text:
False positive in lucas test for primary alcohol due to formation of tertiary carbocation by rearrangement?
From Wikipedia:

The Lucas test in alcohols is a test to differentiate between primary, secondary, and tertiary alcohols. It is based on the differences in reactivity of the three classes of alcohols with hydrogen halides via an $\mathrm{S_N1}$ reaction:
$$\ce{ROH + HCl -> RCl + H2O}$$
The differing reactivity reflects the differing ease of formation of the corresponding carbocations. Tertiary carbocations are far more stable than secondary carbocations, and primary carbocations are the least stable.

So since alkyl halides are insoluble, they give a turbid solution upon formation, which easily happens for tertiary alcohols.
Consider a primary alcohol, like $\ce{(CH3)3-C-CH2OH}$, which forms a tertiary carbocation, $\ce{(CH3)2C+-CH2CH3}$ by an alkyl shift from primary carbocation during the substitution process. So will Lucas test give a false positive in this case, since a tertiary alkyl halide can be easily formed from a primary alcohol?

A similar question with no answer: https://chemistry.stackexchange.com/questions/117827/does-2-2-dimethylpropan-1-ol-give-a-positive-lucas-test

In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions
In chemical kinetics, there are two ways to deal with multiple-step mechanisms

Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate-determining step can be compared to the neck of a funnel. (1)
Steady-state approximation method- This is used when no step is the clear slowest. The steady-state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction is constant. (2)

So, in the Lucas test, the formation of the carbocation is the clearly slowest first step. This governs the reaction rate and the rest of the steps don't really matter. So, although it might seem deceptive at times if the R.D.S. is slow then the whole net reaction in slow.
Hence, the rate of formation of the one-degree carbocation is so less, that the entire rection does not proceed. So what you can theoretically say will be the only difference in say a Neopentyl alcohol and a 1-alkanol would simply be that after the super slow first step, the rest of the reaction will be much faster in the case of Neopentyl alcohol. But, due to the slow first step your given example will give an absolutely accurate negative Lucas test

Hydrochloric Acid ($\ce{HCl}$) reacts in the same way, although often Zinc (II) chloride (a Lewis acid) is added to help compensate for the lower nucleophilicity of chloride ion.
The mixture of $\ce{HCl}$  and $\ce{ZnCl2}$  is called the Lucas Reagent.
Secondary and tertiary alcohols react via the $\mathrm{S_N1}$ mechanism with the Lucas reagent.
The $\ce{ZnCl2}$ coordinates to the hydroxyl oxygen and this generates a far superior leaving group.
Primary alcohols react in a similar fashion except the free cation is not generated, and the substitution is of $\mathrm{S_N2}$ type.
Source: Reaction of alcohols

But since a tertiary carbocation cam be formed by rearrangement, won't there be $\mathrm{S_N1}$ reaction for the alcohol mentioned in the question ?

No @Gokul the carbocation does not form in the first place. Forget rearrangement.

@Gokul carbocation is not formed in case of primary alcohol. Sn2 reaction takes place


text:
Is there any way of testing antioxidant ability of different substances?
For my investigation, I am looking at the antioxidant levels of different types of tea. How could I compare the antioxidating ability of each tea?
For example, I know there is a method calling for DPPH radical and absolute ethanol, which I do not have at my disposal. Do you know any other method I could use?

Title and text do not strictly correspond. To the title the answer is reasonably yes in the sense while having reliable methods for mixtures and solutions might be, as it seems to a non specialist, complicated.

Could you please clarify your point?

The problem is that teas are not made of a single substance. For "food" the orax scale went common but I think its use is now deprecated.

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text:
Does dissociation of molecules into atoms change the volume of mixture?
For example, if $\ce{H2}$ is converted to two $\ce{H},$ does the volume of the mixture change?
If it doesn't, does it imply that the work done is $0$ because work is $P\,\mathrm dV?$

Yes, that is correct.

What's correct?  The volume changes or work done = 0.

Volume doesn’t change and no work is done.

All depends on how the gas is kept, E.g. if it is kept at constant volume or constant pressure. For the former, pressure increases, but not volume, so no work is done. for constant pressure, pressure remains, but volume increases and works is done.

Short answer is "not necessarily". If we isolate V in the ideal gas law, we have $$V=\frac{nRT}{P}.$$
If $n$ doubles and nothing else changes, then $V$ does also double. But the doubling of $n$ could also be offset by a doubling of $P$ or a halving of $T$, either of which would result in no change in the volume.
And yes, if the volume increases, pressure-volume work has been done, and if the volume remains constant, no pressure-volume work has been done.  That isn't the same as "no work has been done", as there are other types of work besides pressure-volume work.

In the question i was working on, it was mentioned that the temperature was constant, and in the solution, it was somehow implied that the volume would also remain constant.

Could you tell me why

In the dissociation of $H_2$ into $2$ $H$, one mole of something yields $
2$ moles of something else. The number of moles increases. As $V$ is proportional to $n$, at constant pressure, the volume $V$ of the gas must increase.

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text:
Electrochemistry - Understanding Diffusion Potential
I have a big problem trying to wrap my head around this for several hours and I cannot seem to make sense of it.
I have to link this scheme with the 3 colored potential profiles on the bottom of the picture. Considering the electrolyte AK, which splits into A- and K+, that it is 10x more concentrated in compartment I (left) and that the 3 different cases are the assumed:
1) Mobility of A- >>> than mobility of K+
2) Mobility of K+ >>> than mobility of A-
3) Both ions have an equal mobility
Now, I know that in the 1) case, anions will diffuse through the membrane and work toward balancing their concentration. However, this will generate a potential difference because K+ cannot move and compartiment I will be left with a higher potential. Now, I can imagine that if we set our reference (left electrode) at an arbitrary number, let's say 0, then we can assume compartiment II being negative potential in respect to I. This leads to the constant value of the Nernst eq. for same metal electrodes and electrolyte with 10x concentration difference of 59 mV. Since II is considered lower potential VS I in case n°1, should I assume that the potential difference at the membrane's level is -59 mV (green profile)? And another +59 mV between the electrodes because of Nernst equation, which would bring to a final measured potential of 0V. Is this a legit reasoning?
Thank you in advance.

Let's discuss the problem in a qualitative way. In such a cell, the metal will get dissolved in the right-hand side of the cell, where the concentration is low, so that $[M^{z+}]$ increases in this compartment, which becomes the anode. In the left-hand side, the ions $[M^{z+}]$ are discharged  and deposited as a metal layer on the electrode, which is the cathode.
The only problem is what is going on in the solutions. If the anion $A^-$ has a high mobility, it will flow from left to right through the membrane to compensate the chang of concentration of the metal ion. And after some time, the two solutions will have the same intermediate concentrations, and the cell will not deliver any current.
But if the ion $A^-$ has a low mobility, the cation $[M^{z+}]$ has a higher mobility and will easily cross the membrane from right to left. It will take a much longer time for the concentrations to be equalized in the two compartments. The reason is that a much smaller amount of ions $A^-$ will cross the membrane from left to right. But at the end, the two solutions will have the same concentrations.

This addresses at most a half of the question.


text:
Heats of formation of neutral molecules and homolytic vs heterolytic bond dissociation in mass spectrometry
I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

Great efforts have been made to generate accurate and reliable ion thermochemistry data. Once such data is available, it can be employed to elucidate fragmentation mechanisms and in addition, it is useful for obtaining some background on the energetic dimensions in mass spectrometry.
Heats of formation of neutral molecules, $\Delta H_{f(RH)}$, can be obtained from combustion data with high accuracy. Bond dissociation energies can either be derived for homolytic bond dissociation
$$\ce{R - H \rightarrow R^. + H^., \Delta H_{Dhom}} \tag{2.12}$$
or heterolytic bond dissociation
$$\ce{R - H \rightarrow R^+ + H^-, \Delta H_{Dhet}} \tag{2.13}$$
The homolytic bond dissociation enthalpies give the energy needed to cleave a bond of the neutral molecule which, in the gas phase is in its vibrational and electronic ground states, to obtain a pair of radicals which also are not in excited modes. Homolytic bond dissociation enthalpies are in the range of $3-5 \ eV$ (table 2.2). The heterolytic bond dissociation energies apply for the case of creating a cation and an anion in their ground states from the neutral precursor which means that these values include the amount of energy needed for charge separation; they are in the order of $10-13 \ eV$ (Table 2.3). Due to the significantly altered bonding situation, breaking of a bond is clearly less demanding in molecular ions than in neutral molecules.

I think this is good explanation. However, since this is my first exposure to these concepts, I'm interested in a higher-level explanation. Specifically,

What does "heats of formation of neutral molecules, $\Delta H_{f(RH)}$", mean in the context of mass spectrometry, and why is it relevant?

What is the relevance of homolytic vs heterolytic bond dissociation in mass spectrometry?

The Wikipedia articles for these two concepts, 1. Standard Enthalpy of Formation and 2. Bond Dissociation Energy: Homolytic Versus Heterolytic_Dissociation, offer more general explanations, but I'm interested in understanding in a general explanation in the context of mass spectrometry.
I would greatly appreciate it if some of the more knowledgeable contributors would please take the time to clarify these concepts.

