between indole and cyclohexanone on chemical abstracts, but I couldn’t find it.

okay thank you for the information!

The Comments have pretty much ruled out the existence of this reaction in the chemical literature. Like many exam questions the proposed reaction is conceptually possible but that doesn't mean that it occurs. The reaction for indole itself (1b), in principle, would be catalytic in base in that alkoxide is regenerated. Your indole (1a) requires stoichiometric base to neutralize the carboxylic acid in addition to the catalytic amount of base required to effect the reaction. In practice, an excess of base would normally be employed.
Indole (1b) has a pKa of ~21 while methanol's pKa is 15.5. This means that the concentration of resonance stabilized anion 2 is quite low. However, it is a better nucleophile than indole itself. Any addition of the nitrogen of 2 to cyclohexanone (3) will be unproductive because the reverse reaction is its only option. Addition of the carbon site to 3 leads to protonation of the alkoxide and regeneration of the indole nucleus as shown in structure 4. During the course of these steps either 4 or 5 may be protonated and deprotonated. Species 5 effects  elimination of hydroxide affording 6, which upon vinylogous deprotonation leads on to the final product 8 via 7. continued

The foregoing reaction is a case of a strong nucleophile (2) and a decent electrophile (3). What about the use of a strong electrophile (9) and a decent nucleophile (1)? An acid-catalyzed route is less problematic having the Vilsmeier-Haack formylation of indole as a precedent. Accordingly, species 10, which is formed from protonated cyclohexanone 9 and indole 1, aromatizes to indole 11. Acid-catalyzed loss of water from 12 affords 8 via 13.
If I were setting out to accomplish this reaction, I would certainly opt for the acid-catalyzed route.

thank you a lot for your reply! I hadnt thought about pKa, but i gave a similar mechanism for the exam so its nice to see that my thoughts are okay, at least for the sake of examination. I would definitely use acidic conditions if i were doing this in a lab


text:
Substituent effect on Aromatic compound (Nucleophilic addition)
The reaction shows a nucleophilic addition:

The reaction on the left is faster because methoxy substituent is on the meta position of the ring, while the right side reaction is located on the para position of the ring. What difference does the 2 different positions come into play here? The -OMe substituent is electron donating and in both cases will donate electrons to the ring. Why is then the left handed reaction preferred or faster?

Left reaction has predominantly inductive effect, which increases the electrophilic nature of carbonyl carbon. Right reaction dominated by mesomeric effect, which reduces the electrophilic nature of carbonyl carbon.

Deaw a resonance structure with $\ce{C+ - O-}$ at the carbonyl group. Then see where additional resonance structures can delocalize the positive charge from the carbonyl carbon. Which molecule ends up with more delocalization and thus weaker positive charge on the carbonyl carbon?


text:
How is Clausius statement violated for reversible heat engines that are assumed to have different efficiencies?

I'm trying to verify that all reversible heat engines have equal efficiencies by using this diagram. Here, the second heat engine (HE2) is assumed to has higher efficiency than the first heat pump (HP1). Since W2 > W1, some work produced by the HE2 can be used for other work, which is denoted as W2 - W1. This concludes that the overall engine will be extracting heat energy from the low temperature reservoir and completely convert it to work (W2 - W1 = Q2 - Q2'), which violates Kelvin-Planck's statement.
What I'm having trouble is, I don't get how Clausius statement (heat doesn't flow spontaneously from cold to hot regions) is violated here? Can someone help?


text:
Cp=Cv for isochoric processes?
According to the first law of thermodynamics, ∆U = ∆H + work done, where magnitude of work done is given by W = PdV.
In an isochoric process work done would be zero (provided the external pressure doesn't change, which is usually the case), and the above equation would give us ∆H = ∆U. If we were to use the values: delta H = nCp∆T and delta U = nCv∆T, we would get Cp = Cv, which is false.
I also know that one of the ways to find the relation Cp = Cv + R would be to substitute the PdV with ∆nRT, but if we were to use the formula for an isochoric process as it is, then the result would be Cp = Cv.
The only counter argument to this that I could think of was that the ∆T in the above formulae were different (with the condition that the ∆T in ∆H pertains to the change in temperature on the condition that pressure remains constant and the condition of constant volume for that of ∆U). But even this argument was squashed by a youtube video I watched where it is mentioned that the above formulae could be used regardless of the process the gas undergoes. So where exactly am I going wrong?
Please note that I'm in 12th grade so my knowledge of thermodynamics and some of the math used in it (such as partial derivatives) is limited.

https://physics.stackexchange.com/questions/410009. Also, YouTube as a platform is designed for entertainment and collecting ad revenue, not rigorous studying.

$\Delta H = \Delta U + p \cdot \Delta V + V  \cdot \Delta p$, so for isochoric system  $\Delta H = \Delta U  + V  \cdot \Delta p$.

Cp is isobaric molar heat capacity. Cv is isochoric molar heat capacity. The system cannot have 2 heat capacities at the same heating process. You heat it up either isobarically with Cp , or isochorically with Cv. ( or neither of both ways, then neither of capacities is applicable.)

@Poutnik so you're saying that the assumptions of those formulae being correct in all processes is the problem?

dH/dT=dU/dT + d(pV)/dT=nCv + nR=nCp. It is valid for ideal gas, not generally. // The fact that dH/dT=nCp does not mean Cp=Cv, as dQ/dT=dU = nCv <> dH for isochoric process and dQ/dT=dH =nCp <> dU for isobaric process.


text:
Enthalpy in an irreversible isobaric process
I know that in a reversible isobaric process, the enthalpy change of the system is the change in heat for that system. What happens if the process is not reversible? I.e only the outside pressure is constant, for example atmospheric pressure?
For example, in a container with a movable piston on the top, a gas phase chemical reaction takes place with ΔH, in an uncontrolled manner so that the process is irreversible. The outside pressure is atmospheric. Is still ΔH=Δq for the gas (assume in the reaction the moles of gas are changed, increases)?

If the initial pressure in the system is atmospheric and the final pressure at equilibrium of the system is atmospheric, then  $q=\Delta H$

@ChetMiller unless there is non-PV work, right?

@Karsten Yes.  I was trying to keep it simple and at the OP's level.

The reversibility of the reaction is irrelevant; any measured isobaric reaction will have ΔH = Δq. Keep in mind that simply knowing a process is reversible does not give any indication as to what equilibrium will actually look like for the process (meaning you can't generalize about how reversible vs. irreversible reactions will behave, as they are all simply reactions); you could easily have a reversible process that produces the same pressure change as an irreversible one. The reactions may have different mechanisms, activation energies, and enthalpies, but the underlying calculations will be the same. ΔH will always equal Δq if the process is in an isobaric environment. It does not matter how the reaction proceeds. The only thing that affects enthalpy in an isobaric environment is the energy changes within the reaction and how much heat is produced. Reversibility itself has no impact on this.

Only if $P_i = P_f = P_{ext}$ as well. Otherwise you can have an "adiabatic" irreversible isobaric process where $P_i \ne P_f = P_{ext} = $ constant and $\Delta H = \gamma W \ne Q = 0$.

I know that in a reversible isobaric process, the enthalpy change of the system is the change in heat for that system. What happens if the process is not reversible? I.e only the outside pressure is constant, for example atmospheric pressure?

I am going to extend a bit to clarify this simple but interesting case.
The key lies when you carry out the mathematical derivation, for a constant-pressure process, in a closed system and apply the 1st law
\begin{align}
  dH &= dU + d(PV) \\
  dH &= dQ + dW + PdV \\
  dH &= dQ \underbrace{- PdV} + PdV \rightarrow \boxed{Q = \Delta H} \tag{1}
\end{align}
it is precisely where the underbraces appear, that you state that the work is $dW=-PdV$. This substitution is what we call a mechanically reversible process, which is equivalent to two propositions:

The system is uniform in pressure, i.e., $P$ is not a function of spatial coordinates.
The pressure of the system is only differentially displaced with respect to the pressure of the surroundings. Thus, we envision the system walking from $P_1$ to $P_2$ in an infinite number of steps, each one characterized by a mechanical equilibrium between the system and the surroundings.

