Is Dalton's Atomic Theory a theory or a postulate? It is always mentioned that Dalton proposed an Atomic Theory. But when you read it, it consists of the following:

  The main points of Dalton’s atomic theory, an explanation of the structure of matter in terms of different combinations of very small particles, are given by the following postulates:

  All matter is composed of indivisible atoms. An atom is an extremely small particle of matter that retains its identity during chemical reactions.
  An element is a type of matter composed of only one kind of atom, each atom of a given kind having the same properties. Mass is one such property. Thus, the atoms of a given element have a characteristic mass.
  A compound is a type of matter composed of atoms of two or more elements chemically combined in fixed proportions. The relative numbers of any two kinds of atoms in a compound occur in simple ratios. Water, for example, a compound of the elements hydrogen and oxygen, consists of hydrogen and oxygen atoms in >the ratio of 2 to 1.
  A chemical reaction consists of the rearrangement of the atoms present in the reacting substances to give new chemical combinations present in the substances formed by the reaction. Atoms are not created, destroyed, or broken into smaller particles by any chemical reaction.

(From the book: General Chemistry by Ebbing and Gammon)

I'm sure many of you might be knowing that postulates and theories are quite different. Can someone explain me whether Dalton proposed a postulate or a theory?

It is most definitely a theory. The 4 points you mention could be construed as hypothesis by themselves. But Dalton went further, he put them to the test.
The results of his exeriments formed yet another component of his atomic theory called the Law of Multiple Proportions. This law served as proof that the other components of his theory were true. I used to spend most of a class period explaining this law to students, it's best you search the web for a complete explaination but I'l give you a brief summary via example.

Say an experiment shows it takes exactly 16 g of oxygen to combine with 12 g of carbon to make carbon monoxide. Now suppose we make carbon dioxide. We would find that it now takes 32 g of oxygen to combine with that same amount, 12 g, of carbon. So, for a fixed amount of carbon, the amount of oxygen required to make the two different compounds can be expressed as a small, whole number ratio. What would explain this better than to surmise that elements come as particles and thes particles combine in small, whole number ratios to form compounds.

When a hypothesis have been proven by repeatable experimentation, we call it a theory.


What is the difference between enthalpy of formation and Gibbs free energy of formation? I'm having trouble (not alone, apparently) understanding the concept of Gibbs free energy, but I'm beginning to. What I still don't understand, however, is how Gibbs free energy of formation differs from heat- or enthalpy of formation. Is it more precise, somehow? given that entropy is accounted for in Gibbs free energy?

Thanks in advance

Gibbs free energy is not more "precise." The two are different but complementary. Enthalpy and Gibbs Free Energy indicate different things. Enthalpy can tell you about the relative stabilities of the products and reactants. Gibbs free energy however can tell you about whether a reaction is spontaneous (whether a reaction will occur) under a set of specified conditions.

A lot of people will confuse exothermic - bond improving - reactions with spontaneous reactions. While it is true that a bond improving reaction is in itself quite favorable - after all - all chemical processes tend toward greater stability and lower potential energy - we must remember that there is another dimension to the problem of reaching maximum stability - namely, entropy (a measure of disorder).

Recall that all physical processes tend toward greater entropy (greater disorder) as well. Just as one's bedroom becomes more and more chaotic despite the pleas from Mom to clean one's room, physical processes tend toward chaos (disorder) as well. Gibbs free energy takes entropy into account with enthalpy and can tell us whether a process will actually go.

Consider this reaction:

$\ce{2NO_2(g) -> N_2O_4(g)}\ $

Consider only the forward reaction. The forward reaction is spontaneous under standard state conditions and 298 K - i.e. we have a negative Gibbs free energy value. Don't believe me? Nitrogen dioxide is brown. Nitrogen tetraoxide however is colorless. Put some brown nitrogen dioxide into a flask and if you observe it, you'll notice that the brown color fades somewhat as the colorless nitrogen tetraoxide is made.

The enthalpy of reaction is definitely negative - the reaction is exothermic because the reaction is a dimerization reaction - we are forming a bond here (specifically, a N-N bond) by using two nitrogen dioxide molecules (which are actually radicals, and remember that radicals are highly unstable). Nitrogen tetraoxide however is not a radical.

The entropy of reaction however is negative as well - i.e. we have an unfavorable change in entropy. Remember that $ΔS^o = S_{final} - S_{initial}$. In this case, it is clear that $S_{final} < S_{initial}$ because on the right hand side, we have one mole of gas. On the left hand (reactant) side, we have two moles of gas. There is definitely a lot more chaos (disorder) with 2 moles of gas molecules bumping around and whizzing around every single way imaginable. So we've gone from a disordered system on the left to a somewhat more ordered system on the right.

We would describe this reaction as (at 298 K and std. state conditions):

1) Spontaneous
2) Exothermic
3) Entropy lowering
4) Exergonic


why do we see characteristic colors of alkali metals in oxidizing flame? I am a high school student and I am a little confused in a topic, My confusion is that:
We know alkali metals and their salts imparts characteristic color in oxidizing flame, for example: Lithium gives crimson red color, sodium gives yellow , potassium gives violet and Caesium gives blue, My school textbook says that this is because the heat from the flame excites the outermost electron to a higher energy level and when it returns back it will radiate in visible range as I mentioned above, but if this is the correct reason then we know the outermost electron  of lithium is the most attracted  among all alkali metals .So, it means that it would take high energy to excite its outer electron so a short wavelength of light is needed i.e probably in violet region, so when it returns back it should emit the same wavelength. Then why do we see these weird pattern of colors? I think this is not the complete reason for why do we see those colors? please explain this in brief only so that I can easily understand.

I don't think it is a good idea to connect ionization energy with the color of flame emission, especially in a Bunsen burner.
Ionization energy means that you have separated the electron out of the nucleus attractive field. However, electronic transitions, corresponding to visible wavelengths, take place while the electron is still bound to the nucleus. Thus the flame color of alkali metals might correlated with the IP  but not the causation.
Let us look at the resonance wavelengths of emission of Li, Na, K, Rb, and Cs. Resonance wavelengths means the transition from the $\mathrm{1s}$ to $\mathrm{2p}$ orbital or $n\mathrm s$ to $(n+1)\mathrm p$ orbital in the case of alkali metals.
$$
\begin{array}{r|ccccc}
\hline
\text{Element} & \ce{Li} & \ce{Na} & \ce{K} & \ce{Rb} & \ce{Cs} \\
\hline
\lambda/\pu{nm} & 670 & 589 & 769 & 794 & 894 \\
\hline
\end{array}
$$
You cannot even see these wavelengths with your eyes after 700 nm which are in the deep deep red. Some people can see the red lines of K 769 nm. However, you can see the energy gap is indeed decreasing as we go down the group. The ionization energy also goes down the group. However, you can see, the colors which you see in the Bunsen burner have nothing to do with ionization energy.
The wavelengths have been extracted from a old book titled Graphische Darstellung der Spektren von Atomen und Ionen mit ein, zwei und drei Valenzelektronen by Dr. W. Grotrian, 1928. I think NIST atomic spectra database will be better today but I am not sure if they list resonance lines clearly. This book has graphically labelled transitions for almost all elements known in 1928.


How can molecularity be the order of the rate determining step? In my textbook the molecularity is defined as(along with the standard definition) the order of the rate determining step(RDS). It is also given that the order can be a fraction but molecularity is always a whole number. But, then if the order of RDS is a fraction molecularity will be a fraction which is a against the definition. So, where is the problem? Is it that the order of RDS can only be a whole number?

What your book, and most other elementary books present, is at best a generalization, not applicable to all cases.

Molecularity, as per the collision theory, is defined only for elementary reactions. One of the things that define an elementary reaction is the fact that the entire transformation from reactants to products occurs simultaneously, and follows from a properly oriented, appropriately energetic collision of the reactant species. If the reaction is not a direct consequence of a single collision, it is not an elementary reaction, and hence molecularity is meaningless.