The heat of formation of a neutral molecule is important in mass spectrometry because in order to measure the characteristics of individual molecules, a mass spectrometer has to convert them to relevant ions. Once ions form, they can be moved about and manipulated by external electric and magnetic fields as depicted in following sketch of mass spectrometer (MS):

The Ion Source, the Mass Analyzer, and the Detector are the three main components of the mass spectrometer. The Ion Source, which can be considered as the heart of the spectrometer, is the one needed to ionize individual molecules. Electron Ionization (EI) is the most common ionization technique used for mass spectrometry (Note: Older literature usually refer to EI as Electron Impact, but Electron Ionization is the currently accepted term; EI works well for many gas phase molecules, but it does have some limitations as well, such as causing extensive fragmentation so that the molecular ion is not observed for many compounds). In this ionization technique, molecules of the sample are bombarded by electrons generated from a heated filament (see the insert red box in the diagram). The electrons are produced by passing a current through the filament. The amount of current controls the number of electrons emitted by the filament. An electric field accelerates these electrons across the source region to produce a beam of high energy electrons. When an analyte molecule (usually organic molecule) passes through this electron beam, a valence shell electron can be removed from the molecule to produce an ion, mostly a positive ion (Note: Ionization does not occur by electron capture, which is highly dependent upon molecular structure).
Since the ionization is produced by a single electron that is accelerated to $\pu{70 V}$, the technique is commonly referred to as $\pu{70 eV}$ EI. This high energy electron collides with a molecule (analyte), it transfer more than enough energy to over come ionization energy of the particular site (usually a electron from a lone pair of oxygen of carbonyl group or alcohol group) of the molecule, often ionizing the analyte by knocking away one of the molecular electrons (either bonding or non-bonding). The resultant ion is a radical cation. This web page describes the possibility of the fragmentation happens in mass spectrometer:

The amount of energy transferred during this process depends upon how fast the electron (from ion source) is traveling and how close it passes to the molecule (analyte). In most $\pu{70 eV}$ EI experiments, approximately $\pu{1400 kJ/mol}$ ($\pu{15 eV}$) of energy is transferred during the ionization process. There is, however, a distribution of energy and as much as $\pu{2800 kJ/mol}$ ($\pu{30 eV}$) is transferred to some molecules. Since approximately $\pu{960 kJ/mol}$ ($\pu{10 eV}$) is required to ionize most organic compounds and a typical chemical bond energy is $\pu{290 kJ/mol}$ ($\pu{3 eV}$), extensive fragmentation is often observed in $\pu{70 eV}$ EI mass spectra. The distribution of energy transferred during ionization and the large number of fragmentation pathways results in a variety of products for a given molecule. Other electron voltages may be used to vary the amount of fragmentation produced during ionization. For most organic compounds the threshold energy for EI is about $\pu{10 eV}$. After ionization of the molecule, the residual energy from the collision may cause the molecular ion to fragment into neutral pieces and smaller fragment ions. This fragmentation can be through either homolytic cleavage or heterolytic cleavage. The molecular ion is a radical cation, but the fragment ions may either be radical cations or carbocations, depending on the nature of the neutral fragment. For example, Fragmentation of Toluene is depicted in following diagram:

Important: In Mass spectroscopy, the electron volt ($\pu{eV}$) is frequently used as an energy unit (for single molecules or atoms). Keep in mind that $\pu{1 eV} = \pu{1.6022 x 10^{-19} J}$ (per molecule or atom). Thus, $\pu{1 eV}$ (per molecule or atom) $= \pu{96.4152 kJ/mol}$.
A recent publication described why most fragmentations in mass spectroscopy preferred homolysis over hererolysis (Ref.1):

Bond strength is an essential parameter for discussion of bonding and reactivity. While bond fission can occur by homolysis or heterolysis, for neutral compounds such as alkanes the term “bond strength” generally denotes rupture by the lowest energy homolytic pathway. For example, the $\ce{C–C}$ bond strength in ethane is listed as $\pu{359 kJ mol–1}$,$^{[Ref.2]}$ representing homolysis, which requires one-third of the energy for heterolysis ($\pu{1297 kJ mol–1}$).$^{[Ref.3]}$

In addition, Ref.4 is a great review on mass spectrometry‐based analysis of protein structure.
References:

Karlee L. Bamford, Saurabh S. Chitnis, Rhonda L. Stoddard, J. Scott McIndoe, Neil Burford, "Bond fission in monocationic frameworks: diverse fragmentation pathways for phosphinophosphonium cations," Chem Sci. 2016, 7(4), 2544–2552 (doi: 10.1039/c5sc03804a).
Donald F. McMillen, David M. Golden, "Hydrocarbon bond dissociation energies," Annual Review of Physical Chemistry 1982, 33(Oct), 493–532 (https://doi.org/10.1146/annurev.pc.33.100182.002425).
Pratim K. Chattaraj, Andres Cedillo, Robert G. Parr, Edward M. Arnett, "Appraisal of Chemical Bond Making, Bond Breaking, and Electron Transfer in Solution in the Light of the Principle of Maximum Hardness," J. Org. Chem. 1995, 60(15), 4707-4714 (https://doi.org/10.1021/jo00120a010).
Frederik Lermyte, Dirk Valkenborg, Joseph A. Loo, Frank Sobott, “Radical solutions: Principles and application of electron‐based dissociation in mass spectrometry‐based analysis of protein structure,” Mass Spectrometry Reviews 2018, 37(6), 750-771 (https://doi.org/10.1002/mas.21560).

It is always molecular ion (called it as parent ion) forms first. Fragmentation occurs as a consequence. Also, this fragmentation in my answer happens in electron impact (EI) mass spectrometers. Advance technologies exist to give parent ion as the base peak (100% intensity). Such a techniques is MALDI-TOF.

Please refer to highlighted portion of my answer.


text:
How does Fe2+ become Fe in sacrificial protection?
I've searched low and high on the internet, but nothing to help me.
I know that zinc is oxidised in preference to iron, and that it "donates" those two electrons to iron.
$$\ce{Zn -> Zn^2+ + 2e-}$$
But how does iron make use of those two electrons to revert back to iron when it turns into iron oxide, like so:
$$\ce{Fe^2+ + 2e- -> Fe}$$
Doesn't the $\ce{Fe^2+}$ in the $\ce{Fe2O3}$ need to be dissolved in water? How can it be affected by $\ce{2e-}$ travelling along? Maybe I am missing something.

I think the point that you're missing is that $\ce{Fe2O3}$ is never formed when there is $\ce{Zn}$ in the system. When the metal gets oxidized, the electrons come from the $\ce{Zn}$, not from the $\ce{Fe}$. This is true even if the physical location of the oxidation is on the iron, since the $\ce{Fe}$ and $\ce{Zn}$ are connected and conductive, so all of the electrons on both are shared.

So wait. Does this mean that the oxygen gets its electrons (which came from Zinc ions) but the zinc does not get attached to it to form zinc oxide? I am talking about when the physical location of the oxidation is on the iron.

Exactly. The oxide formation and the oxidation are actually separate reactions. After the zinc is oxidized to $\ce{Zn^{2+}}$, it can dissolve in water if there is enough present, or it can react with water to give zinc hydroxide ($\ce{Zn(OH)2}$) or zinc oxde ($\ce{ZnO})$. But these oxygen atoms need not be the same ones that were reduced. Salty water greatly increases the rate of reaction because it conducts charge easily between the site of reduction (Fe metal) and oxidation (Zn metal) to offset the electron movement just as in an electrochemical cell.

But what happens to the reduced oxygen atoms? Isn't it necessary for an ion to be part of a compound or be dissolved in order to exist as an ion? Fundamentally I guess it's possible, but the forces of attraction between negative and positive charges in the world would not allow the ion to.

The sacrificial Zn only works if the Fe and Zn are also connected by water (preferably salty). So when the oxygen gets reduced, it picks up protons to form water. That means hydroxide is formed somewhere. Through proton transfers in the water, the negative charge is "conducted" to the location of the zinc ion and the hydroxide is formed there, so the negative and positive charges are united.

Aluminum is used to protect steel in sea water; A complex "calcareous" deposit is built up on the steel which reduces the amount of sacrificial aluminum needed for protection.

@Andrew You're such a legend, bud. I am painfully and slowly getting it now. Much of my misconception is cleared by "The oxide formation and the oxidation are actually separate reactions." However what exactly do you mean by oxygen picks up protons to form WATER?

If you keep adding electrons to $\ce{O2}$, you eventually get two $\ce{O^{2-}}$. Add protons to get $\ce{H2O}$.

@Andrew I just remembered that $\ce{H^+}$ ion is basically a proton. Sometimes you forget this sort of stuff too hehe. So basically the reduced oxygen atoms at the iron surface bump into oxidised hydrogen atoms (protons) somewhere and form water. Is that correct?


text:
Formic acid and carbon monoxide outgassing from latex mattress
Formic acid is used as a coagulant to turn rubber tree sap into latex, formic acid also decomposes to carbon monoxide. Does this mean things like latex mattresses have the potential to off-gas carbon monoxide? I am assuming they don't, but I have had carbon monoxide poisoning twice, so I would like to buy a latex mattress and I don't need $\ce{CO}$ issues for the third time.
If this isn't an appropriate forum for this question and you can point me in the direction of one that is that would be great.

I see a few possible issues with this reasoning. 1) assuming that formic acid is used in the production of latex, is that specific type of latex used for mattresses? And if so, does it still contain any formic acid? Many toxic chemicals are used in the manufacture of certain products, but that does not mean that the products still contain them. The manufacturing process may include steps during which the acid is removed. That is easily checked by looking at the specification of mattress latex. My guess is that they will not allow much formic acid to be in there.

assume that a latex mattress contains formic acid. Formic acid can indeed decompose to CO, but only under very strenuous conditions or in the presence of catalysts. See this article. So the question would be: how fast is CO produced? If very fast, then CO would probably be gone before the mattress reaches the customer. If very slowly, then CO would probably not reach dangerous concentrations during normal use.

formic acid is quite volatile (see wikipedia), so it might actually evaporate before decomposing. 4) CO poisoning in a domestic environment is usually associated with the use of faulty heating equipment. If that is what happened in your case (BTW, sorry to hear that!) perhaps I would advise to check that before going for less obvious options. Having said all that, I once bought a latex mattress, it was plastic-wrapped, and when I unwrapped it at home it smelled so bad I had to air it for 2 weeks before it was bearable! But CO is odourless :)


text:
According to KMT, is the velocity of an ideal gas always sqrt(3RT/M)?
It is a common textbook question to treat the root mean square velocities of an ideal gas as given by the following equation:
$$v_\text{rms}=\sqrt{\frac{3RT}{M}}$$
I was wondering about the validity of this equation, particularly as it applies to gases that are not monatomic. For example, in the following question from Zumdahl et al., it says:

Calculate the root mean square velocities of $\ce{CH4}$ and $\ce{N2}$ molecules at
$\pu{273 K}$ and $\pu{546 K}$.

The answer key gives $\pu{652 m/s}$ for methane at $\pu{273 K}$, $\pu{493 m/s}$ for nitrogen at $\pu{273 K}$, $\pu{922 m/s}$ for methane at $\pu{546 K}$, and  $\pu{697 m/s}$ for nitrogen at $\pu{546 K}$. These were all calculated with the equation $$v_\text{rms}=\sqrt{\frac{3RT}{M}}$$
Why is this equation valid for all ideal gases? Shouldn't it be $$v_\text{rms}=\sqrt{\frac{6RT}{M}} \text{ for methane}$$ (because there are $3$ rotational degrees of freedom in addition to $3$ translational degrees of freedom), and
$$v_\text{rms}=\sqrt{\frac{5RT}{M}} \text{ for nitrogen}$$ (because there are $2$ rotational degrees of freedom)?
If my thoughts are not true, then why is it commonly reported that the average molar specific heat of an ideal monatomic gas (at constant volume) is $3/2RT$, that of an ideal diatomic gas is $5/2RT$, and that of a nonlinear gas is $3RT$?

By velocity i think it's meant to be velocity of centre of mass , so the rotatory velocity does not impart any velocity to centre of mass.