We have three comments at hand now:

Regarding enthalpy: since it is a state function, if you can measure $P_1$, $T_1$, $P_2$, $T_2$, the enthalpy change can be computed directly
$$ \Delta H = H(P_2,T_2) - H(P_1,T_1) \tag{2} $$
independently if your process was highly irreversible or totally reversible.
Regarding heat transfer: if the process was not mechanically reversible
$$ Q \neq \Delta H \tag{3} $$
and we cannot calculate the heat transferred between the boundaries of the system/surroundings. I will emphasize: you cannot calculate it that way, because you cannot substitute $W = -PdV$ in Eq. (1).
Regarding reversibility: imagine the process is still mechanically reversible and thus $ W = -PdV $, this doesn't mean that the process is reversible. Lets give a fast proof right now, and we will see that this isobaric process can be irreversible. For simplicity, we will consider $\ce{Ar}$, since it has a $C_P \neq f(T) = 5R/2 $ and it behaves perfectly as an ideal gas. The compression is illustrated in the following image:

The calculations are straightforward: (1) the heat transfer to the system $Q$, the work done by the system $W$ to the surroundings, and the change of entropy of the universe yields
\begin{align}
  Q &= \Delta H = \frac{5}{2}R(T_2 - T_1) \rightarrow Q = 2079 \; \pu{J/mol} \\
  W &= -P(v_2 - v_1) = -(P_2v_2 - P_1v_1) = -R(T_2 - T_1) \rightarrow W = -831.4 \; \pu{J/mol} \\
  \Delta S_\pu{uni} &= \Delta S_\pu{sys} + \Delta S_\pu{surr} \\
  \Delta S_\pu{uni} &= c_P \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) + \frac{Q_\pu{surr}}{T'} \\
  \Delta S_\pu{uni} &= \frac{5}{2}R \ln\left(\frac{T_2}{T_1}\right) + \frac{1}{T'}
  \left(-\frac{5}{2}R(T_2 - T_1)\right) = 3.646 \; \frac{\pu{J}}{\pu{mol·K}} > 0 \\
\end{align}
The conclusion is: you can apply $W = -PdV$ if you want, but it doesn't mean that the process is reversible.
How to know $Q$ then? For an irreversible process there are many approaches, if you want to still apply thermodynamics. I will present you the simplest one. Consider that the system is mechanically reversible anyways and compute
$$ W^\pu{id} = -\int P \; dV $$
Say the process resulted in a compression (like the example). Therefore, the system has done work and $W<0$. In reality, we are being too optimistic, and the work in fact was less. Thereby we penalize it by an efficiency $\eta < 1$, and calculate the real work
$$ W = \eta W^\pu{id} \tag{4} $$
Finally, the heat is obtained by appling the 1st law. Like enthalpy, internal energy is independent of the path, and by Eq. (4) we have
$$ Q = \Delta U - W \rightarrow \boxed{Q = U(P_2,T_2) - U(P_1,T_1) - \eta W^\pu{id}} \tag{5}  $$
To know the efficiency means to have valuable information of your system, and how it functions. Other approaches, as to calculate $P$ as a function of $V$, can also be applied. This one has again the same consequence, i.e., obtain information of how the system behaves when it expands or compress.

Your diagram doesn't make sense -- if you have an isobaric compression, the temperature has to decrease per $PV = nRT$, so $T_2 < T_1$. Also your point #2 contradicts the other answer here which states "the reversibility of the reaction is irrelevant; any measured isobaric reaction will have $\Delta H = \Delta q$" as well as the comments by Chet Miller in the original question as well...

Likewise since enthalpy $H$ is a state function it is thus path-independent and so in the absence of non-PV work, it has to follow that $\Delta H= \Delta q$ applies equally to all isobaric paths whether they are reversible or not.

@ManRow Hello ManRow. I don't have much time to engage because I have to do some exams. You are not correct. I assure you that $Q = \Delta H$ is only true for isobaric and mechanically reversible processes. I attach to you the pages of the following book that explain this, in here. It is underlined in red. If you want, you can read its Chapter 2, and come back for next discussions.

We can ask @ChetMiller. He is a chemical engineer as me, and has read Smith, Van Ness, and Abbott one million times. What is your answer Chet? If the process is isobaric but irreversible, can we say that $Q = \Delta H$?

@ManRow Here you have another proof, in Chapter 2 of Atkins' book in Physical Chemistry. Pay attention to the underlined lines. If the process is irreversible, can you say $\mathrm{d}w = -p,\mathrm{d}V$? No. If you can't say that, you can't replace that expression and reach $Q = \Delta H$. I await your answer to this discussion. Otherwise, I will flag these comments for words such as 'doesn't make sense' and not backing up your claims.

If you have an isobaric process where $V$ is decreased, you must also decrease $T$ in order to still ensure that $PV = nRT$ (since at constant pressure $P\Delta V = nR \Delta T$, so $V_2 < V_1 \Rightarrow T_2 < T_1$).

Also when internal and external pressures are the same then it always follows that $dW = -pdV$ regardless of whether the process is reversible or not. Here is an answer on the Physics-SE detailing this: https://physics.stackexchange.com/a/760921/168695

@ManRow I agree that the image is wrong, in the sense that the volume seems to "decrease". Unfortunately, I decreased the volume to put that text above that says $Q_\mathrm{surr{ = -Q$. Fortunately, in science and technology, we have authorative sources which I have cited to you. This helps us in the sense that if I write nonsense, they won't publish it. That's why I cited you two books. In that post, he/she statesit, doesn't details it. I won't go for further discussion. Thanks.


text:
∆G or ∆G° to predict spontaneity
I have a doubt about using Gibbs free energy to predict the spontaneity of a reaction.
It is shown that ∆G = ∆G° + RT ln (Q).
That said, in order to predict which direction the reaction spontaneously goes, should ∆G or ∆G° be used? As I understand it, ∆G is what actually allows one to predict the spontaneity of the reaction, that is, whether the reaction proceeds spontaneously from the reactants to the products (in the direction written, ∆G < 0, exoergonic) or whether it proceeds spontaneously from the products to the reactants (in the opposite direction, ∆G > 0, endoergonic), and all of this depends on the value of Q, since ∆G° is constant. Instead, ∆G° should be used to understand how far the reaction is shifted to the right or left when it is at equilibrium (∆G = 0). However, very often I see that ∆G° is used to predict the spontaneity of a reaction, so I see a contradiction (especially in biochemistry, when it comes to reactions in which ATP is involved).
P.S. again referring to Gibbs free energy: can it be said that a process of mixing two substances is (almost) always characterized by a ∆S > 0 and therefore that the spontaneity of the process depends essentially on ∆H?

Prediction of reaction direction by $\Delta_r G^\circ $ is identical to prediction of reaction direction by $\Delta_r G $  at $Q=1$. The former is limited to $Q=1$, the latter is applicable to any $Q$. By other words, the former predicts reaction spontaneity generally, the latter in the particular case.

So, basically ∆G° allows you to see where the reaction is most shifted at equilibrium.
Eventually it can also be used to predict the direction in which the reaction spontaneously proceeds, but only when the concentration of the reactants is equal to that of the products. For all other cases, ∆G is used.

Very often books refer to ∆G°: if ∆G° is positive, then the reaction is endoergonic and vice versa. But this is not entirely true, because the book does not refer to the specific case of Q=1. When I see ∆G° I think directly of the equilibrium situation (∆G=0)

Delta G0 is Delta G from Q=1 to Q=K.

What do you mean @Poutnik? Delta G^0 is Delta G for Q=K?

A process is called spontaneous when reacting in the forward direction would approach equilibrium (no matter whether a reaction in that direction is observed or not).
For a chemical reaction, this depends on the activities (concentrations and/or partial pressures) of reactants and products. To figure out whether a reaction under a given set of conditions (activities, temperature, pressure) is spontaneous in the forward or reverse direction, you have to know $\Delta_r G$.
Sometimes, people ask whether a certain reaction is spontaneous or not. The answer depends on the conditions, i.e. the activities of the reactants and products. If no set of conditions is given, you can answer that question by stating whether at standard conditions (i.e. Q = 1) and a temperature of choice the reaction is spontaneous in the forward or in the reverse direction. In that case, you would look at $\Delta_r G^\circ$ to determine which direction is spontaneous at standard conditions.
When a reaction proceeds in the absence of work, it approaches equilibrium; $\Delta_r G$ will continuously increase until it reaches zero at equilibrium.

Thanks for the answer! I don't know if I'm getting it right. Basically, if you know that the reaction is taking place under non standard conditions, then to predict spontaneity you use ∆G and the equation ∆G = ∆G° + RT ln (Q): if you know Q (i.e., concentration of products and reactants, as you said), then it is easy to trace ∆G and thus spontaneity, since ∆G° is constant. So the reaction will proceed until ∆G cancels. Contextually, ∆G° gives us indications of how far the equilibrium is shifted, because at equilibrium ∆G is zero and ∆G° becomes directly related to Keq. And that's it so far.

