Oxide surfaces develop a charge in H2O. Do we have any, even vague, handle on how long does an indivdual charge last on avg. before being neutralised? We often know, from e.g. titration experiments, what the average surface charge density of a surface is.

How do these charges vary over both time and space (i.e. dynamics)? I have been unable to find much (well, any really!) literature on this.

Example

Taking the example of oxide surfaces, you have the following processes going on:

$\ce{S-O- + H3O+<=>[\text{K1}]S-OH +H2O}$ #Protonation/Autoionisation

$\ce{S-OH + OH- <=>[\text{K2}]S-O- +H2O}$ #Deprotonation/Protonation

$\ce{S-OH + H3O+ <=>[\text{K3}]S-OH2+ +H2O}$ #Protonation/Deprotonation

If you know (from experimental/theoetrical calculation of the Equilibrium Constants K1, K2, K3), that you have 100 negative charges per micrometer squared (i.e we are at high pH). Is it a more realistic picture on a picosecond/nanosecond timescale to consider it being something like 10,000 charges per micrometer squared which are being regularly neutralised over time or simply 100 charges per micrometer which are fixed?

Proton transfer reactions in water such as the ones you wrote down are diffusion controlled under ambient conditions (rate constant about 1e+12 per sec.) A snapshot in time will have a certain distribution of charges on the surface, but these are continually in flux, so any one charged site will have an exceedingly short lifetime. You could reasonably represent that system with a fixed set of charges on the surface, for some types of models (e.g., states at equilibrium).


Latent Heat, Vapor pressure and Equilibrium So after reading about this topic i wanted to share what i learned and i wanted you guys to tell me if what i am writting is correct:

A pressure vessel has propane inside and it is pressurized. The liquid is at atmospheric temperature meaning that it is in equilibrium with the outside temperature. The pressure inside the pressure vessel will change according with ambient temperature.

If I open the valve of the pressure vessel, the gas will come out and the pressure inside the vessel will decrease and the liquid propane inside the vessel will immediately star to boil so it can re-establish the equilibrium. When boiling the liquid goes into the gaseous state and to do this it requires energy in the form of heat (latent heat of vaporization).

Since the liquid has the same temperature that the outside it cannot take heat from the outside and it will need to take this energy from itself, decreasing the temperature of the liquid.

Now there is a ΔT between the liquid and the outside temperature, so the liquid will steal heat from the outside (sensible heat) so it can incease his temeprature in order to have enough energy to go to the gaseous state.

If the flow of gas that is coming out of the vessel is constant the pressure and the temperature of the liquid and of the gas will initially decrease a little but it will stabilize, because the heat that is being taken from the outside is enough to keep the liquid boiling without decreasing more of his temperature and pressure.

If the flow of gas that is coming out of the vessel increases the outside heat may not be enough to maintain the flow of gas required, so the temperature of the liquid will start to decrease (auto refrigeration) (heat absorbed by the liquid < heat used by liquid to go to the gaseous state due to the high gas demand)

If the temperature of the liquid keeps decreasing so will the pressure and if it decreases below the atmospheric boiling point, the liquid will stop boiling and the gas will stop to come out (negative pressure inside the pressure vessel).

All is about  3 basic principles.

Liquid evaporation consumes thermal energy, taken from liquid itself or surrounding. Vapour condensation releases this energy.
Liquid saturated vapour pressure (about) exponentially raises with temperature.
If saturated vapour pressure exceeds external pressure, liquid boils. If vapour pressure exceeds the saturated vapour pressure, vapour condensates.


Doubts Regarding Ceric Ammonium Nitrate Test for Alcohols I read about Ceric Ammonium Nitrate Test for Alcohols and a red colour is produced if the solution has an alcohol functional group however as a side note it was mentioned that this test is only useful for compounds having less than 10 Carbon atoms. Why is it so?

The question does not mention in what the ammonium cerium nitrate is dissolved in, but I guess it is water. If so, one expects that sufficiently large compounds that one wishes to test are not sufficiently soluble in water, such as decanol. Obviously, this is not entirely correct because disaccharides such as lactose do are soluble in water.

I expect a sufficient solubility to be important for the test such that it finishes in a timely manner and does not become inconclusive due to prolonged reactions with air, water etc.


Why is the C-Br bond considered polar? Why is a C-Br bond considered polar? From electronegativity considerations, both carbon and bromine have very similar electronegativities - 2.5 and 2.8 - respectively.

Nonetheless, I am told that such a bond would be polar. We generally classify bond between atoms with EN differences < 0.5 as non-polar; why would C-Br be an exception?

I poked around on the internet and found a lot of sites use the Pauling electronegativity scale (link 1, link 2).  For carbon the value is 2.55 and for bromine 2.96.  Not quite a 0.5 difference, but closer - and this "rule" is somewhat arbitrary to begin with.  Also the dipole moment of bromobenzene is around 1.7 D (link 3), certainly not insignificant.  I too would have said bromobenzene is polar.  It is not strongly polar, whether it is "a little" polar, "mildly" polar, "somewhat" polar, etc. seems to be the question - but that's just semantics.  Maybe the reason many people consider the $\ce{C-Br}$ bond to be polar is because it has a very observable effect on electrophilic aromatic substitution reaction rates.  Specifically, why should the relative rate for electrophilic aromatic substitution at the meta position in bromobenzene be slower that the rate in benzene itself unless you invoke the inductive effects caused by a polar $\ce{C-Br}$ bond?


Would sodium metal react differently with aqueous solutions than with pure water? I know sodium metal reacts violently with water, but what about aqueous solutions (like sodium bicarbonate). Also, what would the reaction kinetics be?

I have watched YouTube videos were a small piece of elementary sodium reacts completely with water in a matter of seconds. Would that also be the case of elementary sodium in aqueous solutions (for example sodium bicarbonate)?

It helps to know what the reaction actually is. When sodium metal reacts with water, it is oxidised and the hydrogen in water is reduced according to the following equation:

$$\ce{2Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$

Note that no other components play a role; the only thing that does is the $\mathrm{pH}$ value of the water (lower $\mathrm{pH}$ would mean a more violent reaction) — this is a spoiler and later shown in equation $(8)$. Since this is a redox reaction, we can use standard half-cell theory to analyse it. Most importantly, thanks to equation $(2)$, only the effective cell potential influences $\Delta G$ ($z$ is the number of electrons transferred while $F$ is Faraday’s constant) which in turn can be separated into two half-cells according to equation $(3)$ which can then be analysed by the Nernst equation $(4)$.

$$\Delta G_\text{cell} = - n_{\ce{e-}}FE_\text{cell}\tag{2}$$
$$E_\text{cell} = E_\text{cathode} - E_\text{anode}\tag{3}$$
$$E = E^0 + \frac{RT}{zF} \ln \frac{a_\text{Ox}}{a_\text{red}} = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\text{Ox}}{a_\text{Red}}\tag{4}$$

The half-cell reactions are shown in equations $(5)$ (oxidation) and $(6)$ (reduction) respectively, along with their standard potentials $E^0$ (under standard conditions, i.e. most importantly $\mathrm{pH\ 0}$).

$$\ce{Na+ + e- <=> Na (s)} \qquad E^0 = -2.71~\mathrm{V}\tag{5}$$
$$\ce{2 H+ + 2 e- <=> H2 (g)} \qquad E^0 = 0.000~\mathrm{V}\tag{6}$$

Let’s take a look at the anodic oxidation first. The initial concentration of sodium ions could be, say, $3~\mathrm{M}$. Remember that the activity is defined as $1$ for solids. With the assumption that $a \approx c$, that would mean our oxidation potential is $-2.68~\mathrm{V}$ as shown in equation $(7)$.

$$E_\mathrm{anode} = -2.71~\mathrm{V} + \frac{0.059~\mathrm{V}}{1} \lg \frac{3}{1} = -2.71~\mathrm{V} + 2.82 \cdot 10^{-2}~\mathrm{V} = -2.68~\mathrm{V}\tag{7}$$

The corresponding calculation for the reductive side, equation $(8)$, is a little harder since we have no real starting value for the partial pressure (and thus activity) of hydrogen gas.

$$E_\text{cathode} = \frac{0.059~\mathrm{V}}{2} \lg \frac{a(\ce{H3O+})}{p(\ce{H2})} = -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\tag{8}$$

But we do know that a reaction will happen if $E_\text{cell} > 0~\mathrm{V}$ and we have a good idea of the anode’s potential. Thus, if we want no reaction we must set the $E_\text{cell} < 0$ (equation $(9)$):

$$\begin{align}0 &> E_\text{cathode} - E_\text{anode}\\
- 2.68~\mathrm{V} &> E_\text{cathode}\\
- 2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\end{align}\tag{9}$$

The second summand is negative if $p(\ce{H2}) > 1~\mathrm{bar}$. (The equation itself requires a dimensionless pressure which is typically defined to have a the same value as if it were in $\mathrm{bar}$). However, that basically requires performing the reaction under a hydrogen atmosphere. This setup is rather hard and not part of the question. Instead, let’s take the natural concentration of hydrogen which is $1~\mathrm{ppm}$ according to Wikipedia. That results in $0.177~\mathrm{V}$ for the second summand (equation $(10)$).

$$-0.0295~\mathrm{V} \times \lg 10^{-6} = (-6) \times -0.0295~\mathrm{V} = 0.177~\mathrm{V}\tag{10}$$

At least we didn’t lose too much. But there is still the first summand, which is also negative by sign. We just need a high enough $\mathrm{pH}$ value to compensate. The calculation is in equation $(11)$.

$$\begin{align}-2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} + 0.177~\mathrm{V}\\
-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH}\\
96.9 &< \mathrm{pH}\end{align}\tag{11}$$

So we would need a $\mathrm{pH}$ value of almost $97$ to prevent the reaction from happening due to too basic conditions. Of course, this value is unobtainable in water or practically any solvent known to us. Even tert-butyllithium, one of the most basic species known has a $\mathrm{p}K_\mathrm{a}$ value of only $\approx 50$.

You could now try to argue that you would just need to increase the hydrogen gas partial pressure appropriately. So how does the value change if the reaction is performed under hydrogen atmosphere? See equation $(12)$ for the answer.

$$\begin{align}-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \mathrm{pH}\\
90.8 &< \mathrm{pH}\end{align}\tag{12}$$

As you may or may not have guessed, a logarithmic difference of $6$ for the partial pressure results in a lowering of the $\mathrm{pH}$ by $6$ units. So to arrive at achieveable $\mathrm{pH}$ values, your reaction would need to happen at a hydrogen pressure of $10^{75}~\mathrm{bar}$ — go figure.

All things considered, the reaction is so violently endergonic that there is basically no way to inhibit it whatsoever. If metallic sodium sees a sufficient amount of water, it will start reacting, end of discussion.


Oxidation number of nitrogen in KCN How can one determine the oxidation number of nitrogen in $\ce{KCN}$?

KCN is a ionic compound consisting of $\ce{K+}$ and $\ce{CN-}$. You can know that the cyanide ion will have a negative charge, because a potassium ion always has a positive. Also the elements in the cyanide ion are more electronegative - that means they will have the negative oxidation number.