$E=\frac 12mv^2 \implies  v=\sqrt{\frac{2E}{m}} $ is valid for translational kinetic energy and the speed of the centre of mass.
Vibrational or rotational energy does not count. An object may vibrate or rotate even if it's centre of mass has zero speed.
As each available degree of freedom has the mean energy $E=\frac 12kT$,
and as there are 3 independent translational degrees of freedom,
the mean translational kinetic energy is $E_\mathrm{transl}=\frac 32kT \le E_\mathrm{kin}$
and the respective
$v_\mathrm{RMS}=\sqrt{\frac{2E_{transl}}{m}}=
\sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$
even for ideal gas consisting of  multiatom molecules.

I am not too sure about your statement. Consider an ideal gas; the total internal energy is defined as $U = \text{PE} + \text{KE}$, where $\text{PE}$ and $\text{KE}$ are the potential and kinetic energies, respectively. Since in an ideal gas, there are no attractive or repulsive forces, $\text{PE} = 0 $. Are you saying that $U = \text{PE} + \text{KE} = 0 + 3/2kt$ for all gases? Surely this can't be true....

@DrPepper No, I do not say that. KE >= translational KE =3/2kT. ( Al as mean energy)

According to the equipartition theorem, energy is shared equally among all accessible degress of freedom.
A monatomic ideal gas has three translational degrees of freedom, so each translational degree of freedom has an energy of $\frac12 k T$, for a total translational energy of $\frac32 k T$.
A diatomic ideal gas has three translational degrees of freedom and two rotational degrees of freedom.  At the same temperature, it thus has the same translational kinetic energy as a monatomoic ideal gas ($\frac32 k T$) and thus, for the same mass, the same average speed.
However, its two rotational degrees of freedom give it an additional $2\times\frac12 k T = k T$ of rotational kinetic energy.  As a result, the heat capacity of a diatomic ideal gas is higher, because to heat it to the same temperature requires more energy, because you have to put energy not only into the three translational modes, but also into the two rotational modes.

so the kinetic energy only takes into account the translational modes?

@DrPepper  No, the kinetic energy is the sum of the translational, vibrational, and rotational kinetic energy.  The speed, however, is due only to the translational (as opposed to rotational or vibrational) kinetic energy.  This is because translational kinetic energy is the energy the molecules have as a result of how fast they move (i.e., translate) through space (and their mass).

Thank you so much! I understand it now :) Your comment was very helpful.


text:
Hydrogen disulfide heterogenous equilibrium

Hydrogen sulfide can be removed from natural gas by the reaction
$$\ce{2 H2S(g) + SO2(g) -> 3 S(s) + 2 H2O(g)}$$
a) Calculate $ΔG^\circ$ and $K$ for this reaction at $\pu{298 K}.$
b) What is the mass of sulfur at the equilibrium (at $\pu{298 K})$ when $\pu{1.00 mol}$ sulfur dioxide are sealed in a vacuumed 1 liter vessel?
$$
\begin{array}{lrrr}
\hline
  & Δ_\mathrm fH^\circ/\pu{kJ mol-1} & Δ_\mathrm fG^\circ/\pu{kJ mol-1} & S^\circ/\pu{J mol-1 K-1} \\
\hline
\ce{H2S(g)} & -20.6 & -33.4 & 205.8 \\
\ce{SO2(g)} & -296.8 & -300.1 & 248.2 \\
\ce{S(s)} & 0 & 0 & 32.1 \\
\ce{H2O(g)} & -241.8 & -228.6 & 188.8 \\
\hline
\end{array}
$$

I was able to calculate the Gibbs free energy and the equilibrium constant, using the formula
$$ΔG = -RT\ln K$$
and got the following answers, though I am not sure whether they are correct:
$$
\begin{align}
ΔG &≈ \pu{-90.2 kJ mol-1} \\
K &≈ \pu{6.28E15}
\end{align}
$$
Next, I tried to solve for the mass of sulfur but was unable to do so. I tried to solve it in the following way, which led to no solutions:
$$\begin{array} {c c c c c c c}
\ce{&2 H2S(g) &+ &SO2(g) &-> &3 S(s) &+ &2 H2O(g)} \\
c_\mathrm{alku}/\pu{mol dm-3}&1.00 && 1.00 && 0 && 0 \\
c_\mathrm{loppu}/\pu{mol dm-3}&1.00 - 2x && 1.00 - x && 3x && 2x
\end{array}$$
$$\pu{6.28E15} = \frac{(2x)^2}{(1.00 - 2x)^2(1.00 - x)}$$
I would like to know whether my answers for step a are incorrect and how I should approach step b.

The equilibrium constant is in terms of partial pressures (in bars), not moles.  Also, the 90.2 should be 90.3

The equilibrium constant is so huge (bigger than $10^{15}$), that you can admit that $100$% of the sulfur atoms are transformed into solid sulfur. So $\ce{x = 0.5}$. The final amount of elementary sulfur is $1.5$ mole.


text:
What is the explanation of the energy gap law in radiationless transitions?
The rate of radiationless transitions such as internal conversion and intersystem crossing decrease with the size of the energy gap between the states, according to
$$k\propto \exp\left(\frac{-\gamma\Delta E}{\hbar\omega_M}\right)$$
The papers by Jortner[1] that I have found which derive this expression involve taking steepest descent integrals of the flux autocorrelation function, but this is not a very intuitive explanation. (And in particular is not suitable for undergraduate chemists.)
How can this expression be understood qualitatively from the perspective of the sum over Franck-Condon factors:
$$k_{sl} = \sum_j \frac{2\pi}{\hbar} |V_{s0,lj}|^2\delta(E_{s0}-E_{lj})$$
References:

Jortner, J. Radiationless transitions. Pure Appl. Chem. 1971, 27 (3), 389–420. Section IX, weak coupling. DOI: 10.1351/pac197127030389. Mirror: pdf via publications.iupac.org, pdf via the Internet Archive

I deleted the answer because there were some discrepancies. However I suggest instead of k (rate constants), first look at the transition probabilities of radiationless transitions. There is 2 paged original paper which derived the expression for radiationless transitions via vibrational levels. Energy Gap Law and Franck–Condon Factor in Radiationless Transitions by S. H. Lin, Citation: The Journal of Chemical Physics 53, 3766 (1970); doi: 10.1063/1.1674571
View online: http://dx.doi.org/10.1063/1.1674571

I will have a read of this, many thanks

see also comments and references in this answer https://chemistry.stackexchange.com/questions/73985/how-to-calculate-the-quantum-density-of-states/74107#74107

also here https://chemistry.stackexchange.com/questions/28883/non-radiative-transitions-or-what-features-of-rhodamine-result-in-it-being-so-hi/55287#55287 . Note that your first equation (energy gap law) is only valid for solution experiments. In the gas phase the opposite effect occurs and both are derived from Fermi Golden Rule as Franck Condon weighted density of states.

@porphyrin I am interested to know what you mean by "In the gas phase the opposite effect occurs". Do you have a reference for this? It is not immediately obvious to me why this should be the case. Of course a rate like description may not be valid, and the assumption present in both equations that the system thermalises more rapidly than all other processes may break down, is this what you meant?

In the isolated molecule gas phase the rate constant goes up with gap not down as in solution.  The difference is due to weighting with Boltzmann distribution.

These are a few 'rough and ready notes' and some data on radiationless transitions.
Intersystem crossing $S-T$ and Internal Conversion $S_1-S_0$ are described by the Fermi Golden Rule:
The rate constant decreases exponentially with energy gap between S and T in solution. In contrast in isolated molecules the gas phase rate increases with energy gap, see figures below.
Molecules in condensed phase

Molecules isolated in the vapour phase

The pure singlet and triplet states cannot interconvert because of the strict spin selection rule. An interaction, which is spin-orbit coupling, mixes the states together to make new states that contain a bit of each initial state. The singlet now contains a little triplet and triplet contains a little singlet. As the S1 and S0 states are orthogonal to one another without an interaction; $\int S_1 S_0 d\tau = 0$ so no transition. States must have different symmetries if coupled by a photon similarly  a vibration with the correct symmetry can also couple S1 and S0 to cause internal conversion.
Transitions occur because an interaction mixes basis states together (singlet and triplet, say) to form new states containing part of each. These new eigenstates are shown schematically below, the thick lines on the lower pair show the amount of each base state in the mixed state.

One state can couple to many others, and when there are very many final states (below) we think of them as a quasi-continuum of levels. The radiationless transition rate is the rate from one state (optically excited) to these many final states.

Equation to solve is $\displaystyle (H_0+\Omega)\psi=i\hbar \frac{\partial}{\partial t}\psi$ to get rate constant $W_{kn}$ between levels $k,n$

Robinson & Frosch simplified the calculation of Fermi Golden Rule with approximations;
(a)Assume that the interaction energy is the same to all $n$ levels ,
(b)The (vibrational) Franck-Condon factors take an average value.
(c) The density of states $\rho$ is made to fit data! ( $\rho$ = number of states / wavenumber)
The approximate formula is: $\displaystyle W_E =\frac{2\pi}{\hbar}|V|^2F_c\rho(E)$  where $W$ is rate constant at energy $E$, $V$ is interaction matrix element between $k$ and $n$ , $F_c$ is the average Franck Condon factor and $|V|^2F_C=\int\psi_1\Omega\psi_2d\tau$. The $\Omega$ is spin-orbit operator for intersystem (singlet to triplet) crossing, kinetic energy operator for internal conversion.  The term $\rho(E)$ is the effective density of states at energy $E$.
This formula is not good for accurate calculation but indicates what is important.
The full and detailed calculation shows that :
(a) In the initial state, say singlet, only certain vibrational modes are effective in causing a radiationless transition.
(b) These modes are separated out and called promoting modes, at least one is present, and can only be identified by experiment.
(c) These promoting modes are the only modes which do not have zero electronic coupling terms.
(d) promoting modes have zero displacement between the two electronic states and are non-totally symmetric
(e) In the final state accepting modes receive the vibrational energy, these are modes of largest frequency so that as few quanta as possible of each are excited. Usually these are CH or CC vibrations.
There are two cases molecules in condensed phase and isolated molecules in gas phase.
(a) Condensed phase
The energy gap law  $\displaystyle K_{ST}=Ae^{-\beta\Delta E}$ is observed experimentally and is obtained theoretically from $W_{kn}$ by taking a Boltzmann distribution over initial levels and then the maximum value of resulting expression. It turns out that this term is greater than all the others combined. (btw, this energy gap law has nothing to do with Arrhenius equation.)
The full expression is
$$W=|V|^2\sum_m \left[\prod_n\frac{X_n^{m_n}}{m_n!}\right] \delta (\Delta E -\hbar \omega_k-\hbar\sum\omega_n m_n)$$
where $\displaystyle X_n=\frac{\mu_n\omega_n}{2\hbar}(Q_n-Q_k)^2$ is the displacement between vibration $n$ between energy surfaces $Q_n\to Q_k$. The term in the square bracket is the 'Franck Condon Factor density of states' assuming harmonic oscillators.  $\omega_k$ is the promoting mode and has no displacement between the two states. The $\delta$ term ensures energy conservation, i.e. $\delta(0)=1$, else is zero.
This equation is very difficult to use, but to a good approximation the complicated $\Sigma\Pi$ term can be replaced by its maximum value and a significant simplification is obtained.
We now take $X$ as being the displacement of the vibration of maximum frequency, $\omega$, usually CH or CC vibration. $\Delta E$ is singlet -triplet energy gap. The expression for the energy gap law is
$$W=|V|^2e^{-X}e^{-\gamma\Delta E/\hbar\omega}, \qquad \gamma= \ln\left(\frac{\Delta E}{\hbar\omega X}\right)-1$$
and $\gamma$ has values in the range 3 to 5 typically and is effectively a constant.
Its success is that it describes fall off with energy gap and also D/H difference in rate constants due to difference in frequency, see first figure above.  The difference to the gas phase is the averaging over the Boltzmann distribution.
(b) Gas phase
In the gas phase with a large molecule, say benzene, we are in the statistical limit with a huge number of final states, effectively a continuum, then the calculation is from one initially excited level $k$ to many levels $n$ in the final state.
$$W_{kn} \sim |V|^2\sum_n \big|\int\psi_n\Omega \psi_k d\tau\big|^2 \delta (\Delta E_n-\Delta E_k )$$
the equations to use in this case can be found in Beddard et al Chem. Phys. Letts. 1973, v 18 p 481, if you cannot get to this paper I can post again.