Instead, when a reaction occurs under standard conditions, ∆G° can be used both to predict spontaneity, both to predict where equilibrium is shifted, taking advantage of the equation ∆G° = -RT ln Keq. For example, if Keq > 1, then ∆G° < 0 and the reaction is both exoergonic, both right-shifted at equilibrium. So in this case ∆G is not considered at all (?). However, previously we said that ∆G° is a constant: so in this case, of a reaction occurring under standard conditions, ∆G° should vary until it reaches zero, because it is now used to predict even spontaneity. Contradiction?

Starting with standard conditions, as soon as the reaction starts, you are no longer at standard conditions. The Gibbs energy of reaction changes continuously as the reaction proceeds. It is the derivative of the Gibbs energy with respect to the progress of reaction.

But doesn't the standard state, in the context of Gibbs free energy, depend only on pressure (1 atm)? It is sufficient to keep the pressure constant for the reaction to always run in the standard state.

No, the standard state has certain requirements for every reactant and product species: https://goldbook.iupac.org/terms/view/S05925. See also https://chemistry.stackexchange.com/q/114651

Okay, I think I understand. I'll bring back an example:
Consider the following reaction and evaluate its spontaneity:
CaCO3 (s) --> CaO (s) + CO2 (g)
∆G = ∆G° + RT ln Q
If no pCO2 value is reported, then we take it as obvious that we are in standard conditions, so pCO2 = 1 atm and, consequently, Q=1 as well (solids have activity equal to 1). Consequently, ∆G = ∆G° and so ∆G° gives us information about where the reaction proceeds under these standard conditions. At equilibrium ∆G=0 and in this case ∆G° gives us information about where equilibrium is shifted.

Conversely, if we are told that pCO2 has a specific value, then we are not at standard conditions, Q≠1 and ∆G ≠ ∆G° and in this case we can only use ∆G to predict spontaneity. However, ∆G° always allows us to see where the equilibrium is shifted when ∆G =0.

You are correct in supposing that $\Delta_r G$ is what is important. A criterion for a spontaneous reaction is $\Delta_r G< 0$ and for equilibrium is when $\Delta_r G = 0$ so that $\Delta G^\text{o} =-RT\ln(K_e)$. Only here is $Q\equiv K_e$. At equilibrium the change of free energy is zero, it must be so, and the entropy is at a maximum.
Suppose that $\Delta G^\text{o}=0$ then the mixture of reactants and products is such that all have unit activity and are at equilibrium, and then $K_e=1$ as required by $\Delta G^\text{o} =-RT\ln(K_e)$ . If $\Delta G^\text{o} <0$ the mixture is not at equilibrium and this is now produced by reactants converting to products and the reaction proceeding to the right. As $\Delta G^\text{o} <0$ then $K_e >1$ which is just another way of saying that the reaction moves to the right. The larger $K_e$ becomes the further to the right the reaction reaches equilibrium.
If $\Delta G^\text{o} > 0$ then $K_e<1$ and the reaction does not proceed spontaneously and the equilibrium lies on the left.
(Your p.s. Mixing always does increase the entropy which via $-T\Delta S$ can have a big effect on the free energy but additionally reactions may increase or reduce the number of different gaseous species which also has a big effect on entropy changes.)

Thank you for your answer.
But ∆G° is constant for a given reaction, how can it take values >, < or = to zero depending on the situation?
Also, the equation ∆G° = - RT ln Keq already refers to an equilibrium situation because the equation assumes that ∆G = 0. How can you say that if ∆G° > 0 then we are not at equilibrium? We may very well be at equilibrium but have a positive ∆G°. The ∆G° should only give information about how far the equilibrium is shifted to the left or right.... However I also refer you to the comment under Karsten's post.

You seem to have misunderstood, of course, $\Delta G^\text{o}$ is fixed for a particular reaction. I gave examples of different values that cover all cases.  It is  $\Delta_r G$ (the slope of $G$ with extent of reaction) that varies until equilibrium is reached then it is zero.  $\Delta G^\text{o}$ is the standard state value and is fixed by what the molecule is and what conditions define it.


text:
Mysterious violet-purple compound in "Gerber Method" & "Banana oil synthesis"
During the determination of fat in milk (Gerber method), in a milk butyrometer are mixed milk + $\ce{H2SO4}$ 90% w/w + isoamyl alcohol (isopentyl alcohol) and heating.
The acidic mixture separates into a black aqueous acidic phase (due to dehydration of the sugars, proteins, ...) and a colorless organic phase containing the partially esterified fat.
https://m.youtube.com/watch?v=vORSXW_imUw&pp=ygUPZ2VyYmVyIGZhdCBtaWxr
A (sometimes pale, sometimes intense) violet-purple precipitate can be seen at the bottom of the butyrometer.

This is how it looks like in my experiment:

The violet-purple precipitate is also formed if milk is replaced with $\ce{H2O}$. This means that the purple substance is formed by mixing isoamyl alcohol + $\ce{H2SO4}$ alone.
The same purple color is seen in the synthesis of isoamyl acetate (banana oil), mixing isoamyl alcohol + glacial acetic acid + a drop of $\ce{H2SO4}$ 96% w/w and heating.
https://m.youtube.com/watch?v=F9fGv9M_zyk&pp=ygUUYmFuYW5hIG9pbCBzeW50aGVzaXM%3D

I would like to know what this purple compound is.
I suppose it is a compound with a large conjugate system, perhaps due to the dehydration of isoamyl alcohol (formation of amylene) and subsequent partial polymerization.
Any ideas/references?
Ref.1 is the only source where the unusual color of the solution is explicitly indicated (the formation of the purple precipitate, which occurs on heating to more than $\pu{65 ^\circ C}$, is not indicated). Accordingly.

II. To exaggerate the action and to see the effect of temperature, mixtures were made containing:

a. Sulphuric acid: 10 c.c.; Amyl alcohol: 1 c.c.

b. Sulphuric acid: 15 c.c.; Amy1 alcohol: 1.5 c.c.

One series was cooled during mixture and kept at $\pu{14 ^\circ C}$  for fourteen hours, and the other series was not cooled and was left at $\pu{37 ^\circ C}$ for fourteen hours; water was then added to make up to 30 c.c., and the separated layer measured. The results were:

At temperature $\pu{14 ^\circ C}$: 0.295 c.c. H2SO4 + 0.590 c.c amyl alchol.

At temperature $\pu{37 ^\circ C}$: 0.424 c.c. H2SO4 + 0.800 c.c. amyl alchol.

The layers at $\pu{14 ^\circ C}$ were nearly transparent, while at $\pu{37 ^\circ C}$ they were deeply coloured.

III. Ten c.c. sulphuric acid, 11 c.c. of water, and 1 c.c. amyl alcohol (all at $\pu{29 ^\circ C}$) were mixed; no separation occurred in centrifugalizing, and the solution only became light brown. One c.c. amyl alcohol, 10 c.c. sulphuric acid, and 11 c.c. water (all at $\pu{29 ^\circ C}$) were mixed and centrifugalized, a liquid layer equivalent to about 0.2 per cent. fat was obtained, and the liquid became violet.

I hypothesize that perhaps at low temperature $\ce{H2SO4}$ can dehydrate isoamyl alcohol to amylene, but its polymerization is slow enough to allow the formation of low weight (with conjugated systems) soluble oligomers. At high temperatures the polymerization is faster and therefore large insoluble polymers are formed instead. In the synthesis of "banana oil", perhaps only the soluble compound is formed (not every times) since only a drop of $\ce{H2SO4}$ is used, so very little isoamyl alcohol is dehydrated to amylene.
The fact that it is not always visible could be due to the different amount of $\ce{H2SO4}$ used by different people (1 drop VS a few drops).

Reference:

H. Droop Richmond  and  F. R. O'shaughnessy, "Note on a possible source of error in modifications of the Leffmann-beam method for estimation of fat in milk," Analyst 1899, 24, 146-148 (DOI: DOI   https://doi.org/10.1039/AN8992400146).

Also note change to brown on neutralization. Good question.

It is very surprising that if we search Gerber's method and purple/violet precipitate, there are no relevant results.

Indeed. I thought the same thing. By heating to a basic pH, the same purple compound is not formed, even if the acidic pH is subsequently restored (unless if the solution is then heated).

In the cited publication (isoamyl alchol + H2SO4 at 29°C)  they say that the purple compound stratifies on the acidic aqueous phase. Therefore, I suppose that the purple compound is nonpolar.

