Why does diamond conduct heat better than graphite? Our teachers made us accept (without any explanation) that diamond conducts heat better than graphite. What is the reason behind this (alleged) fact?

Diamond is one of the best thermal conductors known, in fact diamond is a better thermal conductor than many metals (thermal conductivity (W/m-K): aluminum=237, copper=401, diamond=895).  The carbon atoms in diamond are $\ce{sp^3}$ hybridized and every carbon is bonded to 4 other carbon atoms located at the vertices of a tetrahedron.  Hence the bonding in diamond is a uniform, continuous 3-dimensional network of $\ce{C-C}$ single (sigma) bonds.  Graphite on the other hand is formed from $\ce{sp^2}$ hybridized carbon atoms that form a continuous 2-dimensional sigma and pi bonding network.  This 2-dimensional network forms sheets of graphite, but there is little connection between the sheets, in fact, the sheet-sheet separation is a whopping ~3.4 angstroms.  This might lead us to suspect that heat conduction in the 2-dimensional sheet of graphite would be superior to diamond, but that heat conduction between graphite sheets would be very low.  This is, in fact, an accurate description of thermal conduction in graphite.  Thermal conductivity parallel to the graphite sheets=1950, but thermal conduction perpendicular to the sheet=5.7.  Therefor, when we consider thermal conduction over all possible directions (anisotropic) diamond would be superior to graphite.


Why are regular plastic bags not smell-proof? Due to the wide use of plastic bags as containers for almost everything in everyday life, I naturally use them to wrap and hold dirty diapers as a temporary stop before getting to throw them in the garbage outside. However, these bags do not stop the smell from spreading around the bags at all.

I wonder why is that? These bags seem to be tight enough to hold liquids (when intact), but they do not hold smells at all. I guess the obvious answer is that the plastic\nylon polymers have wide enough gaps between them for the passage of small clusters of molecules in the micro- or nano-level, but I was wondering if there was some other explanation and if anyone could elaborate on the matter.

Well, you basically answered your own question nicely with your "obvious answer" discussion. On a molecular scale, smallish molecules, like the volatile compounds that evaporate and make way to your nose, are able to permeate the plastic. This happens in part for the reason you said; on such a small scale, the plastic that appears fully "solid" and impermeable on a macro scale is actually a mesh of molecules with paths large enough for these small, smelly, molecules to escape.

There are factors at play other than molecular size, like the chemical composition of the plastic and the escaping compounds. If the compounds have a strong affinity for the plastic, it will take less time for them to permeate as they will more quickly absorb through the plastic.

Regarding your observation that the plastic seems to hold liquids, but not smells; this is probably an illusion. I suspect that the liquids you've placed in the bag actually have little or no smell (i.e. water, food oils). For example, in an analytical laboratory setting, water samples or water-based standards stored in various plastic containers have a limited shelf life because the water slowly permeates the plastic, evaporating into it's surroundings. So the water is escaping the plastic just like a smelly compound might do, but of course water has no smell.


Prediction of ionic reduction? Here are some standard electrode potentials:-

$$\ce{SO_4^{2-} + 4H^+ + 2e^- -> SO_2 + 2H_2O}\\E^o=0.17\mathrm{V}\\
\ce{NO_3^- + 4H^+ + 3e^- -> NO + 2H_2O}\\E^o=0.97\mathrm{V}\\
\ce{NO_3^- + 2H^+ + e^- -> NO_2 + H_2O}\\E^o=0.78\mathrm{V}$$
When nitric acid reacts with a metal, either $\ce{NO_2}$ or $\ce{NO}$ is formed? How do you predict which is thermodynamically favourable and which reaction will actually take place? For example, when dilute nitric acid reacts with copper, Nitric oxide is liberated and when it is concentrated, Nitrogen dioxide is evolved? As Nitric acid formation is more favourable, shouldn't it proceed in all cases as increasing the concentration will decrease/increase the potential of both the reaction simultaneousy(Nernst equation)?

In the case of sulfate ion, although the reduction of sulfate is more favourable than that of hydrogen, it is hydrogen ions which are getting reduced when a metal reacts at dilute concentrations? why is this so? Moreover, at high concentrations, once again sulfur dioxide is evolved? How can we understand such occurrences? Is overpotential involved in these observances?

In the case of permanganate ions, there are so many possible redox pathways, but a different one is preferred in a different case based on the conditions. For example, in acidic medium, permanganate is reduced to $\ce{Mn^{2+}}$ while in neutral/slightly alkaline solutions, it reduces to $\ce{MnO_2}$. Why is this so? Here are the redox potential values for permanganate conversions (latimer diagrams)

I want to know the general way to explain all these phenomena, as to what drives the reaction to give one product under certain condition and another in different conditions. Explaining any one example would be enough to set the general method which I can extrapolate to understand the remaining Examples.I am comfortable with basic electrochemical equations and those of thermodynamics.

Fortunately, the explanation is not as complex as it might seem. It is an oft-made analysis, but one must be very careful when determining reaction spontaneity by looking at the standard free energies $\Delta G_r^o$ alone, as most reactions are not performed in standard conditions and therefore the actual reaction free energy change $\Delta G_r$ may be quite different.

You are correct that, in standard conditions, $\Delta G_r = \Delta G^o_r$ and therefore that $\ce{NO}$ is the most stable reduction product of $\ce{NO_3-}$. However, standard conditions entail, among other things, that $\ce{[H+]\ = \ 1\ M}$ (at least approximately). This corresponds to quite an acidic aqueous solution, with pH zero. If one were to examine the reaction in a more dilute acidic medium, then the value of $\Delta G_r$ for each reduction reaction would change. The reaction that produces $\ce{NO_2}$ has a $\Delta G_r$ that depends on $\ce{[H+]^2}$, while for $\ce{NO}$ the $\Delta G_r$ varies with $\ce{[H+]^4}$. It is clear then that, as one goes to a more and more dilute acidic medium, $\Delta G_r$ decreases more strongly for the reaction that produces $\ce{NO}$, and at sufficiently weak acidic solutions the relative stabilities of the products actually switches, so then $\ce{NO_2}$ becomes the most stable product. A very similar argument can be made for the switch between $\ce{H+}$ and $\ce{SO^{2-}_4}$ reduction in media with sulfuric acid. For $\ce{MnO^{-}_4}$, note that the conversion to $\ce{Mn^{2+}}$ implies the consumption of eight protons, and hence if it can happen, it will likely do so in acidic media. As the acidity of the solution decreases, the conversion to $\ce{Mn^{2+}}$ is quickly suppressed, favouring reactions for which the $\Delta G_r$ depends less strongly on $\ce{[H+]}$, and so we find that $\ce{MnO^_2}$ becomes more favourable.


Difference between positional isomerism and metamerism (organic chemistry) I don't want the definition of positional isomerism and metamerism but want what differentiates the two types of isomerism.

I know metamers have different alkyl groups on either side of it, but doesn't that just mean the position of the functional is changing? Hence shouldn't it be a positional isomer.

If functional group carbon is part of principal chain then metamers are also positional isomers. In case of polyvalent functional group like sulphoxide($\ce{>S=O}$), secondary amine ($\ce{-NH-}$) etc. structure formed by change in structure of alkyl groups will be metamers only.  For example, diethyl amine ($\ce{Et2NH}$) and methyl propyl amine ($\ce{MeNHPr}$) are metamers.


When comparing boiling points of two ionic compounds, what bonding factors should you assess? When comparing boiling points of two ionic compounds, should you look at their electronegativity difference, lattice energy, or strength from LDF forces (or all 3 factors)?

For example, between LiCL and NaCl, which one has the higher boiling point?

If we look at EN difference and LDF forces, it would be NaCl, because the EN difference here would be $2.1$ compared to $2.0$ for LiCL; also, since Na has an extra energy level, it has a larger electron cloud and thus stronger LDF's.

However, if we look at it in term of lattice energy or Coloumb's law, LiCl has a larger boiling point because high lattice energy is a combination of large charge and small ions. Therefore, even though Li and Na have the same charge, Li has a smaller atomic radius than Na.

To compare 2 ionic compounds' melting point, there are generally 3 determinants:

Ionic Radii, smaller the ionic radius, closer the ions are to each other, therefore stronger the electrostatic attraction between them. (analogous to stronger attraction when closer to a magnet) A good example would be melting points of cations from the same group with the same anion (NB: LiF does not follow the trend, potentially due to its more covalent nature):

$\ce{NaF}\ -\  \pu{993 ^\circ C ( 1819 ^\circ F)}$
$\ce{KF}\  -  \pu{858 ^\circ C ( 1576 ^\circ F)}$
$\ce{RbF}\  -\  \pu{795 ^\circ C ( 1463 ^\circ F)}$

Charge of ions, larger the charge an ion has, more positive or negative the ion is, therefore there is a stronger attraction between ions. (analogous to stronger attraction with two stronger magnets in opposite poles) An example would be $\ce{NaCl}$ and $\ce{MgO}$:

$\ce{NaCl}\ -\ \pu{801 ^\circ C ( 1,474 ^\circ F)}$
$\ce{MgO}\ -\ \pu{2852 ^\circ C ( 5166 ^\circ F)}$

Ratio of ions, in most cases, closer the ratio is to 1 : 1 between anions and cations, stronger the attraction is. larger or smaller the ratio is, the more the anions or cations will have to 'share' or 'divide' the electrostatic attraction from the oppositely-charge ion, decreasing the melting point. An example would be $\ce{NaCl}$ and $\ce{MgCl2}$:

$\ce{NaCl}\ -\ \pu{801 ^\circ C ( 1,474 ^\circ F)}\ [1 : 1]$
$\ce{MgCl2}\ -\ \pu{714 ^\circ C ( 1317 ^\circ F)}\ [1 : 2]$

The determinants for melting points would generally also apply to the boiling points of ionic compounds. However, it should be noted that other factors would affect the boiling point as well.


Energy Diagram for Propane I've been unable to find an energy diagram for propane either in my book or online via Google. So I resorted to making my own. I suspect that propane isn't covered because it's essentially the same deal as ethane - we can have one staggered conformation and one eclipsed conformation (only in propane it's a methyl eclipsing a hydrogen, not hydrogens eclipsing each other).

So I drew the Newman projections and an energy diagram.

Your diagram is correct.  Here is a link to an on-line energy diagram for propane and here is the diagram itself


Is graphite a better lubricant on the Moon than on the Earth? My book says that it is a better lubricant on the Moon because of absence of gravitational pull on the Moon. However, I also read that it's lubricant nature is because of a film of moisture or gas molecules absorbed on the surface of its layers and in vacuum, it gets dried up and would be a bad lubricant. Which is correct?

Graphite, like most solid lubricants, has a layered structure which makes it a good lubricant.
The atoms of a layer are strongly bonded to each other, while the interlayer (van der Waals) forces are much weaker.
The layers align themselves between the sliding surfaces and the layers slide one over the other providing lubrication.