Now, what usually happens is that out of several elementary steps constituting a reaction, one is very slow as compared to others. So whatever the rate of that elementary reaction, becomes practically the rate of the entire reaction. And since it is an elementary reaction, its order is its molecularity, and to an approximation, the molecularity of the slowest step then becomes the rate of the equation.

But, in most cases, the slowest step is not slow enough for it to entirely determine the rate. It is the slowest step, and it is an elementary reaction and therefore its molecularity is its order. But now, its order is not necessarily the order of the entire reaction, since that was an approximation valid only in cases where the slowest step was so slow that it entirely determines the rate. Then, the order of the entire reaction, due to the influence of the other steps (elementary in itself), can be fractional, while the order of the slowest step, is still its integer molecularity. ( here the term slowest step is preferred over the misnomer rate determining step, as it does not entirely determine the rate).

Also keep in mind, that many a times what people refer to as the "rate determining step" or the "slowest step" is not an elementary reaction. then its order need not be an integer, even if it is slow enough for the order of the net reaction to be equal to the order of that step.


Why is the H−C−H bond angle in сhloromethane larger than 109.5°? Symmetrical tetrahedral molecules (like $\ce{CH4}$) have a bond angle of $109.5$. Those with lone pairs in place of one atom (like $\ce{NH3}$) have bond angles less than $109.5$.

The standard explanation for why is that the lone pairs are "larger" than the hydrogens, creating repulsive forces that push the hydrogens down, decreasing the bond angle.

However, this explanation fails when confronted with $\ce{CH3Cl}$. According to the above logic, the $\ce{Cl}$, which is much bigger than the hydrogens, should repel the hydrogens just like a lone pair, making the $\ce{H-C-H}$ bond angle smaller than $109.5$. However, the $\ce{H-C-H}$ bond angle is actually $110.5$ (see here, I could not find a more authoritative source).

Why is this (or is my source wrong)?

Chlorine is more electronegative than carbon which is further more electronegative than hydrogen and thus chlorine pulls the bond pair towards itself, polarizing the bond. This decreases the electronic repulsion between the $\ce {C-Cl}$ bond pair and the $\ce {C-H}$ bond pair, decreasing the $\ce {H-C-Cl}$ bond angle. Consequently, the $\ce {H-C-H}$ bond angle increases to 110.5°.


Why does chemisorption decrease with increasing temperatures? Why is it that chemisorption decreases with higher temperatures?
The explanations I found online were quite vague and one of them cited Le Chatelier's principle, but I thought chemisorption is an irreversible process (because covalent bonds are formed between the adsorbate and adsorbent) so there is no reversible equilibrium.
Explanations I found online:

"Further increase will increase the energy of the molecules adsorbed and will also increase the rate of desorption. Therefore, the extent of adsorption decreases." - why does the rate of desoprtion increase?

"The initial increase is due to the fact that heat supplied acts as activation energy. The decrease afterward is due to the exothermic nature of adsorption equilibrium."

Let's answer this question with the help of some thermodynamics and kinetics. Chemisorption is a chemical process with $\Delta S<0$ and $\Delta H<0$. Thus, in simpler words it is an enthalpy driven exothermic equilibrium
Every chemical process is an equilibrium with varying extent of completion. So your idea of chemisorption not being an equilibrium is wrong. Every chemical reaction is theoretically reversible in nature. What we call irreversible reactions in general are such reactions which are heavily favoured on one side of the equilibrium.
Now,
$\ln K = -\frac{\Delta H}{RT}  + \frac{\Delta S}{R}$
It is evident from this above expression that the $K$ will decrease when we increase $T$ when $\Delta H<0$ and thus the aforementioned conclusion.
But this is only half the story. Like all other chemical reactions, chemisorption also needs some activation energy to overcome the transition state. At lower temperatures there are very few molecules which have the kinetic energy enough to overcome the activation energy. The fraction of these molecules increases with increase in temperature and thus increasing the extent of chemisorption.
Conclusion
Due to the combination of thermodynamic and kinetic effects the rate of chemisorption first increases with the increase in temperature but after a certain saturation point the thermodynamic effects start dominating and thus the extent of chemisorption starts to decrease.
The graph below might help you understand how the extent of chemisorption changes with change in temperature.

Here,
$x$ = Mass of adsorbate
$m$ = Mass of adsorbent


What is the chemistry behind cleaning silverware with aluminium foil This question has a accepted answer on this site already:

What happens during cleaning silverware?

but the reason I ask the question again is because I found different "recipes" for the technique and also different reaction formulas in other sources.

The "recipe" given in the linked question is

  "Put [the silverware] into a pot with some club soda and a piece of aluminium foil and pour over a hot water."

And the reaction formulas given for "blackening" is the following

$$\ce{3Ag2O(s) + 2Al(s) -> 3Ag(s) + Al2O3(s)}$$

and for the reaction with the aluminium foil

$$\ce{3Ag2O(s) + 2Al(s) +3H2CO3(aq) -> 3Ag(s) + 2Al(OH)3(aq) + 3CO2(g)}$$

But I also found the following "recipe" in a Swedish newspaper

"Put aluminium foil in a bowl. Put some baking powder on the foil. Put in the silverware. Add more baking powder. Pour over hot water and wait."

I assume this recipe is more or less equivalent to the one given earlier since the baking powder essentially turns the water into carbonated water?

The reaction formula for the "blackening" of the silver given in the newspaper is

$$\ce{Ag + sulfurcompunds + O2 -> Ag2S + H2O}$$

and the reaction with the aluminium is described as following:

$$\ce{2 Al(s) + 3 Ag2S(s) + 6 H2O(l) - 6 Ag(s) + Al2O3(s) + 3 H2S(g)}$$

The newspaper also links to a small piece written at a University in Sweden with regards to this question.

where they explain the blackening of the silver in the same way as the newspaper but they give a different reaction formula for the aluminium reaction

$$\ce{2 Al + 3 Ag2S -> 6 Ag(s) + Al2S3}$$

So my question is: Is any one of these reaction more "true" (occurring more often naturally) than the others or is it the case that a little bit of everything is happening?

Is any one of these reaction more "true" (occurring more often naturally) than the others or is it the case that a little bit of everything is happening?

I think the best answer is "a little bit of everything".  Silver sulfide forms faster but requires exposure of the silver to sulfur-containing materials (like human skin, food, etc.).  Silver that isn't exposed to sulfur would still be oxidized by the air, but at slower rates.  So depending on the life history of the silver object in question, different amounts of the sulfide and oxide could be present.


How are partial molar Gibbs excess functions correctly defined? I think I've found a mistake in my thermodynamics textbook, Chemical Thermodynamics for Process Simulation, but as thermodynamics is hard and I'm a relative novice, I wanted to check here.

The textbook, in talking about partial molar Gibbs excess functions of species in solutions, writes that we can express the partial molar Gibbs free energy of a species $\bar{g_i}$ in solution, as the sum of the partial molar Gibbs free energy in an ideal solution, $\bar{g_i}^\mathrm{id}$, and an excess function, $\bar{g_{i}}^\mathrm{ex}$:

$$\bar{g}_i = \bar{g}_i^\mathrm{id}+\bar{g}_i^\mathrm{ex}.$$

It doesn't specify whether we are thinking of evaluating $\bar{g}_i^\mathrm{id}$ at the system pressure or at a reference pressure, but immediately after it 'expands' the above expression into the form:

$$\bar{g}_i(T,P) = \Big(\bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i\Big) + \left(\mathcal{R}T\ln \frac{f_i}{x_if_i^0}\right)$$

Clearly we're comparing $\bar{g}_i$ to the ideal value, $\bar{g}_i^\mathrm{id}$, at standard pressure $P_0$. However, they then go on to define the activity $a_i=f_i/f_i^0$ and the activity coefficient $\gamma_i = a_i/x_i$ and substitute these in to get

$$\bar{g}_i(T,P) = \bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i  +  \mathcal{R}T\ln \gamma_i.$$

Now, I'm all fine up to here. The first part, $\bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i $, is the ideal solution Gibbs free energy at standard pressure, and the final term, $\mathcal{R}T\ln \gamma_i,$ is a correction in two ways: It corrects for the non-ideality of the solution, and it corrects for the fact that we may be at a different pressure $P$.