Color of the gloves must be coincidental. :-P

There's a mistake in your quoting from the paper. The paper says in the "a" scenario, 0.295 cc sulfuric acid and 0.424 cc amyl alcohol are obtained. But your quote says 0.295 cc "a" and 0.424 cc "b" are obtained. While "a" and "b" are in fact depicting two scenarios, not two materials. There is the same mistake in 0.590 and 0.800 cc.


text:
How to make NH3 + H20 favor its products (NH4+ and OH-)
I am putting together a (fake) proposal for a process that would convert ammonia in cow manure into elemental nitrogen. The first step includes the conversion of that ammonia into ammonium so that it can be used by bacteria in a later step. I know that this specific reaction favors the reactants, so I am looking to find the right combination of methods to push equilibrium towards the products. I used le chatelier's principle and noted an excess of H20 already, but I have a hunch that wouldn't be enough to sway the equilibrium as much as I want. I know that reducing pH would also favor the products. My concern is that there would be a reaction with the desired ammonium and any acid (or their conjugate bases) added to the solution.
Would a common strong acid (HCl, H2SO4, etc) would be appropriate for this scenario, or is there another method of removing OH- from the solution that could work?

Be careful not to confuse 0 and O in formulas.

Well, you might want to be a little more precise with what you mean by "favour the products". Extreme dilution will drive the degree of ionisation to almost 100%, and in that sense "favours the products", except if you actually try to count the ions, there will be almost nothing, (i.e. there are no products). Meanwhile, increasing the concentration of ammonia decreases the degree of ionisation, and in that sense "favours the reactants", but up to a point it will increase the actual amount of ions in solution. This second case is probably what actually matters to you.

It does not have to be a strong acid. Acetic acid, citric acid or any soluble carboxylic acid should do the job, and the byproduct is a naturally occurring organic material.


text:
If atomic orbitals are orthogonal, how can one mix then in Molecular Orbital Theory?
If two molecular orbitals are orthogonal, their net overlap is 0. Which means that the mixture of them would result in no observable change in any of the orbitals if they occur. However, this contrasts totally with the Molecular Orbital Theory as well as the Valence Bond Theory, whereby atomic orbitals are often combined.
In this manner, is it not plausible that the orthogonality condition of the wave function hinders the combination of orthogonal wave functions -i.e. atomic orbitals?

Atomic orbitals of one atom are orthogonal, but those of different atoms aren't.

This makes sense to explaining chemical bonds, but not to explaining hybridization in an atom.

In MO theory, you first mix orbitals to fit them to the symmetry of the molecule. That in itself does not explain bonding, but it prepares you for then combining the symmetry-adapted combinations of one set of atoms (e.g. the hydrogen atoms of methane) with the symmetry matched of another set of atoms (e.g. the carbon atom of methane).

Within MO theory, there (strictly speaking) is no hybridisation. For a single atom such a description (without an external potential) doesn't make much sense. And in VB theory there are (strictly speaking) no eigenvalues of the hybrid orbitals. A VB description of a single atom also doesn't make much sense; the theory is not designed for this. You also cannot just mix and match both approaches.

@Karsten thus, in MO Theory, I first mix the present orbitais from the atoms to find if the orbitals can make one MO - with aid of Group Theory - for then populating the MO according to the Aufbau Principle, which in turn enable me to analyse the degree of bonding in the molecule?

@Martin-マーチン and what is the implication of having no eigenvalues for the orbitals? That we cannot analyse quantitatively the bond lengths or angles, since there is no wave functions whatosever or orbitals being precisely described?

There is of course a wave function as the total description is in VB theory the same as in MO theory. Just the parts towards this equivalence can't be interchanged. And conclusions from one theory don't carry over to the other. There are a few questions on the platform. It's not a simple topic, and certainly nothing for the comment section. Here is something to get you started: https://chemistry.stackexchange.com/q/109615/4945

A bond is formed between two atoms, and a pair of AOs from each doesn't share the same  coordinate system. Obviously, because there is a finite distance, the bond lenght, between the (0,0,0) coordinates of each AO.
That means the concept of orthogonality makes no sense for the constituent AOs of a MO. Simply speaking you can freely align the two coordinate systems to give the biggest overlap. Which is what nature does, because that gives the biggest energy gain from the bond.

Thanks to you and the other guys' clarifications, it was clear to me that I was mixing Quantum Mechanics theories with VB theory, which is not applicable.


text:
Does opening the windows increase the rate at which a floor dries?
After my mom mopped the house, she insisted me to open the window doors, so as to allow the floor to dry faster. This was at late evening time, so I told her there was no point as there was no sun to heat the room, because the temperature of the room and the temperature from outside were equal.
However, she argued back saying that the moisture inside would evaporate more, as existing moisture would go outside. A simplified image would be the following:

I quickly caught on she was basically describing the idea of Le Châtelier's principle. I had nothing to rebute with, so I opened the window.
But, I still feel confused since this conversation occurred. Does the principle really holds when there is no energy as form of heat to evaporate the water? This energy would be brought if both parts were not in thermal equilibrium, but in this case, they are.

Consider the relative humidity inside (and how it changes as the floor dries), and outside.

Many students use the Le Chatelier's principle (note the spelling) as a magic wand, applying it on everything.
The principle is also called the Equilibrium law. But evaporation is not an equilibrium state, it is a process. Additionally, at the same temperature, there is not even an equilibrium shift of saturated vapor pressure.
Opening the windows at balanced indoor and outdoor temperature has these major effects:

Horizontal air flow along the wet surface disturbs the air boundary layer of high humidity at the floor, what accelerates vapor transport and therefore evaporation too.
Exchanging indoor and outdoor air masses decreases indoor absolute (g/m3) and relative (%) humidity (increased by evaporation), what accelerate the kinetic rate of evaporation.
As evaporation cools down the surface and air, bringing outdoor air helps decreasing this temperature drop what accelerates evaporation too.

" evaporation is not an equilibrium state". Absolutely critical

When a liquid boils its vapour pressure becomes equal to atmospheric pressure at the boiling temperature and the vapour at one atmosphere is in equilibrium with the liquid and $\Delta G =0$ which means that $0=\Delta H-T\Delta S$. The entropy change is  $\Delta S=\Delta H/T_B$. For water this is $\approx 109$ J/mol/K as $\Delta H = 40.7$ kJ/mol. (btw. $\Delta S$ is larger than  many other liquids due to hydrogen bonding)
To spontaneously evaporate we suppose that the free energy must be slightly negative  then
$$\displaystyle\Delta G =\Delta H-T\Delta S\\ \text{small neg} =  +40700 - 300\Delta S$$
which produces a value $\Delta S \approx 136$ J/mol/K which means that for the process
$$\displaystyle \ce{H2O}\text{ (liq, 1 atm 300K)}\to \ce{H2O}\text{ (vapr in air 1atm, 300 K)}$$
the entropy change must be larger than $136$ J/mol/K. This value can be achieved but only if a sufficiently dilute vapour is produced, ($\Delta S=nR\ln(V_2/V_1),\; V_2\gg V_1$) by opening the windows and letting dry air into and moist air out of the room.
( Evaporation cools the liquid, so damp clothes can freeze on a washing line in winter. In very dry and warm climates evaporation could be used to produce ice)

Hm, using TD to compare two rates is not a typical case.


text:
What is the significance of semipermeable membrane in osmosis?
My teacher said that semipermeable membrane allows passage of only selected particles and then said that it is required for osmosis but I cannot totally understand the main function of semipermeable membrane in osmosis.

When in doubts about statements of lectures or textbooks, always try first to verify the facts in independent available offline or online sources, before you decide to ask. CH SE should not be the 1st choice in looking for the authoritative answer.

If the membrane was not selective, it would allow passage of nothing or of all. Is any of these two cases osmosis?

There is some detail in this answer https://chemistry.stackexchange.com/questions/167189/osmotic-pressure-definition-clarification/167286#167286

@Poutnik I am preparing for an exam and I also have to deal with 3 other subjects also so I don't have much time for verification I know you are correct but please understand my situation also.

You do not have time to verify but you have time to ask. But the former is faster.

A semipermeable membrane is a membrane containing tiny holes. The dimension of the holes is just sufficient to let small molecules like water molecules ($\ce{H2O}$) go through the membrane. Big molecules like sugars or proteins are not small enough to go through the membrane. So if a semi-permeable membrane separates pure water on one side and a sugar solution on the other side, water goes through the holes and sugar does not. In the long range, more water is going through the membrane in the direction water -> solution than in the inverse movement. As a consequence, the aqueous solution gets slowly more and more diluted. This is osmosis.

Thanks sir for such a clear explanation but can you tell me one more thing that, why                    π is directly the to molar concentration of solute (π = osmotic pressure) ?


text:
Band gap voltammetry sensors
I'm new in electrochemistry and I am leading with modification of sensors and analytes. I would like to know why the analyte voltammetric signal should be between the band gap of the modifier. If I run a DPV passing through the band gap limit do I lose my electrode? or modification?