So we know that since potassium has an oxidation state of +1, the cyanide ion has an overall oxidation state of -1. As stated above, ions have oxidation states corresponding to their charge. But you want to know what the oxidation state of the NITROGEN atom has.

There are rules you can use for polyatomic ions (ions with multiple atoms in it), such as $\ce{CN-}$. Nitrogen is more electronegative than carbon. Thus it will have the negative oxidation state (think: it's better at grabbing electrons, which are negatively charged). It is 3 columns away from being a noble gas, thus its oxidation state it -3. The overall oxidation state of the ion is -1, thus carbon must have an oxidation state of +2.

To sum up: K = +1, C = +2, N = -3. If you're trying to balance an equation though, you don't need to look at C and N individually. You just look at the $\ce{CN-}$ as a whole and the $\ce{K+}$.

How did I know $\ce{CN-}$ was an ion? Since potassium is a metal, and only forms ionic bonds. C and N do not form ionic bonds, they bond covalently, thus they act as a single ion in ionic compounds. Does that make sense? Other than that it's just something you learn to recognize with time!

Read rules for determining oxidation states here:
http://www.tmcleod.org/cgi-bin/chem1/wiki.cgi?action=browse&diff=3&id=OxidationNumber

There's a comment to this answer explaining what causes electronegativity. A general rule of thumb is that electronegativity increases as you go towards higher group number and low period number. You can study it for yourself in the picture below:

(source: wikispaces.com)


Relationship between molar mass and density in solids In ideal gases, it's pretty clear that 1 mole of gas occupies 22.4 L at STP. By knowing this, it's easy to calculate density given molar mass. However, this isn't true for solids. If I know molar mass of a solid, I cannot derive it's density.
I want to clarify whether my understanding of why we can't derive density in case of solids is correct. For solids, since each solid has its own lattice structure, knowing how the molecules are arranged is necessary, because if there is more space between molecules, it will have a lower density, which isn't really an issue in (ideal) gases where the arrangement of molecules/atoms is uniform.

If your sample is solid and crystalline, knowledge of the unit cell's dimension and symmetry (e.g., fcc) and density allows you to determine the molecular mass of your compound.
In fact, in crystallography, you may determine the macroscopic density of your sample (e.g., find a liquid which i) wets your crystals and ii) lets your crystals float) in first place.  If you then know the dimensions of the unit cell by the diffraction experiment, knowledge of the chemical formula per unit cell may tell you the number of formula units per unit cell.  See e.g, this demonstration about Ne.
The example above about Ne may be extended to crystals consisting of molecules for that you may assign atomic volumes to each element that add up to a molecular volume. An early approximation for organic compounds[1] was to assign $\pu{18 Å^3}$ for any atom of $(\ce{C, N, O})$ and $\pu{2 Å^3}$ for any atom of $\ce{H}$.  Decades later,[2] backed by much more experimental data, it was possible to refine these assignments, e.g., to average volumes of $\pu{13.87 Å^3}$ ($\ce{C}$), $\pu{11.8 Å^3}$ ($\ce{N}$), $\pu{11.39 Å^3}$ ($\ce{O}$), and $\pu{5.08 Å^3}$ ($\ce{H}$), respectively.  Large deviations from these guesses from the then experimentally determined molecular volume may be caused by solvent-accessible voids in the crystal structure.[3]
References:
1 Kempster, C. J. E.; Lipson, H., A rapid method for assessing the number of molecules in the unit cell of an organic crystal in Acta. Cryst. B28, 1972, 3674; doi: 10.1107/S056774087200857X.
2 Hofmann, D. W. M., Fast estimation of crystal densities in Acta. Cryst. B58, 2002, 489-493; doi: 10.1107/S0108768101021814.
3 Florence, A. J., The use of supplementary information to solve crystal structures from powder diffraction in International Tables for Crystallography (2019). Vol. H. ch. 4.4, pp. 433-441; doi: 10.1107/97809553602060000959.


How can the outermost electrons in silicon be 2s and 2p electrons? My book says the number of outermost electrons in silicon is $4$. Yes, right. But then it says those are $\ce{2s}$ and $\ce{2p}$ electrons. The electronic configuration of silicon is $\ce{1s^2  2s^2  2p^6  3s^2  3p^2}$.

How are the outermost electrons $\ce{2s}$ and $\ce{2p}$ electrons. Those belong to third shell.

Then it says the maximum number of electrons in the outermost shell is $8$. Correct. But then it says those  are $\ce{2s + 6p}$ electrons. How?

The book means to say that there are 2 electrons in the  $\mathrm{3s}$ subshell and and 2 electrons in the  $\mathrm{3p}$ subshell. You can see this in the electron configuration as

  $\mathrm{...3s^2 3p^2}$

and that the maximum configuration has 2 electrons in the  $\mathrm{3s}$ subshell and 6 electrons in the  $\mathrm{3p}$ subshell which would be

  $\mathrm{...3s^2 3p^6}$


Atomization enthalpies of transition elements So, my book says that transition elements have higher enthalpies of atomization than other elements (say s- or p- block) because of stronger metallic bonding, primarily due to large number of unpaired electrons in the d-orbitals.

Now, if that is the case, then the atomization energy for $\ce{Ni}$ should be less than for $\ce{Fe}$ and $\ce{Co}$, as $\ce{Ni}$ has only 2 unpaired electrons compared to $\ce{Fe}$ and $\ce{Co}$ (4 and 3 respectively).

Actually, it is just the opposite.

$$\begin{array}{cc}
\text{Metal} & \Delta_\text{atom}H(\ce{M})\text{ / }\mathrm{kJ~mol^{-1}} \\ \hline
\ce{Ni} & 430 \\
\ce{Co} & 425 \\
\ce{Fe} & 416
\end{array}$$

Why could this be?

The sorts of generalities always sweep aside other considerations.

$\ce{^26Fe}$ has a body-centered cubic crystal structure.

$\ce{^27Co}$ has a hexagonal close-packed crystal structure.

$\ce{^28Ni}$ has a face-centered cubic crystal structure.

Here is a nice table showing the overall trends.


Is using oxidized aluminium baking forms safe? My apologies in advance if this is not the best forum for this kind of question.
Recently, I've bought a few non-anodized aluminium bread-baking forms, such as this one:

As you can see, there is already some discoloration on them when they come from the shop. My knowledge of chemistry is rudimentary but I presume it is an oxidized aluminum layer (do correct me if I am wrong here).
Now, my wife decided it is a good idea to wash these forms in a dishwasher. The forms became even darker, becoming rather gray over their entire area. There are a lof of posts on various forums and blogs telling you to not put aluminium cookware in dishwashers, and most of them go on about how to remove the oxidized layer.
For my purposes, I do not care how it looks. I would like to know whether it is safe to bake bread in those (post-dishwasher and without removing the oxide layer). There is very little information about this, as the online discussions revolve mostly about removing the gray layer. I've read someone saying that the oxide layer can even be beneficial as it might prevent aluminium leaking into food. This does make sense, presuming that the oxide itself is not harmful and does not leak. Could someone comment on this please? Any reference to scientific literature would be a bonus.
As a backstory, the reason I bought these forms is because they are traditionally used for making specific sorts of rye sourdoughs in the Baltic region. So I'd like to give them a go before buying more modern (and more boring) stainless steel forms.
Thanks in advance!

In accord with this forum's policy on not providing health advice, I  will just comment on the chemistry surrounding any created Al3+ interacting with the superoxide radical anion (which is incidentally, present in the human body). As a source, here is a 2014 article by Tushar Kanti Das, et al, published in Archives of Neuroscience, available here.
The authors explicitly references, in Figure 4, “Formation of Aluminum Superoxide Semi reduced Radical Ion and Aluminum Superoxide Complex (43)”, with described reactions proceeding as follows (also adopting notation and water complexing from yet another article):
$$\ce{[Al(H2O)4](3+) + O2•− <-> [Al(O2•−)(H2O)4](2+) }$$
$$\ce{[Al(O2•−)(H2O)4](2+) + Fe(3+) --> O2 + [Al(H2O)4](3+) + Fe(2+)}$$
And, in acidic conditions:
$$\ce{[Al(H2O)4](3+) + O2•− + H+ <-> [Al(O2•−)(H+)H2O)4](3+)}$$
$$\ce{[Al(O2•−)(H+)H2O)4](3+) + [Al(O2•−)(H+)H2O)4](3+) --> 2 [Al(H2O)4](3+) + H2O2 + O2}$$
where the above chemistry suggests a role for Al3+ in possibly assisting with the subsequent production of miscellaneous powerful radicals associated with a Fenton reaction  as sourced from an iron ion presence (as in blood).
Now, as to the possible originating source of Al3+ itself, in the particular context of baking with Aluminum pans, and further washing said aluminum ware in a dishwasher also employing Baking Soda, is that the latter NaHCO3 on warming in aqueous solutions, apparently will attack Al metal, commencing with the reaction:
$$\ce{ NaHCO3 + H2O + Heat --> NaOH + H2O + CO2 }$$
And further, to quote Wikipedia:

Sodium aluminate is also formed by the action of sodium hydroxide on elemental aluminium which is an amphoteric metal. The reaction is highly exothermic once established and is accompanied by the rapid evolution of hydrogen gas. The reaction is sometimes written as:
$$\ce{ 2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2 }$$
however, the species produced in solution is likely to contain the [Al(OH)4]− ion or perhaps the [Al(H2O)2(OH)4]− ion.[6]

This problem with Aluminum ware is well known (see warnings, for example, here, and also here).
I hope this, in part, advanced chemistry presentation nevertheless is of assistance in answering your question "Is using oxidized aluminium baking forms safe?".


Entropy Change of Resistor A very large swimming pool filled with water of temperature equal to $20\ \mathrm{^\circ C}$ is heated
  by a resistor with a heating power of $500\ \mathrm{W}$ for $20$ minutes. Assuming the water in the
  pool is not in any contact with anything besides the resistor:

  Is the change of entropy of the resistor positive, negative, or zero?

My Attempt

We know that entropy is given by the formula:

$$\mathrm dS = \frac{q_\text{rev}}{T}$$

My first thought was that since the resistor is giving off heat energy to the water, $q$ must be negative and therefore the entropy change must be negative. However, does the fact that there is a continuous current of electrons passing through the resistor complicate this simple calculation? I am thinking that the current of electrons increase the entropy of the resistor and offsets the decrease in entropy due to heat loss. Am I correct?

  The answer in the book is that the entropy change is $0$.

Here entropy change will be zero because resistor convert the electrical energy into heating so it take as work not heat so entropy change will be zero


Why was the prefix 'bi' used in compounds, such as for bicarb of soda? According to the Wikipedia article IUPAC nomenclature of inorganic chemistry,

  he prefix bi- is a deprecated way of indicating the presence of a single hydrogen ion

A very common example is the commonplace 'bicarb of soda', or sodium bicarbonate (or using its correct chemical name sodium hydrogen carbonate). On the Wikipedia page for sodium bicarbonate there is a note that states that:

  The prefix "bi" in "bicarbonate" comes from an outdated naming system and is based on the observation that there is two times as much carbonate ($\ce{CO3}$) in sodium bicarbonate ($\ce{NaHCO3}$) and other bicarbonates as in sodium carbonate ($\ce{Na2CO3}$) and other carbonates.