All very informative, I will have to study in more detail. But for now I am still not clear on why the gas phase dependence should be opposite. Is it related to a difference in the initial density operator, i.e. not being allowed to assume the system thermalises before transitions occur? If so can you give more detail on what this difference is? Thanks

Also I am not sure that I agree with your claim that the energy gap law has nothing to do with the Arrhenius expression. My interpretation is that this is all equivalent to being in the inverted regime of Marcus theory except that tunnelling means the exponent no longer behaves as a quadratic of the energy gap but instead becomes linear. Since in Marcus theory the quadratic term is precisely the energy of the TS it seems that it is all very much related to the Arhenius expression. Does this sound correct?

J.L. no I don't think that what you say is correct. There is no chemical reaction involved in radiationless transitions, just an interaction between one electronic state and another within a molecule.  Marcus theory involves reorganisation of the solvent as the fundamental step. This is nothing like this.


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Why O and N promote SN1 but C=O inhibit SN1 although both are electron-withdrawing?

Source:
Jonathan Clayden, Nick Greeves, Stuart Warren, Organic Chemistry 2th Edition, 339.
The general idea I got so far, is that, carbocation will be more stable if the + charge at central C is dispersed.
How can a very electronegative atom bonded to C bearing leaving group promote SN1? That electronegative atom should intensify the + charge, making the central C more electon-hungry, making the carbocation less stable, no?
C=O inhibits SN1 because it is electron-withdrawing, but why not O and N inhibit SN1 too? They too are electron-withdrawing.

Those carbocations are termed as 'stable' as the O/N atom establishes a '2p-2p back bond' with the empty p orbital on the carbon atom carrying the positive charge. This helps the carbocation stabilize as the O/N atom shares its electron density to the starving cation. This effect is FAR more dominating than the pulling inductive effect of the O/N atom.

C=O inhibits SN1 because it is electron-withdrawing, but why not O and N inhibit SN1 too? >They too are electron-withdrawing.

I am not quite sure what you are referring to, but I think you are talking about a carbonyl group on the alpha (just adjacent) carbon to the carbocation. That destabilizes the cation as the Oxygen atom cannot back-bond now with the cation. Now it pulls electrons owing to its negative inductive effect, or its electronegativity, you might say.


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Can rate constant for forward reaction be equal to rate constant for backward reaction?
For a particular reaction, can we have
$K_c = 1$?
My confusion arised when I was studying the effect of temperature on $K_c$ . I see that the temperature must affect forward and backward rate constants, differently for a reaction so that it changes $K_c$ (which is ratio of the two rate constants). So I thought for a reaction with $K_c$ equal to one, both the rates must be equal, so both will get equally affected by temperature, and thus no change in $K_c$ , or does it have something to do with the Arrhenius equation?
I know that there is a lot of lacking in my intuition as I have just began studying Le Chatelier's principle and all the stuff. So a detailed explanation would be much welcome.

Does this answer your question? Same Activation Energy for Forward and Backward Reactions

A part of it perhaps,so that I conclude that Kc will not be affected by temperature then(if Kc =1).

No. If you read the linked answer, you'll see that the requirement is that the enthalpy change is 0. K=1 only means that $\Delta G=0$. That only requires that $\Delta H = T\Delta S$. In the case where $\Delta H = \Delta S = 0$, there will be no temperature effect, but otherwise there will be a temperature effect. (And K=1 will only occur at one temp).

When forward and backward reaction have the same rate, $K_c$ is equal to $1$ in the case of a reaction where the total number of reactants is equal to the total number of products in the chemical equation describing the change. If the total of these stoichiometric coefficients is not the same before and after the horizontal arrow, $K_c$ may be equal to $1$, but must have a dimension, like mol/L or L/mol, or any combination of these units.
As the forward and backward rections have different activation energies, their rates change differently with increasing the temperature. So the equilibrium constant $K_c$ can be equal $1$ at only one temperature.

But this is not true for many reactions, is it? For instance what about reaction    2ICl <-->Cl2+ I2     Kc is 0.001 even though the reaction has same stoichiometric coefficients

The important factor is not the coefficients, but the relative contribution of enthalpy and entropy. Also K is always dimensionless because it is technically calculated from activities not concentrations.


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Is there any material that can prevent silver from tarnishing?
I am designing a machine that uses paper-thin sheets of silver. I am trying to find a material that can cover the silver to prevent it from tarnishing over time. The material does not have to be abrasive-resistant, as the silver will be placed in a controlled environment. Because of this, appearance does not matter. Perhaps aerogel can work?

Does the covering need to be abrasion resistant? Chemically inert under specific conditions? Nice looking? Without more information, don't expect a useful answer.

Something that blocks diffusion of oxygen or sulfur-containing species from the air and doesn't produce any of its own. What about a very thin layer of gold? It can be evaporated or electroplated, but I don't know if the electroplating process is compatible with silver. Now that would be a good question to ask separately. It's also possible that a different but related question could be asked in Home Improvement SE just to see if there are existing methods. You could try https://en.wikipedia.org/wiki/Varnish#Polyurethane and see what happens.


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Is levomenthol superior to Menthol as a pain killer in humans and what are the ways to determine this?
I desire to learn whether levomenthol is superior to Menthol as a pain killer in humans.
I didn't find a wiki article about levomenthol and the article Menthol doesn't mention this molecule.
The chemical formulas are the same but I understand that the only difference is the spatial position of chemical bonds between the atoms, as can be learnt in pubchem:

https://pubchem.ncbi.nlm.nih.gov/compound/Menthol
https://pubchem.ncbi.nlm.nih.gov/compound/l-Menthol

My problem
I don't know what are the main ways for chemists to decide which one should work better in humans but hope for a simple explanation for non chemists.
I thought about this after reading that Strepsils, the famous British sore throat lozenges, contain levomenthol.
I think about this because I know some other medicines are "levo" versions of others (for example, cetrizine and levocetrizine).
My question
Is levomenthol superior to Menthol as pain killer in humans and what are the ways to determine this?

Oookay, I can see you try there, but menthol isn't exactly a painkiller and l- (levo) is just a description of isomer, another being d-  (dextro)  https://en.wikipedia.org/wiki/Dextrorotation_and_levorotation

@Mithoron Here is a starting point to read on the analgesic properties of menthol, and the channel it binds to: https://www.ncbi.nlm.nih.gov/pubmed/29524352

Another paper: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6600640/

@KarstenTheis Tell that to House M.D. Strepsils isn't exactly Vicodin.


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What causes the differences in price in home testing kits for lead in water?
The price differences between home testing kits to test tap-water for lead are huge.
20 dollar will either buy 1 test kit or 100 test strips. Both are intended to test drinking-water safety.
What does this mean? Is one of the tests 100 times less sensitive? Or 100 times less accurate?
How can i as an average customer know which test is good enough?

You should shop somewhere they know what they´re selling.

A test kit means a set of solutions for repeated use until the are spent. How many analysis can be done should be noted on the kit package.
Paper strips are easier to use, but may not be better for result.

Quantity is noted on the package: there is a package of 1 and a package of 100, both are roughly the same price. So the question is, what causes the 100 fold price difference and is this difference likely reflected in test reliability?


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What exactly is the requirement (in terms of saturation) for a reaction to be at equilibrium?
A saturated solution is a solution that contains the max amount of solute able to be dissolved. However, can't a solution still be at equilibrium if you add more solute to a saturated solution as long as it doesn't reach supersaturation? What exactly is the requirement (in terms of saturation) for a reaction to be at equilibrium?

If you add solid sodium chloride to a saturated sodium chloride solution, it is still at equilibrium. If you increase the sodium ion concentration or the chloride ion concentration (by e.g. adding a high concentration of NaOH or LiCl), then the solution is supersaturated and will not be at equilibrium.

What exactly is the requirement (in terms of saturation) for a reaction to be at equilibrium?

A reaction is at equilibrium when the reaction quotient is equal to the equilibrium constant. For a precipitation reaction, the solid species is not part of the equilibrium constant expression, so adding more solid does not change the situation. Changing the concentration of the ions in solution, though, will change the reaction quotient, and the reaction will no longer be at equilibrium.


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Independant variable for Asprin pill titration with NaOH solution
I am assigned to do an independently designed chemistry experiment for school. I want to do titration of Aspirin pills with NaOH solution using Phenolphthalein as indicator. I am struggling to find and independent variable for the experiment. I thought about doing heat, but is unsure about the mechanisms of that. If anyone could help me or guide me in a direction I would greatly appreciate it!
Thank You!

Independent variable means the variable which you can change easily. Direct titration aspirin is not a good idea as aspirin hydrolyzes into salicylic acid. There are a lot of non-sense high school/college level experiments doing direction titration and ignoring the side reaction of hydrolysis.
The proper procedure of aspirin analysis is by back-titration with NaOH and taking hydrolysis into account. Ask your teacher about it by showing him/her this procedure Back titration of aspirin
Coming back to independent variables...Just now you have another independent variable...comparison of two different methods of titration (direct vs. back). Think of other independent variables and share in your comments.