Since it are the shear forces that come into play, gravity has no influence on the system.
From that point of view, there is no difference between its use here or on the Moon.
But, there are also extrinsic factors which influence its performance.

Graphite works best in a humid environment.
This is probably due to the weakening effect of the water molecules on the interlayer forces.
So, graphite should not work well in vacuum.

Another solid lubricant, $\ce{MoS2}$, which has also a layered structure like graphite, works much better in vacuum and should be the lubricant of choice on the Moon.


Clogged drain, commercial products are ineffective I have a clogged drain from my shower. Commercial products like Savo or WC Net totally don't work. Is there any chemical substance which can be bought without permission or which can be made easily to clear the drain?

I live in Czech Republic, it's part of European Union, just for your information about the law system here (can be helpful to decide what's legal).

My first idea is sodium hydroxide, is it powerful enough?

I don't know what the products you mention contain, but you basically have four chemical weapons, and two physical weapons against a clogged drain. Choosing depends on the nature of the clog, which is hard to guess.

Chemistry first:

Acidity: sulfuric acid. ’nough said.
Basicity (causticity): lye, a.k.a. “caustic soda”, a.k.a. sodium hydroxide.
Oxidation: bleach or peroxide.
Enzymes: take time to act, but can be very powerful. This you should find at an eco-friendly store or supermarket aisle.

Now, on to physics:

Hot water: if the pipe is clogged by solidified fats, boiling water can melt them and unclog. Hardly typical for a shower drain, but very often useful for kitchen sink.
Last, but not least: mechanical action. Either a drain snake, or a plunger. Most often, the clog is a localized problem, and by breaking it into smaller parts, you will restore the flow.

For a typical shower drain, hair and other organic matter is most likely the issue, so I would choose drain snake first, then sodium hydroxide, then enzymes.


What is the rate-determining step in free radical halogenation? Recently, we were learning about free radical halogenation in school. A set of notes given to us wrote that the rate-determining step is the propagation step. This seems to go against my understanding of the chemistry of this reaction. The step that likely has the highest activation energy should be the initiation step as it is the only step that involves bond-breaking solely. I have looked up various sources and they also seem to suggest that the rate-determining step ought to be the first step.

Advanced Organic Chemistry (5th ed.) by Carey and Sundberg states that:

  Free radicals are usually highly reactive and the individual steps in a chain reaction typically have high absolute rate constants. However, the concentrations of the intermediates are low. The overall rates of reaction depend on the balance between the initiation and termination phases of the reaction, which start and end the chain sequence.

It does seem odd as a blanket statement. However, the bond dissociation energies (BDEs) for Cl2 and Br2 are quite low, 243 and 193 kJ/mol, respectively, while BDEs for the C-H bond in alkanes are much higher (eg. 435 kJ/mol for methane). So, it's not necessarily the case that events involving only bond breaking will be the rate-determining step, particularly since, as Zhe points out, the propagation steps are bimolecular.

Bromine radicals are significantly less reactive than chlorine radicals, leading to interesting selectivity differences between chlorination and bromination reactions - https://youtu.be/c5enwHzAwmA

Edit: BDEs are taken from Klein, "Organic Chemistry", 3rd ed., Chapter 10.


What are the signs for work done by (on?) a gas during isothermal reversible compression? For the expansion of a gas, isothermally and reversibly, what is the sign taken?
Do I use this equation
$-nRT\ln  (Vf/Vi)$
or this one
$nRT\ln  (Vf/Vi)$ ?
Moreover, when I'm asked about the 'work done', what exactly does the work? The gas itself? The surroundings? I'm confused about this because my text says one thing and the guidebook, another.
Thanks.

The easiest way to remember the sign conventions is to think of the gas particles as "wanting" to get away from each other. To reduce the volume of the gas, you have to do work on the gas - the surroundings are doing the work. In chemistry we always take the point of view of the system. If we do work on the system, we add energy to it, and therefore, the sign of work is positive.

This means the opposite is true as well. When the volume of the system increases, the sign of the work term must be negative - the system is losing energy and doing work on the surroundings.

For your equations, all of this holds for the first one:

$$-nRT\ln  (Vf/Vi)$$

This is because a decrease in volume will give you a fraction in the $\rm{ln}$ term, which leads to a negative sign. The leading negative sign reverses that. The result is that any compression will have a positive work term, and any expansion will have a negative work term.

You can see how the correct sign is a result of the derivation in the wikipedia article.


Polymerization Isomerism In Transition-Metal Complexes Not True Isomerism? At school we've just finished learning about Coordination Compounds.

Our textbook has defined Isomers as

  Compounds having same molecular formula but different arrangement of atoms, due to which they differ in one or more physical properties are called isomers.

It further goes on to discuss the various kinds of structural isomerism seen in (transition-metal) complexes:

Ionization Isomerism
linkage isomerism
ligand isomerism
solvate/hydration isomerism
coordination isomerism
coordination-position isomerism
polymerization isomerism

Polymerization isomers, according to our textbook, are

  those complexes that have the same empirical formula, but differ in molecular mass by integral multiples of empirical mass.

e.g., $\ce{[Pt(NH3)2 Cl2]}$ and $\ce{[Pt(NH3)4][PtCl4]}$ are polymerization isomers.

Yet another book I've read, simply states that polymerization isomerism is not isomerism in the true sense of the word, without elaborating on this statement.

But looking at the definition for isomerism, the way I see it, polymerization isomerism fits the bill, so what exactly do the authors of the book imply by that statement? Are they wrong?

I’m pretty sure that polymerisation isomerism should mainly be considered a historical term nowadays. Going back in the history of science, all people knew at some point was which elements a specific compound is made out of. That worked well for main group metal salts but not too well for transition metal salts. Later, quantitative analysis established the law of multiple proportions, which basically says that if two compounds are made up of the same set of elements, the ratio of element mass ratios will always be an integer ratio. Sounds hard to understand, here’s an example:

$\ce{CuCl}$ has one chlorine per copper, $\ce{CuCl2}$ has two. So for every $x~\mathrm{g}$ of copper, one of the two will have $1~\mathrm{g}$ of chlorine, the other will have $2~\mathrm{g}$, and $1$ and $2$ form a simple integer ratio.

And all of a sudden there were compounds that were clearly distinct but gave the same ratio of elements, a simple example being $\ce{NO2}$ and $\ce{N2O4}$. To understand this, the term polymerisation isomer must have been introduced.

Nowadays, isomers are defined as having the same elemental composition on a per-fragment basis. So examples such as $\ce{N2O4}$ (which is a different molecule than $\ce{NO2}$) or the platinum complexes you give would no longer be considered isomers — neither constitutional ones nor stereoisomers. Hence why so many people said ‘your second book is right.’


Chemical equilibrium and Le Chatelier's principle $$\ce{energy + 2KClO3 (s) <=> 2KCl(s) + 3O2 (g)}$$

The reaction will be more active in the reverse direction if:

$\ce{O_2}$ is added
more $\ce{KCl}$ salt is added.
More $\ce{KClO3}$ is added
Increasing the temperature.

My thoughts:

I think (1) and (2) are both right. But the book states that only (1) works and I can't find reason for that searching google was useless.

$K_c = [\ce{O2}]^3$, so if we increased $\ce{KCl}$ or decreased $\ce{KClO3}$ it won't affect the equilibrium?

$$\ce{Fe(OH)(s) <=> Fe^{2+}  + 2OH^-}$$ The reaction will be more active in the reverse direction if : ... is added

$\ce{KOH}$
$\ce{Na_2S}   $
$\ce{Fe(OH)_2}  $
$\ce{Fe(NO3)2}$

Won't it be (1) and (4)? The book says it's (4) only.

1) Increasing Oxygen pressure will increase the reverse rate because to reestablish equilibrium some of the oxygen should be consumed.
2) Out of system due to precipitation.
3) Out of system due to precipitation.
4) This is endothermic reaction. So increasing temperature will increase the forward rate to consume the added heat.


What is the oxidizing agent for Fe in the following redox reaction? Redox: $\ce{Fe + 2HCl  ->  FeCl2 + H2}$

Oxidation:  $\ce{Fe -> Fe^{2+} + 2e-}$

Reduction: $\ce{2H+ + 2e- -> H2}$

The $\ce{HCl}$ and $\ce{H2}$ are formed by covalent bonds. The $\ce{FeCl2}$ forms an ionic bond.
The iron actually loses 2 electrons to the two chlorines, so the oxidation equation makes sense to me. But the hydrogen is always in a covalent bond with the chlorine or another hydrogen atom, it is always sharing electrons, never losing or gaining, so the reduction equation doesn't make sense to me. When the $\ce{H}$ is bonded with the $\ce{Cl}$, it "has" two electrons because it is sharing a pair with the $\ce{Cl}$. After the reaction, when the $\ce{H}$ is bonded with another $\ce{H}$, it still "has" two electrons because it is sharing a pair with another $\ce{H}$, so no electron was gained as the reduction equation indicates.

After a little a thinking I came up with an answer, which raised even more questions:

Since the covalent bond between the $\ce{H}$ and the $\ce{Cl}$ is a polar bond and it behaves more like an ionic compound because of that, when $\ce{HCl}$ is dissolved in water, the $\ce{H}$ will lose an electron to the $\ce{Cl}$, making $\ce{H+}$ and $\ce{Cl-}$ ions, so there is no covalent bond when $\ce{HCl}$ is in water. The $\ce{H+}$ ions can gain one electron and then make a covalent bond with each other, making $\ce{H2}$ molecules.

But what about the $\ce{Cl-}$ ions in the solution? they have already formed an octet in their outer shell by getting an electron off the hydrogen when they got separated, so they are satisfied. Then why $\ce{Fe}$ atoms oxidize? who is the oxidizing agent? The $\ce{H2}$ molecules are already comfortable with their covalent bond, and the $\ce{Cl-}$ ions are also satisfied with their octet.

I'm very confused.

To answer your initial question:

The oxidizing agent is the reagent that gets reduced. Conversely: The reducing agent is the reagent that gets oxidized.

Now, addressing your trouble with the concept:
You are right that hydrochloric acid is found in the ionic form when in aqueous solution, so:
$$\ce{HCl ->[][\ce{H2O}] H+ + Cl-}$$

Now we have the reagent that gets reduced, which is the proton. Together with another proton and two electrons, they form hydrogen gas that then evolves from the solution:
$$\ce{2H+ + 2e- -> H2 ^}$$
This is basically our electron sink, because that's where they go "to die", i.e. to vanish from our solution.

But where do these electrons come from? Well, we have two potential electron donors here: The chloride anion and the iron. Which one will give up his electrons more easily?

The answer we already know: Iron.