However, the statement above is not particularly clear, especially as carbonates are not the only compounds that used the 'bi' prefix, such as:

Bisulphite $\ce{HSO3-}$

Bisulphate $\ce{HSO4-}$

Why was the prefix 'bi' used to indicate the presence of a single hydrogen ion in compounds?

According to the Online Etymology Dictionary, the word bicarbonate is derived as follows:

  1814, bi-carbonate of potash, apparently coined by English chemist William Hyde Wollaston (1766-1828), from bi- + carbonate.

The question then becomes, what is the meaning of the prefix bi- here?  We know it means two, but it also can mean double or twice.  If that is the case, then sodium bicarbonate is sodium double carbonate.

Consider a given quantity of acid, let's say $\ce{HCl}$, that we are going to neutralize with sodium carbonate, $\ce{Na2CO3}$, and an equal quantity of acid that we are going to neutralize with sodium bicarbonate, $\ce{NaHCO3}$.  We add the sodium carbonate to the acid, and as the neutralization reaction goes forward, it produces some quantity of carbon dioxide.  The same process is repeated for sodium bicarbonate, and the carbon dioxide is again collected.

$\ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O}$

$\ce{NaHCO3 + HCl -> NaCl + CO2 + H2O}$

In the first reaction, one unit of sodium carbonate is able to neutralize two units of acid.  In the second reaction, one unit of sodium bicarbonate can only neutralize one unit of acid.  Another way of looking at this is that double the quantity of carbon dioxide is produced from the sodium bicarbonate neutralization.  That is, the compound is a double carbonate.

Similar explanations make sense for bisulfate and bisulfite.  The acid-base reactions that a sulfate or sulfite ion might undergo are already partially completed in the bi- versions of the compound, so it takes twice as much sulfate or sulfite to do the same job as the non-hydrogen form.

The hydrogen phosphates don't fit this pattern as nicely.

Additionally, if the cation in these salts is a relatively light weight one, then the required mass of the bicarbonate or bisulfate is substantially more than that of the non-hydrogen form of the compound, though it's not quite double.  53g of $\ce{Na2CO3}$ could neutralize one mole of $\ce{HCl}$, but it takes 84g of $\ce{NaHCO3}$ to do the same job.  This is closer to accurate for the more massive sulfur and phosphorus based oxyanions.


How did Joseph-Louis Proust know what gases weighed? Proust's work led him to conclude that compounds always contain the same ratio, by weight, of their constituent elements. How did he know (or deduce?) what gases weighed?

Precise answers appreciated - I can find nothing on the net that explains how the weight of gases was measured at that time - all accounts of Proust & Dalton's work skip over this & other accounts just refer to modern tech.

I am not a chemist - I just want to know this.

He knew because he weighed them

This might seem ridiculously obvious but the early pioneers of modern atomic theory based the idea on actual results they got from weighing things before and after reactions.

It is perfect possible to weigh a container full of gas if you have an accurate enough balance (the buoyancy of the air cancels out if you use the same vessel for different gases). Note that the idea of a vacuum was developed in the 17th century so was well-known by the time of Proust and he could also have measured the empty weight.

The idea is simple: fill a vessel with some gas and weigh the vessel very accurately. A litre of oxygen weighs about 1.5g which is large enough to measure and with good equipment and technique large enough to measure accurately. A liter of hydrogen would be a little under .1g, a big difference.

A series of careful experiments like this with a range of gases would reveal something about the weight-ratios and carefully counting weights and volumes before and after reactions would tell you more (2 volumes of hydrogen react with one of oxygen, for example).

This is what people like Proust did, though with equipment we would now regard as primitive. The reason why it isn't mentioned a lot in modern descriptions is probably because it seems too basic and simple. Nowadays we just teach the theory not so much about the careful experiments that came first and actually led to the theory.


How does this brominated intermediate lose its methyl group? The problem is such:

So far my working has been:

I'm concerned about the last step where the ester forms a ring. The methyl group doesn't look like a plausible leaving group, but it does not appear in the product so how can I get rid of it?

A methyl cation definitely won’t leave by itself. However, you can consider bromide ions floating around in solution. These bromide, being nucleophilic can attack the methyl group in an $\mathrm{S_N2}$ manner, because any positively charged oxygen is a good leaving group. Therefore, the side product would be bromomethane.

Other than that, your mechanism is looking good.


CIF fields that are available for some atoms, but not others I've been working on another CIF file, this time for (-)-threo-methylphenidate based on Pharma Res, 1995 (sadly the relevant crystallography data isn't supplied in the free preview at Springer). I have several questions relating to this file, the most pressing is that so far I have this loop:

loop_
_atom_site_label
_atom_site_chemical_conn_number
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
_atom_site_adp_type
_atom_site_B_iso_or_equiv
_atom_site_type_symbol
N1   1    -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2) N
H1a  2    -0.9650000 0.08500000 -1.0010000             H
H1b  3    -0.8800000 -0.0800000 -1.0330000             H
C2   4    -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2) C
H2   5    -0.7460000 -0.0800000 -0.9360000             H
C3   6    -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3) C

that is generating this error in enCIFer:

Too many or too few data values in the loop

which I'm going to take a stab in the dark and say that it's because two fields (Biso & $B_{eq}$ value) aren't supplied for each atom in this loop (namely it is missing from the hydrogen atoms). The problem is that the paper I cited didn't specify these for hydrogen atoms. Should I use separate loops for hydrogen & other atoms (i.e., one loop for hydrogens, another for the other atoms)? I ask because I fear that this might make molecular modelling software treat the hydrogens & the other atoms as being part of separate molecules.

Your loop defines 8 records. That works fine for heavy elements. However, the lines with hydrogen atoms have only 6 items. That's the reason why this error is reported (all you records are not multiplier of 8).

Usually in refinement the hydrogen atom's $U_{iso}$ is 1.2–1.5 bigger than this of atom to which the H is bound (smaller 1.2 for more constrained movements like on aromatic carbons, bigger 1.5 for aliphatic)

Try something like this:

N1   1    -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2) N
H1a  2    -0.9650000 0.08500000 -1.0010000 Biso 4.     H
H1b  3    -0.8800000 -0.0800000 -1.0330000 Biso 4.     H
C2   4    -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2) C
H2   5    -0.7460000 -0.0800000 -0.9360000 Biso 4.     H
C3   6    -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3) C


Why does protic solvent suggest E1 mechanism rather than E2? If I understood correctly, having a protic solvent increases chance of E1 mechanism. But I can not rationalize this because carbocation intermediate is a positively charged molecule, and by surrounding that with a lot more positive protons will make the whole system in an unstable state. Hence it should make it less likely for E1 to occur. Is my understanding wrong or did I miss some important fact?

First of all, a protic solvent does not mean a highly acidic solvent; it mainly means a solvent whose molecules contain protons that can reasonably be expected to act as an acid. Take for example methanol: its $\mathrm pK_\mathrm a$ is somewhere around $15$, so calling the hydroxy proton acidic is a stretch by most definitions but there are sufficiently basic compounds that can abstract said proton.

Secondly, the question of how a protic solvent affects the E2 versus E1 ratio has less to do with the intermediate cation and more to do with the leaving group. In an E1 reaction, the leaving group dissociates from the main molecule first taking the bonding pair with it. That typically means this leaving group is now anionic or has a somewhat nucleophilic lone pair on it. The protons of the solvent can ‘block’ that lone pair from reattacking the intermediate cation, preventing the backwards reaction and increasing the odds of proceeding. The resulting cation can then be stabilised by the polar bit of a polar protic solvent: the more negatively polarised part of the solvent molecules will interact favourably with the cation.

In an E2 reaction, the base must interact with the starting material for both the starting material’s proton and the leaving group to be displaced in one step. In this case, having a protic solvent which is capable of protonating the base (reducing its basicity) is harmful meaning the reaction will proceed slower. In E1 reactions, weaker bases can be used that are not able to deprotonate the solvent molecule so this pathway is not much of a problem.


Why is my solution for the partial pressures of two different gases incorrect? I have the chemical equation: $\ce{NH4NO3(s) -> 2H2O (g) + N2O (g)}$. There are $12.8$ g of ammonium nitrate, and it decomposes completely in an empty container at $25.00\ ^\circ\pu{C}$ until the total pressure is $3.70$ atm.

This is what I tried:

Because there are 2 more $\ce{H2O}$ than $\ce{N2O}$ I said that there are $\pu{9.06g}$ of $\ce{H2O}$ and $\pu{3.02g}$ of $\ce{N2O}$.

$$\pu{9.06g}\times \frac{\pu{mol}}{\pu{18.016 g}} = \pu{0.503 mol}\  \ce{H2O}$$
$$\pu{3.02g}\times \frac{\pu{mol}}{\pu{44.02g}}= \pu{0.0686 mol}\ \ce{N_2O}$$

I then found the mole ratio of both compounds:
$$X_{H_{2}O}={0.503\over 0.572}=0.879 $$
$$X_{N_{2}O}={0.0686\over 0.572}=0.1199$$

Then I used the relationship between mole ratios and partial pressure $X_i={P_i \over P_{tot}}$, where $P_{tot}=\pu{3.70atm} $:

$$P_\ce{H2O}=(0.879)(\pu{3.70atm})=\pu{3.25atm}$$
$$P_\ce{N2O}=(0.1199)(\pu{3.70atm})=\pu{0.444atm}$$

According to the answer solution for this problem $P_\ce{H2O}=\pu{2.47atm}$ and $P_\ce{N2O}=\pu{1.23atm}$. What am I doing wrong? I can't figure out how to get the correct solution.

The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).

$$P_{\ce{N2O}} = \frac{1}{3} \times 3.70\ \mathrm{atm} = 1.23\ \mathrm{atm}$$
$$P_{\ce{H2O}} = \frac{2}{3} \times 3.70\ \mathrm{atm} = 2.47\ \mathrm{atm}$$

The bottom line is that chemical equations (i.e. stoichiometry) describe the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.


Is it safe to condense water into a liquid nitrogen trap? I'm trying to remove water from an imidazolium salt (i.e. dry the salt).  However I want to know if it is safe to remove the water under vacuum.  Is it safe to condense water into a liquid nitrogen trap?  Wouldn't the water freeze in the trap since the freezing temperature of water is 0 degrees Celcius?  Just curious b/c I was informed that it was ok to do it this way.

One use of a liquid nitrogen cold trap is to collect organic and water vapors that are removed from a system under vacuum.  In this case, the vapors are collected as ices (solids).  See this reference which has some useful information on Schlenk line safety and use.

Where you want to be cautious is when your collection is finished.  You do not want to keep a liquid nitrogen cold trap exposed to the atmosphere because oxygen will condense and cause an explosion hazard.  Typical procedures would include shutting of the vacuum pump, dismantling of the vacuum apparatus and allowing the trap to warm to room temperature (in a fume hood if organic vapors were collected).

Some additional references might be helpful, including this one that not only describes proper Schlenk line technique but also provides some good references; and you can see how liquid nitrogen traps are used for several industrial purposes here and here.


What is the mechanism by which lactones are hydrolyzed during a Wolff-Kishner reaction? I've been away from orgo for a few years but lately someone asked me a question about lactones during a Wolff-Kishner reduction.  I understand the basic mechanism of a Wolff-Kishner but I cannot figure out how a lactone gets hydrolyzed.  Can anyone help me with that step?