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How to find the value of $K_c$?
This is part of my homework. Given the chemical equation :
$$\ce{Sb2S3_{(s)} + 3H2_{(g)} <=>2Sb_{(s)} + 3H2S_{(g)}}\text{        }K_{p}=0,429\text{ at }713\text{K}$$
What is the value of $K_c$ for the reaction?
I used the formula : $K_p = K_c(RT)^{\Delta n}$, but I know that $\Delta n$  considers only the coefficients of the reactants and the products of gas forms, so $\Delta n = 3-3=0 \Rightarrow K_c=K_p$?
Am I right? Thanks for any help!

Looks correct to me. Note, however, that such questions regarding proof-reading are rarely off-topic on Chemistry.SE: Policy on AMIRITE questions.

Ups...I am sorry, this is my first post actually... thanks for the notice anyway.


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How many π-electrons are there in 2-methylene-2H-pyran?
I thought that the structure will be aromatic and stabilised because one of the lone pairs being delocalised with the help of conjugated double bond system but still only 3 π-bonds will be present at a time and 2 lone pairs, so the answer should be 8.
But the book says the answer is 10. Please explain.

Get a better book. Why? It should be 8, as only one pair can be delocalised.

Please quote and cite the book you have used. This is a typical case where the [tag:erratum] should be applied.

Yes, the textbook is wrong. With one lone pair from the oxygen capable of conjugation (the other lone pair is a sigma pair in the plane of the ring) plus the three carbon-carbon bonds, we have eight pi electrons.
Possibly the confusion arises because eight pi electrons "should" make this molecule antiaromatic, thus highly unstable which does not match with experiment. But the 4n+2/4n rule applies only when the conjugated electrons are in a single cycle and most strictly with all conjugated atoms identical. The methylene-pyran molecule shown, with its pendant pi bond, does not meet either criterion.
Protonating the CH2 group in 2-methylene-2H-pyran leaves six pi electrons in a single cycle and thus a (pyrylium) cation with some aromatic character, so we may expect the molecule to be basic. Evidence for the existence and, indirectly, the basicity of this molecule may be seen in Zvezdina et al.[1], where the protonated molecule as a perchlorate salt is reacted with a sufficiently strong base to generate 2-methylene-2H-pyran.
Reference
1.
Zvezdina, É.A., Zhdanova, M.P., Bren, V.A. et al. "Reaction of methyl-substituted pyrylium salts with tertiary amines." Chem Heterocycl Compd 12, 1222–1226 (1976). https://doi.org/10.1007/BF00470218


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Is the kinetic energy change of a system (q) equal to the enthalpy change of a system (∆H)?
When silver nitrate and hydrochloric acid are put together in a container, they form silver chloride. If you're given information about the reactants (including temperature change, heat capacity, and mass of reactant aqueous molecules), then you can find the kinetic energy change of the system.
What I want to know is: does the $q$ value you find for the system equal to the $\Delta H$ value for the system? Worded differently, does the kinetic energy change of the system equal to the enthalpy change of the system? Is this one of those obvious questions that most people understand instantly but people like me who lack common sense don't?

does the q value you find for the system equal to the ΔH value for the system?

It seems that the reaction is undergoing at constant pressure. In that case, q value is indeed equal to the ΔH value for the system.

Worded differently, does the kinetic energy change of the system equal to the enthalpy change of the system?

q value is not the same as 'kinetic energy change'. q is energy exchanged in the form of heat.
When system intakes heat, its internal energy can change or remain the same.
Internal energy is actually Total Energy = Kinetic Energy + Potential Energy. So, when Internal Energy changes, any or both of KE & PE can change. KE remains the same in an isothermal process.

Under the restrictions of constant $p$, and no non-$pV$ work, we have:
$$q_p=\Delta H$$
q is the heat flow into the system (postive for flow in, negative for flow out).  However, q is not necessarily the change in the system's kinetic energy.  For instance, suppose you have an endothermic chemical reaction at constant temperature.  Heat will flow into the system.  However, some or all of that heat flow could go into increasing the potential energy of the chemical bonds.  And if the endothermic chemical reaction is accompanied by an increase in volume, some of that heat flow will go into the work needed to expand the system against the external pressure.
Indeed, if the temperature is constant, and the heat capacity of the reactants and products are the same, then there will be no change in the kinetic energy of the system.  I.e., in that case, none of the heat flow will go into changing the kinetic energy of the system.

It is right if you consider that kinetic energy may have different forms: it can be either vibration, or rotation or translation energy.


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How to name the following compound?
I know that when both carboxylic acid and ester group are present then the carboxylic acid takes priority and ester is treated as side chain . It is denoted by prefix alkoxy carbonyl. But the arrangement is different in the given figure. The ester group is attached with single bonded oxygen connected to main chain .Please explain how to name such compounds. Answer to given problem is option d.

The IUPAC name of the following compound is:
(a) 1-Acetoxy acetic acid
(b) 2-Acetoxy acetic acid
(c) 2-Ethanoyloxy acetic acid
(d) 2-Ethanoyloxyacetic acid

Acetoxy is not accepted by IUPAC nomenclature as it is a trivial name. Ethanoyloxy is the proper name of the substituent attached at C-2. Also, there is no space (Which you should know), which is why (d) is the correct answer.

Related: Naming ester as a substituent to carboxylic acid

You have already found out that the carboxylic acid group is the principal characteristic group. Therefore, the ester group has to be expressed as a prefix. The corresponding rule in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows.

P-65.6.3.2.3 Esters cited as prefixes
When, in an ester with the general structure $\ce{R-CO-O-R'}$ or $\ce{R-S(O)_x-O-R'}$, another group is present that has priority for citation as the principal group or when all ester groups cannot be described by the methods prescribed for naming esters, an ester group is indicated by prefixes as ‘acyloxy’ for the group $\ce{R-CO-O-{}}$, and ‘alkyloxy(alkanyl)…oxo’ or ‘alkyl(alkanyl)oxycarbonyl’ for the group $\ce{-CO-OR'}$.

In this case, the ester group has the form $\ce{R-CO-O-{}}$, which is expressed as a prefix of the form ‘acyloxy’.
Note that some acyl groups that are derived from carboxylic acids have retained names used as preferred IUPAC names. In this case, the preferred prefix for the acyl group $\ce{CH3-CO-{}}$ is the retained name acetyl rather than the systematic name ethanoyl. The preferred name for the corresponding acyloxy group $\ce{CH3-CO-O-{}}$ is acetyloxy. Also note that the name acetyloxy is preferred to the contracted name acetoxy that may
be used in general nomenclature.
The prefix acetyloxy is a compound substituent group. It consists of a simple substituent group (oxy) to which is attached one more simple substituent group (acetyl). Parentheses are used around such compound prefixes. Therefore, the complete name for the compound that is given in the question is (acetyloxy)acetic acid.

Note that locants are omitted for parent compounds when all substitutable hydrogen atoms have the same locant. Therefore, the preferred IUPAC name is (acetyloxy)acetic acid and not 2-(acetyloxy)acetic acid.


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Is there a cheap and effective way to make PdZn-Ag foam?
I am trying to design a machine that can convert carbon dioxide into methanol and/or hydrocarbons. My design requires PdZn-Ag foam (nanoparticles), which is made by depositing palladium-zinc vapor onto a silver base, but this seems to be very costly. So is there is a cheap and effective way of producing a measurable amount of PdZn-Ag foam?

Palladium is not cheap!

Perhaps an alloy would help?

I need around 3-4 g of palladium.


text:
Why is Quinine more soluble in an acidic aqueous solution than an basic aqueous solution?
How do I find out if Quinine or in general if a compound is more more soluble in an acidic aqueous solution than an basic aqueous solution?
Here is the structural formula for quinine:

Investigate the properties of amines. This ? may be marked down without some effort on your part.

Observe whether a compound has basic centres. The protonated form has a much greater chance of being water souble.


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Nucleophilicity Of Amino Acid Side Chains
Out of the side chains of aspartic acid, methionine, threonine, and tyrosine, which is the least nucleophilic?

Highly dependent on solvent, electrophile etc

Aspartic acid < Tyrosine < Threonine < Methionine is the order of nucleophilicity of the amino acid side chains. So, Aspartic acid (neutral form) is the answer. Every side chain has a Oxygen or Sulphur (Methionine) group available for electron pair donation, but in Aspartic acid the Oxygen group has lesser inclination for donation of electron density (due to resonance). Moreover, the solvation layer surrounding the acid group of Aspartic acid makes the job of nucleophilic attachment even more tougher.

Do you have a reference for this?

Nope, just arguments. How can one measure nucleophilicity of a side-chain anyway! Nucleophilicity is a property of a molecule or ion.


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How will the dielectric constant of water change when ethyl acetate is added?

How will the dielectric constant of water change when ethyl acetate is added?

The $\mathrm{p}K_a$ value of weak acids in aprotic solvents will be lower than in protic solvents because the dielectric constant does not favor ionization. If I increase the $\mathrm{\text{%}}$ concentration of ethyl acetate in water and gradually increase it to nearly $\mathrm{100\text{%}}$ ethyl acetate, how will the $\mathrm{p}K_a$ values of a weak acid in this solvent change?
I want to determine the endpoint with an indicator dye and a suitable titrant of a strong base such as sodium hydroxide. My general aim is to calculate pKa in an easier way.

Note that AcOEt solubily in water is just 8.3g/100ml https://en.wikipedia.org/wiki/Ethyl_acetate what is less than 2% molar fraction.

...and note that ethyl acetate will decompose into ethanol and acetic acid in water.


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Organometallic and oxygen
I often read that reactive organometallics must be handled under an inert atmosphere because they react with oxygen in the atmosphere. This applies to grignard reagents, organolithium reagents, and organozinc reagents.
What is the reaction of these organometallics with oxygen? I don't see how oxygen can be electrophilic. Maybe it forms a peroxyacid?

Part of the trouble is that the reaction with oxygen is usually very exothermic, so it tends to take place all over your fume hood. I would expect to get a whole range of reactions.

A lot of organometallics are sensitive to oxygenation which oxygen is quite good at.

Grignards react with molecular oxygen to form hydroperoxides. According to this paper  here the yield for alkyl hydroperoxides can be as high as 90%, yields from aryl Grignards are lower (10-20%).


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Is sodium percarbonate the safest and by weight the most efficient oxygen generator?
I have been reading about $\ce{Mn}$-catalyzed reaction of sodium percarbonate and water to produce gaseous oxygen in medical emergency situations.
This seems safe due to its reagents and products, however I am wondering if there are other reagents that would be suitable and safe enough to fulfill above use.
I did read about lithium perchlorate being used in so-called oxygen candles and capable of a high ratio of oxygen per weight output, however they hardly seem as safe as the above reaction.