But the book then says:

  For an ideal mixture the excess part is equal to zero, since the activity coefficient is $\gamma_i=1$.

I think this is ridiculous: we can only make this conclusion if we have an ideal solution at $P=P_0$.

Is this an error in the textbook? Also, I can see two possible routes to an error: Either $\bar{g}_i^\mathrm{id}$ is actually to be evaluated at system pressure, in which case their last statement is correct but the formula they give is wrong, or else their definition of excess functions is the standard one (actual value minus ideal value at ref. pressure) but they made a mistake in concluding $\gamma_i=1$ for all ideal solutions, independent of the pressure.

The confusion is perhaps not uncommon. However, whichever book it is has jumped from discussing the free energy of pure materials, and their variation with various things (T, P, V, ...), to discussing mixtures.  In this case, all variation of the 'pure' substances is present in the ideal terms - that is because those variation exist without regard for any mixing terms.  What you are focused on here are the impacts of making a solution (or binary alloy, ...).

So, to make the mixture, you take a certain amount of substance A, and a certain amount of substance B, and mix them together. Before mixing, the total free energy of the system is $x_A G_{A} + x_B G_{B}$ where $G_{A,B}$ are the Gibbs free energies of those substances at whatever the conditions you specify before mixing. So, what does mixing do? Well, at the very least you get the increase in entropy, $RT(x_{A}\ln{x_{A}} +x_{B}\ln{x_{B}}$. If there is any interaction between A and B, you now have non-ideal terms which can be represented in any number of different ways. However, the point is that they are "all the rest that happens" during the mixing, and those things may be functions of concentration, temperature, pressure, ...

But, to come back to your question - no, this is not an error in the textbook (whichever it is). But, the did not make it clear that they were leaving behind the explicit dependencies of free energy on the various parameters - they figured you had that part down.

I would also note that, when one calculates a phase diagram (ultimately one hopes the book might get to that, and miscibility gaps and whatnot), you are free to choose the reference points or phases of A and B yourself - that choice makes no difference in the phase boundaries and tie lines. And, this comes back to just what $G_{A,B}$ means.


What is the difference between ΔG and ΔG "with an o on top"? What is the difference between $\Delta G$ and $\Delta G^\circ$? I couldn't figure it out.

Ex:

$\Delta G^\circ = -RT\ln K$
$\Delta G = \Delta G^\circ + RT \ln Q$

What is the difference, is it that $\Delta G^\circ$ is the value predicted by the standard free energies and that $\Delta G$ is just the actual value?

A simple explanation would be that the "o on top" denotes standard state and the one without the "o on top" denotes conditions that are not standard state. However, that may not allow you to "get it" so let's look at the equations.

$\Delta G^{\circ} = -RT\ln K$
$\Delta G         = \Delta G^{\circ}  + RT \ln Q$

The first one allows us to find the Gibbs free energy at the standard state, as was mentioned. However, it would be useful to review what exactly the standard state is:

For the standard state of formation:

  The partial pressure of any gas involved in the reaction is 0.1 MPa.
  The concentrations of all aqueous solutions are 1 M.
  Measurements are generally taken at a temperature of 25 °C (298 K).
  The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions.
  (source)

Or, very simply:

Partial Pressure of gases = 0.1 MPa
Concentration of Aqueous solutions are = 1 M (Molar)
Temperature usually 298 K

So if it's so nice and well, why would we need the one without "o on top"? Well, if we meet a problem or situation without the concentrations or pressures as stated above, we need to use our handy equation to relate the standard state which may have been given to the actual situation, with perhaps 0.1 M or some other concentration. This is powerful because it allows you to view different states,perhaps with more or less reactants or products, and realize how one can affect equilibrium.


Shouldn't enthalpy be constant measured at constant volume? Shouldn't enthalpy be measured at constant volume? Then the work done on and by the gas is 0 meaning that the temperature change is caused solely by heat released or absorbed by the system. You'd be able to calculate
$$\mathrm dQ = \mathrm dT \cdot C_\mathrm s$$
(specific heat).
However, since enthalpy is at constant pressure the substance will do or have work done on it and thus there is no easy way to relate the enthalpy change to the change in temperature. Am I mistaken somehow?

Enthalpy is heat flow associated with a constant pressure process.  Here is how:

Start with the familiar first law of thermodynamics for energy change:

$\Delta E=q+w$

$E$=energy

$q$=heat flow into the system

$w$=work done on the system

The work done is rendered thusly:

$w=-\int_\text{start}^\text{end}P\,\mathrm dV$, $P$=pressure, $V$=volume

Pressure is deemed constant so:

$w=-\int_\text{start}^\text{end}P\,\mathrm dV=-P\Delta V=-\Delta(PV)$

And then our first law balance equation becomes:

$\Delta E=q-\Delta(PV)$

$q=\Delta(E+PV)$

So heat flow is the change in a combination of state variables involving energy, pressure and volume.  This combination is what we define as enthalpy and usually label $H$.  Thereby, this enthalpy variable is designed to automatically include (mechanical) work inherent to a constant pressure process.


NMR Spectra Oddities I've noticed that as the chain length of n-alkanes increase, their number of carbon environments tend to 5 peaks, and no more, no less. Why is this? I've looked on multiple NMR (C13 NMR, I might add) databases and they show that there are 5 carbon environments. The outer carbons are all distinct environments like you'd normally expect, but the inner part of it (e.g. the innermost carbons in tridecane, C13H28) is all one environment. I hope somebody could shed some light on it. It's stumped both me and my chemistry teachers.

The environments are the same only to within the limits of the spectrometer's resolution and other experimental details; remember that every measurement comes with uncertainties, some of which are baked into the very fabric of the universe.

If you had an "infinitely precise" NMR spectrometer, all the peaks would be distinct (except when symmetry demands they be the same). In fact, every single possible compound would be uniquely identifiable by its atoms' specific shifts, shared with no others. In principle, you'd be able to assign a carbon peak at 123.6207426[...]964109454164[...] ppm to riboflavin, whereas a peak at 123.6207426[...]964109454165[...] ppm would indicate monobrominated hexabenzocoronene intercalated into a strand of DNA, and so on. Of course this is just a thought experiment and would never be possible in practice, even though it would make characterisation of complex molecules and mixtures a dream...


Enthalpy in the van't Hoff equation In the van't Hoff equation, why do we say that the change in enthalpy at standard pressure is constant? My book derives the equation for a mixture of ideal gases reacting with each other. The proof itself is good, until the authors declare that the equation says that the graph of $\ln K$ with $\frac{1}{T}$, where $K$ is the equilibrium constant and $T$ is the temperature of the mixture, will be a straight line. This would only be true if the change in enthalpy is constant (since the slope of the curve is proportional to this change). But isn't the enthalpy of an ideal gas proportional to temperature? So why is the change in enthalpy constant in $T$, then?
I've attached an example of this effect, as presented by the book. See the image below.

Good question. Enthalpy of reaction dependence on temperature is given by Kirchoff's equation (you can check it) and it is true that enthalpy of reaction changes with temperature. For most reactions if temperature change isn't too big, enthalpy of reaction won't change much so we can regard it as constant in that temperature span on which we are plotting lnK vs 1/T for sake of simplicity.