Here is a link to the basic mechanism that mentions lactones.

This is a general feature of the hydrazine reagent that is employed in Wolff-Kishner reductions. When attacking an ester’s (includes lactone’s) carbonyl function, the generated tetrahedric intermediate can break down to form a hydrazide as shown in the scheme below.

The carbonyl function is not reduced but the ester (or lactone) is hydrazinolysed.

One could assume that the displace alcoholate could reattack the carbonyl group to regenerate the ester/lactone. However, apart from hydrazine being much more nucleophilic than alcoholates, the resulting hydrazide is also stabilised much like an amide, and is therefore less reactive.


"Adjust pH of the solution to 5.0±0.1 with acetic acid (by potentiometry)" In a procedure description I'm translating, there's this sentence (I here quote it literally, word for word, as it is in Russian):

  Adjust the pH of the solution to 5.0±0.1 with acetic acid (by potentiometry).

The meaning of "(by potentiometry)" is that the personnel should use a potentiometric pH meter diring the adjustment process.

However, I'm not sure what the custom is of indicating this in English.

Maybe "(control using potentiometry")? Or maybe one should add the word "potentiometrically":

  Adjust pH of the solution potentiometrically to 5.0±0.1 with acetic acid.

There must be some commonly used turn of phrase for this.

I suspect this is very likely a sentence adapted from of one of the methods from the Russian State Pharmacopoeia (RSP), which has numerous entries for the adverb «потенциометрически» (Eng. “potentiometrically”).
For instance, there is a nearly identical match in the normative section for the preparation of acetate buffer solution, ОФС.1.3.0003.15-1.27 Ацетатный буферный раствор рН 5,0:

К 120,0 мл 6,0 г/л раствора уксусной кислоты ледяной прибавляют 100,0 мл 0,1 М раствора калия гидроксида и 250,0 мл воды, перемешивают. Доводят рН до 5,0 потенциометрически с помощью 6 г/л раствора уксусной кислоты ледяной или 0,1 М раствора калия гидроксида и доводят объём раствора водой до 1000,0 мл.

RSP underwent a vigorous clean-up since the Soviet era when it was primarily original, and currently it is mostly translated from the European Pharmacopoeia (EP), so I decided to take a look at the corresponding entry in  EP 8.0 (currently invalidated, but one can at least find bits of it in the internet), Section 4.1.3. Buffer solutions:

Acetate buffer solution pH 5.0. 4009100.
To 120 mL of a 6 g/L solution of glacial acetic acid R add 100 mL of 0.1 M potassium hydroxide and about 250 mL of water R. Mix. Adjust the pH to 5.0 with a 6 g/L solution of acetic acid R or with 0.1 M potassium hydroxide and dilute to 1000.0 mL with water R.

EP omits the details as to how the pH is controlled, but there are several phrases in the prior text

Adjust to pH … with …, monitoring the pH potentiometrically

referring to the section 2.2.3. Potentiometric Determination of pH, so I think the phrase that you have proposed

Adjust pH of the solution potentiometrically to 5.0±0.1 with acetic acid.

is perfectly fine and will be correctly understood as a more literate alternative to something like "the pH was determined with a pH meter."
A couple of similar textbook-usecases:

Concise Encyclopedia Chemistry [2, p. 803]:

The pH of a solution is determined experimentally with a suitable pH color indicator (see Indicators) or potentiometrically, with a pH meter (see Glass electrode).

Fundamentals of Electrochemistry [3, p. 590]:

The concentrations of hydrogen ions (solution pH) and of a number of other inorganic ions … are determined potentiometrically.

References

Council of Europe; European Directorate for the Quality of Medicines & Healthcare. European Pharmacopoeia 8.0; European Directorate for the Quality of Medicines & Healthcare, Council of Europe: Strasbourg, 2013.
Concise Encyclopedia Chemistry, English language ed., 2nd ed.; Scott, T., Eagleson, M., Eds.; de Gruyter: Berlin; New York, 1994.
Bagot︠s︡kiĭ, V. S. Fundamentals of Electrochemistry, 2nd ed.; The Electrochemical Society series; Wiley-Interscience: Hoboken, N.J, 2006.


Does "higher symmetry" mean any symmetry group of higher order? What is the precise definition of a higher symmetry in group theory? I see this term scattered around, such as in Cotton's Chemical Applications of Group Theory:

  ...In the cyclic groups each of the operations, Cn, Cn2, Cn3, ..., , Cnn-1, constitutes a class by itself and we continue to use this notations. However, in all other groups of higher symmetry, the number of classes spanned by these operations will be reduced in the following way...

Does higher symmetry strictly mean a supergroup? Or is it a quantitative statement about something like the number of symmetry elements or number of irreducible representations (which are not always ordered the same way, as in the case of C3v and S4).

I am failing to find intuition because, if it does mean strictly supergroups, we can't compare the "amount" of symmetry in C2 compared to C3 (or Td compared to Ci even!).

"Higher symmetry" per se does not mean anything, unless you define it to mean something.

Depending on the context, it usually means one of the following things:

Any symmetry that is a supergroup of the group in question. In this case $C_{4v}$ is higher than $C_4$, while $C_6$ is incomparable to both. Bear with it. Partially ordered sets are quite a thing in math.
(rarely used) Any symmetry group having more elements than the group in question, so that $C_{4v}$ is higher than $C_6$, which in turn is higher than $C_4$.
(apparently, the meaning used by Cotton) Any symmetry group which is not a cyclic rotation group. In this case $C_{4v}$ is higher, while $C_4$ and $C_6$ are lower.

See, there is an intuition, after all. Worse yet, there are quite a few different kinds of intuition.

So it goes.


The Salt Room, is there salt in the air? I work in a restaurant where we dry age our uncut steaks in a cooler. Dry aging is a wonderfully complex process that improves the flavor of richly marbled steaks. Our particular type of aging cooler has a wall of himalayan pink salt (NaCl with other minierals and enough iron oxide to make a pink hue). Our process is even patented by a famous chef. The patent claims that the salt improves the flavor of the beef, and that the closer the beef is to the salt wall, the more flavorful the beef becomes.

I have a limited understanding of NaCl's basic properties (this is necessary in my line of work). It seems to me that no appreciable amount of salt is making its way to the beef through the air. Air simply circulates around the cooler keeping the beef dry and mitigating microbial grotwh. There is also moisture control that keeps the humidity within a certain range, somewhere in the 60% ball park (relative humidity, i think).

When I assert to some of my co-workers that the salt does nothing besides pull some moisture from the air (via hygroscopy), they contest. They cite the gradually diminishing volume of the salt and the smell of sea air (by the sea, not in the room) to support the claims that the salt is present in the aging room atmosphere.

I would counter that the salt is simply ending up on the floor as it is dissolved by the water it pulls from the air via hygroscopy (The floor is cleaned daily). And on the sea air: I know that near the ocean, salt, along with other minerals and organic material, can become suspended in the air in microscopic water droplets and these droplets can deposit these salt, minerals, etc on surfaces. However, the formation of these "sea salt aerosols" involves the churning of the sea water, formation of bubbles, and the action of wind, two of which are not present in the salt wall.

I guess the best summary of my question would be the following: Can damp salt with an air current flowing around it transfer any amount of salt, via the air, to another surface?

In short, no. The vapor pressure of salt is trivial compared to the grams of salt you would put on a steak when cooking it. I think your analysis is good- the most likely impact is that the salt modulates the humidity.


Why do all radioactive decay series terminate at lead? Thorium series:

Uranium series:

Actinium series:

Why do all the radioactive decay series terminate at lead isotope? Why not hydrogen?

(All pictures were taken from Wikipedia.)

There are four main decay chains for actinides and superheavy elements. This is a simple consequence of the fact that one of the main processes to increase a heavy nucleus' stability is the emission of alpha particles, which have a mass number of 4 ($\ce{^4_2\alpha}$); notice that if you take the isotope's mass number and divide it by 4, the remainder of this division (0, 1, 2 or 3, corresponding to the $4n$, $4n+1$, $4n+2$ and $4n+3$ mass number decay chains respectively) stays constant under alpha, beta or gamma decay. There are decays which change the remainder (neutron emission, proton emission, spontaneous fission, etc) and therefore allow hopping between the main decay chains, but for simplicity these can be approximately ignored for the isotopes which are not too unstable or too heavy.

Only three out of these four main decay chains "stop" at lead, so already the statement in the question is incorrect. The $4n$, $4n+2$ and $4n+3$ chains "stop" at $\ce{^208_82Pb}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$, respectively. The $4n+1$ decay chain reaches $\ce{^209_82Pb}$, but this lead isotope is quite short-lived ($t_{1/2}=3.25\ \mathrm{h})$ and decays further to $\ce{^209_83Bi}$ where it "stops" (or at least seemed to until 2003).

Note that I said the chains "stopped" at these isotopes. That's because in reality they don't. It just happens that $\ce{^208_82Pb}$, $\ce{^209_83Bi}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$ are all very long-lived isotopes, with half-lives billions of times greater than the current age of the Universe. This means that any continued decay is severely bottlenecked at these points, so for almost all practical purposes, the decay chains stop there. However, rigorously speaking, if you were really patient there would be further decays in the sequence. In fact, theoretically the heaviest isotope which is not susceptible to any known mode of spontaneous radioactive decay (except proton decay) is an isotope of zirconium, $\ce{^92_40Zr}$, so the decay chains actually go on well past lead or bismuth. It just takes an immense amount of time for the decays to happen.


Density calculation with cubic equation of state I want to calculate the density of CO2 using a cubic equation of state in order to plot the P-v diagram. From theory I know that I get one real root when the substance is in single-phase region and two when it is in two-phase region. In the latter case the smaller volume is associated to liquid while the other to vapor. I'm using Matlab to solve the cubic eos, incrementing the pressure and the temperature at fixed intervals.  How can I merge the two solutions to represent the actual Pv plot?

First off, in the two-phase region you have three roots not two.

Smallest volume = liquid

Largest volume = gas

Middle volume = an unstable state you can't really reach

To determine whether a given temperature and pressure  gives the liquid or the gas you must compare the Gibbs free energy, $G$, between the liquid and gas roots as determined from solving the cubic equation.

The change in Gibbs free energy from liquid to gas is given by

$\Delta G = \int_{liquid}^{gas}{V dP}$

where $P$ and $V$ are pressure and volume along a path of constant temperature.  Note that the need for constant temperature; you can't just hold pressure constant and change the volume because that is not constant temperature.

The proper way to do it is to render

$dP=(dP/dV) dV$

where the derivative $dP/dV$ is obtained as a function of temperature and volume by differentiating the equation of state at constant temperature.  Then you have

$\Delta G = \int_{liquid}^{gas}{V(dP/dV) dV}$

with $dP/dV$ known as a function of volume at constant temperature.  You carry out the integral, which requires knowing how to integrate rational functions, and check whether $\Delta G$ is positive, negative, or zero:

$\Delta G>0$ means liquid has lower free energy, so liquid is more stable.  You plot the point for the liquid phase root.

$\Delta G<0$ means gas has lower free energy, so gas is more stable.  You plot the point for the gas phase root.