Under what conditions? Safest in an operating theater? Aboard an airliner? To carry in an ambulance? In what quantity or rate?

Careful, as sodium percarbonate (basically $\ce{3Na2CO3.2H2O2}$) is usually marketed as a chlorine-free bleach. The latter usually has an additive, TAED, which acts as a key activator. See, for example, Wikipedia on OxiClean, where not even listed on the label is TetraAcetylEthyleneDiamine, which is apparently commonly employed as a bleach activator in many laundry products, see this, for example.
TAED reacts with H2O2 in alkaline conditions (referred to as perhydrolysis) creating, in part, Peracetic Acid. The latter PAA is actually preferred over sluggish H2O2 for bleaching. In other words, it is not the H2O2 as the active bleaching agent, but PAA created in situ.
So, other than possibly misleadingly implying that eco-friendly hydrogen peroxide is the active agent, it is, in reality, PAA, a probematic compound with associated health concerns (see Marquand, E. C.; et al. , 2007. "Asthma Caused by Peracetic Acid-Hydrogen Peroxide Mixture". J. Occup. Health. 49 (2): 155–158.). Here is an MSDS on an actual PAA mix, ( 76-61% water, PAA 20-35%, 3% Acetic acid and 1% H2O2), which is far from friendly but does eventually totally decompose into harmless O2, CO2, H2O,...
As such, anyone using a commercial percarbonate product should at least be aware of potential safety issues that one may likely be employing a Peracetic acid mix.
Note, there is an indicator that you do not have pure sodium percarbonate. It is the presence of some dire warning labels.


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Distance Dependence of Weak Forces
I am currently taking a Biochemistry course and one of the topics on midterm covers various weak forces and its distance dependence. We covered:

charge-charge (ionic)
hydrogen bonds
van der waals
hydrophobic effect

I am not sure how to put them in order and how they interact with each other once distance comes into play.
If anyone can explain in details, it would be very helpful.
Thank you in advance.

Does "the book cover various weak forces and its (sic) distance dependence" or not?

If you look up 'intermolecular  forces' on Wikipedia you will find answers to your question.


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Ozone layer: why shouldn’t there be holes at the poles?
Just read "What property allows ozone to absorb uv light" together with excellent answers.
The driver here, i.e. the source of both the short and long wave UV, is of course sunlight. This is notably absent for six months of the year at the poles, so can anyone explain why one would "not" expect holes in the layer there?
Also, is it true that there have never been measurements of the layer that did "not" reveal holes at the poles?

Apart from anything else the holes were big enough to affect non-polar areas and were expected to lead to significant increases in hard UV at ground level in inhabited places.

I sense this has been interpreted as a climate debate platform - not true. There is good reason to suppose there is a complex dynamic at work here, for example the layer thickens as one moves towards the arctic circle, then thins towards the equator. This is surely counter intuitive! I am curious to know if this dynamic has been modelled and if the CFC influence is still regarded as the dominant mechanism.


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Can dissipative work be circumscribed as irreversible heat transfer
My question refers to an article I read yesterday in which the author states that for the calculation of the entropy change for a process (in that case friction, during an irreversible adiabatic gas expansion), we could circumscribe the dissipative work as irreversible heat transfer, find a reversible pathway for the process and calculate the entropy change.
I had never read something like that before.
What is your opinion to this statement
$$\mathrm{d}S = \int \frac{W_\mathrm{diss}}{T}$$

If you want to learn how to determine the entropy change for a system that undergoes an irreversible process, see the following primer:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

'Dissipative work' or energy dissipation from friction can occur through heat, sound or light generation. Since, all forms of energy are equivalent, one can circumscribe it as heat transfer only.
Dissipative work is considered irreversible as there is no way one can reverse the act of friction.
In such an adiabatic expansion entropy changes due to volume increase and 'irreversible heat transfer'. Since, entropy is a state function its change in value remains the same in a reversible as well as an irreversible process. So, it is instructed to 'find a reversible pathway for the process and calculate the entropy change'.


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Importance of zinc blende and wurzite crystal structures for semiconductors?
I am currently studying the textbook Physics of Photonics Devices, Second Edition, by Shun Lien Chuang. In a section discussing the basic concepts of semiconductor band and bonding diagrams, the author gives the following description:

The basic idea is that for a semiconductor, such as $\ce{GaAs}$ or $\ce{InP}$, many interesting optical properties occur near the band edges. For example, Table 1.1 shows part of the periodic table with many of the elements that are important for semiconductors, including group IV, III-V, and II-VI compounds. For a III-V compound semiconductor such as $\ce{GaAs}$, the gallium ($\ce{Ga}$) and arsenic ($\ce{As}$) atoms form a zinc-blende structure, which consists of two interpenetrating face-centered-cubic lattices, one made of gallium atoms and the other made of arsenic atoms (Fig. 1.1).

This seems to be the relevant Wikipedia article for zinc-blende (Zincblende) structure.
I was interested in understanding the relevance of this structure to semiconductors, so I did further research. I then came across this website, which says the following:

Here are the two most important crystal structures for semiconductors.
They are often referred to by the historical names "Zinc blende" from the German "Zinkblende" $= \ce{\alpha-ZnS}$, a rather ubiquitous mineral. The name "Sphalerite" also comes form the German: "Sphalerit", which, as was the custom of the time, stems from the Greek "sphaleros" meaning treacherous or malicious because it is easy to confuse it with other minerals.
Wurtzite was and is the name of the $\ce{\beta-ZnS}$ modification - the hexagonal high-temperature variant. The named after the French chemist C. A. Wurtz (* 1817, † 1884), which gives us an idea of how old those names are.
Zinc blende or sphalerite or Diamond structure

Wurtzite or hexagonal structure
Red lines are not showing bonds.

However, this website does not actually explain why these are the two most important crystal structures for semiconductors. I was wondering if someone would please take the time to explain this.
EDIT:
Thinking about Justanotherchemist's comment and Oscar Lanzi's answer, I wonder if all group $\ce{III-V}$ semiconductors have these $\ce{ZnS}$ structures? Because, if so, then that could very well be the answer to my question here, since, as I understand it, group $\ce{III-V}$ semiconductors are the most important semiconductors for practical purposes? As Oscar Lanzi said in his answer, not all semiconductors have a $\ce{ZnS}$ structure, so the answer cannot be that they are the most important because all semiconductors have a $\ce{ZnS}$ structure.

I don't know if the structure has something to do with it's ability to be semiconducting but these III-V structures adopt the ZnS structures due to something we call 'Grimm-Sommerfeld' rule. Same applies for I-VII as well in for example AgI.

@Justanotherchemist Interesting. Thanks for the information.

@Justanotherchemist Thinking about your comment and Oscar Lanzi's answer, I wonder if all group $\ce{III-V}$ semiconductors have these $\ce{ZnS}$ structures? Because, if so, then that could very well be the answer to my question here, since, as I understand it, group $\ce{III-V}$ semiconductors are the most important semiconductors for practical purposes? As Oscar Lanzi said in his answer, not all semiconductors have a $\ce{ZnS}$ structure, so the answer cannot be that they are the most important because all semiconductors have a $\ce{ZnS}$ structure.

As suggested I can try to summarize the basic idea in an answer although I wouldn't say it solves the bounty. I'm also unsure why you even set the bounty for it.
For this I'll quote the book that I found a few days ago. As I own the missing pages now as well the answer seems a bit more elaborate then before. As the book is in German I will translate the text.

The formation of the band-structure of typical element semi-conductors like diamond (C), Si and Ge was already shown in chapter 7.3. By mixing s- and p-wavefunctions we'll get a tetrahedral bonding orbital ($sp^3$), that will lead to a splitting into a bonding and an anti-bonding orbital in the range of the equilibrium-bonding distance. The bonding orbitals form the valence band and the anti-bonding orbitals form the conducting band. The distribution of the four s- and p-electrons into bonding and anti-bonding orbitals leads to a completely filled valence band while the conducting band remains empty.
  The result is therefore an insulator like diamond or, if the energy gap between the valence and conducting band is smaller, to semi-conductors like Si or Ge.

I'll skip the next pages that deal with the formation of the band-structure and the symmetry aspects as I have no experience in these topics and continue with the $III-V$-semi-conductors.

As the formation of $sp^3$ hybrid orbitals in the chemical bonding of Si and Ge seem to be the a substancial part of the semi-conducting properties it seems reasonable that other materials with a tetrahedral crystal structure, i.e. $sp^3$ hybridisation, should show semi-conducting properties as well.

At this point I will switch to basic solid state chemistry as the book doesn't elaborate this part well enough. As solid state chemist I often have examples like these with my students. A while ago we were discussing the silver-mobility in high-temperature $AgI$, which shows quite some conductitiy through defects and the low temperature modification. At this point I would typically ask for reasons why $AgI$ might be problematic. Typical answer that I expect are:

HSAB-concept  with soft $Ag^+$ and soft $I^-$
A covalent three-dimensional tetrahedral network with ZnS-structure

For the second part they either check these facts before we talk about it or they can use the things they have learned in lectures. One of them being the Grimm-Sommerfeld rule. Basically, compounds, that form between two elements to the left and to the right of another element will likely adopt the structure of said element. So for a group $IV$ element like the silicon structure, it can be $III-V$ or $II-VI$ or $I-VII$. It's one of these rules that are taught in the very first semester but people forget about it although it's super useful for compounds you don't know.
Therefore my idea would be here that if, as the book mentions, $sp^3$ hybridisations and covalent, tetrahedral structures as the main source for the semi-conducting properties it's only natural that $III-V$ compounds show a similar effect. Those bands are basically molecular orbitals. If you start from a di-atomic molecule and increase the amount of bonding partners you will get more and more orbitals that move closer and closer together until you have bands. If let's say Ge has a certain band-gap and you change to GaAs, or course the single contributions to the HOMO and LUMO and later to the bands will be different but not that far away from Ge, so we expect a comparable band-gap.
Source: H. Ibach, H. Lüth, Festkörperphysik: Einführung in die Grundlagen, 7th edition, Springer, 2008.

Thanks for taking the time to post this excellent answer! So just to clarify something that I wasn't clear on from the textbook, but that seems to be implied: Group III-V compound semiconductors all form a zinc-blende structure?

That's a good question, at least hat high pressure they will go into the NaCl structure. Some lead compounds might have problems as the lone-pair will require some space. There is an entry somewhere here where I explained how the PbO structure is created from the CaF2 structure. There it should become obvious why no ZnS is formed.

But the author of the website says that the two $\ce{ZnS}$ crystal structures are the most important for semiconductors. And we are told that silicon and germanium semiconductors form the $\ce{ZnS}$ structure, and that they have $\ce{sp^3}$ hybridization. But I'm struggling to understand what the connection is between these two. The former seems to be a statement about all semiconductors, whereas the latter seems to be a statement about silicon and germanium semiconductors.