Why? Because as you correctly stated, the chloride anion has all the shells filled and is very happy because of that filled octet state. So the energy to get out an electron from it is higher than to rip it away from an unsuspecting iron atom.1)
$$\ce{Fe^0 -> Fe+ + e- -> Fe^{2+} + 2e-}$$

The electrons then vanish (because of the formation of hydrogen, remember?) which leaves us with some iron cations and chloride anions. They will then start to mix with each other. Don't judge them, they don't know any better!
$$\ce{Fe^{2+} + 2Cl- <=> FeCl2}$$

Since ferric chloride (II) is quite soluble in water, it will most probably not precipitate.

Footnotes:

Actually, you could also compare the standard reduction potentials of the reactions $\ce{Fe -> Fe^{2+} + 2e-}$ and $\ce{Cl- -> Cl^{.} + e-}$. The one that has the lower standard reduction potential is the winner (because the energy of a system is always minimized). This is also what you would usually do, but in this example it wasn't necessary because you could get to the solution via chemical concepts.


How do I calculate the grams of a product after a reaction? If you mix $50\,\mathrm{mL}$ of $2.5\,\mathrm{M}$ $\ce{HCl}$ with $100\,\mathrm{mL}$ $2.5\,\mathrm{M}$ $\ce{NaOH}$, you will get sodium chloride. If such reaction occurs only at $50~\%$, how many grams of sodium chloride will you get?

Well, I guess that the reaction is like

$$\ce{HCl + NaOH \to NaCl + H2O}$$

Since the concentration is the same for both reagents ($2.5\,\mathrm{M}$), I imagine that the concentration will be the same for the products?

So we got $2.5\,\mathrm{M}$ of $\ce{NaCl}$. However, to get the grams I first need to know the moles of $\ce{NaCl}$. I know that the concentration is $2.5\,\mathrm{M}$, but what of it? What can I do to get the moles of the resulting $\ce{NaCl}$?

The concentration $c$ of a solution is the molar quantity of the solute $n$ devided by the volume of the solution $V$:

$\begin{equation}
c = \frac{n}{V}
\end{equation}$

as you state, for example for HCl 2.5 mol/L (or, equivalent, 2.5 M).  Inversely, if you know volume and concentration of your solution, you may access the molar quantity of your solute.  Actually, there is such an equation for each of the solutions, indepent from each other; you may index the variables accordingly, too.

Yet watch, there is a catch: mixing the two solutions together increases the total volume, and simultaneously decreases the concentration of the specis in question -- without altering the molar quantity of $\ce{Na+}$, nor $\ce{Cl-}$ ions present.  Hence "2.5 M of NaCl" (as measure of concentration) in the solution in common is incorrect.

Because of the initially asked question, one of the solutions contains the limiting reagent.  Figure out which one -- this includes the stoichiometry in the equation of chemical reaction. Knowing the molecular amount $n$ of it, your are able to determine the amount of sodium chloride formed; initially as moles, later (via molecular mass) in grams.


Heavy Metal Toxicity Ok, so I am working on a project where I have to make a tax for manufacturers using non recyclable materials. So for this tax you multiply the weight (in kg) by how toxic the material is, this gives you an answer in pence. So I have a list of heavy metals that are dangerous to the environment but I don't know what one is the most toxic and the least toxic.

This is what I think the order from most dangerous to least dangerous is:

Selenium (most harmful to the environment)
Arsenic
Mercury
Copper
Lead
Cadmium
Nickel
Chromium
Antimony
Zinc (least harmful to the environment)

Could someone please share their opinion on this order and suggest a better order. Also could someone suggest a suitable number to multiply the weight by to get price in pence.

Thanks

This is probably a more complex problem than your instructor intended. In addition to the various metals having different toxicities, different compounds of these metals have different toxicities.

For example, nickel metal has low toxicity, but the compound $\ce{Ni(CO)4}$ is a poisonous, flammable, volatile liquid.

Likewise alkyl mercury compounds, e.g. $\ce{(CH3)2Hg}$ are vastly more toxic than the low solubility mercury salts like $\ce{Hg2Cl2}$.

If you just want to know about the pure metals, then this can be easier. Look up a Safety Data Sheet for each metal. It would be better if you pull all your SDS from the same source, that way they are all using the same data sources.

For example, here is a link to the SDS for mercury from Fisher Scientific. Section 12 contains ecological information, including LC50 (concentration that kills half of the population in the time period given) data for various aquatic organisms. Section 11 contains toxicological information, including LD50 (dose that kills 50% of the population) for various organisms.

The best way to do this is to try to find one set of data for all the metals with the same organism for the best comparison. Then, devise your scale.


Solubility of methanol and ethanol in water Which substance is the most soluble in water?

  $\ce{CH3OH}$
  $\ce{CH4}$
  $\ce{C2H6}$
  $\ce{C2H5OH}$

I wasn't able to decide between 1 and 4, as both have hydroxides, but the answer is 1. How?

For primary alcohols, the trend is the longer the chain, the less soluble.

However, at room temperature, methanol, ethanol and propanol are all completely miscible with water.

At lower temperature, below the freezing point of water, there is finite solubility.

See the phase diagram that is Fig. 1 in CRYSTAL GROWTH OF WATER ETHANOL MIXTURE and that in Fig. 1 of Solid-Liquid Phase Diagram of the System Methanol-Water

Which alcohol would you choose for windshield deicing solution?


Where does Free Energy go in a Redox Reaction? This doubt came to my mind while answering the following question:

  Why does a voltaic cell not operate unless the two separate compartments are connected by an external circuit?

The answer seems to be that only on connecting the two halves of a Redox reaction via an external circuit, does one allow the electrons to be transferred from one species to another. And while the electrons move towards their goal (to a higher potential), they do work in the way (light a bulb or a motor or something).

Another way to view this process is that the free energy of the Redox reaction is used to do useful electrical work.

But then my brain started to wonder and there came up a question - what happens to all this energy when none of it is actually used to do electrical work, i.e, when the Redox reaction is let to occur spontaneously, where does all the free energy go? Does it get converted to heat, thus warming the solution? Or does it simply stay stored in the solution in some form?

Free energy is a state function. No matter how you run the reaction, if you start with a certain state (set of concentration) and end with a certain state (set of concentrations at equilibrium), the change will be the same.

The free energy of reaction is the maximal amount of non-pV work (electrical work in this case) the reaction can do. If it does not do any work, that energy will be released as heat instead. For example, if you short the battery, the wire and the electrochemical cell will heat up while the reaction is going on.

If you want an argument from first principles, you should look at the first law of thermodynamics. The change in energy of a system is equal to the sum of the heat and the work exchanged with the surrounding. When making optimal use of the free energy, you maximize the work (even sometimes transferring heat from the surrounding to do even more work).


For what reasons, is isopropyl alcohol widely used as a cleaning solvent? I have worked in electronics and aeronautics industry 30 years ago, and I have always wondered, what were the specific properties of isopropyl alcohol, which make it used instead of other alcohols like common methyl or ethyl alcohol ?

Edit:

I know that it's less toxic than other solvent and that it is easier to separate from water (by adding salt) than other alcohols (reduce producing costs ?), but some reasons might be less pertinent than others.

From this Wikipedia article

  Isopropyl alcohol dissolves a wide range of non-polar compounds. It also evaporates quickly, leaves nearly zero oil traces, compared to ethanol, and is relatively non-toxic, compared to alternative solvents. Thus, it is used widely as a solvent and as a cleaning fluid, especially for dissolving oils.

 You should try Googling first :-)

Also, the antiseptic properties of isopropyl alcohol are an added benefit (over the other lower alcohols).

As you pointed out, methanol is notoriously toxic, so using that as an industrial cleaning agent is a no-no.

Ethanol for industrial purposes is often available as denatured alcohol; which has added impurities to render it toxic and/or alter its taste and smell.
These impurities may interfere with the equipment (you mention the electronics and aeronautics fields) you clean it with.

However commonly available isopropanol isn't denatured (then again, there's no real need to denature it). Since it is lacking in added impurities, it presents itself as a better choice over ethanol for cleaning such equipment.


What is it meant by two different groups in stereoisomerism? I understand there are some properties in an organic compound that allow stereoisomerism.

double bond can't rotate
two double bonded carbon atoms each have two different atoms or groups attached

I'm unsure by the meaning of two different atoms/groups around on each side of the double bond.

With respect to but-2-ene. From my understanding, both sides of the carbon atom have $\ce{-CH3}$ on either side of the double bond so how does this comply with the rule that "Two different atoms or groups attached around each carbon of the double bond"

Stereoisomers are different due to the spatial arrangement of their atoms.

The picture that you gave has two $\ce{CH3}$ groups on one side. That is called cis - as they are on the same side. There is another isomer, called trans. That is when one of the $\ce{CH3}$ is on the top and the other $\ce{CH3}$ is at the bottom. The arrangements of the $\ce{CH3}$ and $\ce{H}$ is different in cis and trans.

This link explains it well: Chem.Libretexts


Molar concentration calculation with absolute uncertainty A solution of $\ce{F-}$ is prepared by dissolving $(0.0933\pm0.0004)\ \mathrm g$ of $\ce{NaF}$ $(M= (41.989\pm0.001)\ \mathrm {g/mol})$ in $156.00\pm\ \mathrm{mL}$ of water. Calculate the concentration of $\ce{F-}$ in the solution and its absolute uncertainty.

First of all, uncertainty aside, I got $0.0141$ for the concentration of $\ce{F-}$. However, this is not the correct answer (according to Sapling Learning).

Furthermore, what technique should I use to find the absolute uncertainty?

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematic operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. The same applies to the equation for the uncertainty of the desired result.

In this case, the concentration $c$ shall be calculated. The mass $m=0.0933\ \mathrm g$, the molar mass $M=41.989\ \mathrm{g\ mol^{-1}}$, and the volume $V=156.00\ \mathrm{ml}$ are given. The uncertainty of the mass $m$ and the molar mass $M$ are given as $u(m)=0.0004\ \mathrm g$ and $u(M)=0.001\ \mathrm{g\ mol^{-1}}$, respectively. The uncertainty of the volume $V$ is not given; however, when a number is given without any further information, it is generally interpreted so that the last digit is rounded. Therefore, the given value $V=156.00\ \mathrm{ml}$ is assumed to represent a value between $155.995\ \mathrm{ml}$ and $156.005\ \mathrm{ml}$, or $V=\left(156.000\pm0.005\right)\ \mathrm{ml}$. Thus, the assumed uncertainty of $V$ is $u(V)=0.005\ \mathrm{ml}$.