Calculate the standard oxidation potential from given potentials The standard electrode potentials for the following redox couples are given as: >$E_0(\ce{Fe^{2+}/Fe}) = x~\mathrm{V}$ and $ E_0(\ce{Fe^{3+}/Fe^{2+}}) = y~\mathrm{V}$
  What will be potential $E_0(\ce{Fe^{3+}/Fe})$ (in $\mathrm{V}$)?

My approach: $\ce{Fe^{3+} + e- + Fe^{2+} + 2e- -> Fe^{2+} + Fe}$

So $E_0(\ce{Fe^{3+}/Fe})=(x+y)~\mathrm{V}$

You can write down the three half reactions:

\begin{aligned}
 \ce{Fe^{2+} + 2e- &-> Fe}      & E_1&= x\,\mathrm{V}\\
 \ce{Fe^{3+} + e-  &-> Fe^{2+}} & E_2&= y\,\mathrm{V}\\
 \ce{Fe^{3+} + 3e- &-> Fe}      & E_3&= z\,\mathrm{V}
\end{aligned}

The third half-reaction is simply the sum of the first two. However, we cannot just sum $E_1 + E_2$ in order to obtain $E_3$. The problem is with the meaning of the potentials, and their relation to the stoichiometry.

An half reaction with potential $E$ in volts means that the elecrons have some energy assigned to them. The meaning of volts is joules per coulomb. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).

Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry

$x$ joules per coulomb, or
$96500x$ joules per mole of electrons, or
$2\times 96500x$ joules per mole of reaction.

In the same fashion, the second half reaction has the electrons carrying

$y$ joules per coulomb,
$96500y$ joules per mole of electrons,
the same $96500y$ joules per mole of reaction.

The third half reaction is the sum of the first two, so its electrons carry

$2\times 96500x + 96500y$ joules per mole of reaction, which is
$3\times 96500 z$ joules per mole of reaction, or
$96500z$ joules per mole of electrons, or
$z$ joules per coulomb.

Now you can simplify this to $2x + y = 3z$ and calculate the potential you want.


Why is bipolar resonating structure less stable than monopolar structure? In my book, it is written:

  Resonance energy of a system involving monopolar resonance structures is greater than that involving bipolar resonance structures.

The book cites example:

  Carboxylate ion which contains negative monopolar charge is in fact more stable than the corresponding carboxylic acid which contains both negative and positive charge.

But what's the reason?? What is going on?? Please help me with a math-free explanation.

In the diagram below, the 'bipolar resonance structure' refers to the resonance structure with both positive and negative partial charges (left), while the monopolar resonance structure refers to the anionic form in the right. I don't think this terminology is appropriate, and I have never encountered it before. I will refer to the structures as 'protonated' (left) and 'deprotonated' (right).

The diagram above illustrates the scenario. While it is true that the resonance stabilization is greater for the deprotonated form, the protonated form is still overall more stable. The book is inaccurate in that regard.

The "resonance stabilization" refers to electrons being donated into empty orbitals within the molecule. In this case, there is a lone pair in the hydroxide oxygen atom being donated into the antibonding pi orbital of the C = O. A way to rationalize the effect is that, when the proton is attached to the oxygen, you have both the H and the O pulling on the electrons. When the H is removed the electrons will more readily flow into the C = O pi*. Since both O atoms are equivalent, you end up forming an orbital that looks like an extended pi bond.

The stabilization due to the resonance of the deprotonated form is of greater magnitude than the stabilization due to resonance of the protonated form. However, the proton-oxygen bond is more stabilizing than the resonance, so the protonated form is overall more stable.

Diagram soure: http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Physical_Properties_of_Carboxylic_Acids


Hiyama vs Sonogashira coupling Why is Sonogashira's coupling more feasible than Hiyama's coupling when both could occur? I have seen many examples in syntheses, here is one (the penultimate reaction):

So I will answer my question. In Organic Chemistry Portal it is written that Hiyama coulping requires an activating agent such as fluoride ion or a base because crucial for the success of the Hiyama Coupling is the polarization of the Si-C bond. Activation of the silane with base or fluoride ions (TASF, TBAF) leading to a pentavalent silicon compound is a first necessary step. The whole mechanism is given here:

On the other hand, Sonogashira coupling doesn't require any activation and it happens preferentially in the above mentioned synthesis.


What is the extent of a reaction between an acid and its conjugate base? What is the extent of say, this reaction?

$\ce{H_2CO_3 + HOCO_2^- <=> HOCO_2^- + H_2CO_3}$

K for the reaction appears to equal 1, and given the stoichiometry of the reaction, this implies that the reaction reaches an extent of 50% at equilibrium.

However, is this actually true?

The reaction you've shown is termed an "identity reaction", the products are the same as the reactants so $\ce{K_{eq}}$ must equal one.  In the particular case you've drawn the reaction involves a simple proton transfer and will likely be diffusion controlled.  Other examples of identity reactions are shown below.  That they are actually occurring can be demonstrated using isotopically labeled substrates.  The last one is one of my favorites, the degenerate rearrangement of bullvalene into itself via a Cope rearrangement.  My drawing of bullvalene doesn't do it justice, just google bullvalene and prepare to be fascinated.  If you number each of the ten carbon atoms in bullvalene you'll find that there are over 1.2 million possible arrangements.  Sequential Cope rearrangements will convert each of these arragments into every other possible rearrangement.  Connectivity in bullvalene is lost, each carbon atom is rapidly bound to every other carbon atom.  At a bit over 100 °C, both the proton and carbon-13 NMR of bullvalene show only 1 sharp peak.


Why is volume inversely proportional to pressure? If temprature is directly proportional to volume (Charles's law) and
temperature is directly proportional to pressure (Gay-Lussac's law), then why is pressure and volume are inversely proportional?

Given you know and understand Charles' and Gay-Lussac's laws, it's not about chemistry, rather, simple ratios:

$$
\begin{cases}
T \propto V\\
T \propto p
\end{cases}
\implies
p \propto \frac 1 V
$$

which, as Zenix commented, is a math form of Boyle's law.


Why would a reaction favor one isotope over another? In my instrumentation class last semester, we were asked to read a short paper. The paper described a technique for determining whether dinosaurs were cold-blooded or warm-blooded based on the amount of a certain isotope in a sample of dinosaur bone.

What would cause a reaction to favor one isotope over another? In my general chemistry course, I learned that isotopes act similarly in chemical reactions. Of course, I have found over the years that pretty much everything taught in general chemistry was approximations probably so students don't feel inundated.

Chemically speaking compounds containing isotopes are very similar but not exactly similar

Broadly when compounds contain different isotopes of the same element (say different isotopes of oxygen) their reactivity is the same. But this is a little over broad. There are subtle and (usually) small differences and these can show up as small differences in chemical reactivity.

To get an idea why consider the simple physical properties of water and heavy water. In heavy water the two hydrogens are replaced by deuterium (hydrogen with an extra neutron in the nucleus). A heavy water molecule is about 10% heavier than a normal water molecule (molecular mass ~20 rather than ~18). As a result the physical properties are a little different: bp is about 2°C higher, mp about 4°C higher and the liquid is about 10% denser. So careful techniques can separate the two (by repeated fractional distillation, for example).

The slight differences also appear in chemical reactions. The reasons why are more complex than the explanation of simple physical effects. But one hint that they exist is simply that the vibrational frequency of a bond depends on the mass of the atoms making up the bond. Isotopes have different masses so there will be some difference with different isotopes. And bonds vibrating are part of the mechanism of reactions involving that bond so the reactions will have different rates. The detailed theory is pretty complex (see this) and the size of the effect depends on the detailed mechanism of the reaction.

Reaction rates for some reactions involving bonds to hydrogen can differ by large factors (because deuterium is twice as heavy as hydrogen) and are sometimes 10 fold. Most other isotopes have only small relative mass differences compared to their parent element and the net rate differences are often only small single-digit percentages. This results in some reactions concentrating some isotopes over others (eg 18O vs 16O. But these can be large enough to detect by mass spectroscopy.