Half of success of obtaining useful answers is providing sufficient relevant context along the questions.


text:
Are configuration interaction (CI) and configuration interaction method (CI-method) the same thing?
I am trying to study molecular quantum mechanics and I am quite confused because the material that I'm reading has a separate section called "Configuration interaction" and another separate section called "Configuration interaction calculations".
In the CI section, it is explained that in $H_2$-molecule, if the basis set consists of 1s-orbitals of the hydrogen atoms ($\phi_A$, $\phi_B$), the two electrons can occupy the molecular orbitals (MOs) according to four different configurations:
$1\sigma_g(1) 1\sigma_g(2)$
$2\sigma_u(1) 2\sigma_u(2)$
$1\sigma_g(1) 2\sigma_u(2)$
$2\sigma_u(1) 1\sigma_g(2)$
The last two configurations can form antisymmetric and symmetric linear combinations:
$\frac{1}{\sqrt2}(1\sigma_g(1) 2\sigma_u(2) - 2\sigma_u(1) 1\sigma_g(2))$
$\frac{1}{\sqrt2}(1\sigma_g(1) 2\sigma_u(2) + 2\sigma_u(1) 1\sigma_g(2))$
The four possible total wave functions are the first two configurations multiplied by spin wave function and the linear combinations multiplied by spin wave functions. The term symbols associated with these wave functions are  $^1\Sigma_g$, $^1\Sigma_g$, $^3\Sigma_u$, $^1\Sigma_u$, respectively. As you can see, two wave functions have the same symmetry, and as the internuclear distance $R \rightarrow \infty$, the states of the same symmetry converge to the same dissociation limit. This means that the energies of the states of the same symmetry never cross at any $R$ due to the non-vanishing matrix element between the states in the Hamiltonian. Due to the off-diagonal matrix elementrs, configurations of the same symmetry mix. This is called the Configuration interaction.
Now, the other section is the Configuration Interaction calculations, where it is explained that CI-method is a method to approximately solve the electronic SE. In CI-method, the exact wave function for any electronic state can be expressed as a linear combination of all possible n-electron Slater determinants arising from a complete set of spin orbitals. This is called the CI wave function. In practice, e.g. CISD is used, where only singly and doubly excited determinants are used in the calculation.
I kind of understand both of the things explained above, but I am very confused since I'm not sure if the two things explained above are the same thing? Or are they completely different topics but have the same name? Can someone explain, are these the same thing? Or is it that the CI-method just takes the Configuration interaction into account?

CI is within its limitations an exact theory. CISD is truncated and will have shortcomings. The latter is an approximation to the former. So the method and the theory are definitely not the same thing.


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Relative Concentrations of Reactants and Products at Equilibrium
I cannot understand what makes a reaction at equilibrium lie far to the right (large K) or far to the left (small K). What exactly makes the product or the reactant concertation larger than the other when the reaction producing the products (forward reaction) and the reaction producing the reactants (reverse reaction) have the same rate, and rate is change in concentration over time.

When I was reading to find an answer, I came across the definition of the equilibrium constant as a ratio of the rate constants of the forward reaction to the rate constant of the reverse reaction. Does that have some effect on determining the concentration which reaction species will be greater at equilibrium, given that at the end of the day, the rates of the reactions are the same. If the rate constant does have an effect, then what is the factor determining how large the rate constant at a given temperature is.

I just can't wrap my head around it, if the rates of the reactions are the same, (and obviously the initial concentrations of the reaction species have no effect on K), then there must be another factor affecting the concentrations of the reactants at equilibrium, but what is it?

The point is, the rates of the forward and backward reactions are equal at and only at the reaction equilibrium. If for $\ce{A <=> B}$ and $[\ce{A}] = [\ce{B}]$ isf the forward rate  (much) bigger than the backward rate, then K>1 ( K >>1 ) and vice versa. $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k_f[\mathrm{A}] - k_b[\ce{B}]$$ and similarly for $\ce{A}$, assuming the reaction is elementary.


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How to choose the solvent gradient for the flash column chromatography of a complex mixture?
When performing a flash column chromatography, it is advised to select an eluent in which the desired compound has an $R_\mathrm{F}$ of about $0.25$ in TLC.
However, if I want to separate and isolate all the compounds present in the mixture, or I do not know which is the desired product, it is advised to use an eluent gradient for the column. But what eluent should be selected to start with? Maybe an eluent in which the compound with the highest $R_\mathrm{F}$ (least polar in direct phase chromatography) has $R_\mathrm{F} = 0.25$? What should I consider when performing such chromatography?
I do a gradient from hexane to ethyl acetate until everything is separated. Is there a more efficient method that saves time and solvents?

That would be good start since you don't know which product is the desired one. To save time and chemicals, I'd buy an authentic sample and follow through.

I would do a complete gradient from 100% A to 100% B where A is something non-polar (hexane maybe) and B is something polar (e.g. EtOAc) to ensure I separated everything, and then a final wash with 10% MeOH in EtOAc


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Can cyclopropane be synthesized from propan-1-ol?
I know the dehydration of propan-1-ol derives in prop-2-ene by removal of the hydroxyl group and a hydrogen atom from the second carbon, but could the hydrogen come from the ones bonded to the third one?

And why would it?

At first look at how cyclopropane can be produced from a haloalkane

This is the creation of cyclopropane by Zn metal reduction of terminal vicinal dihalide. Actually cyclopropane is a highly strained molecule and it's production is quite unfavorable, it is anyways produced in this reaction due to the presence of Cl at both ends.
When Zn is introduced to 1,3-dichloropropane, it abstracts the two Cl atoms and produce two $\ce {Cl^-}$ ions. This means there is a lack of electrons in the propane chain as each Cl atom takes away an extra electron. Due to this there can't be unsaturation in propane to make propene and thus the reaction proceeds toward cyclization to make cyclopropane.
But cyclopropane cannot be made from   propane-1-ol. See this reaction:

This reaction occurs in the presence of a protic acid e.g. $\ce {H2SO4}$ or $\ce {H3PO4}$. The $\ce{H^+}$ from these protic acids attacks oxygen of propane-1-ol to leave behind a carbocation and elimination of $\ce{ H2O}$. The carbocation, which is quite unstable  and it will get to be stabilize by picking up electrons from the nearby $\ce {C-H}$ bond. This creates a double bond between the two C atoms and elimination of $\ce H^+$. So the formation of carbocation is what makes the conversion of propane-1-ol to propene and rules out the formation of cyclopropane.


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How does stirring demonstrate a diffusion-controlled reaction?
I was reading about diffusion-controlled reactions, and on the Wikipedia page (Diffusion-controlled_reaction) it says the following:
"One classical test for diffusion control is to observe whether the rate of reaction is affected by stirring or agitation; if so then the reaction is almost certainly diffusion controlled under those conditions."
How does stirring affect the rate of reaction in this scenario?
I was taught that if a reaction was homogeneous in nature, then stirring should not affect the rate of reaction. I would have thought that if a reaction was homogenous and occurring in a viscous medium, that the rate would not change therefore. So I do not see how stirring a viscous system increases the rate of reaction if it is already homogenous.
UPDATE: Thank you for all the comments so far!
I have done some reading, and I see that stirring can affect the effective diffusion coefficient. This has been demonstrated with the Belousov-Zhabotinsky reaction. According to the paper that I was reading, the effective diffusion coefficient is the sum of the turbulent diffusion coefficient and the molecular diffusion coefficient, and stirring affects the turbulent diffusion coefficient. But what does this actually mean?
Does stirring lead to more collisions and therefore increase in the rate of reaction, or is it to do with changing the shear rate which then influences the observable rate of reaction?

I think that I agree with you, I cannot see how stirring would affect a normal (i.e. homogeneous) solution phase reaction that mostly occurs on first contact of reacting species, i.e. one whose activation energy is close to $k_BT$. Diffusion controlled rate constants can be $\approx 10^{10},\mathrm{dm^3,/mol/s}$ in mobile solvents such as hexane.

Hmm, what is homogenous at macroscale, is not homogenous at nanometer scale.

Stirring in a heterogeneous reaction lowers the width of the diffusion layer, making the reaction faster. I updated the wikipedia article to make the statement less general.

@Poutnik, re homogeneous, in chemical kinetics we consider that the discrete nature of matter is unimportant since kinetics is an ensemble measurement (unlike those on single molecules). The theory by Smoluchowki  and Einstein that leads to ther eqn. for diffusion controlled rate constants supposes that the solvent is a structureless fluid in which the reactants diffuse.