$\Delta G=0$ means both phases have the same free energy, you hit the equilibrium between liquid and gas. You plot both points and connect them with a line segment corresponding to both phases in equilibrium.  Jackpot!


Boiling point of: Halogenoalkanes vs alkanes vs alcohols We know that Boiling point (alcohol) > Boiling point (halogenoalkane) > Boiling point(alkane) with same number of $\ce C$ atoms, but is the following, good explanation of that?:

Since e.g. $\ce{CH_3CH_2-O^{\delta_1 \textbf{-}}-H^{\delta_1}+}$, similar with $\ce{ClCH2CH3}$ and with $\delta_2\textbf{-}(\text{at}\ \ce{Cl})\Longrightarrow \delta_1 \textbf{-}\stackrel{\text{bigger, electronegative}}{>}\delta_2 \textbf{-}$, thus electron pair is closer to oxygen atom, so more energy is needed to move that electron in alcohol than in halogenoalkane, so higher temperature is needed to reach boiling point.

Does this have any sence, if not then could somebody explain why does it not have sence?

The cause of this is the hydrogen bond that forms in the liquid state between the molecules, and debye bonds that form between polar molecules.

A hydrogen bond can only occur between an electronegative atom and a hydrogen atom, himself linked to an electronegative atom  (F, O or N).

In order to evaporate the liquid, you have to provide enough energy to break the hydrogen bond. So the more bonds there are, the higher the boiling temperature. Soft debye bonds can also occur between two polar molecules and have to be broken too.

The alkanes can not participate in any bonds, thus have the lowest boiling temperature.
The halogen alkanes may not form hydrogen bonds, but they are polar, thanks to the very electronegative halogens. They form debye links, that are not very strong but sufficient to raise the boiling temperature.
The alcohols have -OH extremities, very likely to associate themselves in hydrogen bonds, that are stronger than debye links, and thus alcohols have the highest boiling temperature.


Why does decreasing the pressure of the system increase the relative volatility of a binary solution? In our teaching lab, we were posed with the following question as an exersize:

  If the boiling points of two compounds differ by $\pu{50 ^\circ C}$ at atmospheric pressure, what will be the effect on the relationship by decreasing the pressure to $\pu{1 mm}$ (no number, just a general process)?

I did not find much information on the subject in our lab manual or book but, but based off of some reading I've done around the internet (see below), it seems that for most binary solutions, decreasing the pressure increases the relative volatility.  However, asking some of my peers, they seems to believe that while both boiling points would decrease, the difference in their new boiling points would remain about the same (i.e. $\pu{\approx 50 ^\circ C}$), To make matters slightly more confusing, using a pressure-temperature nomograph, picking arbitrary numbers, I found that the relative volatility decreased (see Figure 1).

Who is more correct?  Does decreasing the system pressure increase the relative volatility or does it stay the same?
If the relative volatility increases when system pressure is decreased, why is this the case?  I would actually guess the opposite, because as you decrease the pressure towards $0$ (a vacuum), it would seem that all boiling points would approach $\pu{0 K}$, thus be getting closer together.
If the relative volatility increases when system pressure is decreased, does that mean it is always more efficient to perform fractional distillation under a partial vacuum?

Figure 1  Two nomographs measured via Pressure-Temperature Nomograph Interactive Tool by Sigma-Aldrich decreasing the system pressure to $\pu{1 mm}$ for two arbitrary substances with $\pu{50 ^\circ C}$ boiling point disparity at $\pu{1 atm}$.

References:

Effect of temperature or pressure on relative volatility chemengineering.wikispaces.com via the Internet Archive (under "Effect of temperature or pressure on relative volatility")
Industrial-scale applications of vacuum distillation at Wikipedia

Let's assume that the vapor pressure of both components can be modeled with the Antoine equation.  This gives us a convenient way to address the question algebraically.  The vapor pressure of most chemicals is excellently modeled by the correct Antoine equation for that chemical.  The equation is:
$$\log_{10}{p} = A-\frac{B}{C+T}$$

where the the parameters $A$, $B$, and $C$ are compound specific parameters.  The equation can be solved for temperature explicitly yielding $T = \frac{B}{A-\log_{10}\, p} - C$.

So at $\pu{1 atm}$, the equation for pure species $i$ and also a different pure species $j$ that is higher boiling by $\pu{50 ^\circ C}$ is:

\begin{align}
T_1 &= \frac{B_i}{A_i-\log_{10}\, p_{\pu{1 atm}}} - C_i\\
T_1 &= \frac{B_j}{A_j-\log_{10}\, p_{\pu{1 atm}}} - C_j - \pu{50 ^\circ C}
\end{align}

where $T_1$ is the boiling point of $i$ at $\pu{1 atm}$.  To make the math easier, let's suppose the equation is in units of atmospheres, in which case $\log_{10}\, p_{\pu{1 atm}} = 0$.

\begin{align}
T_1 &= \frac{B_i}{A_i} - C_i\\
T_1 &= \frac{B_j}{A_j} - C_j - \pu{50 ^\circ C}
\end{align}

The question is, what happens when $p$ changes, in this case, when it is lowered by ~760-fold, let's call that 1000-fold to keep the math easy.

\begin{align}
T_{0,i} &= \frac{B_i}{A_i-\log_{10}0.001} - C_i = \frac{B_i}{A_i+3}-C_i\\
T_{0,j} &= \frac{B_j}{A_j-\log_{10}0.001} - C_j = \frac{B_j}{A_j+3}-C_j
\end{align}

Now we have four equations but six unknown parameters, $A_i$, $A_j$, $B_i$, $B_j$, $C_i$, and $C_j$.  So in general, there is no generally valid constraint for the temperature difference $T_{0,i}-T_{0,j}$:  It could be higher, lower, or about the same as the difference at room temperature.

However, we could make some further assumptions.  A good one here could be that both species in question have a constant heat of vaporization, which means that $C_i=C_j=0$.  If that assumption is true, then the four equations become:

\begin{align}
T_1 &= \frac{B_i}{A_i}\\
T_1 &= \frac{B_j}{A_j} - \pu{50 ^\circ C}\\
T_{0,i} &= \frac{B_i}{A_i+3}
 =\frac{\frac{B_i}{A_i}}{1+\frac{3}{A_i}}
 =\frac{T_1}{1+\frac{3}{A_i}}\\
T_{0,j} &= \frac{B_j}{A_j+3}
 =\frac{T_1+50 °C}{1+\frac{3}{A_j}}
\end{align}

Now, the temperature difference of interest $T_{0,j}-T_{0,i}$ is

$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{1+\frac{3}{A_j}} - \frac{T_1}{1+\frac{3}{A_i}}$$

To keep the formulas easy let me define new parameters $a_i$ and $a_j$ such that $a_x=1+\frac{3}{A_x}$.  Then the temperature difference is

$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{a_j} - \frac{T_1}{a_i}= \frac{a_i(T_1+\pu{50 ^\circ C})-a_j T_1}{a_i a_j}=\frac{(a_i-a_j)T_1 + a_i(\pu{50 ^\circ C}) }{a_i a_j}$$

Now we must introduce further assumptions about the Antoine $A$ parameters.  They are positive and thus the $a$ parameters must also be positive, and further positive $A$ implies  $a>1$.  If we further assume $a_j=a_i$, then

$$T_{0,j}-T_{0,i} =\frac{(\pu{50 ^\circ C}) }{a_j}$$

Since we know already that $a_j>1$, then the temperature difference (i.e. the difference in boiling points) at the lower pressure is less than the temperature difference of $\pu{50 ^\circ C}$ at \pu{1 atm}.  Now, even if $a_i$ is not precisely equal to $a_j$, then the temperature difference will still be less than $\pu{50 ^\circ C}$ as long as the $a_i$ and $a_j$ parameters are similar enough to keep $(a_i-a_j)T_1 << a_i(\pu{50 ^\circ C})$.  It would take a very unusual choice of chemicals $i$ and $j$ for that to happen.  I don't think I could identify such a pair.

So, in summary, to get to the result that I think your instructor was asking for, we required many assumptions:

Both chemicals obey the Antoine equation in the temperature range of interest (probably an OK assumption).
Both chemicals have approximately constant enthalpy of vaporization in the T range of interest so that we can neglect the Antoine $C$ parameters.
The Antoine $A$ parameters for both chemicals must be positive.  (A good assumption but it still must be made.)
The chemicals have Antoine $A$ parameters (and thus also $a$ parameters as I defined them here) that are not too different from each other.

It might be possible to weaken some of the assumptions and still find that the lower-pressure boiling point difference must be lower than $\pu{50 ^\circ C}$, but I'm confident that the result is not necessarily generally true.  It isn't a thermodynamic law or anything.  It definitely depends on several key assumptions about how vapor pressure "usually" or "typically" behaves for most chemicals.


Trans-hydroxylation of alkenes In the trans-hydroxylation of alkenes to give alcohols, there is a formation of epoxide.

Does the reaction stop here if we don't add water?

Even though if we add water, why does the product depends on which medium the reaction is carried whether acidic or basic?
My text says the ring opens differently for basic and acidic medium, but both finally gives us a trans-vicinal-diol.

This answer has technically been given in the comments, but it bears repeating.

Yes, you can stop the reaction at the epoxide stage through careful choice of conditions. If you prevent both acidic and basic aquaeous environments, then epoxide rings are stable. They are present in a wide range of molecules and even in natural products such as epothilone.

You could also open epoxides with other nucleophiles to create different 1,2-substitution patterns — with predefined stereochemistry, if the epoxide was stereoselective.

And finally, there are even methods out there to turn vicinal diols into epoxides (usually via sulphonylation, e.g. to a tosylate followed by intramolecular $\mathrm{S_N}2$ reaction).


Why does HCO₂⁻ exhibit delocalized bonding while CO₂ does not? $\ce{HCO_2^-}$ exhibits delocalized bonding and $\ce{CO2}$ does not?

What I am confused about is, that both of them have resonance structures. Therefore, why doesn't $\ce{CO2}$ exhibit delocalized bonding?

Here is an orbital diagram for $\ce{CO2}$.

Note how the red pi orbitals on the terminal oxygens are rotated 90 degrees with respect to one another; they are said to be "orthogonal" to one another.  This means that the p-orbitals on the terminal oxygens can only interact with the p-orbitals with which they are aligned on the central carbon; there is no way for an electron on one oxygen to interact with the orbitals on the other oxygen.

Here is the analogous drawing for the formate anion ($\ce{HCO2^-}$).  Note how the p-orbitals on the terminal oxygens are now both aligned parallel with the p-orbital on the central carbon atom so that continuous overlap (delocalization) is now possible.

For completeness, we can restate the drawings from above in terms of resonance structures.  In the top row of the following picture I've drawn the possible resonance forms for $\ce{CO2}$.  The only charge-separated structures we can draw involve charge separation on each of the two orthogonal $\ce{C=O}$ groups.