So, despite their chemical similarity, some reactions and some physical processes will favour one isotope over another (though often only very slightly). If you understand the key reaction mechanism, the small differences in the concentrations can tell something about the conditions of the reaction (temperature dependence, for example).


Chemical Equation: Sulfuric acid and powdered sodium carbonate I was asked to write a balanced chemical equation for the reaction between sulfuric acid and sodium carbonate. The balanced equation isn't the problem. The question asks to also show all the state symbols.

All the equations I find online have sodium carbonate as a solution, not a solid. I'm also aware there are two forms of the balanced equation:
$$\ce{H2SO4(aq) + Na2CO3(s) -> H2CO3 + Na2SO4 -> H2O + CO2 + Na2SO4}$$
So do I say the sodium sulfate is a solid or in solution?

Try to follow this reasoning. (mouse over for answer)

Which parameter is mandatory to determine if a substance precipitate or not?

   Solubility.

How does this parameter behave with sulphates?

   Many sulphates are soluble in water. Generally the sulphates become less soluble as you go down the group. So sodium sulphate is surely soluble, confirm with Wikipedia.

Additionally think about others parameters that can have some sort of influence in this to improve your understanding of chemistry. So which parameters can affect this behavior?

   In fact as best practice you should always take in account the temperature and the amount of substance, and if the solution is saturated the excess will precipitate.

What's the answer?

   (aq)


Can aluminum and hydrochloric acid be used to generate hydrogen gas? I am looking to produce hydrogen gas in a small amount with supplies I currently have. Most of the methods I have seen involve reacting zinc or magnesium with $\ce{HCl}$ to obtain the gas, but I do not have zinc or magnesium at the moment.

Is there any reason why aluminum cannot be used just as effectively as zinc? According to the reactivity series, aluminum should be even more reactive than zinc.

Yes. See this video for a demonstration.

The reaction is:
$$\ce{ 2Al (s) + 6 HCl (aq) -> 2AlCl3 (s) + 3H2(g)}$$

You can capture the hydrogen and proceed with your experiments.


Shouldn't MSDS data on products be publicly disclosed? I am wondering whether companies are required to disclose MSDS data on their products: toxicity, $\mathrm{LD_{50}}$, etc. Not only that, but I am wondering whether they are required to disclose the methodology that they used. I know that this isn't a legal forum, but some companies that I asked are unwilling to give details on how the toxicity was tested. What are some laws that are in place (regardless of country)? I thought that all MSDS data should be public.

There is a bit of a misconception here.

Of course, the material safety data sheet (MSDS) is publicly available. Just go to your favourite chemicals supplier, type in your substance and download the corresponding MSDS in your language for your jurisdiction. At least in Europe that is mandatory for all chemicals a company sells; I would be very surprised if there is any developped country where making MSDS publicly available be not mandatory.

On the other hand there is the question of how a value was determined. That is not necessarily public information. Of course, there is a set of standardised tests that companies would perform, but they don’t have to exactly mention whether they used method A or method G. They also have a reason to be nondisclosive about this: imagine an animal rights activist calling you pretending to just be an interested scientist and then using the information they get on the phone to start a name-ruining anti-ChemicalCompany campaign. You don’t want that, do you? And then, of course, there is the general secrecy issue. If you just invented a compound you want to put on the market as a drug, you don’t want anybody knowing anything about it until you’ve started selling it.

Of course, government bodies have a certain reason to know how the tests were perfomed and what the raw results were (otherwise the company could just print a number on their MSDS which the responsible person invented on the toilet). But you’re not government. You’ll have to somewhat trust the government to properly monitor the companies in question.

Of course, the chemist who works safely doesn’t need to know how dangerous something is. They will work with it in a way that they are not affected.


Oxygen production from thermal decomposition of metal oxides Does thermal decomposition of metal oxides always produce oxygen?

A lot of metal oxides decompose under liberation of $\ce{O2}$ when subjected to sufficiently high temperatures. For example, $\ce{HgO}$ decomposes at temperatures above $\pu{500^\circ C}$ into elemental mercury and oxygen:

$$\ce{2HgO ->~ 2Hg + O2}$$

However, disproportionation is another possible reaction pathway. An example is the thermal decomposition of iron(II) oxide to elemental iron and iron(II, III) oxide at temperatures below $\pu{575^\circ C}$ ($\ce{FeO}$ is more stable above that temperature; sources: 1, 2):

$$\ce{4FeO \rightleftharpoons Fe + Fe3O4}$$

Another example is the decompostion of $\ce{SnO}$ at temperatures between $700\mathrm -\pu{1000K}$, which proceeds via an intermediate $\ce{Sn3O4}$ mixed oxide. Tin disproportionates from the oxidation state +II in $\ce{SnO}$ to $0$ in $\ce{Sn}$ and +IV in $\ce{SnO2}$ (1, 3).

\begin{aligned}
\ce{4SnO ->~& Sn3O4 + Sn} \\
\ce{Sn3O4 ->~& 2SnO2 + Sn}
\end{aligned}


How can every atom have unique spectral lines? Why does every element have unique spectral lines?

I have studied that spectral lines help us to identify the properties of an element. This is because after excitation when the excited electron(s) fall back to their original energy level(s), photons of certain frequency corresponding to the difference between the energy levels are emitted.

Now I have a question,  every substance consists of atoms, and each atom consists of electrons, electrons are fuzzy clouds and are described by wave-functions given by the solutions of the Schrodinger equation, so what makes the spectral lines different although every substance has the same thing?

I searched on the site and discussed with various members of the community and some of the conclusions I drew are mentioned below:

Electron-nucleus attraction and the electron-electron repulsion.
Hydrogen is a special case because it has only a single electron so there is no electron-electron repulsion.
Helium has two electrons so now we have some e-e repulsion, Lithium has three electrons so there is even more e-e repulsion and so on...
Every atom has a different number of electrons and a different nuclear charge, so the balance of nuclear attraction and e-e repulsion is different. So every atom has its own set of atomic orbitals that are specific to it and are different from every other atom.
The spectrum comes from transitions between orbitals, and since the orbitals are different for every atom the spectrum is different for every atom.

Are my conclusions reasonable or is there something else I missed to note?

But this answer creates one more question in my mind,

Do all noble gases or alkaline earth metals have similar spectral lines considering the above points I mentioned?

Do all noble gases or alkaline earth metals have similar spectral
  lines considering the above points?

The question is interesting after you modified it. The basic set of reasoning you provided is the main story. Each element has a different nuclear charge and the outermost electron(s) is responsible for the atomic emission spectrum. Since the nuclear charge is different, those outermost electrons experience a different potential energy. Their kinetic energy is also different from element to element.

The key question is what is meant by similarity? The atomic spectra of all the elements is visualized as bright lines on a dark background. The reason they appear as lines is just because of the instrument used to observe the atomic spectrum. There is nothing fundamental in the "line"spectrum. The atomic emission appears as lines because the slit in the monochromator is shaped like a very narrow rectangle. This is the image of the slit. If I made a very narrow circular opening, the images will appear as bright points rather than lines. Consider that line spectrum is the conventional way of looking at the atomic spectrum.

For the visible region, let us look at the visible spectrum of alkali and alkaline earth metals.

or look at the visible spectrum of noble gases

Are they similar in any way? Visually there is no similarity at all in the spectrum. Look at the spectrum. The reason we cannot see any similarity is because the dispersion of the lines is on the "wavelength" scale. However, very bright, I mean really genius spectroscopists of the 18th and the 19th century were able to find out a patterns (or call it as mathematical series). They figured out that each set of lines forms a pattern as Sharp series, Principal series, Diffuse Series and Fundamental series and there they found similarities in patterns-in terms of mathematical series. In short, our eyes cannot interpret those series. Another complication arises because we are just looking at the visible region. There is a ultraviolet region and then there is an infrared region. We cannot see it without using more sophisticated instruments.