@Porphyrin Hmm, but that may lead to deviations from theory. ( I still remember the lecture of the Smoluchowski theory of fast coagulation in 1987.  funny stuff )

@Poutnik, Just to confound my comment above I recalled this paper, G. Beddard, T. Doust & J.Hudales, Nature (1981), 294(5837), 145-6. which shows that rotational diffusion rate constants are different in some solutions of the same viscosity but different composition, showing that Stokes-Einstein eqns. are not always correct.

What is meant by homogeneous? Consider a monomer and catalyst system such as ethyl 2-cyanoacrylate and water in the ration 1:10-6. In effect, though water molecules are evenly distributed, most cyanoacrylate molecules are not in contact with water. If only a tiny fraction of the mixture is water, even if it was evenly distributed at the start, as soon as a molecule of water initiates polymerization, it is effectively walled off from the mixture by the polymer, so the mixture is no longer homogeneous. Stirring could then be effective.
A reaction might proceed in steps, so that at each reaction site, a buildup of intermediate product limits the contact with rest of media, which would no longer be homogeneous.
Another example is the initially homogeneous mixture in the Belousov-Zhabotinskii reaction, where stirring does affect the reaction.

In biochemistry, a diffusion-limited reaction is assumed when the (enzyme-catalysed) reaction is extremely fast. There are some reactions that are even faster than the theoretical diffusion limit. In these cases, a long-range electrostatic interaction is invoked to explain why (charged) reactants are entering the enzyme active site faster than theoretically expected.
In general, a diffusion-limited reaction should have an activation energy of zero. You can test for that by measuring the temperature-dependence of the reaction rate (while considering that diffusion gets a bit faster at higher temperature, but reactions with a non-zero activation energy get a lot faster because the proportion of productive collisions increases exponentially). For an enzyme operating at diffusion-limit, the expectation is that every single binding event leads to formation of product, so an increase in temperature will not increase the proportion of productive collisions (it is already 100%).


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Understanding the definition of enthalpy
My chemistry text describes Internal Energy (U) of a substance as the sum of all categories of energy, i.e., Kinetic (temperature), potential, bonding, etc.
Then, Enthalpy is defined as:
$$H=U+PV$$
But, it seems to me that $PV$ (pressure x volume) is a form of potential energy, even though $U$ presumably already includes potential energy. Why does it look as though potential energy is being included twice?

When in doubts about statements or explanations of lectures or textbooks, always try first to verify the facts in independent available offline or online sources, before you decide to ask.
// This way you prevent creation of redundant info  if existing answer or info could be easily found.

Funny fact: Gas expansion work is not release of gas potential energy,but it is conversion of kinetic energy, at least for monoatomic gases. (Or of mostly kinetic energy for other gases.)

In my judgment, trying to assign a physical interpretation to enthalpy is a waste of your valuable time.  Enthalpy is just a convenient shorthand parameter of U and PV to work with in solving many kinds of thermodynamics problems.

@ChetMiller You could have linked to your earlier answer, instead of repeating yourself. I don't see any problem with interpretation. U is a property of an object itself and PV is a correction for pressure, ambient, or otherwise, that adds up to U.

@Mithoron I didn't know how to look up my previous answer.

Umm, it was linked right there, in related / https://chemistry.stackexchange.com/questions/49701/enthalpy-change-vs-change-in-internal-energy-in-a-system now is linked thrice ;)

But, it seems to me that PV
(pressure x volume) is a form of potential energy, even though U
presumably already includes potential energy. Why does it look as though potential energy is being included twice?

PV is the work you have to do on the surrounding if you start with a volume of zero and expand it reversibly to the volume of the system V. The reason we add it is for convenience, and once we add this term, the value no longer reflects the inner energy, but something new called enthalpy. While the meaning of enthalpy is less clear than that of internal energy, it turns out to be useful nonetheless.
Specifically, the change in enthalpy $\Delta$H happens to be equal to the heat exchange under certain conditions (same temperature and pressure before and after, no work except for expansion work p $\Delta$V).


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Protecting Thiol group in presence of alcohol
What protecting group can I use that will protect the thiol in the presence of an alcohol? Is there a set of conditions, or a group that I can use that is selective to the thiol over the alcohol?
I am converting an acid halide into an ester whose R group's terminal is a thiol. I will be using 3-Mercapto-1-propanol. I want to make sure that the acid only reacts with the alcohol side, so I can use the thiol group in the next step.
EDIT: I added the general reaction scheme for reference

How about oxidising it to the disulfide then running the (di)ester formation then reducing back to the thiol?  The disulfide is commercially available https://www.alfa-chemistry.com/cas_177854-95-4.htm

2-Hydroxyethyldisulfide is commercially available. You can use it as substrate for esterification and then reduce the resulting ester to thiols as Waylander suggested.


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Confusion about how to interpret magnitude of time constant in a paper
When looking at this paragraph in Midorikawa's work for a fluorescence review I was confused, specifically around a time constant of ~4s being compared to ~300 ms.

It is not explicitly mentioned so I assume the 300 ms is a time constant correct?
How is the conclusion drawn that it is the step after tethering, priming, by the discrepancies in time constant? I am not hoping any readers know why biologically it is this way but rather how I can understand a 13x larger time constant can lead to conclusions that it is not rate-limiting?
(emphasis mine)

By comparing the time course of fusion events and tethering events, it was suggested that tethering of new vesicles could refill the empty release sites with a time constant of ∼4 s. This is also surprising since the replenishment of the readily releasable vesicles (RRVs) at the calyx of Held terminal was estimated to be ∼300 ms by the electrophysiological experiments (Neher and Sakaba, 2008; Hallermann and Silver, 2013). Given that a depolarization elicit fusions of only a small fraction of already-tethered vesicles in TIRFM imaging, the discrepancy suggests that the replenishment time course was determined by the step after the tethering, i.e. priming.

Reference
Midorikawa, M. (2018) Real-time imaging of synaptic vesicle exocytosis by total internal reflection fluorescence (TIRF) microscopy. Neuroscience Research, Volume 136, (2018), pp. 1-5, ISSN 0168-0102, https://doi.org/10.1016/j.neures.2018.01.008. (https://www.sciencedirect.com/science/article/pii/S0168010217307551)
Thank you for the help!

@Maurice I don't think you can talk for everyone on the site. Still, the question could be borderline off-topic... and somehow feels homeworky. Some links to provide context wouldn't hurt, not to mention title is pretty impressively unclear.

I linked the paper. I am writing a review of this total internal reflectance fluorescence microscopy paper which falls squarely into chemical biology and I was confused by the wording around results involving a time constant.  Hope that provides enough context

(1) yes. (2) The extract suggest that they are in fact measuring different events as a way of explaining the discrepancy. Only you can know this  as I'm not familiar with how good these techniques are. I presume time constant is the reciprocal of rate constant and not half-life.


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Boiling of liquids and their vapour pressure
Hello I was wondering about the boiling of liquids, I will use water as an example.
Liquids boil when their vapor pressure function p(T) exceeds the ambient pressure, for example water boils at 100 °C at 1 atm atmospheric pressure because at this pressure, its vapor pressure function p(T) has the value of 1 atm.
However, how is this atmospheric pressure defined? Is it just the gas pressure without the water vapor?
Suppose the water is heated up slowly until 100°C. It will form its own vapor, alongside the ambient pressure of 1 atm. So at the boiling point, does the water need to overcome the 1 atm ambient pressure + its own vapor pressure in the surrounding atmosphere to boil (would be over 1 atm tho) or just the pressure of the other gases?

It need not to be atmospheric pressure. Just pressure. It is hydrostatic pressure in the bulk liquid versus pressure of vapor. If the former is stronger, any microscopic bubble of vapor is smashed back to be liquid. If the latter is stronger, vapor pushes liquid away and bubbles grow.

Maybe think if you're using pressure cooker in this, or not?

Vapor is a gas like all others. Higher partial vapor pressure in air means lower partia pressure of permanent gases, maintaining locally the same total air pressure, unless balancing pressure is restricted.

The water vapor partial pressure in the gas phase plus plus the partial pressures of oxygen, nitrogen, and trace species adds up to atmospheric pressure.

So it doesn’t matter if it is its own vapor or other gases, just sum it up and you get the pressure that needs to be overcome?

Yes, that’s right.

@Mäßige What made you think otherwise? A flying water molecule hits  water surface by the same way as any other gas molecule, participating on the total pressure.