In the bottom row of the picture I've drawn 4 resonance structures for the carboxylate anion (when R=H this is the formate structure in the title of your question).  Structure II corresponds to the charge separated structure I drew above for $\ce{CO2}$.  But notice that I have drawn 2 new resonance (III and IV) where the negative charge is delocalized over the O-C-O 3 atom array.  This is because the p-orbitals on the terminal oxygens are now parallel and can overlap with the p-orbital on the central carbon.  Just like the "particle in a box" problem you may have studied in physics, it turns out the longer the path the electron has to roam over, the more stable the electron.  So resonance structures III and IV, where the electron can roam (delocalize) over 3 atoms are very important contributors to the description and stability of the carboxylate anion.  Carbon dioxide has no comparable resonance structures due to the orthogonal alignment (90 degree rotation) of the orbitals on the terminal oxygens


How does one determine the probability of H⁺ to H₂O in pure water? The ratio of $\ce{H+}$ to $\ce{H2O}$ is something around $10^{-8}$. How did people determine that?

You cannot use the $K_\mathrm{w}$, $\mathrm{p}\ce{H}$, $\ce{[H+]}$ ion concentration value to answer because you get all that from the probability.                                                                   Is there any independent way to find the probability. Please elaborate on the steps.
I do not want to know why the ionic product is what it is. how was the ionic product found ?

The answer to this question was surprisingly difficult to find because all modern ways to measure pH use reference solutions of known pH and thus rely on prior knowledge of the $H^+$ activity of said reference solutions.

Finally I found this on wikipedia:

  it is possible to measure the concentration of hydrogen ions directly, if the electrode is calibrated in terms of hydrogen ion concentrations. One way to do this, which has been used extensively, is to titrate a solution of known concentration of a strong acid with a solution of known concentration of strong alkaline in the presence of a relatively high concentration of background electrolyte. Since the concentrations of acid and alkaline are known, it is easy to calculate the concentration of hydrogen ions so that the measured potential can be correlated with concentrations. The calibration is usually carried out using a Gran plot.[9] The calibration yields a value for the standard electrode potential, $E^0$, and a slope factor, f, so that the Nernst equation in the form
  $$E = E^0 + f\frac{2.303RT}{F} \log[\mbox{H}^+]$$
  can be used to derive hydrogen ion concentrations from experimental measurements of E. The slope factor, f, is usually slightly less than one. A slope factor of less than 0.95 indicates that the electrode is not functioning correctly. The presence of background electrolyte ensures that the hydrogen ion activity coefficient is effectively constant during the titration. As it is constant, its value can be set to one by defining the standard state as being the solution containing the background electrolyte. Thus, the effect of using this procedure is to make activity equal to the numerical value of concentration.

This means that the precise cocentration of $H^+$ is actually unknown what is known is it's chemical activity which has been defined as equal to the concentration for a certain standard solution.


Why are the dispersion forces in CS2 stronger than the dipole-dipole forces in COS? London dispersion forces supposedly have the least strength out of all the intermolecular forces. But $\ce{CS2}$, which has only dispersion forces, has a higher boiling point (and thus stronger intermolecular forces) than $\ce{COS}$, which has dipole-dipole attraction in addition to dispersion forces. Why is this?

I suppose that it has something to do with $\ce{CS2}$ having a thicker/more inducible electron shell, but then a new question arises: how would you know if the dispersion forces in one molecule are stronger that the dipole-dipole forces in another?

(Theoretically, without using boiling points or other experimental data. Also, this is based on question 4a from the 2018 AP chemistry free response.)

Although individual dispersion forces are weak, they are cumulative, and increase with molar mass.  As a general rule, boiling point increases with molar mass.

Polar molecules will have higher boiling points when compared to molecules with similar molar masses.  For example, ethanol($\ce{CH3CH2OH}$) has a higher boiling point than dimethyl ether ($\ce{CH3OCH3}$).

$\ce{CS2}$ is ~16 g/mol heavier than COS.


What are disadvantages of using sublimation vs. recrystallization as a method of purification? I can only think that a small sample can be sublimed while a bigger sample size can be recrystallized. Are there more?

The most notable difference is that sublimation requires a volatile solid, whereas most solid can be recrystallized from a solvent system that is not too exotic.
Recrystallizations tend to recover more pure product.
In regards to green chemistry, sublimations are preferred as they generate no solvent waste.


How do I identify the compound(s) that formed from a failed synthesis of another compound? A general answer to the question is great, but specified to my problem is very much welcomed.

My team had a lab project — to synthesize chrome alum crystals.

We first dissolved $0.3 \, \rm g$ of potassium dichromate on $4 \,\rm mL$ of $1 \,\rm M$ sulfuric acid, then added $2 \,\rm mL$ of $99\%$ ethanol and let it rest for a day. There were some crystals formed, but the color was orange (it should've been purple).

From the instruction that we followed, we should heat up the solution, so we did that, and carelessly let the temperature go up past $70 \,^{\circ} \rm C$. It become greenish in color and doesn't form any crystal after a week.
We resorted to evaporating the liquid part and obtained some leftover greenish yellow powder at the bottom which is very much far from the desired chrome alum.

Theoretical part

Chrome alum $\ce{KCr(SO4)2 * 12 H2O}$ is synthesized according to the following reaction (average oxidation numbers are shown above the elements undergoing redox reaction):

$$\ce{K2\overset{+6}{Cr}_2O7 + 3 \overset{-2}{C}_2H5OH + 4 H2SO4 →[15-40~°C] K2SO4 + \overset{+3}{Cr}_2(SO4)3 + 3 \overset{-1}{C}_2H4O + 7 H2O}$$

after joint crystallization of both $\ce{K2SO4}$ and $\ce{Cr2(SO4)3}$ with water forming a complex which is better described as aquacomplex $\ce{[K(H2O)6][Cr(H2O)6](SO4)2}$ with no sulfate(VI) anions in neither of the first coordination spheres.

Note:

It's absolutely crucial to control the temperature during the addition of alcohol to acidified dichromate. If the temperature below 15 °C, the reaction is too slow, but if it's above 40 °C, ligand exchange starts to take place: $\ce{H2O}$ ligands in $\ce{[Cr(H2O)6]^3+}$ are partially substituted by $\ce{SO4^2-}$, which is reflected in color change from blue-violet to green. Also, evaporation of alcohol also happens quicker at elevated temperatures, depleting the reducing agent at higher rate.
Whereas you need stoichiometric amounts of dichromate and sulfuric acid, alcohol is usually taken in excess to ensure complete reduction of $\ce{Cr^6+}$ to $\ce{Cr^3+}$ and to salt out crystals of chrome alum from the solution.
Working with such small quantities (few mL) of pretty abundant and cheap chemicals seems unreasonable to me as it increases the error of measurement, increases the negative impact of impure chemicals or erroneous concentrations and makes the experiment less demonstrative.
I'd advise to work with larger quantities (e.g. 1-2 g of dichromate) in the future.

Experimental results and discussion

  There were some crystals formed, but the color was orange …

It's extremely likely that the orange crystals are unreacted potassium dichromate(VI), so it looks like the reduction was incomplete.
That's why I doubt that alcohol was ever oxidated further to acetic acid, hence I see the formation of chromium(II) acetate as suggested in the comments as improbable.
Also, chromium(II) acetate is very unstable and would've decomposed under these conditions.

There is also a rudimentary method of checking the oxidized product by its smell.
If it smells like apples, then it's acetaldehyde $\ce{C2H4O}$; if it smells like vinegar, then the oxidation took further and alcohol has been oxidized to acetic acid $\ce{CH3COOH}$.

  … (it should've been purple).

Since you haven't dissolved potassium dichromate in water and just added sulfuric acid to the powder, the concentration of sulfate(VI) was too high.
Plus, you've added all alcohol at once, which might've resulted in temperature spike. Both factors effectively prevent formation of chromium(III) aquacomplex, leaving no chance for alum to crystallize. (see Addendum for the reference)

  From the instruction that we followed, we should heat up the solution, so we did that, and carelessly let the temperature go up past 70 °C. It become greenish in color and doesn't form any crystal after a week.

At this stage you pushed the reduction further and probably evaporated both ethanol and acetaldehyde completely, and the solution was consisting of very concentrated chromium(III) sulfate, still no chance for alum to crystallize due to above mentioned reasons. (see Addendum)

  We resorted to evaporating the liquid part and obtained some leftover greenish yellow powder at the bottom

Green powder is likely a mixture of potassium chromium(III) sulfate $\ce{KCr(SO4)2}$ and also minor amounts of chromium(III) oxide $\ce{Cr2O3}$ formed along the way.
Now yellow powder is with a very high confidence just potassium chromate(VI) $\ce{K2CrO4}$ leftovers from the dichromate(VI) once the acid was used up:

$$\ce{Cr2O7^2- + H2O <=> 2 HCrO4^2- <=> 2 CrO4^2- + 2 H+}$$

One might also claim that it's elemental sulfur formed by reduction of sulfate(VI) after chromium(VI) was completely converted to chromium(III), but I find it rather implausible.

Addendum

From Ullmann’s Encyclopedia of Industrial Chemistry [1, p. 4441]

  Potassium chromium(III) sulfate [10141-00-1], [10279-63-7], [7788-99-0],
  potassium chrome alum, $\ce{KCr(SO4)2 · 12 H2O}$, $M_\mathrm{r}~499.11$, $\pu{1.813 g/cm3}$, crystallizes in the cubic system forming violet regular octahedra
  which decay in air and melt at 89 °C, the color changing to green; the enthalpy of formation is − 5788 kJ/mol. The solubility in water at 25 °C is 11.1 wt %. The solution is violet when cold, but becomes green above 50 °C, this change being accompanied by a decrease in the molar conductance. For 0.125 M solution at 50 °C, the molar conductance is $\pu{221 Ω−1 cm2 mol−1}$ for the violet form and $\pu{202 Ω−1 cm2 mol−1}$ for the green form. This change is reversible and its rate is increased by acids. The green form always occurs as an amorphous solid and crystals
  are unknown.

  […]

  For the preparation of potassium chromium alum, a saturated potassium dichromate solution is reduced with sulfur dioxide in the presence of sulfuric acid.
  During the reaction, the temperature must be kept below 40 °C by cooling to prevent the green modification being produced. Apart from sulfur dioxide, such organic compounds as formaldehyde, methanol, or starch are suitable as reducing agents. Crystallization starts after sulfuric acid has been added and the temperature is kept
  further below 40 °C. If the solution is allowed to settle in vats, large crystals are produced, whereas ﬁne ones result if the solution is stirred.

References

Ullmann’s Encyclopedia of Industrial Chemistry, 40 Volume Set, 7th ed.; Wiley-VCH, Ed.; Wiley-VCH: Weinheim, 2011. ISBN 978-3-527-32943-4.


Is it possible to have negative corrorion rate and weight loss, meaning specimen [steel] gained weight after immersion in electroyte solution? Regarding this article, I have read that the corrosion rate and weight loss of a steel specimen after immersion in different corrosive media is negative. I'm surprised that this is really possible. Is there a chemical explanation for this?

In the data cited in your reference, corrosion is measured by weight loss or gain. In the case of the steel coupon immersed in sodium carbonate, the weight gain is not due to "negative corrosion", but likely due to a tightly-adhering film of oxides that was not removed in cleaning. The oxides are made from the iron of the coupon plus oxygen from water, which is greater than the mass of the iron alone. (Normally, cleaning the coupon removes loosely adhering oxides, so there is a weight loss.)