Well it's ZnS crystallizing in the Silicon structure rather than the other way but yeah. I have to guess here. The most common ones are in fact all ZnS-type. There may be some that do not form this structure (although I cannot find any) but still this wouldn't matter. The bands form from a combination of the two elements orbitals. If the energy gap is within a certain range it's semi-conductive, if they overlap and delocalize it's conductive, if the gap is really high it's an insulator. Hence other structures besides ZnS can be semi-conducting as well.

Interesting. Thank you for taking the time to do all of this research and explain it to me. This has been very illuminating.

Zincblende and wurtzite structures are not the only ones that afford semiconducting materials.  This short list of narrow-gap semiconductors involves compounds with a variety of crystal structures.  Aside from the fact that several examples clearly have different structures because they do not have 1:1 stoichiometry, the examples lead sulfide (PbS) and lead selenide (PbSe) have halite structures.

Interesting, but that doesn't tell us why, if at all, these are the "two most important crystal structures for semiconductors".

It may have nothing to do with semiconduction as such.  It may have more to do with what has been applied in practice.

I think that's what they're saying: These two crystal structures are the most important for semiconductor applications. Or am I misunderstanding what you're saying?

Thinking about your answer and Justanotherchemist's comment, I wonder if all group $\ce{III-V}$ semiconductors have these $\ce{ZnS}$ structures? Because, if so, then that could very well be the answer to my question here, since, as I understand it, group $\ce{III-V}$ semiconductors are the most important semiconductors for practical purposes? As your answer says, not all semiconductors have a $\ce{ZnS}$ structure, so the answer cannot be that they are the most important because all semiconductors have a $\ce{ZnS}$ structure.

Unfortunately I'm just a solid state chemist and no physicist. I can only tell you that most things like I-VII, II-VI, III-V will adopt either the hexagonal Wurzite or the cubic Sphalerite type ZnS structure. It was mentioned in a book somewhere that in Silicon and Germanium the sp³-hybrids formed in these structure might actually be responsible for the semiconducting properties. Unfortunately Google Books does not show me the rest of the pages.

They also argumented that as most of these III-V compounds like GaAs or so are somewhat covalent we can still think of sp³-hybrids but with different band gaps. That might actually be the answer here. Your original IV structure like Silicon has interesting properties. If you change the elements but remain similar bonding situations the different orbital energies overlapping will lead to new band structures.

You can see something similar with Graphite. Graphite can be seen as big system of connected benzene rings. Which means there are orbitals all pointing orthogonally where current can be transported. If we switch to BN however, though it's the same structure, the nitrogen will draw more of the electron density and the thing turns white and non-conductive.

@Justanotherchemist Ahh, you make some very interesting points. Indeed, what you mentioned may very well be the answer to my question. Thanks for the clarification!

@Justanotherchemist Do you want to post an answer? The bounty expires in 3 days, and so far you're the only one who seems to have a somewhat reasonable idea of what the answer might be, so it would be better to have the option to give you the bounty than let it expire and go to waste.


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Producing hydrogen sulfide at home - how dangerous is it?
When I was a kid, I used to go to the pharmacy to buy sulfur and iron powder, then mix it together and heat it until it starts to react into iron sulfide. Put a chunk of iron sulfide into some vinegar, and you get a room smelling badly of rotten eggs. This experiment was also described in some "chemistry for kids" boxes from the 80's.
Now I'd like to do this experiment with my son, but in the meantime I learned that hydrogen sulfide is a very potent poison, even more poisonous than hydrocyanic acid. For this reason, I'd like to set up some precautions.
Of course I won't do it in a closed room, and I won't let my son breathe in much of the gas.
The gas can be smelled long before it starts to get dangerous, so I think it should be pretty easy to use an amount of sulfur and iron that will cause the characteristic smell, but at the same time is guaranteed to not cause any danger. So, I'm looking for some quantitative estimation:
How much iron sulfide would be enough to create a dangerous amount of hydrogen sulfide? To make the calculation easier, let's assume we grind the iron sulfide and use a stronger acid than vinegar, so that all of the gas will be created almost instantly.
If possible, it would be great to have two estimations: one of the beginning of the "danger zone", and one of the beginning of the "lethal zone". According to Wikipedia, the gas starts to smell at $\pu{0.00047ppm}$, the danger zone starts at about $\pu{10 ppm}$, and at about $\pu{320 ppm}$ it can be lethal.

Not a proper answer, but as long as you do it outdoors, I have a hard time seeing now you will reach any lethal amounts. You'll likely puke your guts out way before that happens, and if you start puking your guts out, it is about time to abort mission.

Can you tolerate the smell of rotten eggs?

In our lab practicals (where we had to work with hydrogen sulfide), we have been told not to rely on our sense of smell, which could become dull over time. Take care to get ventilated.

@StianYttervik Note that you can't count on the smell of the gas, since at high concentrations, the gas paralyzes your olfactory receptors, so you won't smell anything when it starts to become dangerous. Yes, I also don't expect to reach lethal amounts, I've done the experiment many times, but anyway, I want to have some quantitative estimation.

@StianYttervik "... if you start puking your guts out, it is about time to abort mission." I wish someone had told me that when I started university.

@cheesussaysstopfiringmods Uh, yes you can. You most certainly can. If you choose to stay there, revolted and alarmed, then over time your sense of smell will disappear. Or, if you suddenly walk into a room with >100ppm you could theoretically also not smell it. But rest assured, all the way up to olfactory paralysis will be a foul smelling fight against your desire to leave. And as I said, as long as you are outdoors where lethal concentrations are unlikely to hang around.

The main rule of chemistry for kids boxes is to use the minimal amount required to demonstrate an effect. Something littler than a a little pinch will suffice so go ahead. I guess prank vials also contain H2S. They were common when I was a kid :)) for carnival

You might look for the paper strip mentioned here for a round elegant :) experiment - evolution then capture of the gas. https://chemistry.stackexchange.com/questions/126728/indicator-for-h2s

Not dangerous to me at all, because I wouldn't do it.

On the question as to how dangerous is H2S let me detail some facts that I find problematic. Here is a good paper.
I summarize some of my findings:
First, H2S after initial exposure can deaden your sense of smell (10-minute exposure at 50-100 ppm impairs the sense of smell). So, one may not really have an accurate conception of the magnitude of one's time exposure to the gas.
Second, skin exposure may be a path to ingestion, especially if moist I suspect. Note, some sources actually disagreed on skin exposure poisoning of Hydrogen sulfide with one study alluring to skin toxicosis from high concentrations over a long period.
Third, following a significant exposure event, the very fact that you are still alive should not be comforting. People having incurred a lethal dose may be a de facto walking dead with a slow hourly progression to death.
Fourth, the ppm considered fatal for H2S is not that far removed from some known very toxic gases (like HCN). However, my opening reference doubts if one can even assign accurate figures on ingestion via the lungs. Here is an extract addressing dosing level ranges:

Findings in the canary, cat, dog, goat, guinea pig, rabbit and rat (2) are consistent as to the effects of hydrogen sulfide: at 150-225 mg/m3 signs of local irritation of eyes and throat after many hours of exposure; at 300-400 mg/m3 eye and mucous membrane irritation in 1 hour and slight general effects with longer exposure; at 750-1000 mg/m3 slight systemic symptoms in less than 1 hour and possible death after several hours; at 1350 mg/m3 grave systemic effects within 30 minutes and death in less than 1 hour; at 2250 mg/m3 collapse and death within 15-30 minutes, and at 2700 mg/m3 immediate collapse, respiratory paralysis and death.

Viewing all these together, if one has a choice to avoid working or experimenting with the gas, I would do so.

Thank you. Not sure if I understand your last sentence. As long as you safely stay within some reasonable quantities which are far from the lethal ones, why would you want to avoid experimenting with the gas? It might be even a safety advantage if students know how the gas smells. Not everyone knows how "rotten eggs" actually smell.

To answer your question, some professionals (with safety gear), including, for example, sewer workers, still manage to die from toxic exposure. Hence, my speculation of moist skin (especially someone working in hot humid conditions) being a not clearly specified route to ingestion. Also, read my cited reference on the attack of organs including the brain,  now that is troubling.

If one decides that working in targetted safe ranges with H2S is the best and safest alternative, that is keeping, I believe, with working with good judgment. However, at times experiments may involve errors (like in quantities employed) and thus may result in an accidental exposure. With that in mind as a forethought, was there another path not employing H2S?

According to Wikipedia, at sea level, 20°C warm air has a density of $1.20 kg/m^3$. 78% of it is Dinitrogen, which under the same conditions has a density of $1.17 kg/m^3$. To simplify the calculations, we will assume that the air is 100% dinitrogen, which is a reasonable simplification because the density is so similar. So a cubic meter of air weighs $1.17kg$, and because $N_2$ has a molar mass of $28.014$, we have $41.76$ moles of dinitrogen in a $m^3$.
To produce enough hydrogen sulfide for $1000$ ppm in one cubic meter, we would thus need about $0.041$ moles of $H_2S$ (neglecting the density of $H_2S$, since there is so much more $N_2$). With a molecular mass of $34.08g / mol$, this means $1.39 g$ of $H_2S$ in one cubic meter of air is about $1000$ ppm.
$88$% of the weight of $H_2S$ comes from the sulfur, so we need roughly $1.23$ grams of sulfur to create this amount of $H_2S$.
I hope this is more or less correct.


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Would the same equilibrium concentration of carbon dioxide, carbon monoxide and oxygen always result for a given initial atomic carbon fraction?
Question: Starting with a gas mixture of some combination of gases containing carbon and oxygen with a total carbon atomic fraction $f$, would it always reach the same equilibrium proportions of $\ce{CO2 + O2 + CO}$?
For example if I had an equal number of $\ce{CO2}$ and $\ce{O2}$ molecules then $f=0.2$, but so would two molecules of $\ce{CO2}$ for every three molecules of $\ce{O2}$. If I started with either mixture in this example, would I end up with the same equilibrium mixture of $\ce{CO2}$ and $\ce{CO}$ and $\ce{O2}$?
Simplifying assumptions: Assume $\pu{273K}$ and $\pu{610Pa}$ if necessary, the answer may be as simple as "yes it would" or "no it wouldn't, because...", a supply of ionizing radiation from space (hard UV and energetic cosmic ray particles) and a billion years, and ignore transmutation of elements by those particles.

Background and motivation is from my Space Exploration SE question Are Mars' atmospheric CO₂, O₂ and CO in equilibrium? Are sunlight or chemical reactions involved?, which begins:

According to the NASA JPL video linked below the top five gases comprising the martian atmosphere include $\ce{CO₂, O₂}$ and $\ce{CO}$. Do the proportions reflect some chemical equilibrium? Are sunlight or chemical reactions involved in maintaining these proportions?