The concentration $c$ is defined as
$$c=\frac nV\tag1$$
where $n$ is amount of substance, which is unknown.
Since the molar mass $M$ is defined as
$$M=\frac mn\tag2$$
the concentration $c$ can be calculated according to
$$\begin{align}c&=\frac m{M\cdot V}\tag3\\[6pt]
&=\frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}}\\[6pt]
&=1.42437\times10^{-5}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&=0.0142437\ \mathrm{mol\ l^{-1}}
\end{align}$$

In this case, we may assume that the uncertainties of $m$, $M$ and $V$ are not correlated. Therefore, the relative uncertainty of the concentration $c$ may be estimated as
$$\begin{align}
\left(\frac{u(c)}{c}\right)^2 &= \left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2\tag4 \\[6pt]
\frac{u(c)}{c} &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2}\tag5
\end{align}$$

Thus, the absolute uncertainty of the concentration $c$ is

$$\begin{align}
u(c) &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2} \cdot c\tag6 \\[6pt]
&= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2} \cdot \frac m{M\cdot V}\tag7 \\[6pt]
&= \sqrt{\left(\frac{0.0004\ \mathrm g}{0.0933\ \mathrm g}\right)^2 + \left(\frac{0.001\ \mathrm{g\ mol^{-1}}}{41.989\ \mathrm{g\ mol^{-1}}}\right)^2 + \left(\frac{0.005\ \mathrm{ml}}{156.00\ \mathrm{ml}}\right)^2} \cdot \frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}} \\[6pt]
&=6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&\approx0.00006\ \mathrm{mol\ l^{-1}}
\end{align}$$

Note that the same result may be obtained using the general formula

$$\begin{align}
u(c) &= \sqrt{\left(\frac{\partial c}{\partial m}\right)^2\cdot u^2(m)+\left(\frac{\partial c}{\partial M}\right)^2\cdot u^2(M)+\left(\frac{\partial c}{\partial V}\right)^2\cdot u^2(V)}\tag8 \\[6pt]
&= \sqrt{\left(\frac1{M\cdot V}\right)^2\cdot u^2(m)+\left(-\frac m{M^2\cdot V}\right)^2\cdot u^2(M)+\left(-\frac m{M\cdot V^2}\right)^2\cdot u^2(V)}\tag9 \\[6pt]
&= \sqrt{\left(\frac1{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}}\right)^2\times \left(0.0004\ \mathrm g\right)^2\\
+\left(-\frac {0.0933\ \mathrm g}{\left(41.989\ \mathrm{g\ mol^{-1}}\right)^2\times 156.00\ \mathrm{ml}}\right)^2\times \left(0.001\ \mathrm{g\ mol^{-1}}\right)^2\\
+\left(-\frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times \left(156.00\ \mathrm{ml}\right)^2}\right)^2\times \left(0.005\ \mathrm{ml}\right)^2}\\[6pt]
&=6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&\approx0.00006\ \mathrm{mol\ l^{-1}}
\end{align}$$

The numerical values of the concentration $c$ and its uncertainty $u(c)$ should not be given with an excessive number of digits. In this case, it suffices to quote $u(c)=0.00006\ \mathrm{mol\ l^{-1}}$ and accordingly $c=0.01424\ \mathrm{mol\ l^{-1}}$; i.e. $c=\left(0.01424\pm0.00006\right)\ \mathrm{mol\ l^{-1}}$ or $c=0.01424(6)\ \mathrm{mol\ l^{-1}}$. In some cases, however, it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.

Note

The large equations for the uncertainty $u(c)$ may be confusing and susceptible to errors.

To verify the result, you should check whether the calculated quantity and its absolute uncertainty have the same unit (here: $\left[c\right]=\left[u(c)\right]=\mathrm{mol\ ml^{-1}}$).

Furthermore, you should compare the relative uncertainty of the result with the relative uncertainty of the input data. Usually, the relative uncertainty of the result is greater than or equal to the relative uncertainty of the input data. In this case:

$$\frac{u(m)}{m}=\frac{0.0004\ \mathrm g}{0.0933\ \mathrm g}=0.0042872\approx0.43\ \%$$

$$\frac{u(M)}{M}=\frac{0.001\ \mathrm{g\ mol^{-1}}}{41.989\ \mathrm{g\ mol^{-1}}}=2.38158\times10^{-5}\approx0.0024\ \%$$

$$\frac{u(V)}{V}=\frac{0.005\ \mathrm{ml}}{156.00\ \mathrm{ml}}=3.20513\times10^{-5}\approx0.0032\ \%$$

$$\frac{u(c)}{c}=\frac{6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}}{1.42437\times10^{-5}\ \mathrm{mol\ ml^{-1}}}=0.0042874\approx0.43\ \%$$

We find that, in this case, the uncertainty of the concentration $c$ is mainly caused by the uncertainty of the mass $m$.


Why is the voltage produced by a voltaic cell affected by temperature as described by the Nernst Equation? I understand why changing the concentrations of the electrolytes in a voltaic cell affects the voltage. Given a zinc-copper cell for example, if I were to increase the concentration of the zinc sulphate electrolyte, the equilibrium between zinc ions and zinc metal would shift towards the side of the zinc metal. This would cause the reduction potential for zinc ions to increase, decreasing the overall cell potential. Please correct me if I’m wrong as this is how I understand it!

However, I don’t quite understand why temperature affects the voltage the way it does. According to the Nernst equation, if the reaction quotient is greater than one (meaning that the zinc sulphate concentration is greater than the copper sulphate concentration in this example), increasing temperature decreases voltage. Conversely, if the reaction quotient is less than one, increasing temperature increases voltage.

I am familiar with Le Chatlier's principle, but if I were to apply it to this scenario, given that the redox reaction taking place is exothermic, it would suggest that increasing temperature would always decrease the voltage produced since it would cause a shift towards the reactant side. This, of course, does not correspond with the relationship predicted by the Nernst equation.

I have looked all over the internet but I could not find any answer to this seemingly straightforward question. Any help is greatly appreciated!

Consider that even when exist a "driving force" to move ions to electrodes, it is necesary that particles stay not static, they need some capacity to move to change direction and take way to electrodes, and what gives this capacity to move, Energy, and in any particle system this energy distribution follows Maxwell-Bolzmann and this means that temperature counts.. e^-(1/kT)


How does the volume of a mole of ideal gas relate to pressure and temperature? The volume of a mole of gas is:

$\ce{22.4 l}$
directly proportional to pressure and Kelvin temperature.
directly proportional to pressure, inversely proportional to Kelvin temperature.
inversely proportional to pressure, directly proportional to Kelvin temperature.
inversely proportional to pressure and to Kelvin temperature.

I said the answer is 3 because the ideal gas law is $PV=nRT$ and "proportional to" is the same as saying which ones are on opposite sides. So $n$ and $P$ are on opposite sides and temp is on the same side as $n$.  But I'm wrong and the answer is not 3, but I don't know why.

Your starting point of reasoning is good, but you misread the equation. The question is about the volume of 1 mole of gas so let's rewrite the ideal gas law to give this relation:
$$P V = n R T \rightarrow \frac{V}{n}=\frac{R T}{P} $$
Now we have the volume of a mole of gas $\frac{V}{n}$ on the left hand side so we can see the proportionalities on the right hand side.

The volume of a mole of gas is inversely proportional to the pressure and directly proportional to the temperature, i.e. 4 in your list of answers


Unusual hydrolysis of methyl 2,4,6‐trimethylbenzoate Problem 223 from Conceptual Problems in Organic Chemsitry by Singh [1, p. 367]:

The given answer is b and I don't get why.
Usual ester hydrolysis would've given a.
I somehow managed to know that the mechanism is SN2, but why?
Also, if someone could provide the reaction mechanism, it would be wonderful.
Reference

Singh, D. K. Conceptual Problems in Organic Chemistry: For Engineering and Medical Entrance Examinations, 3rd ed.; Pearson India, 2013. ISBN 978-93-325-8207-1.

In the typical reaction of hydroxide ion with a methyl ester, the hydroxide attacks the carbonyl carbon of the ester, resulting ultimately in elimination of methoxide. Given the basic conditions, the carboxylic acid formed immediately deprotonates to a carboxylate, and the methoxide is protonated to methanol. In older texts particularly, this reaction mechanism is referred to as BAc2, indicating "bimolecular base-catalyzed acyl-oxygen cleavage" (*) because the bond that is broken is between the acyl group and the oxygen of the methoxyl group [see for example Ingold, C.K. (1953) Structure and Mechanism in Organic Chemistry, Cornell Univ. Press].
An alternative mechanism is the BAl2 mechanism, designating "bimolecular base-catalyzed alkyl-oxygen cleavage". As the name suggests, in this mechanism, the bond between the alkyl group and the oxygen is broken. To do this, the hydroxide attacks the methyl carbon in an SN2 reaction, directly yielding the carboxylate and methanol products. This mechanism is also described in detail in Ingold's book and many other sources.
In the years since this mechanism was proposed, there has been some debate about how often it really occurs. Certainly for most esters, BAc2 is the dominant mechanism, and even when reaction by that mechanism is very slow, only a fraction of molecules, if any, react by the BAl2 option.
There seem to be two requirements for the BAl2 type reaction. First, the alkyl group must be methyl or a very ring-strained lactone. Other alkyl groups react much more slowly, as is observed in any SN2 reaction. Second, the carbonyl carbon must be sterically hindered to the point that the BAc2 mechanism is essentially impossible.
I am not certain that the tri-methyl benzoic acid in your question is hindered enough to react exclusively by the BAl2 mechanism, but it is certainly hindered enough that BAc2 would be substantially diminished.
You can find a more thorough discussion of these mechanisms in many textbooks, including the Ingold reference above. Another paper that may be useful is Mueller, P. and Siegfried, B. (1974) SN2 Reactions with Carboxylic Esters. Selective Cleavage of Methyl Esters. Helv Chim Acta 57:987.
*Note that since the net reaction in my example is $\ce{R-CO2Me + HO- -> R-CO2- + MeOH}$, the base is a stoichiometric reactant rather than a catalyst, so "base-catalyzed" is a misnomer. That terminology is used because the mechanism is the same if a different nucelophile is used, and the product is a non-acidic acyl derivative rather than a carboxylic acid. In that case, the base is indeed a catalyst.


Metering Device decreases pressure yet produces saturated liquid? This question pertains to the change of state of liquid using a Metering Device as the refrigerant enters an Evaporator Coil of an HVAC system:

The Metering Device takes liquid and then reduces the pressure and then spits out 75 percent liquid and 25 percent vapor. I don't understand why decreasing pressure, thus decreasing temperature will create a saturation point where liquid is both water and vapor. Shouldn't it be the other way around? Decreasing pressure should change saturation point so the liquid is both water and ice.

This is my rationale behind what I say:

In terms of changing of state of a liquid, as you add more heat content (molecules in motion or BTU), such as 340 BTU all the way up to 1310 BTU, all the liquid is converting into gas. This is latent heat. So adding heat changes liquid to gas. Conversely, when we remove heat it changes state of liquid to ice. At 0 degrees F and 0 BTU, you have ice.

In terms of gas laws, if you compress a gas to reduce its volume, both the temperature and pressure will increase. If you expand a gas to increase its volume, both the temperature and pressure will decrease.