According to Toyin, "the corrosion inhibition efficiency of $\ce{Na2CO3}$... indicates that a protective film is formed on the metal surface. The passivation of the mild steel may be caused by the formation of $\ce{Fe2O3}$  and $\ce{Fe3O4}$."

That said, it does appear that high carbon concentrations may be effective in greatly reducing the corrosion of mild steel.


Change of temperature upon dilution Consider a concentrated dye solution, say 10µM of Rhodamine 6G in water, which is rapidly diluted down to 10nM in an open container. I'm trying to estimate the (presumably minute) temperature change, assuming both solutions (dye and water) are initially at the same room temperature. I believe dilution increases entropy ever so slightly, but I'm struggling to turn this statement into a quantitative temperature change. Specifically, how do I calculate the change in entropy? Once I know this, the change in temperature should be straight-forward.

At such low concentrations, can’t you just assume an ideal solution? Then the  change in entropy from mixing would be $\Delta S = -R(x_A\ln x_A + x_B \ln x_B)$. But because the solution is so dilute, you can see that both terms are going to be essentially zero. (For a 10 nM solution, the contribution from entropy of mixing is on the order of $10^{-8}$ J K$^{-1}$ mol$^{-1}$.)

The dilution process would thus lower the total entropy, since there’s less uncertainty in the system. (You’d get the value you compute at 10 nM, the final state, minus the value at 10 µM, the initial state.)


Calculating ΔG at the extremes of reaction extent This question follows on from a great post on comparing $\Delta_\mathrm{r} G^{\circ}$ to $\Delta_\mathrm{r} G$ answered here

The G vs ξ graph that is used in the answer is an excellent tool to understand how $\Delta_\mathrm{r} G$ changes with the extent of reaction:

If $\Delta_\mathrm{r} G$ is found on this graph by the slope of the line, what happens at the two extremes, where we have pure reactants on the left ($Q = 0$) and pure products on the right ($Q ⟶ ∞$)?

The equation to calculate $\Delta_\mathrm{r} G$ is: $$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln Q$$

This would give  $\Delta_\mathrm{r} G ⟶ -∞$ (vertical slope asymptotic to the y axis) when $Q=0=ξ$ (i.e., when we have pure reactants). Logically, that would mean that absolutely pure reactants have infinite G (which doesn't seem right).

A similar problem occurs as we approach pure products: $\Delta_\mathrm{r} G ⟶ ∞$ when $Q⟶∞,ξ⟶max$ (vertical slope asymptotic to a line at ξ = max).

Can anyone reconcile this graph with the equation to calculate $\Delta_\mathrm{r} G$?

After a lot of help, I have the following to suggest as an answer:

Imagine a reaction with $\Delta_\mathrm{r} G^{\circ} = -1000Jmol^{-1}$ at 298K. Using the following equations: $$\Delta_\mathrm{r} G^{\circ} = - RT\ln K$$ $$e^{-\frac{\Delta_\mathrm{r} G^{\circ}}{RT}} = K$$
This would give a K value of 1.50, which indicates the Q value on the curve where G is at a minimum (favoring products slightly in the equilibrium mixture). Using another equation (valid e.g. for a reaction $\ce{A(aq) <=> B(aq)}$): $$ξ = \frac{Q}{1+Q}$$

Gives the ξ value (1.50/2.50 = 0.600) which is the x coordinate of the minimum on the graph, where the x axis runs from ξ = 0 (pure reactants) to ξ = 1 (pure products). This graph is pretty similar to the one posted in the question:

Using the equation:$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln Q$$

And substituting in:$$Q = \frac{ξ}{1-ξ}$$

Gives us:$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln \frac{ξ}{1-ξ} $$

or:

$$\frac{\delta G}{\delta \xi} = \Delta_\mathrm{r} G^{\circ} + RT\ln \frac{ξ}{1-ξ} $$

$\Delta_\mathrm{r} G^{\circ}$ is a constant even as G changes as the value of ξ changes, so are R and T. This differential equation can be separated and integrated to gives us the original function of how G depends on ξ:

$$\begin{align} \int \delta G = \Delta_\mathrm{r} G^{\circ}ξ + RT \int \ln \frac{ξ}{1-ξ}\delta \xi \end{align}$$

$$\begin{align} G = \Delta_\mathrm{r} G^{\circ}ξ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1 + C)\end{align}$$

Graphing this on desmos gives us the correct graph for how G varies with ξ.

The graph shows the correct ξ value for the equilibrium point and also the difference in G between the "pure" reactants and products. All of this is as predicted by the theory.

If you trace the line along the domain of x towards zero (or 1) the y values increase until the value becomes undefined, but I have zoomed in until my brain hurt and the value never seems to shoot upwards (as I thought the maths predicted). L'Hopital's rule seems to prevent this from happening.

I can't pretend that I fully understand the maths (I don't), maybe others' comments will help here. If anyone wants to use the desmos calculator to produce their own plots (you can change T or $\Delta_\mathrm{r} G^{\circ}$ to other values to see the effect) please do so, I may use this in teaching the material in the future.

I credit @orthocresol @Night Writer and @Karsten Theis with helping me. Also Darrell Bach and my students Nick Kleiderer and Sylver Riddell.

****** Addition to deal with limits ******

The limits of G at each end of the domain of ξ are below.
(For ease of reading I will use x = ξ)

$$\begin{align} G = \Delta_\mathrm{r} G^{\circ}ξ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1 + C)\end{align}$$
$$ G(x) = \Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)$$

Limit of G when $x \to 0 ^+$ :

$$ \lim_{x \to 0 ^+} G(x) = \lim_{x \to 0 ^+} [\Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)] $$

$$\lim_{x \to 0 ^+} G(x) = \lim_{x \to 0 ^+} [\Delta_\mathrm{r} G^{\circ}x - RT + RTC] + \lim_{x \to 0 ^+} [RT(x \ln (x) + (1-x) \ln (1-x))] $$

Note (1):   $ \lim_{x \to 0 ^+}[(1-x) \ln (1-x)] = 0 $

Note (2):   $\lim_{x \to 0 ^+} [x \ln (x)] = 0$   (by l'Hopital's rule)

$$\lim_{x \to 0 ^+} G(x) = - RT + RTC + RT(0+0) $$

$$\lim_{x \to 0 ^+} G(x) = - RT + RTC $$

$$\lim_{x \to 0 ^+} G(x) = RT(C-1) $$

Limit of G when $x \to 1 ^-$ :

$$ \lim_{x \to 1 ^-} G(x) = \lim_{x \to 1 ^-} [\Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)] $$

$$\lim_{x \to 1 ^-} G(x) = \lim_{x \to 1 ^-} [\Delta_\mathrm{r} G^{\circ}x - RT + RTC] + \lim_{x \to 1 ^-} [RT(x \ln (x) + (1-x) \ln (1-x))] $$

Note (1):   $ \lim_{x \to 1 ^-}[(1-x) \ln (1-x)] = 0 $ (by l'Hopital's rule)

Note (2):   $\lim_{x \to 1 ^-} [x \ln (x)] = 0$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} - RT + RTC + RT(0+0) $$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} - RT + RTC $$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} + RT(C-1) $$

These limits to G show that there is a value to G as you approach pure reactants and products along the curve of G vs ξ. The algebra shows some interesting points, which agree with other parts of the theory of equilibria and state functions:

The curve of G vs ξ can be thought of as the sum of a simple line between the G of reactants and the G of products $$G = \Delta_\mathrm{r} G^{\circ}ξ + RT(C-1)$$

for $C = 0$

$$G = \Delta_\mathrm{r} G^{\circ}ξ - RT$$

and a second function (which is always negative) that deals with the Gibbs free energy of mixing $$ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1)$$

This is seen as a dip from the line, reducing the Gibbs free energy in the mixed system. This disappears as ξ approaches 0 or 1, as the system is becoming pure reactants or products:

For a reaction where K > 1 and $\Delta_\mathrm{r} G^{\circ} <0$ (such as the graphs in this example) the pure products are lower in G than the reactants, but the equilibrium mixture is lower still
$\Delta_\mathrm{r} G^{\circ}$ refers to the tangent of the curve at standard conditions ($Q=1$ and $ξ = 0.5$) and also to the slope of the straight line that connects $G_{reactants}$ and $G_{products}$. As ξ changes by 1 from reactants to products, $\Delta_\mathrm{r} G^{\circ}$ is also the absolute $\Delta G$ between pure reactants and products


What can be done to increase the yield of a product in a chemical reaction that has reached equilibrium? Let's take the following reaction that has reached equilibrium

$$\ce{CaCO3 (s) <=> CaO (s) + CO2 (g)}$$

What can be done to increase the yield of calcium oxide ?

My initial thought was to remove the carbon dioxide produced in the reaction mixture. I understand that this is correct since, according to Le Chatelier's principle, the reaction would shift to increase the yield of the products.

Why wouldn't adding more calcium carbonate lead to a shift to the products and hence lead to a greater yield of the products? Doesn't more reactant mean more product?

I agree with your initial starting point, that the thermal decomposition of calcium carbonate into calcium oxide and carbon dioxide may be treated as an example of an heterogeneous equilibrium

$$\ce{CaCO3 (s) <=> CaO (s) + CO2 (g)}$$

that still is of relevance today

(source, Hughes et al. in Ind. Eng. Chem. Res. 2004, 43, 5529-5539).

and actually at around $\ce{900 ^\circ{}C}$, the equilibrium pressure of $\ce{CO2}$ over $\ce{CaCO3}$ and $\ce{CaO}$ would equal to $\pu{1 atm}$.

Your first reasoning, while keeping the other parameters of this equilibrium fixed ($T, p$), a removal of $\ce{CO2}$ may increase the yield of $\ce{CaO}$ since this eventually (macroscopically) leads to an exhaustive decomposition of $\ce{CaCO3}$ is correct.

For the second part, you have to balance the general idea about chemical equilibria (which in this case indeed means, keeping other parameters fixed, adding more $\ce{CaCO3}$ as starting material yields more $\ce{CO2}$) with the fact that this particular reaction is heterogeneous.  Looking up the data for molar mass and density, (1, 2, and 3) allows you to estimate the volume one mole of each pure compound under standard conditions occupies:

| compound | molar mass |  density | molar volume |
|          |    [g/mol] | [g/cm^3] |   [cm^3/mol] |
|----------+------------+----------+--------------|
| CaCO3    |     100.09 |    2.711 |        36.92 |
| CaO      |      56.08 |     3.37 |        16.64 |
| CO2      |      44.01 | 0.001977 |     22261.00 |
|----------+------------+----------+--------------|

The value of the molar volume of $\ce{CO2}$ may remind you to the value of the volume occupied by the Ideal Gas at standard conditions ($\pu{22.7 L/mol}$) and demonstrates that the influence of $\ce{CO2}$ on the position of the equilibrium is much more important than the two other of the solids with their -- in comparison -- often negligible molar volumes. Hence, to underline that this equilibrium is this much pressure dependent, it is described by $K_{\text{p}}$, rather than by a normal $K$ value.

The general idea of influencing a chemical equilibrium along "an increase of  partial volume (or partial pressure) of starting material(s) will increase the yield of the product(s)" better full-filled in the case of homogeneous reactions, like the synthesis of ammonia

$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}. $$


Effect of inert gas on Q Adding an inert gas will not affect Q if that reaction is held at a constant volume. But, if the reaction is held at constant pressure, then adding inert gas will change Q bc the partial pressures of the gases will change.