What is the difference between a radical and a neutral alone atom? I've done a bit of googling but can't find a clear answer to this question. Is there any difference to e.g. an alone neutral chlorine atom and a chlorine radical?

A chlorine radical and a chlorine atom are effectively the same thing. This does not mean that "radical" and "atom" are interchangeable as jargons. They are used in fairly particular contexts, namely

Saying an "atom" is a reference to an instance of a particular element. They are often discussed abstractly, in terms of their constituent parts (nuclei and electrons) without coming to discuss their  surroundings or conditions within a system.
It is natural to say molecules are made of atoms, because they conserve the total number of nuclei and electrons with respect to the individual atoms, so we can rationalize them as bonded atoms. But saying molecules are made of radicals is weird.
Radicals can (very often) be polyatomic, like a methyl radical $\ce{CH3.}$
A radical has an odd number of electrons at its outer shell, it is chemically unstable, although electrically neutral. Atoms can be neutral and chemically stable, as in the noble gases, which fully comply to the octet rule.

Thus an atom being a radical and vice-verse is more of a "chemical coincidence" than anything else, kind of like a hydrogen cation is a proton. But they are definitely not equivalent.


Memorizing the periodic table Several people have said that the key to understanding chemistry is through memorizing the periodic table.

I want to ask if there is a simple technique to learn it, or if I just have to remember every element as it is?

There isn't really much sense in memorizing the periodic table. The elements you often use you will know them by heart after a while. And you can always use a table when you need it for the others.

That being said if you really want to do it, mnemonics are probably the best solution to memorizing the whole table. You can find some here :

https://www.mnemonic-device.com/chemistry/periodic-table/mnemonic-device-for-the-periodic-table/

or here :

http://en.wikipedia.org/wiki/List_of_chemistry_mnemonics#Periodic_table


Can Ferrofluids vaporise to gaseous state and Condense back to liquid state and do it many times over? Here is the Link for Ferrofluids for basic idea. After going through this Wiki link, probably many people will come to a conclusion like when Ferrofluid boils, its carrier liquid vaporises and nano particles gets settle down. Therefore Ferrofluid looses its properties(destroyed). I come to this conclusion but not 100% sure. There may be some wild science going over because "nano" particles and its size. So I decided to ask here and get some help from some bright minds across the world.

Part of the ferrofluid could theoretically be vaporized. As ferrofluid is made from tiny magnetic particles(eg. iron powder coated with a surfactant to inhibit clumping) suspended in a fluid. The fluid can be organic solvents, water and most commonly ink. Some of these fluids can evaporate and recondense. However, as a whole ferrofluid, cannot be vaporized. You can imagine ferrofluid as a cup of mixed sand and water that wouldn't settle(suspended).


Where to find data for Gibbs energy, enthalpy, and entropy? Where can you find the enthalpy and Gibbs free energy of formation and entropy of compounds either as a data file or web API? Technically, I only need two since I can figure out the other one from two of them.

NIST webbook does have a lot of data, though they are not in any kind of an API form as far as I know.

http://webbook.nist.gov/chemistry/


Why isn't this a compound? I don't know what to call it I'll just provide a link. From what I can tell it is impossible, but why?

http://www.chemicalize.org/structure/#!mol=structureId%3A1609384473364&source=calculate

This structure is impossible as either the molecule is planar and then the C-C bonds on the outer ring are extremely long (and therefore weak), or you distort from the sp2 geometry (120 degree bond angles) of the central ring carbons so that the outer ring is beneath the inner one, which would also require a very high energy state. The only stable circumscribed ring structures you'll find are macromolecular.


Why doesn't water wash off the coronavirus alone? So the coronavirus has a lipid bilayer where the hydrophilic heads are facing outward. So why can't water just stick to these ends and flush out the virus particles from our hands? Is it because our hands stick better to the virus? If so, why and how does the virus stick better to our hands than water?

Because your hands are covered in a thick crust of dirt, dust and fats, in which the virus can hide.

But anyway still nobody has been able to show that this virus is regularily (or even rarely) transmitted via smear infection. (If someone knows about any evidence to the contrary, please come forward. And yes, some other diseases, especially bacterial infections, do even mainly take this route. That´s how you usually get the "shits".)

It can be detected on surfaces after hours via PCR, however PCR is as sensitive to broken down pieces of a virus particle as it is to the fresh, contagious specimen. Obviously the RNA pieces also of a destroyed virus would not evaporate, so the result of these investigations proves nothing but the extremely obvious. Newspapers are spreading it as a piece of clickbait and tinder, causing people to buy lots of "sanitizers".

Anyway, use soap, leave the disinfectants to the health care people who absolutely, direly need them, and be done.


How to Determine An Element's Colour Different elements have different colours. Some elements may have similar colours like Sodium (Na) and Potassium (K), so that's why i'm asking "is there a way to calculate or even simply deduce the actual colour (not spectral lines) of an element based on its properties?"

Consider the energy required to elevate electrons from one "shell" to another, and the frequency of visible light that has this amount of energy.  The substance is more likely to absorb wavelengths of light matching the energy difference of that band gap, and reflect or transmit others.  The colors NOT absorbed can reach your eye as an observer, so (assuming a white light source) you see white (which is all the colors together) minus the colors absorbed by the material (matching the band gap).  For example, gold (Au) absorbs the blues and greens and leaves just the colors adding up to what you now think of as "gold."  If your light source is not white, what your eyes see will also be missing any colors your light source was missing.
The crystal structure can also make a difference, e.g. compare the colors of carbon (C) as graphite vs. diamond.

Useful links: See also this same question on ResearchGate, WebExhibits, and Quora.


Why does paper crumble on getting wet? Why does paper crumble on getting wet?
This is something I have noticed hundreds of times but cannot think of an explanation.

Does it have anything with water disturbing the intermolecular forces between cellulose micro fibrils in paper or something else?

Also, one peculiar observation is that it does not crumble immediately but only after the water evaporates? Why?

Polymer properties in water often relate to the relative strength of polymer-polymer interactions versus polymer-water interactions. For example, the solubility of amylopectin > amylose > cellulose is attributed to how strongly molecules of each of these polymers interact with water molecules compared to other molecules of themselves.

In the paper making process, the cellulose fibers of paper dry out and interactions with water tend to get replaced by interactions with nearby polymer molecules. The resulting structure is strong because of polymer-polymer interactions as well as purely physical effects (e.g., "matting" of fibers, and surprisingly strong surface-surface interactions), all of which is enhanced by the compression and tension forces placed on the paper during its manufacture (google it to see photos).

When paper is later made wet, the entrance of water eventually disrupts these interactions, leading to local regions of what you might think of as micro solubility.  Then, when the paper dries out again as you describe, it probably does so without any compression/tension forces on the paper, thereby reducing the extent to which the matting of the fibers and chemical interactions between polymers takes place, and so it crumbles because of those micro regions of "solubility".

(Disclaimer: The above is just rationalizing off the top of my head. But since it is based on expertise as a PhD Chemist for nearly three decades, I'd give it a 70+% chance of being accurate ;)

The real science begins when you test the above ideas: For example, one test would be drying out paper under different degrees of compression and/or for different time periods and then testing the crumbling behavior.  Another might be to wet the paper while it's under different degrees of compression.  But the best test of the above ideas would be if there were some way to assay the extent of polymer-polymer versus polymer-water interactions -- no doubt such interactions result in very different vibrational modes. So there, I've outlined a Ph.D. thesis for you.


Is carbonyl-enol tautomerization only intermolecular or can it be intramolecular? During workup of a conjugate addition reaction, it is likely that protonation occurs at the oxygen of the enolate formed. See discussion here: Do enolates get protonated at the carbon or oxygen atom? In that discussion, it is essentially said "The oxygen is probably protonated, but it really doesn't matter." However, I think this could be important in select instances, specifically in the protonation of enolates with diastereotopic faces (which isn't all that rare). But this mostly depends on if tautomerization can occur intramolecularly.