We can use PSI or PSIA (absolute) to measure pressure per square inch of area. If you look at a Pressure Temperature Chart for Water, 212 Degrees F and 14.7 PSIA is at saturation where there is both liquid and vapor. If we add any heat, it will change state from liquid to vapor.

If we reduce the atmospheric pressure (e.g. from 14.7 PSIA to 0.33 PSIA) and if we put the water into a vacuum (e.g. from 29.6 inHg to 0.68 inHg), then we can change the state of the water at 70 degrees F. Why? Because decrease in pressure means decrease in temperature.

Consequently, how is reducing pressure creating vapor in the case of the Metering Device?

I'm a bit confused by your question -- your diagram shows the path of refrigerant through the cooling system, but you seem to be discussing water's state changes.

Liquid refrigerant under pressure enters the metering device. As it leaves the device at a lower pressure, it boils, producing some vapor, but not evaporating completely. The rest of the liquid evaporates as it passes through the evaporator coil, absorbing heat as it does so.


Which is stronger, PET or PVC? I have been given the task of designing a new wheel rim (the metal part of the wheel) but instead of metal, it has to be made out of a polymer.

I'm thinking of using either PVC or PET but I don't know if either of those plastics will be able to support the weight of a car and am not sure how to find that out.

If anyone can think of a different polymer that I could use instead that would be better suited feel free to include it as an answer. This is only theoretical so cost/difficulty to manufacture is not a big hassle.

You need to define what you mean by "stronger".  Are you talking about resisting deflection (and thus rolling losses)?  Yield?  Break?  Impact?  Abrasion?  Lifespan?  Ability to withstand heating from its deflection as the vehicle rolls?  Also, any material can "withstand the weight of a car" if you use enough of it, do you know what sort of geometry you're talking about?

Lastly, the material properties of polymers depend a lot on the manufacturing processes and the details of the raw materials (chain lengths, branching, monomer ratios, etc) and additives.  A good example is PET.  Check out the difference in tensile strength between simple thermoformed PET and BoPET (Mylar), for example; they're vastly higher for BoPET.  And the reason is because it's elongated to near breaking at lower temperatures - thus orienting the polymer - and then given a heat treatment to relax it into its new shape.

I think you need to define your problem statement better.  Think about what properties you want, your geometry, and how you plan to manufacture it.  Also, what are the parameters you're trying to optimize?


How is the pKa of extremely weak acids determined? According to Wikipedia, the conjugate acid of LDA has a pKa of about 35 - which is in line with figures I've seen elsewhere.

How then can such a weak acid's pKa be determined? If the determination is made in water, how is that even possible? Water would be a stronger acid by far and couldn't it easily mask the acidity of LDA's conjugate acid?

Is LDA's pKa value even determined in water?

Similarly, how are extremely strong acids' pKa values determined? Wouldn't the leveling effect come into play as one introduces, say, tosic acid into water? How then can we determine the strength of an acid stronger than hydronium ion in water?

Standard pKa measurements

pKa is, as you mentioned, commonly measured in water. This worked well initially, but as chemists got more ambitious in their measurements, they began to come up against the levelling effect, whereby the acidity/basicity of the compound being measured was limited by the acid/base properties of the solvent being used to measure it.

To a certain extent, this was mitigated by moving from water to a solvent that is unchanged over a wider range of pKa (the range of pKa is known as the discrimination window), DMSO being the example chosen. This change to DMSO allowed the pKa of far more compounds to be measured. DMSO has been used to measure the pKa of LDA, the example you gave.

Even in cases where DMSO is unsuitable pKa may be measured in other solvents. There is however an issue with using different solvents to measure pKa in that the values are not directly comparable since different solvents affect the ease of protonation/deprotonation. The discrimination windows of several solvents are given below (Source: Wikipedia).

Standard pKa measurement is carried out by dissolving the compound in the solvent and titrating in acid/base slowly, whilst measuring the pH change, which can be done in many ways. This gives a curve from which the pKa may be read off (pKa is the point at which the compound is 50% ionised).

Measuring pKa of weak acids

Whilst pKa has been determined for many common chemicals, there are instances where pKa will be extremely high/low. In these situations, we can abandon the direct measurement as described above, and look for some other way of doing it.

-A brief interlude- To give an analogy that may (or may not) be more familiar. In medicinal chemistry, logP is an incredibly useful property. It was traditionally measured by partitioning the drug molecule in-between octanol and water and seeing how much of the compound ended up in each layer. This was laborious, and so medicinal chemists moved from directly measuring logP to carrying out other analysis from which logP could be derived. The common way of doing this is to use a column (similar to chromatography, but specially adapted). The chemist first runs a set of samples with known logP through the column, followed by the compound being studied. By looking at retention times and then comparing and doing some maths, a calculated logP can be obtained which is generally fairly reliable.

With pKa measurement, the situation is a similar in that we can indirectly measure the pKa and extrapolate.

There are, however, two separate issues here. A first problem is that pKa measurements at the extreme are just difficult (i.e. very small or very large) due to the titration curve becoming very difficult to measure accuratly, it isn't impossible however. This is really the main issue.

The other problem is related to levelling, but as you can see from the table above , we do have solvents available to measure over pretty much as wide a range as you could ever want (note:  if you're considering measuring a pKa in hydrofluoric acid, you should probably reconsider why you need to measure it in the first place).

The following paper (free to view) details the most common direct and indirect methods of determination.

From my experience, there are three methods I think are worth discussing here as alternatives to direct pH change measurement. . .

HPLC measurement in which the pH of the mobile phase is varied. As the pH changes, the compound may ionise, which changes its retention time on the column. This may be used to extrapolate pKa by comparison to samples of known pKa. The HPLC columns used now are able to tolerate the use of highly acidic/basic mobile phases for pKa determination.
NMR measurement in which we 'titrate' in the NMR tube, adding acid/base, and watching as the chemical shift changes. This has allowed some super accurate pKa's to be measured.
Comparative measurement in which a compound with a known pKa is reacted with our compound for analysis, and we see the extent to which they protonate/deprotonate one another. By looking at this equilibrium constant we essentially can derive the pKa.

Last, but not least...

Computational determination of pKa

Although not a 'measurement' of pKa, in silico computational methods are becoming increasingly accurate. This is especially true within a normal pKa range, but has also found applications with highly acidic/basic molecules. Some common packages for this are ACDpKa, Epik and Marvin.


Why does a pickled gherkin glow only at one end when a current is passed through it? I recently saw this Periodic Video.

The Professor (Martin Polyakoff) notes that,

The Gherkin normally glows only at one end due to excitation of $\ce{Na+}$ ions.
The end which glows can either be the neutral end or the positive end. It is impossible to predict which one will glow.
During the video, both ends glowed together only for a few seconds.

Even the Professor admitted that probably glowing at one end happens due to pure chance.

  Why should the gherkin glow only at one end? Is there any reason? Normally one would expect it to glow symmetrically around the center, that is, at both ends.

The question has been quite well received. However no one has commented or answered. Does the question need to be re-worded or better explained?

Do let me know in the comments.

It looks like an electric arc to me, more precisely, a single-phase arc -- it seems like AC current was applied, as "ground" and "neutral" ends were used. Hence the light emission, high observed current density, and high temperature. As you noticed, the yellow color is most likely due to the presence of $\ce{Na+}$ ions.

Any conducting (or not so, if the voltage is high enough) object put in contact with a couple of separated electrodes and applied current (DC, single-/three-phase AC) can undergo arc discharge. For example, with the current from the standard household outlet one can do the same with lemons (it's going to smell better though than the gherkin, at least at the beginning due to evaporating essential oils); one can also light a tip of a cigarette moistened with saliva; or put a piece of paper with a pencil drawing on fire within seconds (starting the ignition from the graphite layer) and so on. Even a piece of wood can also  glow when arc discharge paves the way in a form of graphite path after dielectric breakdown  in between electrodes of HV-generator ($\pu{5 .. 10 kV}$ depending on wood and moisture).

It is not the particular end of the gherkin that glows, it is an area where arc discharge takes place. One can also observe arc(s) hopping along between two elongated parallel electrodes in the very same gherkin:

Video courtesy of YouTube channel "ПРОСТАЯ НАУКА" (Russian; English "SIMPLE SCIENCE")

Predicting the path of the arc is a rather non-trivial physics problem. To put it simply, it partially depends on the contact area, which constantly changes mainly due to vigorous boiling substance, as well as the local pressure and composition of formed gases, which results in equilibrium or non-equilibrium plasma formation in positive column (luminous region in the environment between the electrodes).

The process of forming an electric arc can be simplified down to three distinguished stages:

Arc ignition, when, as a result of shock ionization and emission of electrons from the cathode, an arc discharge begins and the ionization intensity is higher than deionization.  As the electrodes are diverged from each other, the contact pressure and the contact surface are decreased due to increased transient resistance in addition to the resistance of the media (mostly vinegar/salt solution in this case), overheating  starts. High temperature subsequently contributes to thermionic emission, when the electron velocity increases so they can be eliminated from the electrode surface. The voltage on the contact gap is quickly restored, and since the distance between the contacts is small, an electric field of high intensity arises, under the influence of which more electrons get accelerated and escaped from the electrode surface. When they struck a neutral atom or molecule, they give away their kinetic energy. If this energy is sufficient to remove at least one electron from the shell of a neutral atom, then the ionization process takes place.
Stable arc combustion maintained by thermal ionization in the arc itself, when the ionization and deionization intensities are approximately equal. Free electrons and ions form plasma arc content, that can be represented as an ionized channel, in which the burning act takes place and continuous motion of the particles is ensured. Negatively charged particles (primarily electrons) move toward the anode, and positively charged particles (deprived of one or several electrons atoms and molecules of gases) in the opposite direction (toward the cathode). This contineous flow makes the conductivity of plasma comparable with the conductivity of metals. So the voltage drop along the length of the arc is small due to the intensive thermal ionization.
Arc extinction, when the deionization intensity is higher than the ionization. In addition to ionization, there is also deionization of atoms and molecules. The latter occurs mainly through diffusion of charged particles to the environment, and the recombination of electrons and positively charged ions, which are reunited into neutral particles releasing energy in a form of heat.

Video bloggers all over the internet popularized the heck out of this effect, which is, strictly speaking, has more to do with physics, rather than with chemistry.


Acidity of Carbonyl group I m a high school chemistry learner. I have learnt that acids are substance that can liberate H+ on hydrolysis. For that they must be connected to an electronegative atom.
In carbonyl ( mainly aldehyde and ketone) the H atom is not directly attached to Oxygen. So I may not consider it as an acid.
But today while performing a chemistry practical when I added Blue litmus solution to aldehyde it turned red.
This leads me to a confusion about whether they are acidic or neutral or What can I infer from the colour change of blue litmus.

Might as well make an answer of the comments.