I was just thinking that maybe because it is water vapor to the corresponding liquid phase, it somehow contributes to boiling point but that was wrong


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Are there examples of publications using DFT instead of HF as a coupled-cluster reference?
I have been searching for a while, and only managed to find an article by M. Saitow,[1] where CAM-B3LYP orbitals are used. I was wondering if this has been done elsewhere.
It was recommended to me by a colleague, and it is an easy enough thing in practice, but I wanted to find some stronger motivation in order to be able to trust the results.
References:

Saitow, M.; Becker, U.; Riplinger, C.; Valeev, E. F.; Neese, F. A new near-linear scaling, efficient and accurate, open-shell domain-based local pair natural orbital coupled cluster singles and doubles theory. J. Chem. Phys. 2017, 146 (16), 164105. DOI: 10.1063/1.4981521.

I remember that Head-Gordon used DFT orbitals for MP2 and MP3 and it looks like it's been used for CCSD(T) vibrations


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Drying Li2CO3 without vaccum
I am looking for drying Li2CO3 without a vacuum oven, simply because we do not have one.
I did not find anything in the literature, or I am overthinking it? Just 100 °C overnight in a normal oven, and afterward simply store it in a vacuum desiccator.

What about sufficiently strong desiccants?  $\ce{P4O10}$ or $\ce{Mg(ClO4)2}$. Even weaker should do. $\ce{CaCl2}$ ? $\ce{KOH}$ or $\ce{NaOH}$ ?

I would be careful with P4O10 as a Lux-Flood acid-base reaction could occur. The others look good.


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Is cleaning silverware with tinfoil and salt dangerous?
Premise: I'm not a chemist and I've already had a look at this and this but they don't fully answer my question.
I have read that you can clean silverware using tinfoil, salt and water, though from what I understand, $\ce{H2S}$ is (or could be?) freed. Again, from what I understand, this a dangerous compound to breath. Now, is it still safe to clean silverware using this method? Is the amount of $\ce{H2S}$ freed dangerous?
I don't want to come across a "bleach + vinegar situation", which is why I'm asking here.

Many things are unsafe if you do them incorrectly, unfortunately. If you perform the reaction in a well ventilated place you reduce the risk effectively to near zero. The answer presumably  also depends on the amount of silverware and the state of the silverware.

I've used this method several times. You'll need to make sure your silver is in very good contact with the tinfoil so wrap it quite tightly. I do not recall ever smelling H2S from the system

The amount of sulfur in the tarnish is microscopic. 2. Much of it reacts with the Al, and is not released as a gas. Consider that there is cyanide bound in almonds and apple seeds. Eating them in reasonable amounts is usually not fatal. There have been exceptions.


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Why causes the curvature of the line in pressure-composition graph?
I have been studying about liquid solutions and came across these graphs.

From the first graph it is evident that the solution considered has been assumed ideal as there is no deviation from ideal behavior and the graph is straight. However, I wonder what causes the curvature in the second graph. Both Raoult's law and Dalton's law describe the proportionality between the mole fraction and the vapour pressure. So, why does the graph deviate from its proportionality in graph 2 despite the ideal nature of the vapours?

So, why does the graph deviate from its proportionality in graph 2 despite the ideal nature of the vapours?

It does not deviate, both curves are the result of the ideal behavior. Lets do some math.
For an ideal binary mixture where: (1) the gas phase behaves as an ideal gas, and (2) the liquid phase behaves as an ideal solution; the LV equilibrium reduces to Raoult's law
$$ y_j P = x_j P_j^\pu{sat} \tag{1} $$
where $P_j^\pu{sat}$ is the saturation pressure of chemical species $j$ at a temperature $T$.
Bubble Curve Summing twice Eq. (1) for both components yields
\begin{align}
  &y_1 P + y_2 P = x_1 P_1^\pu{sat} + x_2 P_2^\pu{sat} \\
  &(y_1 + y_2) P = x_1 P_1^\pu{sat} + (1 - x_1) P_2^\pu{sat} \\
  &\boxed{P(x_1, T) = [P_1^\pu{sat}(T) - P_2^\pu{sat}(T)]x_1 + P_2^\pu{sat}(T)} \tag{2}
\end{align}
Thus, at a fixed temperature, the bubble curve is a linear function of the liquid composition $x_1$. This is the scenario in the left-hand side.
Dew Curve First we put Eq. (1) in a more suitable form to eliminate $x_j$
$$ x_j = \frac{y_jP}{P_j^\pu{sat}} \tag{3} $$
Summing twice Eq. (3) for both components yields
\begin{align}
  &x_1 + x_2 = \frac{y_1P}{P_1^\pu{sat}} + \frac{y_2P}{P_2^\pu{sat}} \\
  &1 = P\left(\frac{y_1}{P_1^\pu{sat}} + \frac{1 - y_1}{P_2^\pu{sat}}\right) \\
  &1 = P\left(\frac{y_1P_2^\pu{sat} + (1 - y_1)P_1^\pu{sat}}
       {P_1^\pu{sat}P_2^\pu{sat}}\right) \\
  &\boxed{P(y_1,T) = \frac{P_1^\pu{sat}(T)P_2^\pu{sat}(T)}{y_1[P_2^\pu{sat}(T) -
  P_1^\pu{sat}(T)] + P_1^\pu{sat}(T)}} \tag{4}
\end{align}
Thus, at a fixed temperature, Eq. (3) has a mathematical equation in the form of $a/(bx + c)$. This is the scenario in the right-hand side.
Therefore in LV equilibrium, as indicated by Raoult's law, the two-phase region is confined between an upper straight line and a lower hyperbolic curve for the $Pxy$ diagram.

Thank You for your explanation. It is very clear to me. I have one more question, it might sound silly, but why do we converge the two graphs to show that there exists equilibrium between the straight line and the hyperbolic curve? How is it possible to converge both graphs if the x-axis represents completely different things? In one graph it represents the mole fraction component B in the solution and in the other it represents mole fraction of it in the vapor phase. xB and yB aren't equal.


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Why isn't my sodium acetate demo working?
So I'm trying to do the sodium acetate tower demonstration that I've seen a ton of videos on.  I don't have sodium acetate and I am making it from a liter of 5% distilled white vinegar and pure baking soda.  The vinegar is expired according to the package (though I'm not sure if this makes any difference). I've seen a ton of videos on this and it is not going as smoothly as I expected.
Although the vinegar is past the "best if used by" date it is still reacting with the baking soda.  Once I hit the point where it's no longer reacting with additional baking soda I have around a liter of volume.  One website recommended to slowly heat it up until it reaches about 100 - 150 mL.  They recommended not boiling it with high heat as it would appear burnt.  So i gently heated it up without boiling.  Once it got down to around 600 ml the color started becoming noticeably yellow.  Once it got down to around 400 ml the color become more of a brownish tea looking color.  When it got to under 300 ml the crystals started rapidly forming.  I then attempted to pour out the liquid into a new container so it could cool and I tried to filter out the existing crystals with a paper towel.  I got some brown liquid but quickly the entire mixture turned into a crystally sludge.
Am I doing something wrong or is this just trial and error and I need to find the exact volume at which the crystals just barely start to form and then stop heating?  I know what's supposed to happen but it doesn't seem to be doing that for me.  Is there a reason it's yellow/brown and not clear and does that make a difference?
Thanks for any advice you can give me!

It looks like vinegar contains some decomposing organic matter, aside of acetic acid, like sugar undergoing caramelization.  Try spontaneous evaporation.

Please give us numerical information about the quantities of soda and of vineggar used, and about the concentration of vineggar.

Maurice, it was 1 liter of 5% vinegar.  The baking soda I was less exact about and just put in about 4 tablespoons full and then went a lot slower with adding until the fizzing finally stopped.  I would say it was about 6 tablespoons of baking soda.  I bought some new distilled white vinegar 5% acidity and I'm going to calculate exactly how much baking soda should react with it and try the whole thing again today

Went a little differently with fresh vinegar and with more exact number for amount of baking soda.  The previous batch with expired vinegar quickly turned yellow and then brown.  This one started to have yellow tint by time previous one turned brown.  I'm down to solution being about 200 mL (started off with a little less than 1000 mL) and  little flakes started forming around 400 mL, i reduced the heat and it kinda lessened.  I'm gonna try to get it close to 150 mL and see if i can see a more hardened crystal forming on the top and then turn off heat, pour in a little vinegar and give it a go


text:
Probability distribution for the momentum of a particle undergoing a collision?
Background
I haven't seen this mauver done before but let's say, I have a thermal gas and I have a molecule with momentum $\vec p$. It undergoes a collision and now has momentum $\vec P$.
Now, due to molecular chaos it should be possible to get the probability distribution function $P$ given $\vec p$ what is the chance it will become $\vec P$ without explicit consideration of the oncoming particle.
My guess
We consider the ansatz:
$$ P(\vec p, \vec P) = A e^{- \beta |\vec P \cdot \vec p|/m}$$
where A is a normalization constant, $m$ is the mass and $\beta = \frac{1}{k_B T}$ is the standard inverse temperature. Let us assume each molecule undergoes a collision then using the Maxwell distribution my new distribution is:
$$ \tilde P = A' \int_{- \infty}^\infty \int_{- \infty}^\infty \int_{- \infty}^\infty e^{- \beta |\vec P \cdot \vec p|/m} e^{- \beta \vec p^2 /2m} d^3 p \propto e^{-\vec P^2/2m} $$
which is as expected a gaussian/Maxwellian gaussian. ($A'$ is another normalization constant).
Thus, even after collisions the distribution remains the same as it would be worrisome that a different distribution was produced due to collisions.
Question
Is this the correct distribution for the momentum of a particle undergoing a collision?