I don't understand this. If the reaction is held at constant pressure, this means that you changed the volume when adding the inert gas. Wouldn't the partial pressures of all the reactants change by the same factor, so Q wouldn't change?

Wouldn't the partial pressures of all the reactants change by the same factor, so Q wouldn't change?

If the number of gas particles is the same for reactants and products, then Q would not change. However, if you have a reaction like

$$\ce{2N2O(g) <=> N2O4(g)}$$

even though the reactant and product partial pressure changes by the same factor, the changes don't cancel out because the coefficients are different.

$$ Q = \frac{[\ce{N2O4}]}{[\ce{N2O}]^2}$$


Why doesn't the SeO2 oxidation of ketones stop at the hydroxyketone stage? When $\ce{SeO2}$ is used to oxidise aldehydes and ketones, $\ce{SeO2}$ becomes hydrated in water, and the active reagent is selenous acid $\ce{H2SeO3}$ which oxidizes aldehydes and ketones to furnish 1,2-dicarbonyl compounds such as 9 in the scheme below.1 The α-hydroxyketone 10 is not observed. Why is that so?

1 Sharpless, K. B.; Gordon, K. M. Selenium dioxide oxidation of ketones and aldehydes. Evidence for the intermediacy of .beta.-ketoseleninic acids. J. Am. Chem. Soc. 1976, 98 (1), 300–301. DOI:10.1021/ja00417a083

As noted in the cited paper by Sharpless and Gordon$^{[1]}$, the problem with the Corey$-$Schaefer mechanism is the existence of the selenous ester (7) which, due to its instability, would undergo hydrolysis prior to any elimination reaction.
A study by Qingjiang Li and Tochtrop$^{[2]}$ supports Sharpless and Gordon's counterargument. The former pair give a good overview of the hypotheses put forward up to that point (i.e., 2014).
$\hskip 1 in$

$\beta$-hydrogen transfer (Speyer mechanism)

Based on the logic of the proposed mechanism, we reasoned that replacement of $\beta$-hydrogens of the substrate with deuterium would result in a ﬁrst-order
isotope eﬀect in the reaction rate.

Consequently, a $\beta$-deuterated fatty acid was prepared ... and converted to triﬂuoromethyl ketone to react with selenium dioxide. The reaction rate was monitored by NMR and compared with that of the analogous fatty acid without deuterium.$^{[2]}$

Based on this experiment, Li and Tochtrop conclude$^{[2]}$,

No signiﬁcant diﬀerence in reaction rates was observed, which suggested that the removal of $\beta$-hydrogen to form a carbon−carbon double bond is not likely to be the rate-limiting step, and we therefore excluded the $\beta$-hydrogen transfer mechanism.

Verdict: not the best explanation.

Sharpless$-$Gordon

While solving the selenous ester controversy, Li and Tochtrop found other problems introduced by Sharpless and Gordon.

[A]dding acid should accelerate the Sharpless mechanism by accelerating the enol−ketone tautomerization, as the rate limiting step is electrophilic attack of selenium dioxide to the enol. Since no signiﬁcant increase of the reaction rate was observed when adding acetic acid to the reaction mixture, the Corey−Schaefer mechanism seemed more reasonable.$^{[2]}$

To additionally verify this result,

$1.2$ equivalents of $\ce{Ti(O^iPr)4}$, a strong Lewis acid to coordinate with the carbonyl group, was added to the triﬂuoromethyl ketone for $1\ \mathrm{h}$ prior to selenium dioxide addition. This experiment resulted in no reaction, which subsequently served as further evidence that the reaction initiates with the ketone instead of the enol.$^{[2]}$

The Sharpless$-$Gordon mechanism also fails to explain some products detected in Li and Tochtrop's work with carboxylic acids.$^{[2]}$
Verdict: not the best explanation.

Corey$-$Schaefer

Verdict: not the best explanation due to the selenous ester intermediate.$^{[1]\ [2]}$
New mechanism proposed by Li and Tochtrop
Li and Tochtrop came up with a fourth mechanism to account for experimental evidence. Here, the reaction initiates via the ketone. The further pathway is dictated by whether $1$,$4$-elimination is restricted.$^{[2]}$
Li and Tochtrop examplify their mechanism with a trifluoromethylketone that has varying radicals attached to the non-fluorinated $\alpha$-carbon.$^{[2]}$
$\hskip 1 in$

As shown in Scheme $5$, we propose that high valence selenium reacts with
triﬂuoromethyl ketone to form the enol ester $(\bf{17})$. The enol ester undergoes $1$,$4$-elimination directly instead of nucleophilic attack of the carbon−carbon double bond to form selenium ester.$^{[2]}$

[---] When $1$,$4$-elimination is restricted, nucleophilic attack at the $\alpha$-position occurs and the subsequent hydrolysis of selenium($\mathrm{II}$) ester $(\bf{18})$ aﬀords alcohol as the product instead of enone.$^{[2]}$

In the context of the Li$-$Tochtrop mechanism, we can thus answer OP's original question.

The $\alpha$-hydroxyketone (10) is not observed. Why is that so?

The intermediate (4) has a small hydrogen attached to the original $\alpha$-carbon. In other words, $\ce{R2 = H}$. The $1$,$4$-elimination is not restricted. Hence, nucleophilic attack of the $\ce{C=C}$-bond does not take place.
The mechanism by Li and Tochtrop is also useful because it explains, to an extent, the dependence of the end product upon the starting material. Solvent and $\mathrm{pH}$ effects still have to be considered, and the pathway that yields diketones is not covered.

The nature of the starting material seems to have a great influence on the course of the reaction but usually one of the two types of compounds can be specifically obtained by using the right solvent. For example, dioxane favors the formation of the dione whereas t-butanol favors the unsaturated carbonyl compound.$^{[3]}$

The amount of diketone (or even triketone) will further depend on $\mathrm{pH}$ and the quantity of $\ce{SeO2}$.$^{[2]\ [3]}$ Depending on the mixture (e.g., with $\ce{H2O2}$), carboxylic acids and $\alpha$-ketocarboxylic acids are also possible$^{[3]\ [4]}$ with various degrees of saturation$^{[2]}$.

Some information on how to eliminate sidereactions that lead to diketones can be found here:
0 Hans J. Reich, James M. Renga, Ieva L. Reich. 'Organoselenium chemistry. Conversion of ketones to enones by selenoxide syn elimination'. Journal of the American Chemical Society 1975, 97(19), 5434$-$5447. DOI: 10.1021/ja00852a019.

1 K. Barry Sharpless, Kenneth M. Gordon, 'Selenium Dioxide Oxidation of Ketones and Aldehydes. Evidence for the Intermediacy of 8-Ketoseleninic Acids,' Journal of the American Chemical Society 1976, 98(1), 300$-$301. DOI: 10.1021/ja00417a083.
2 Qingjiang Li, Gregory P. Tochtrop, 'New Methodology toward $\alpha,\beta$-Unsaturated Carboxylic Acids from Saturated Acids,' Organic Letters 2014, 16(5), 1382$-$1385. DOI: 10.1021/ol500127u.
3 Alain Krief, Laszlo Hevesi (1988), Organoselenium Chemistry I. Functional Group Transformations, pp 115$-$180. ISBN: 978-3-642-73243-0 (print) 978-3-642-73241-6 (online).
4 Thomas Wirth (2012), Organoselenium Chemistry: Synthesis and Reactions, p 212. DOI: 10.1002/9783527641949.


Saytzev and Hofmann elimination in E1 From what I have read, E1 largely produces the more substituted alkene because of the fact that more substituted alkenes are more stable as a result of hyperconjugation. However, in E2 the situation is different for a number or reasons. One of these reasons is steric effects of potentially large substituents on the $\beta$- carbon which would promote formation of the less substituted alkene. This effect does not seem to be mentioned for E1 but surely it must still be valid, right? Do large subtituents or bulky bases promote the formation of less substituted alkenes in E1? If not why not?

In the E2 reaction, the size of the base can affect the selectivity of the deprotonation, because the steric interaction of a large base with a hindered proton will go over a higher transition state than the interaction of a large base with a less-hindered proton. This is kinetic control.

I'm not aware of any practical case where the selectivity of an E1 reaction can be controlled by the size of the base. The reason is that the deprotonation of the carbocation intermediate will have a transition state that is very close in energy to the carbocation itself. I agree with you that the size of the base should have some influence on the transition state energy, but in practice, it seems that it has less influence than the stability of the resulting alkene.

Also note that often E1 is run under equilibrating conditions, where even if a less-substituted (less-stable) alkene is formed, it can be protonated to reform the carbocation, and the material will funnel to the most stable product. This is thermodynamic control.


Defintion of Solubility in Different cases How do we define solubility (in moles per liter) in cases when concentration of cation is not equal to the concentration of anion?

e.g. When we put AgCN in an acidic solution (Leading to formation of HCN) or when AgCl is put in CH3COONa (Leading to precipitation of CH3COOAg)

Do we take concentration of Ag dissolved as solubility or concentration of CN or Cl as solubility?

Please answer for both cases.

There are two ways of dealing with this.

Solubility product

The more general way is to state the solubility product. Taking your first example, this would have the same value no matter how much acetate is present in solution before adding the silver ions. The disadvantage is that for highly concentrated solution (in fact, in the presence of ions, even for moderate concentrations) this gets inaccurate unless you switch from concentrations to activities.

Solubility (of a solid)

The other way is to specify the solubility of $\ce{AgCl}$ in pure water, in 0.1 M sodium acetate, in 0.5 M sodium acetate and so on. In this case, because silver chloride is much less soluble than silver acetate, no silver acetate would precipitate (the silver ion concentration will be sufficiently low in solution so that silver acetate will not be exceeding the solubility product).

Similarly for AgCN, you could specify its solubility in pure water, in aqueous solution buffered at pH 7, or buffered at any other pH. For the latter example, you should probably also specify ionic strength or the entire makeup of the solution prior to adding AgCN. In this case, there will be a strong dependence of the solubility on the pH because HCN is soluble and its formation will decrease the concentration of cyanide ions while the concentration of silver ions will remain the same up to the point where it precipitates.

In both of these cases, there is no ambiguity in terms of concentration because there are no sources of the ions forming the precipitate other than what was added as a (stoichiometric) solid.

On the other hand, if you add NaCl to a solution already containing LiCl, the apparent solubility of NaCl will be lower than in pure water because of the common ion effect. You could still talk about the solubility of NaCl in 2 M LiCl, and the quantity would refer to the concentration of sodium and chloride ions due to the addition of NaCl, not due to ions already present before addition.

Concentration of ions vs. concentration of added solid

With any salt that does not have 1:1 stoichiometry, we give the concentration of the salt, not that of the ions. For example, we talk of a 1 M $\ce{MgCl2}$ solution even though it is a strong electrolyte containing 1 M magnesium ions and 2 M chloride ions. The concentration given is based on the formula of the solid added, not on the concentration of ions after dissolution. In a similar way, we would say we added acetic acid to a concentration of 1 M even though we know full well that some of it will be ionized.