Aldehydes are not only able to be oxidized to acids, they are quite reactive towards this oxidation.  Exposure to air would be enough to trigger enough oxidation and acid formation to get litmus to react, especially if the litmus is moist.  Aldehydes are not alone in this respect; for example iron(ii) salts can be oxidized by air too, especially if you try to store them as a solution or after breaking the original seal on the bottle.

Moist litmus paper is a good qualitative test for acidity or basicity because it is highly sensitive to small amounts of acid or base.  Reaction with sodium carbonate might not be as sensitive if you are looking for enough carbon dioxide to visibly form bubbles, especially as carbon dioxide has some solubility in most solvents.

Yes, you need to follow good practices inspecting, testing and if necessary, purifying sensitive chemicals if you want high precision or accuracy.


Heat transfer dependence on temperature gradient? Is the rate of heat transfer between two bodies brought into thermal contact dependent upon the different in temperature between them, assuming they each have the same mass?
Imagine two bodies of water connected to each other by an impermeable glass barrier. If the difference between the temperatures is 10 degrees, will the system reach its equilibrium state faster per degree than if the difference is one degree?

Yes, it is very much dependent on temperature gradient. The stefan-boltzman law is only for radiative dissipation of heat and not for general heat transfer by conduction(contact).
The equation for steady state heat flow is
$$\vec h=-k \vec{\nabla T}$$ where $\vec \nabla$ is the gradient operator which is based on the temperature difference between the bodies and $k$ is the thermal conductivity. $\vec h$ is the heat per unit area.

If the system is not in steady state, then the heat diffusion equation becomes relavent:-
$$\frac{dT}{dt}=\frac{k}{c}\nabla^2 T$$ where $c$ is the specific heat of the body.
Lastly, a to a crude approximation, the rate of cooling in atmosphere at low difference of temperature of atmosphere and the body is given by Newton's law of cooling (partly by radiation):-
$$\frac{dT}{dt}=-K(T_{body}-T_{surrounding})$$ where $K$ is a constant.

In all these equation, it is the temperature gradient between the bodies in contact, which drives the rate of heat transfer (or temperature change).


GTOs of higher quantum numbers? I recently realised that the STO-nG basis functions were fit to 1s/2s/3d STOs of various exponents, instead of being fit to STOs of higher principal quantum numbers (such as 2s/3p orbitals etc.).

Has a Gaussian expansion of STOs of higher quantum numbers (for example, 2s or 3p STOs) been performed, and are there any GTO integration schemes for Cartesian GTOs of higher principal quantum numbers?

Has a Gaussian expansion of STOs of higher quantum numbers (for example, 2s or 3p STOs) been performed

I am unaware of such fitting. Typically, basis sets using higher angular momenta (cc-pV$n$Z, def2-$n$ZVP, pc-$n$) are fitted on total energies or correlation energies. See https://www.basissetexchange.org/ for examples and references.

  Are there any GTO integration schemes for Cartesian GTOs of higher principal quantum numbers?

Yes, there are. I have seen the following approach in action before:

hand-crafted algorithms are used for common, low-angular-momentum cases
more general schemes, but still optimized schemes for higher cases
a catch-all, not very efficient, unspecific algorithm above a certain threshold


How can we find the eV value of molecular orbital diagrams for a given atom? It seems like sometimes the electron volt value of the molecular orbital is given but most of the time it just shows relative values of s p d orbitals. Is there some lookup table for such values?

Exact values of MO energy are rarely given as they are difficult to obtain in anything other than relatively trivial systems. When they are given, it is often not even clear whether they are experimentally derived (Photoelectron spectroscopy, as mentioned by orthocresol), or computationally derived (by calculations to approximate the wave functions and their properties). For this reason, most MO diagrams presented are qualitative, derived from our knowledge of symmetry, orbital energy etc.

There is a book, The Organic Chemist's Book of Orbitals (Salem and Jorgensen) which gives the MO's of many simple, and slightly more complex  molecules along with computationally derived energies (in eV).

For anything more complicated, you could also try searching the literature, however this would probably just turn up very specific examples that people have calculated to support their research.

If you need specific energies for molecules, you might look into computing them yourself. Simple software such as Spartan does an acceptable job at computing MO energy, with a variety of computational methods available depending on how much computing power you have.


How exactly is ammonium sulfate formed in a Kjeldahl analysis? When conducting a Kjeldahl analysis,  ammonium sulfate is formed from nitrogen which protein consist of.

How in a picture of formation of ammonium sulfate there is involved no nitrogen but only some compounds containing nitrogen.

Ammonia is basic so:
$$\ce{NH3 +H+<=>NH4+}$$
Sulphuric acid is acidic(since it's name is itself so acidic!):
$$\ce{H2SO4<=>H+ +HSO4- <=>2H+ +SO4^2-}$$
Now the final reaction would be like:
$$\ce{2NH3 +H2SO4<=>NH3 +NH4+ +HSO4- <=>2NH4+ +SO4^2- <=>(NH4)2SO4}$$
And remeber since proteins are polymers of peptides which itself are molecules consisiting of both a carboxylic ($\ce{-COOH/-COO-}$) and amino ($\ce{-NH3/-NH4+}$) group, therefore with strong heating ammonium sulfate is expected.


Difference between "physical quantities" and "units"? My textbook, Atkins' Physical Chemistry, 11th edition, by Atkins, de Paula, and Keeler, says the following:

  Although it is good practice to use only SI units, there will be occasions where accepted practice is so deeply rooted that physical quantities are expressed using other, non-SI units. By international convention, all physical quantities are represented by oblique (sloping) letters (for instance, $m$ for mass); units are given in roman (upright) letters (for instance m for metre).

I'm confused about what the difference here is between "physical quantities" and "units"?

I would appreciate it if people would please take the time to clarify this.

Mass is a physical quantity. Its unit may be gram or kilogram.
Volume is a physical quantity. Its unit may be the liter or the cubic meter


Angular dependence of x-ray and neutron scattering factors Why are neutron scattering factors essentially independent of $2\theta$, whereas X-ray scattering factors drop off with increasing $2\theta$ ?

Scattering factors describe scattering of an atom

To predict how an object will diffract, you essentially have to do a 3D Fourier transform of the scattering density (https://chemistry.stackexchange.com/a/115815/72973). The scattering factor is derived from the scattering of a single atom (placed at the origin). For xray diffraction, the relevant density is that of the electrons. For neutron diffraction, it is that of the nucleus. See slide 11, https://www.embl-hamburg.de/biosaxs/courses/embo2012/slides/neutron-scattering-basics-trewhella.pdf

The fuzzier the scattering density, the higher the angular dependency

The Fourier transform of a Gaussian is a Gaussian again; a sharp Gaussian results in a broad transform, and vice versa. For X-ray diffraction, the electron density is fuzzy (same magnitude as the wavelength), so there is a large angular dependence. For neutron diffraction, the scattering density is very compact, leading to almost no angular dependence of the scattering factors.


Crystallisation problem I have an aqueous solution of N,N-dimethylphenylalanine methyl ester hydrochloride and I need to crystallise it as a salt. I'm afraid of increasing the temperature or the concentration of acid due to the possibility of ester hydrolysis.
What crystallisation method would be the most efficient in this situation?
Any help would be greatly appreciated.

This molecule is an ammonium anion with an ester side chain as a chloride salt.

The first option you can choose to undertake is the vacuum evaporation of the solvent at room temperature or 30 degrees should prevent any heat degradation. I do understand your point about increasing the acid because in effect you have a buffer solution.

Your second alternative is to partition into a  warm amine-ophilic non-aqueous/organic layer and crystallize from there.
You can read more about that here: https://books.google.co.uk/books?id=YY3pCAAAQBAJ&pg=PA8&lpg=PA8&dq=partitioning+of+quaternary+organic+salts+into+solvent+layer&source=bl&ots=O0iyoLglNF&sig=jQb5hDP0SvhA2kg7CXGTXsIE4bg&hl=en&sa=X&ved=0ahUKEwjAydrc5pjaAhXCIMAKHQI4ANgQ6AEITjAF#v=onepage&q=partitioning%20of%20quaternary%20organic%20salts%20into%20solvent%20layer&f=false>

Also, see  https://onlinelibrary.wiley.com/doi/pdf/10.1002/jps.2600530804


What is the difference between melting and dissolving? What is the difference between melting and dissolving? I am looking some general features.

The answer should be adaptable to the melting/dissolving of ice cube (water) in a class of pure alcohol (ethanol) just below (or at the) melting point of ice, or similar phenomena.

I am now assuming that the ice is dissolving and melting at the same time.

In other words which reaction energy is higher in the following reactions:

$$\begin{align}
\ce{H_2O(s) &\to H_2O(l)\\
H_2O(s) + n EtOH(l) &\to H_2O \centerdot (EtOH)_{n}}
\end{align}$$

Or, are there substances that release more energy in dissolvation than consume in melting?

Juha invited me to write a summary (see comment on my previous answer ) of the differences between melting and dissolving. I’ll try to outline this roughly in the same order as his. I'm giving this as a new answer since my last answer was quite long as it was.

Differences:

Melting and dissolving are completely different processes on the molecular/atomic level that could not be mistaken for each other if you could observe what was happening at that scale.
If you cool the liquid that arose from melting the solid, to a temperature below it’s melting point, you would see the entire sample solidify. If you cool the solution (dissolved solid + solute) to below the melting point of the solid (solute) in it, you would see no change. (Unless you had a saturated or nearly saturated solution.)
Melting requires only a single substance and energy input while dissolving requires a solvent and a solute that are compatible (“like forces”). (This is actually a pretty huge difference and would potentially affect all of the physical and chemical properties.) Dissolving a solid can be either endo- or exo-thermic.
"Phase change" (wording from the previous summary)  In each case you end up with a liquid. Melting caused a phase change (composition of the substance didn’t change) while dissolving a solid in a liquid is not considered a phase change since a change in composition occurred.

Similarities:

In each case, forces between the particles that comprise the solid are disrupted and that takes energy. (Whether it’s chemical bonds or intermolecular forces depends on the process and on the solid and on your definitions. (See this question.) But melting (rare exception noted in previous comments) is endothermic and dissolving can be either endo- or exo-thermic.
In each case you end up with a liquid. Macroscopically, if you walked into a room and saw the liquid on the table, it would be difficult to say whether this liquid came from a solid that had melted or a solid that had dissolved in a solute and made a solution.
But it would be very easy to determine which you had experimentally in a “dozen” different ways.
Both melting and dissolving require interaction among groups of atoms, molecules, or ions.

There are probably more differences that could be given (how to handle thermodynamic calculations, complexity of the system etc.), and possibly more similarities, but that's enough for me on this topic.