Cross-posted: https://physics.stackexchange.com/questions/761400/probability-distribution-for-the-momentum-of-a-particle-undergoing-a-collision

Can there be a need of the 3rd integral, addressing distribution of momentum after collision of 2 particles with given m and vector p?

@Poutnik Its not clear to me why there would be a third integral? Or what it would represent? Btw due to conservation of momentum if the change in one particle is $\Delta p $ the other one will be $- \Delta p$

The $\Delta \vec p$ can have many possible magnitudes and directions, depending on collision geometry, even for the same initial $\vec p_1$ and  $\vec p_2$ .

I revoke the 3rd integral note, it was confusion due the way how my phone displayed it. But the note about momentum dispersion stays. For 2 molecules, it is determined also by distribution of collision geometries.


text:
What's the formal definition of a ceramic in chemistry?
Dictionary.com defines a ceramic as:

noun (used with a singular verb) the art or technology of making
objects of clay and similar materials treated by firing. (used with a
plural verb) articles of earthenware, porcelain, etc.

This definition seems very narrow. There are numerous types of materials including aluminum oxide, uranium oxide and a variety of super conductors that are also classified as ceramics by chemists. What is the formal definition of a ceramic in the field of chemistry? Would teeth, for instance, fit under that definition of a ceramic?

Teeth are usually not put in ovens nor furnaces. Also, it may be a composite structure with proteins like mollusk shells.

I don't think the dictionary definition is very useful here. Chemists would mostly describe teeth as a ceramic and might even be inspired by nature to help design better ceramics. The dictionary is biased towards how most ceramics are made not what they are.

The enamel on the outside would be a ceramic material -- but inside teeth is organic matter.

To answer the question;
"Do teeth fit the definition of a ceramic material?" Let's look at the question again, it speaks of "a" material. Since teeth are mosty made up of a bone-like material called 'Keratine' (if i'm correct, hair and nails are as well.) And only the outer, protective layer is made of a ceramic-like material. I would say that they most likely do NOT fit that definition. ...but i could be wrong.

As they speak about thermally processed manmade material, I do not see how teeth could fit.

A keramics is an opaque substance made from different amounts of different oxides in non stoichiometric amounts and homogenized by heating to a high temperature. The temperature must be high enough to produce finally a compact solid, even if no melting points are really achieved.
The definition of keramics has recently been extended and includes mixtures containing organic substances or polymers that do not require a high temperature for the compact solid to be created.


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Reduction of static in lyophilized materials that have been powdered
I have recently been working with lyophilized products mostly food based items and sometimes medicinal plants (not cannabis) but I keep running into an issue when I grind the items into a powder the powders develop a static charge is there a way to reduce this?
Should I manually grind them using a mortar and pestle instead of an electric grinder. Or is there another method that does not involve adding moisture back to the powders as they need to stay as dry as possible.

Adding a sprinkle of graphite. :-)  // Perhaps optimal moisture is nonzero moisture.

I thing you should grind your row material while they are wet and then lyophilize it.


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What would the adiabatic and isothermal curves of a system look like for the same initial and final pressure and volume?
Usually, when I see a graph of reversible adiabatic and isothermal processes it's something like this.
Which makes sense mathematically since
$yx^{\gamma} = k_1$ and $yx = k_2$ should intersect only once.
However, let's say if we have a system at $(P_1, V_1)$, then surely we can take it to $(P_2, V_2)$ adiabatically let's say in a thermoflask. We should also be able to take it from $(P_1, V_1)$ to $(P_2, V_2)$ isothermally.
So, my questions are:

How does that work mathematically? If the two graphs never intersect how can they reach the same position?
How would the $P-V$ graph look?
Does this have something to do with the temperature acting as a third dimension?

To get from the same initial state to the same final state, one of the process paths cannot be just isothermal or just adiabatic. E.g. for the curves on the chart,both curves may rejoin by isochoric heating or cooling.

The state doesn't need to be the same. Only the pressure and volume need to be the same. Temperature can vary, well it can vary for adiabatic not for isothermal of course. Would it be possible then that it happens in one single process?

If P and V are the same, then T is the same. Your very basic premise is false: you can't get from P1,V1 to P2,V2 adiabatically. Instead, you get to P2 and some V that you don't get to choose. The word "adiabatic" sets the direction of the process, much like a highway. So does "isothermal", and that's a different highway. Interstate 10 from Houston will take you to San Antonio and El Paso. Can you get to Dallas? No. Or rather, yes, but not by using highway 10. Same thing here.

Who says you can take it back from P2,V2 to P1,V1 isothermally?

However, let's say if we have a system at $(P_1,V_1)$, then surely we can take it to $(P_2,V_2)$ adiabatically let's say in a thermoflask. We should also be able to take it from $(P_1,V_1)$ to $(P_2,V_2)$ isothermally.

The first sentence is correct. However, the second sentence violates the 2nd law! The latter is also in contradiction with your previous statement:

mathematically since $yx^γ=k_1$ and $yx=k_2$ should intersect only once.

This is more than mathematics, it has to do with a principal law of thermodynamics. Lets suppose that you are able to do it, and find that it contradicts the law. I illustrate the situation in the following figure, where the isothermal evolution crosses with the adiabatic evolution twice, and the adiabatic coefficient is $\gamma = 1.4$ (note that the isothermal evolution suffers an unrealistic twist near State 2, but this is because I am trying to nevertheless unite the paths):

Adiabatic evolution: in absence of heat transfer between the system and the surroundings, and given that due to the expansion the temperature is decreasing, we have
$$ Q_{12} = 0 \hspace{1 cm} W_{12} = \Delta U < 0 \tag{1,2} $$
Isothermal evolution: if we assume an ideal gas behavior, then the change of internal energy is zero since $U=U(T)$. The gas is being compressed, so there is work done by the surroundings to the system, and the same amount of heat transfer from the system to the surroundings
$$ W_{23} = -\int_{V_2}^{V_1} \; P \; dV > 0 \hspace{1 cm} Q_{23} = -W_{23} < 0 \tag{3,4} $$

By the $PV$ diagram, the work is the area below the curve of the particular evolution. If we sum them we clearly have that $W_{12} + W_{23} > 0$. However, there is only one heat exchange, and it happens in the isothermal process with $Q_{23} < 0$.
Conclusion: this is a cyclic device, that is capable of converting the heat rejected by the system completely into work done by the surroundings. This the inverse of the Kelvin-Planck's statement of the 2nd law. The process is impossible.

How would the P−V graph look? Does this have something to do with the temperature acting as a third dimension?

We have demonstrated that it impossible to go back by an isothermal path. We need something more, and there are many ways to accomplish this. I will use the proposal of Poutnik: (1) adiabatic cooling, (2) isochoric heating, (3) isothermal compression. We illustrate as follows:

Now the isothermal evolution behaves correctly, look how it cannot collide with the adiabatic evolution.
The work is the same as before, since an isochoric evolution has no work transfer. However, now we are absorbing heat from the surroundings in the isochoric evolution, and thereby we are not violating the 2nd law.

I have another question kind of related to this which is also quite silly. So, if we go from one state to another adiabatically while pressure is constant then we know $TV^{\gamma - 1} = const.$ but also since pressure is constant then $\frac{V}{T} = const.$ Am I missing something here? Surely they can't be related by both of those equations?

@dotmashrc Hello dotmashrc. Are you asking if you can go from state $1$ to state $2$ adiabatically and at constant pressure, at the same time?

Yeah pretty much. Or if a process is isobaric can it also be adiabatic?

@dotmashrc As a rule, when you demand two constraints in ideal gases regarding this type of processes, you don't "move". If it is adiabatically reversible $P_2^{(1 - \gamma )/\gamma}T_2 = P_1^{(1 - \gamma )/\gamma}T_1$, but if $P_2=P_1$, then $T_2 = T_1$. Also, this means that $V_2=V_1$. Thereby if you ask for those two things, there is no evolution.