An example of what I'm getting at:

This is an example I just made up. This reaction likely would not actually happen like this. In actually, I could see some γ-deprotonation happening.

In this reaction, one chiral center is formed, and an enolate is formed, and the most likely atom to be protonated upon addition of an acid is the oxygen:

That enol could go on to form one of two products. The intramolecular reaction, I think, would likely give much more of the thermodynamic product (shown below) than an intermolecular reaction, though I'm not certain on that. The reactions here are simplified; I realize both would most likely make some mixture of the two products:

My thoughts are that the four-membered intermediate required for the intramolecular reaction will just be too strained for this to occur. I realize that in basic or acidic solution, catalysis will occur, as the Wikipedia page mentions. I guess the best way to phrase my question is this: In the hypothetical situation that a single ester enol molecule (like that shown as the product of O-protonation above) is dissolved in pure hexane, would that molecule be stuck in in the enol form or would it be able to rearrange? I realize this is more feasible with an dienol, but I'm specifically trying to figure out if the enol can do this.

For the sake of this question, ignore the basicity of the nitrogen. Maybe I should have used the diphenylmethane anion or something. Doesn't matter, it wouldn't reach the proton, either.

tl;dr: The intramolecular variant is an impossible rearrangement. Only intermolecular proton transfer can be observed.

This is actually a very nice example to discuss sigmatropic rearrangements, because the desired [1,3]-rearrangement would be one. In general, these can proceed in four possible ways, depending on the configuration of the migrating group and whether it remains on the same side of the π system or not. We would have to draw orbital schemes to determine whether or not the orbital of the migrating group is able to interact both with the point of origin and the point of arrival favourably.

With regards to the configuration, the two possibilities are:

migration under retention
migration under inversion

A migration under retention is always possible because the corresponding orbital’s lobe is already pointing towards the π system it is migrating along. A migration under inversion is only possible if the (atomic) orbital whose bond is broken and later reformed has lobes on two different sides; typically for migrating atoms, this means a $\mathrm p$ orbital.

In the case of a proton shift, we must consider a hydrogen atom. Its only occupied orbital through which it bonds to the carbon chain is $\mathrm{1s}$. This orbital does not have two lobes, therefore a rearrangement mechanism under inversion is impossible; we need only consider the retention mechanisms.

With respect to the π system, the two possibilities are:

suprafacial (remaining on the same side)
antarafacial (switching sides)

Again, a suprafacial migration is practically always possible. For an antarafacial migration to be considered, the chain must be flexible enough so that both sides can interact with the migrating group; ideally at the same time.

Since the thought experiments are rather hard to perform, here are some depictions to make it easier.

Figure 1: Suprafacial [1,5] (left) and antarafacial [1,7] (right) hydrogen migrations. The depicted π orbital corresponds to the HOMO of a full π system including the $\ce{C-H}$ bond.

Generally, the full π system and the neighbouring $\ce{C-H}$ bond that is to be broken must be considered as I have done in figure 1. This π system can be drawn up with the standard rules for linear π systems. In the [1,5]-rearrangement, the system is penta-2,4-dien-1-yl while in the [1,7] case it is hepta-2,4,6-trien-1-yl. The $\ce{C-H}$ bond is considered part of the π system with the proton being the migrating group. Therefore, the HOMO (the orbital of importance) will be the one in question. (If the $\ce{C-H}$ bond is considered part of hydrogen, we must consider the π system’s LUMO, meaning that the orbital will remain the same. We must always have an occupied and an unoccupied orbital for consideration.)

The image shows us that the [1,5]-migration — over two double bonds — can only occur in a suprafacial manner with hydrogen. If we had a group capable of inversion, an antarafacial manner would be possible given the correct geometry. In the [1,7] case — over three double bonds —, hydrogen can only migrate in an antarafacial manner under retention; the only possible suprafacial manner would be under inversion which hydrogen cannot do.

This has been turned into a general rule of thumb: If your migration is [1,$4n+1$], it may proceed suprafacial under retention but if it is [1,$4n+3$] it must proceed in an antarafacial manner. In the [1,3]-case, only antarafacial migration would be possible, but there is no way that the π system can twist so it would allow for antarafacial migration. The π orbital in question is shown in figure 2 below.

Figure 2: The π orbital relevant for a [1,3]-migration.

Therefore, an intramolecular proton transfer is impossible and an external source would have to protonate.

One thing to note: the molecule itself is able to deprotonate/reprotonate another molecule of its type intermolecularly. So the compound would not be stuck in the enol form but a second molecule would provide the much desired proton to allow the keto form to be regenerated, even in pure hexane (if it is even soluble in that).


Any examples of liquids volatile at room temp but non-flammable? I'm thinking of something similar to the liquid used in the classic dipping birds. Temperature differential in two "bulbs" at each end of a tube should cause vapor to expand and push liquid to the top, shifting the weight and tipping it. The closer to room temperature the boiling point is, the better. Thanks.

Yes, there are lots. Start with Wikipedia's list of refrigerants; while it doesn't call out flammability, any fully-halogenated compounds will be effectively non-flammable. Trichlorofluoromethane has a standard boiling point just above standard room temperature.

You didn't say anything about pressure in your question. If you don't mind pressures higher than one standard atmosphere, and don't mind toxicity, candidates like ammonia, sulfur dioxide, and chlorine come into play. If you don't mind low pressure, you could even use water, although it might freeze as it boils off.


Determining Charge of Protonated Alkaloids During the extraction of many alkaloids, you lower the pH in order to protonate the compound, making it polar and water soluble. With Mescaline, you must precipitate it as a salt.

What I don't understand, is how can you figure out the mole ratio between the Mescaline and the conjugate base of the acid (for example acetate). My understanding is you'd have to figure out the charge of the protonated mescaline. How? Thanks

Edit:

The reason is not so I can ionize all of the mescaline. I want to know this so I can calculate the molar mass of the mescaline salt. That way I can take the weight I have of product, and calculate (using the molar mass) the weight equivalence of freebase mescaline, for dosage purposes. Also, it's just interesting knowledge that I'd like to know.

Are you sure you're trying to calculate the molar mass of the mescaline salt? Isn't that known, depending on the acid you use?

It seems to me the key you need is the mole ratio. For a weak acid-base conjugate pair, you can use the Henderson-Hasselbalch equation:

$pH = pK_a + log(\frac{[B]}{[A]})$

where [B]/[A] is the ratio of concentrations of conjugate base to conjugate acid. If calculated for a single solution containing both forms, the total solution volume cancels and this is the mole ratio. In a more convenient form,

$\frac{B}{A} = 10^{pH - pK_a} = \frac{K_a}{[H^+]}$

(The first equality is more practical, the second is useful for theoretical manipulations.)

Note that when $pH = pK_a$, $B = A$ (the analyte will be 50% protonated and 50% deprotonated).

For mescaline (more precisely, its conjugate acid), $pK_a =$ 9.56, typical of an amine base/alkaloid. So at a pH of 9.56, approximately half of your mescaline should be protonated. You could extract this, measure the weight of the salt, use the known molar mass to convert to moles of the protonated mescaline (I recommend using a monoprotic acid such as acetic if possible to simplify the situation). By the Henderson-Hasselbalch, the moles of protonated mescaline you extracted at its $pK_a$ will equal the moles of freebase mescaline remaining in your organic solvent.


Chemical Kinetics- Net Number of Particles I have a box of 5 types of particles, $A$, $B$, $C$, $x$ and $y$. They react in the following way.

$\ce{$A$ + $B$ -> $C$ + $x$}$

and

$\ce{$C$ + $y$ -> $A$ + $B$}$

If only the first reaction was happening then $N(C) = k \cdot N(A)N(B)$. Here $N(I)$ is the number of particles of $I$ in the box.
Can I write the total number of $C$ in the box as $\text{number of $C$ created} - \text{number of $C$ destroyed} = N(C) = k \cdot N(A)N(B) - k' \cdot N(C)N(y)$?