Should I be worried about the waste from "sugar rockets" and other potassium nitrate fireworks? "Sugar rockets" or sparklers made of metal fillings, potassium nitrate and some carbon compound are easily made, and there is a lot of tutorials and videos on the internet suggesting them as relatively safe and non toxic DIY fireworks.
However reading on the potassium nitrate reaction, it decomposes to potassium nitrite when heated. This compound is toxic, and is suspected to be mutagenic as well. It is also an oxidizer on its own.
So I have safety concerns and questions:

Do the charred remains of potassium nitrate fire experiments need to be handled as toxic waste?
If potassium nitrite is an oxidizer, what does it decompose to when burned along with a carbon compound?

Potassium nitrate is transformed into potassium nitrite only when heated alone. And it is not really toxic, as it is added in ham to prevent its too early oxidation in air.  When heated with charcoal, sulfur or any other oxydable substance, potassium nitrate is transformed into $\ce{K2CO3}$ or $\ce{K2S}$. For example, black powder is a mixture $\ce{KNO3 + C + S}$ that reacts violently according to ; $$\ce{2KNO3 + 3 C + S -> K2S + 3 CO2 + N2}$$ And if the explosion is made in contact with air, the hot $\ce{K2S}$ is quickly oxidized into sulfate according to $\ce{K2S + 2 O2 -> K2SO4}$
Sugar rockets are made of ordinary sugar $\ce{C_{12}H_{22}O_{11}}$ and $\ce{KNO3}$. This mixture reacts according to the two following equations : $$\ce{C_{12}H_{22}O_{11} -> 12 C + 11 H2O}$$  $$\ce{4 KNO3 + 5 C -> 2 K2CO3 + 2 N2 + 3 CO2}$$
So the ashes of any combustion reaction are mostly made of $\ce{K2CO3}$ which was the main constituant of the wood ashes. These ashes was used in the Middle Age for washing clothes by our grand-grand-mothers. They are not considered as toxic


Can two positive ions be next to each other in a formula? Let's say I want to find the oxidation numbers of the atoms in a chemical formula.

For example: In $\ce{H3AsO3}$,

Oxygen has a $-2$ charge. They're $3$ of them, so its $-6$

Hydrogen has a $+1$ charge. They're $3$ of them, so its $+3$.

So arsenic must have a charge of $+3$ for all the charges to add up to be $0$? Correct?

However, my professor keeps telling me to keep the positive oxidation numbers away from each other. I thought it only applied when drawing Lewis structures or resonance structures.

Is that a rule when writing a formula?

Can two positive ions be next to each other in a formula?

Yes, there are many examples of this, $\ce{HNO3}$ for instance

  So Arsenic must have a charge of 3+ for all the charges to add up to
  be 0? Correct?

Yes, that's correct

  However, my professor keeps telling me to keep the positive oxidation
  numbers away from each other. Is that a rule when writing a formula?

It's not a rule that I'm familiar with.  Here is a link to a nice, concise listing of the rules for writing formulas for inorganic compounds.

  I thought it only applied when drawing Lewis structures or resonance
  structures.

Yes, it certainly applies in those two cases


Green and red crystals of [Ni(Ph2EtP)2Br2] From the reaction of $\ce{NiBr2}$ and $\ce{Ph2EtP}$, it is possible to isolate green and red crystals of $\ce{[Ni(Ph2EtP)2Br2]}$?
A) coordination number and geometries of the complexes?
B) possible isomeric structures for the green and red crystals.

I know that d orbitals effect the color but I don't know how to apply it to isomers.

The coordination number of $\ce{[Ni(Ph2EtP)2Br2]}$ is four;
thus, the complex may be tetrahedral

or square planar

(figures taken from here)

The complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains $\ce{Ni^2+}$, which has eight d electrons.

When filled with eight electrons, the tetrahedral structure is paramagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively small (4.4 Dq).

The corresponding square-planar structure is diamagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively large (10 Dq).

Therefore, the square planar complex absorbs light of higher energy (lower wavelength) than the tetrahedral complex.

The green crystals absorb red light, whereas the red crystals absorb green light. Thus, the red crystals absorb light of higher energy than the green crystals. Therefore, the red crystals correspond to the square-planar complex and the green crystals correspond to the tetrahedral complex.

Furthermore, since the complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains two different ligands, there are cis and trans isomers of the square-planar structure.


Half-life equation for 2nd order kinetics My friends and I were doing some problems from this year's IChO Preparatory Problems (PDF from the 49th International Chemistry Olympiad (2017)) when we stumbled upon a question which we had some confusion with.

  Task 8. Decomposition of Nitrous Oxide
  Nitrous oxide decomposes exothermically into nitrogen and oxygen, at a temperature of approximately $\pu{565 ^\circ C}$.
  $$\ce{2N2O (g) -> 2N2 (g) + O2 (g)}$$
  This reaction follows the second-order kinetics when carried out entirely in the gas phase.

  8.1) If the reaction is initiated with $[\ce {N2O}]$ equal to $\pu{0.108 mol dm-3}$, what will its concentration be after $\pu{1250 s}$ have elapsed at $\pu{565 ^\circ C}$? The rate constant for the second order decomposition of $\ce{N2O}$ is $\pu{1.10\times10^-3 dm3 mol-1 s-1}$ at this temperature.

Task 8 details the kinetics of the decomposition of nitrous oxide. We were confused with part 8.1 of the task which required us to find the concentration of the reactant after a particular duration of time has elapsed, given the temperature, initial concentration and rate constant of the reaction. We approached the question using two different methods:

I approached it using the method, as suggested in the solutions manual, of using the integrated rate law for 2nd order kinetics and substituting the values provided into that equation. My answer was exactly that stated in the solutions manual.

However, they approached it by first finding the half-life using the equation for the half-life for 2nd order kinetics and then, finding the concentration after knowing how many half-lives have passed. Their answer was slightly off.

After substituting random values into the equations for both methods, I realised that the "half-life method" which my friends used only gives a reasonable approximation when the time elapsed < half-life, gives the exact value when the time elapsed = half-life and goes completely off when the time elapsed > half-life. Why is that so? Since the half-life equation for 2nd order kinetics is derived from the integrated rate law (as shown on Chemistry LibreTexts), shouldn't they give the same answer?

Roughly speaking, half-life is not a thing at all for the second order (or any order other than first, for that matter).

What would they do after finding the number of half-lives? Raise 2 to that power? Too bad, because that only works for the exponential decay, which is the solution of the first-order kinetics and no other order. Really, what if the number of half-lives equals 2? After spending the first half-life, we now have a different half-life ahead of us, because it is concentration dependent, and the concentration has changed.

Following your link, we see the same statement, except they don't put enough emphasis on it, as to my taste:

  For this reason, the concept of half-life for a second-order reaction is far less useful.


Trend in the melting point down the group We know that the atomic radii increases down the group. So, less energy is required to pull out the outermost electron as we go down the group, hence,  gradually melting point decreases down the group.  The topmost element has large melting point because of its small atomic radii and so more energy is required to pull out its outermost electron.

But, it is observed that the melting point slightly increases in case of the bottom-most element of group as compared to the previous element. Why?

Examples:

Group 13 - indium - $\pu{156.6^\circ C}$ but  thallium - $\pu{304^\circ  C}$
Group 14 - tin - $\pu{231.92^\circ  C}$ but lead - $\pu{327.46^\circ C}$
Group 17 - iodine - $\pu{113.5^\circ  C}$ but astatine - $\pu{302^\circ C}$

There's a counter effect in play here: van der Waals forces. As you become larger and larger, induced dipole-induced dipole interactions become stronger and stronger.

You'll notice that, in halogens (where abstracting an electron is a fairly tall order), the trend is that, the bigger your atom, the higher your boiling point, regardless of where you are in the group. The boron group loses the conflict of ease of abstraction versus van der Waals forces at around the gallium/indium leap, and for the carbon group, the conflict is lost to van der Waals forces only between tin and lead.


How does LeChatelier’s principle (which makes qualitative predictions) relate to the quantitative changes observed when an equilibrium is disturbed? Le Chatelier formulated his principle in 1884; it predicts the direction in which equilibrium is re-established after it has been disturbed by an external cause. How does this qualitative principle relate to the known quantitative changes that occur when an equilibrium is disturbed by a change in temperature, concentration or pressure? Are there any limitations of Le Chatelier's principle as it is stated by modern sources?

How does this qualitative principle relate to the known quantitative changes that occur when an equilibrium is disturbed by a change in temperature, concentration or pressure?

Equilibrium constant K and reaction quotient Q
One way to quantitatively describe equilibrium processes is to define an equilibrium constant K and a reaction quotient Q. The equilibrium constant is defined by an equilibrium constant expression which contains all concentrations or partial pressures that are affected by the reaction.
For example, if we look at the following reaction:
$$\ce{A(aq) + B(aq) <=> C(aq) + H2O(l)}$$
The equilibrium constant expression for this reaction is:
$$K = \frac{\ce{[C]_{eq}}}{\ce{[A]_{eq}}\ce{[B]_{eq}}},$$
with $\ce{[X]_{eq}}$ designating the equilibrium activity (or fugacity for gas state) of a species. At a given temperature, the equilibrium constant will have the same value when the reaction is at equilibrium, no matter what the starting conditions.
The reaction quotient is defined in the same way as the equilibrium constant, but using $\ce{[X]_{t}}$ (the current activity) instead of the activity at equilibrium $\ce{[X]_{eq}}$. For the example reaction above, it is:
$$K = \frac{\ce{[C]}_t}{\ce{[A]}_t\ce{[B}_t},$$
Q can have arbitrary values from zero to infinity, depending on what reaction mixtures is found at a given time. Q will change over time if there is a net forward or reverse reaction.
Criterion for equilibrium based on Q vs. K
When Q equals K, the reaction is at equilibrium because the current activities $\ce{[X]_{t}}$ plugged into the equilibrium constant expression match K. Any net reaction in the forward direction would typically increase Q, while any net reaction in the reverse direction would typically decrease Q. This relies on the assumption that the activity of reactants decreases and that of the products (of the species included in the equilibrium constant expression) increases as the reaction goes forward (and vice versa as it goes in reverse).
Change in temperature
When, after the reaction has attained equilibrium, the temperature changes (through an external cause) and consequently the equilibrium constant changes, Q no longer equals K. Concentrations that had been constant over time will change now. In the example, we are changing the temperature at $t = 5$ (arbitrary units) so that the equilibrium constant doubles. The only way to make Q equal K again at this temperature is to have a net forward reaction (this will increase product activities and decrease reactant activities in the reaction quotient expression). An example is shown below:

How temperature affects the equilibrium constant is given by the van't Hoff relation. Stated in qualitative terms, the equilibrium constant of exothermic reactions decreases with an increased temperature, while that of endothermic reaction increases.
Change in activity
If we change the activity of a single species (for example by adding or removing solute) in a reaction that has already attained equilibrium, Q will change. If the temperature is maintained, K will remain constant, and Q will no longer be equal to K. As a result, a net reaction will be observed.
For example, if we double the activity of A in the reaction above, Q will be halved, and the reaction has to move in the net forward direction to attain equilibrium again:

After equilibrium is attained, Q will have the same value as it did before the external manipulation, but the set of activities will be different (necessarily, because there is more matter now).
Limitations and Criticism

Are there any limitations of Le Chatelier's principle as it is stated by modern sources?