Component     Fraction
   CO2          95.9%
   Ar            2.0%
   N2            1.9%
   O2            0.14%
   CO            0.06%

The NASA JPL video Crazy Engineering: Making Oxygen on Mars with MOXIE is linked in How many square meters of solar panel to power a MOXIE-like oxygen source per person?.

I'm guessing that the answer is "yes it would" but I'm not a chemist.

As soon as the mixture is allowed to reach equilibrium in the first place, yes. (I guess a billion years is quite enough.) Short of that, no.

Obviously not, because mars rotates and has a climate, with seasons. ;)

@SurArthur7 thanks for the edit! I've never seen \pu{} used before but it looks great. I didn't find it in this tutorial so I'm wondering 1) what it's called, and 2) if it should be added there somewhere?

@Karl this question is not about Mars. That was included as "Background and motivation" but for this question in Chemistry SE I've stipulated a constant temperature and pressure for this gas mixture.

I would add that you should assume in addition that oxygen is present in the same atomic ratio to carbon. By the definition of an equilibrium constant you would expect the one equilibrium composition at constant T,P (nitrogen being assumed inertial). However if you are irradiating the system with high energy particles you might see substantial fluctuations (or perhaps steady-state but not true equilibrium). Something like what happens in our own ozone layer.

Interesting aside, Wikipedia says: "Venus also has a thin ozone layer at an altitude of 100 kilometers above the planet's surface.[5] " Venus' atmosphere is mostly CO2.

@BuckThorn I think Venus' atomosphere is going to turn out to be a really interesting case, much more complicated than Mars'. There are now three approved missions (NASA + ESA) and there will be detailed atmospheric studies... https://www.nature.com/articles/d41586-021-01634-3 and ESA's and NASA's


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Why does hydrogen have a lower ionization energy than fluorine?
I found here that the ionization energy of hydrogen is $\pu{1312kJ/mol}$ and for fluorine, it is $\pu{1681kJ/mol}$. Now clearly, from the data, we can see that hydrogen has a lower ionization energy when compared to fluorine.
I did not expect this because I saw here that fluorine is twice as big as hydrogen (I am referring to the empirical values of the atomic radii). This means that it would be significantly easier to remove an electron from a fluorine atom than from a hydrogen atom. That's why I expected hydrogen to have a higher ionization energy. But I found that it's the other way round. Why is it so? What is wrong with my reasoning? Is this an exception?

@SirArthur7 btw, you don't need to typeset everything in MathJax, so you can save yourself some effort; 1681 kJ/mol is typographically just as correct as $\pu{1681 kJ/mol}$, and H2O (try it on a post) is just as correct as $\ce{H2O} $\ce{H2O}$. The most important thing is to be consistent. In this case, it arguably looks better without the Jax, because then it fits into the surrounding text better. See: https://chemistry.meta.stackexchange.com/a/2935/16683

@orthocresol You know what? Your linked post was for long an ideal doubt of mine, Thanks a lot for clearing out this, this'd surely save myself from needlessly editing units, but if it's really not a problem could I use \ce for $\ce{H2O}$, is it the buffering time that you want me to avoid? If so then H2O could be fine.

Fluorine is bigger, but there are $9$ protons in the nucleus, compared to $1$ in case of hydrogen. So, the outermost electron in fluorine experiences more attraction than what hydrogen's single electron experiences, in spite of larger size of fluorine.
However, the outermost electron in fluorine doesn't experience attraction from all the $9$ protons (nuclear charge). Due to improper shielding, the effective nuclear charge experienced by one $2p$ electron of fluorine becomes -
$\mathrm{ [9 - {(6 \times 0.35) + (2 \times0.85)}]= 5.2}$.
In comparison, the single electron in hydrogen experiences effective nuclear charge of $1.0$
Note: For hydrogen, there is no other electron to shield nuclear charge. So, effective nuclear charge = actual nuclear charge.
Source: To know how to calculate effective nuclear charge visit this Wikihow link.

But then the ionization energy should increase down the group due to increasing nuclear charge. Why does that not happen?

According to Coulomb's law, attractive force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
So, it is basically a competition between the two as Ionization Energy estimation involves calculation of attractive force between the nucleus & the outermost electron.

In the same group, the size factor dominates the charge factor. And that is pretty regular in most cases. But the comparison is tricky when you compare between different groups & periods.
(Note: If you calculate the Effective Nuclear Charge for each members of a group, you will get a clearer idea about why the size factor is getting dominated.)
Li has lesser ionization energy than H. But, along a period Ionization energy increases gradually, with some exceptions, and becomes the maximum at the Noble Gas element.

There is confirmation of the ionization values being for Flourine $1680$ $\pu{kJ/mol}$ and Hydrogen $1312$ $\pu{KJ/mol}$ from the Theoretical Chemistry Textbook (online link).  Theoretically, if you have a chemical like $\ce{HCl}$ it is very understandable that it forms an ionic bond, because the the out-most electron is held with less force for Hydrogen than with Chlorine which takes the electron from Hydrogen in an ionic bond. And Florine is in the same column in the periodic table as Chlorine, so its behavior is similar.
At ValenceElectrons.com the Aufbau electron configuration of Florine is listed as  $1S^2$ $2S^2$ $2P^5$:

It is then important to understand the Aufbau principle further as it represents hybrid energy levels because of a combination of Theoretical Hydrogen charge distributions.  The $S$ $orbitals$ and $P$ $orbitals$ are somewhat combined in atoms aside from Hydrogen, and their energies can be shown in the picture as hybrid $P$ $Bonds$:

In the referenced diagram, just above, due to the Aufbau principle and hybrid orbitals, the energy of the $2$ $S$ $Orbitals$ are lowered, making the energy-well deeper because of the hybrid bonds from the $2$ $S$ and $2$ $P$ levels.  That means that going across a row in the periodic table (with $S$ and $P$ levels to it) that generally it becomes more difficult from left-most (easiest) to right-most (most-difficult) element to remove the outer-most electron, as a general trend across the row.
In the far-field, the $2$ $S$ electrons are not fully shielded and are attracted with a partial $+2e^+$ charge, resulting on a tighter hold of these electrons and a greater bonding energy then normally expected for Hydrogen which in the far-field has no-shielding and only a $+1e^+$ to hold the electron in its shell.
The Theoretical Chemistry Textbook further explains:

The Effects of Electron Shells on Ionization Energy
Electron orbitals are separated into various shells which have strong
impacts on the ionization energies of the various electrons. For
instance, let us look at aluminum. Aluminum is the first element of
its period with electrons in the 3p shell. This makes the first
ionization energy comparably low to the other elements in the same
period, because it only has to get rid of one electron to make a
stable 3s shell, the new valence electron shell. However, once you've
moved past the first ionization energy into the second ionization
energy, there is a large jump in the amount of energy required to
expel another electron. This is because you now are trying to take an
electron from a fairly stable and full 3s electron shell. Electron
shells are also responsible for the shielding that was explained
above.

For Florine, before ionization, it has seven outer shell electrons.  Before ionization, the outer shell structure can be described as s s p p p p p (since the inner $1$ $S$ shell stays basically the same in before and after ionization). The shell has more of a s characteristic then after ionization.
Before ionization, the outer shell structure of Florine can be described as s s p p p p p.  After ionization, the outer shell structure can be described as s s p p p p which has a greater s influence.
The general trend versus row position can be viewed at the below diagram:

I have calculated this effect of the ionization energy of Helium being greater than that of Hydrogen, with some accuracy.  The calculations for Helium are referenced my the free online-book here. The same concept of incomplete shielding for Florine is what is dominating its higher biding energy than that for Hydrogen also.

Your intuition is spot on : the issue is with the definition of atomic radius, which according to that same Wikipedia page "is not a well-defined physical entity, for which there are various non-equivalent definitions". This ambiguity finds its roots in the quantum mechanical description of the atom.
If you use the column with the "calculated" value, you'll see that the radius for hydrogen is 53 and for fluorine 42, i.e. hydrogen is bigger than fluorine. Correspondingly, the ionization energy is smaller, roughly by a factor $\frac{42}{53}$, in line with your reasoning.
Mills* classical model of physics allows us to calculate these radii and ionization energies fairly precisely, as it describes electrons as having fixed distances from the nucleus. The classical Bohr radius (52.91pm) for hydrogen, and 0.7807 times the Bohr radius (41.39pm) for fluorine. The approximate ionization energy, in Rydberg atomic units, is then simply given by $\frac{Z}{r}$ i.e. 1 for hydrogen and $\frac{Z-8}{r}$, or 1.281 for fluorine, with each electron shielding exactly one proton. (Multiply by 1313 to convert to kJ/mol)

*Randell L. Mills, The Grand Unified Theory of Classical Physics, 2018 edition (ISBN: 978-0-9635171-5-9) see equations 10.182 and 10.183


text:
Number of possible isomers (including optical/geometric ones) with C4H8O molecular formula
I found 32 isomers but I think I made a miscount.

Chain alcohols: 6, chain ethers: 5, and chain carbonyl compounds
(ketones/aldehydes): 3
Three-member ring alcohols: 7, three-member ethers: 6 (including ones
that have the ether-bond outside the ring)
Four-member ring compounds: 4, five-member compounds: 1

(Note: Enols are not counted because they're unstable)

It would have been nicer if you included the isomers what you have been counting. However, based on the isomer counts, I assumed you are in correct track but originally thought you might have counted meso-isomer twice (as two stereoisomers) for the total count. Thus I drew all possible isomers and listed accordingly:

Consequently, I also counted 32 isomers (same as yours). When compared to your count, I realized your count of chain alcohols (6), ethers (5), and carbonyl compounds (3) are all correct.
However, your count of three-membered ring alcohols (7) and three-membered ring ethers (6) are incorrect. There are only 6 possible three-membered ring alcohols (one less than your count) while 7 three-membered ring ethers are possible  (one more than your count; see the chart). Therefore, it might not be the miscount of meso-isomer (if this is the reason, you should count 8 possible three-membered ring ethers), but be an error on structure drawing.
Your final count of for-membered ring (4) and five-membered ring (1) ethers are also correct. Precisely, only 1 four-membered ring alcohol is possible.
Conclusively, you and I both have the same structure count but wrong counts on different kind of structures. You can compare your drawing with attached chart here and find out where you made the mistake.

Easy to understand and to the point...as always :)

@Sir Arthur7: Thanks for edit. Good catch.

Is 1-methyl-1-cyclopropanol impossible?

@Harry K: Good catch Harry,  Yes, it is possible. It's just a mental erro on my side. Thanks for pointing, I'll clean this up soon. :-)