The main limitation is that there are multiple non-equivalent ways the principle is stated, so different ways of defining the principle might lead to different counterexamples.
Historically, Le Chatelier wrote (Comptes-rendus de l’Académie des sciences, 99 (1884), p.786-789.) :

Tout système en équilibre chimique stable soumis à l’influence d’une cause extérieure qui tend à faire varier soit sa température, soit sa condensation (pression, concentration, nombre de molécules dans l’unité de volume) dans sa totalité ou seulement dans quelques-unes de ses parties, ne peut éprouver que des modifications intérieures, qui, si elles se produisaient seules, amèneraient un changement de température ou de condensation de signe contraire à celui résultant de la cause extérieure.

This google-translates into:

Any system in stable chemical equilibrium subjected to the influence of an external cause which tends to vary either its temperature or its condensation (pressure, concentration, number of molecules in the unit of volume) in its totality or only in some of its parts, can only experience internal modifications, which, if they occurred alone, would bring about a change in temperature or condensation of a sign contrary to that resulting from the external cause.

Wikipedia states the principle as:

When any system at equilibrium for a long period of time is subjected to change in concentration, temperature, volume, or pressure, (1) the system changes to a new equilibrium and (2) this change partly counteracts the applied change.

Chemistry OpenStax has two versions:

[in text]if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.
[Overview]An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium.

ChemPrime by Moore et al states:

if a system is in equilibrium and some factor in the equilibrium conditions is altered, then the system will (if possible) adjust to a new equilibrium state so as to counteract this alteration to some degree.

Limitations of the principle are discussed, for example, in this letter in the Journal of Chemical Education, and in the literature cited within. Another article in the same journal discusses how applying Le Chatelier's principle leads a convenience sample of educators to incorrectly answer questions such as :

The reaction
$$\ce{2NO2(g) <=> N2O4(g)}$$
is at equilibrium in a syringe. If the volume is decreased at constant temperature by moving the plunger, will the concentration of NO2(g) be higher or lower than the original concentration when equilibrium is re-established? Give reasons for your answer.

The misconception is that while Le Chatelier's principle correctly predicts that the reaction will form product after the pressure is increased, this does not completely offset the increase in partial pressure from lowering the volume.
Some authors argue that it makes sense not to teach Le Chatelier's principle to avoid such confusion, and instead treat equilibrium in a quantitative matter, calculating the new equilibrium state rather than predicting it qualitatively.


Calculating ppm of an ion in a solution Calculate the ppm of $\ce{Fe^2+(aq)}$ if $\pu{0.0055 g}$ of $\ce{Fe(NO3)2}$ is dissolved in $\pu{2.0 kg}$ of water.

I have tried converting g of iron nitrate into moles, using mole ratio between iron nitrate and iron(II) ions and then converting to grams of iron(II) ions. I then took these grams and divided by grams of solution ($0.0005 + 20000$).

You are wrong. Assuming that you are doing ppm in terms of mass: There are $0.0055$ grams of $\ce{Fe(NO3)2}$. Its molar mass is $179.8548$ grams/mol. Thus, you have $3.058022\!\times\! 10^{-5}$ moles. Multiplying this by the molar mass of iron gives you $0.00170775$ grams of $\ce{Fe}$. Thus, you have $0.0037922$ grams of $\ce{NO3}$. The concentration of iron is
$$\mathrm c = \ce{\frac{Fe}{Water+Fe(NO3)2}} = \frac{0.00170775}{0.0055+2000}= 8.538726518 \times10^{-7}$$
Thus, $\ce{Fe}$ has a $\pu{ppm}$ of $0.85$


Effect of direct exposure to sunlight on various drugs Today, we were discussing optical activity in class and our teacher said that some drugs are kept in dark coloured bottles to avoid direct exposure to sunlight. Same is the case with some other drugs in which it's written that the direct exposure to sunlight should be avoided.

The reason he gave for the above was that the one form of isomer can get converted to other form in direct sunlight.
But as far as i know, photolytic decomposition can happen. Is the conversion of d to l isomer and vice versa possible in sunlight?

Conversion of D to L isomer or vice versa is about the last thing to worry about in this situation, even if it may be facilitated by sunlight to a significant extent (which I doubt). Also, storage of chemicals in dark coloured bottles is by no means limited to compounds that have optical isomers, or even to drugs in general. Other examples are numerous; permanganate solutions and nitric acid come to mind.

Photons of visible light have pretty decent energy of a few eV. That's way higher than the energy of thermal motion at room temperature. When a molecule is hit by a photon, it is hit hard. If a molecule is fragile or strained in any way, it will probably break. Photons come in droves, and after a while your precious organic sample turns into a brown goo. (It would probably do so even in the dark, but that would take a bit longer.)

That's what's wrong about sunlight in the first place.


How to compare dipole moment in thiophene, furan, and pyrrole? This is what I found (source):

I wonder about several things here.

First, why is tetrahydrothiophene's dipole moment bigger than that of tetrahydrofuran, when considering that tetrahydrofuran has an atom of stronger electronegativity?

Secondly, why thiophene has weaker dipole moment when the S atom has larger p obitals, and therefore, has weaker conjugation ability than O to aromatic ring. The result seems that S in thiophene has better conjugation effect. (that's why it has less stronger dipole.) and I don't seem to explain why.

So what are the reasons for these dipole moment orders?

As another (downvoted) answer points out, the key to this comparison involves the heteroatom donating an electron pair to the ring in the aromatic compound.  This shifts the negative charge away from the heteroatom's corner into the ring.  Thus we either do not get as much of a negative end at the electronegative heteroatom's corner, or with a stronger interaction we can reverse the dipole.  In the examples shown, pay attention to the directions as well as the numbers of the given dipole moments!

Clearly nitrogen has this stronger interaction, while oxygen and sulfur give the weaker effect, with a reduced but not reversed dipole moment.  Oxygen is so electronegative that even in forming the aromatic ring it donates an electron pair only weakly.  The large sulfur atom has lower electronegativity, but overlaps poorly with the smaller carbon atoms. (In the nonaromatic tetrahydrothiophene, the larger size of the sulfur atom makes the dipole longer and thus may give a larger dipole moment with less charge separation.)  Nitrogen in pyrrole has the moderate electronegativity and strong overlap which, together, enable the stronger interaction.


Handling vectors for double-bond graphics I'm working on a small molecular graphics tool for visualising ball-and-stick style molecules. I'm having trouble handling the linear algebra to position the two bonds in the double bonds. My linear algebra is, unfortunately, very weak. I can follow notation but struggle with working my own way through logical algebra.

If I'm overcomplicating this significantly then I'd appreciate some guidance on how so. Here is my working so far:

I currently have the atomic coordinates of each atom. All of the above atoms are in a plane due to the $sp^2$ nature of atom $i$, though the orientation of the plane is arbitrary.

$\textbf i$, $\textbf j$ and $\textbf k$ are atomic coordinates and $\textbf a$ and $\textbf b$ are vectors between them.

The orientation of the double bond must mean that both of the lines in the double bond are positioned in the plane, and I have quantified the plane as:

$\textbf{a} \times \textbf{b} \cdot (\textbf{r} - \textbf{r}_0) = 0$

where $\textbf r$ is any vector in the plane, and $\textbf{r}_0$ is any other point in the plane.

I can also say that the points which I am looking for on each atom make vectors with the centre of respective atoms at $90 ^\text{o}$ to $\textbf b$, and that the radius at which these points are positioned is constrained to a constant value (and therefore so is the length of said vector). If we call this vector $\textbf l$ we therefore have:

$(\textbf{a} \times \textbf{b}) \cdot (\textbf{l} - \textbf{i}) = 0$

$| \textbf{l} | = \text{const}.$

$\textbf{l} \cdot \textbf{b} = 0$

where $\textbf{a}$, $\textbf{b}$ and $\text{i}$ are all things which are numerically available to me.

I feel as though, based upon geometrical constraints, these algebraic relationships are enough to find only the 4 points which I want. But my amateur self has no idea how to do that.

I'm not sure I exactly know what you want, but such code is implemented in Avogadro and in other places.

The first thing to realize is that the unmarked atom in your plane is also your friend. You can calculate the plane of best fit using all 4 atoms.

You're making your work too hard, though. If you had a single bond, you simply draw a line or cylinder between atom i and j.

So the question at hand is how to displace two cylinders slightly in the plane of the bond. There are a few ways to do this. (Avogadro computes the plane of best fit for the molecule. Some people project towards the plane of the screen.)

My recommendation is to realize that atom k in your diagram is in one direction, and the unmarked atom is in the other. Let's call that other atom l. You can compute the vector between them: k - l. In practice, atom i is midway along this vector. So you want to displace the b vector in your diagram by a small amount, say 0.1 in one direction along that k - l vector, and 0.1 in the opposite direction.


Proof of the chemical potential formula My textbook as well as Wikipedia both state the formula $\mu_A = \mu_A^0 + RT\ln\{A\}$ which gives the chemical potential $\mu_A$ of a chemical species $A$ as function of the standard chemical potential $\mu_A^0$ and the activity (relative to standard conditions) $\{A\}$.

This formula can also be used to derive the Gibbs free energy formula as $\Delta_rG_{T, p} = \Delta_rG_{T, p}^0 + RT \ln Q_r$ where $Q_r$ is the reaction coefficient.

Both formulas appear to be widely used in chemistry, but I have not found a decent explanation for why they are correct. The texts I have seen did not provide sources for the claim and omitted explanation.

Is this an empirical formula, or is it a consequence of statistical mechanics? I would like to find a source with a clear proof or explanation for the origin of this equation. I know that some will define the activity precisely using this equation, but then the question becomes why the activity so-defined corresponds to concentration for ideal solutions. That is, why in an ideal solution, we have $\mu_A = \mu_A^0 + RT\ln c_A$ where $c_A$ is concentration of $A$ relative to standard concentration.

The Sackur-Tetrode equation can be written as $$S^{Trans}=R\ln \left( \frac{e}{C\Lambda^3} \right)$$ where $C$ is the concentration and $\Lambda$ the thermal wavelength. You can rewrite this as $$S^{Trans}=R\ln \left( \frac{e}{\Lambda^3 C^\circ\frac{C}{C^\circ}} \right) = R\ln \left( \frac{e}{C^\circ\Lambda^3} \right)-R\ln\left( \frac{C}{C^\circ} \right) = S^{\circ,Trans}-R\ln\left( \frac{C}{C^\circ} \right)$$ where $C^\circ$ is an arbitrary standard concentration. When you combine it with enthalpy and the other entropy terms you get $$G = G^{\circ}+RT\ln\left( \frac{C}{C^\circ} \right) $$