Actually, the rate law for the species $\ce{C}$ is written as:

$$\frac{\mathrm{d}\ce{C}}{\mathrm{d}t} = k \ce{A} \ce{B} - k' \ce{C} \ce{y}$$

If you integrate this, you would get an equation in terms of the number of particles, as you require. However, this is a very typical integral, and there would be a number of other arbitrary constants, which would make the final result not equal to the one you've shown.

What you've written is a mistake on replacing the rate of formation of $\ce{C}$ with its number of particles itself. Both are not equivalent.


Computational Chemistry (Ab Initio), what should I study? I'm interested in ab initio computational chemistry. I'm trying to figure out what I should study. I know that I should study physical chemistry, mathematics, and programming.

However, I am not sure in what I should be focusing on. It seems to be a fairly new field and books on the topic are rather specific.

Can anyone working in computational chemistry research provide me with a list of (math, computational chemistry, physical chemistry) textbooks they studied or any additional resources to get me started?

Computational chemistry is not a new field, but yes it can be difficult to find the right approach to it, because it is so diversified.  And that means to some degree what you need depends strongly on what you want to study.  Most of the time, you're basically solving the Schrodinger equation all kinds of different ways.

It will be better if you can specify what level of knowledge you do have at this time. Assume that you've just got into college, then there's a list of things you should study:

Basic theory

Math.

You need linear algebra and mathematical analysis.  Any good old textbook will do.

What you use the most everyday would be linear algebra, so you better be very very good at it, along with a good expertise in vectors, matrices and in some cases tensors.  There may also be some functional analysis involved but for most people it really just boil down to linear algebra.

Mathematical analysis is basically just differentiation, integration, and all kinds of tricks related to it.

Quantum mechanics.

You some of it from P-Chem II, but if you take it from the physics department it will be more helpful.  I recommend the Cohen-Tannoudji textbook.  Make sure you start from the beginning, go through every chapter to the very end.

Quantum chemistry.

With a background you can now proceed to understand the conventional methods of quantum chemistry.  For this I would recommend Modern Quantum Chemistry by Szabo and Ostlund.

Courses to take:

In addition to what was already mentioned above, many universities will also have quantum chemistry and computational chemistry courses. I also recommend you to take group theory and calculus of variation.

Programming skills

And that prepares you for the theoretical part.  Whether you need comprehensive programming skill depends on what you want to do.  Some may make do with existing programs, some may need to write there own.  But even if you use an existing program, it requires no less amount of understanding of the quantum chemical theories you're using.

For programming skills, you will certainly benefit by taking numerical methods course, and mastery over one programming language. Take statistics and machine learning if you have time. Any one will do, it doesn't matter. For what we do, its straight forward to switch to a different language whenever you want.  Computational chemists are not usually very big on the algorithms themselves. Most of the times, you just take something right out of a numerical analysis text book or program package and adapt a little bit.  It is much more about the underlying physics, i.e., what form of wave function you choose, what approximation you make, etc.

The most important thing about programming, I feel, is hands on experience.  If you can understand quantum mechanics, programming is not gonna be hard.

Chemistry

As for other branches of chemistry, you don't have to be the expert, but you definitely need a good understanding because those are the problems you're trying to solve.  You need to understand very well the experimental procedure you're trying to study.


What does reaction exothermicity say about the stability of molecules If a reaction is exothermic it means that the energy state of the products is lower than that of the reactants so this will be the state that 'nature' will naturally strive for. Often you see that stability of molecules is argued in this way (e.g. in one of the answers here), saying that the higher the exothermicity of the decomposition reaction, the more unstable the molecule and thus the more unlikely that it will exist in nature.

My question is whether this reasoning is correct?

The reason I am doubting this, is that the reaction enthalpy $\Delta H_r$ is not the only parameter of importance in the energetics of chemical reactions: the activation energy $\Delta E_a$ can also play a major role. To clarify my point, take a look at the diagram for 2 hypothetical reactions I have made. Reaction 1 has a low activation energy and a low exothermicity, whereas 2 has a high activation energy and a high exothermicity.

Judging solely by exothermicity you would conclude that the reactants of reaction 2 are more unstable, but if you look at the activation energies for the reactions than I would argue that the reactants of reaction 1 are more unstable because they more easily react to form their products. Of course having a high activation energy would at best make the reactants meta-stable, but metastability could last for decades as well given the right conditions.

So my general question is: can reaction exothermicity be used to say something about the stability of molecules and what is the role of activation energy in this story?

Good question. There exists two types of stability - thermodynamic stability (determined by $\Delta G_{reaction}$, or $\Delta H_{reaction}$ approximately) and kinetic stability (determined by $\Delta G ^{\ddagger}$, or $E_{activation}$ approximately). The reactant compound in your first reaction is thermodynamically more stable than the second, since it sits at a lower potential energy, but it is kinetically less stable, since there is a smaller barrier for decomposition.

Usually context is sufficient to determine what kind of stability is being talked about, though in general I think stability is more often mentioned in the thermodynamic sense than the kinetic, since thermodynamic stability is related to the reaction free energy (a sum of its enthalpy change and a factor containing its entropy change) and is more commonly tabulated and more readily measured than activation free energies.

For example, all explosives are quite thermodynamically unstable (otherwise they wouldn't be very good at exploding!), but we still speak of "stable explosives", such as TNT or RDX and "unstable explosives", such as $\ce{NI_3}$ or $\ce{Hg(CNO)_2}$. Clearly kinetic stability is what is taken into account.

In constrast, when studying acidity and basicity of substances, it's common to justify reactions based on the "stability" of the acids/bases/conjugate species, but in this case, we're talking about thermodynamic stability, as it defines the acidity/basicity constants.


What is palindromic DNA? What is palindromic DNA and why isn't every complementary strand palindromic?

I know AGCT is palindromic, but what is an example of a strand that isn't?

A palindromic stretch of DNA is a strand whose reverse complement is itself.  So 5'-AAAT-3' is not palindromic.  It's reverse complement is 5'-ATTT-3'.  Those two pieces of DNA are not identical.  However, 5'-GGATCC-3' is palindromic, because the reverse complement is identical.


Why is a trialkylamine a better leaving group, in nucleophilic reactions, than a Cl group? Why is a trialkylamine a better leaving group, in nucleophilic reactions, than a $\ce{Cl}$ group? In the same vein, isn't a trialkylammonium a much weaker acid than $\ce{HCl}$?

$\ce{R-NR3+ -> Nuc-R + NR3}$ (Stronger leaving group)

$\ce{R-Cl  -> Nuc-R + Cl-}$    (Weaker Leaving group)

Compared to:

$\ce{H-NR3+ -> Nuc-H + NR3}$    (Weaker acid, NR3 thus weaker leaving group)

$\ce{H-Cl -> Nuc-H  + Cl-}$     (Stronger acid, Cl- thus stronger leaving group)

Leaving group ability refers to how well a leaving group can stay off whatever it came from.

Consider the following prototypical nucleophilic displacement reaction:

$\ce{R-LG + Nuc- <=> R-Nuc + LG-}$

There are two issues at play here:

1) Thermodynamics. How stable is the reactant with the leaving group, $\ce{R-LG}$, and the nucleophile, $\ce{Nuc-}$? How stable are the products, $\ce{R-Nuc}$ and $\ce{LG-}$? And when examining stability, we must look at not only the reactants and products but also their environments - i.e. the solvent(s).

2) Kinetics. What's the physical ability of the nucleophile to actually displace the leaving group? What's the physical ability of the leaving group to displace the nucleophile again?

Usually we're taught that leaving group ability has to do with acidity and basicity. The less basic the leaving group is, the better its ability to leave. This isn't a perfect rule of thumb, but it's acceptable.

That's the thermodynamic part of the question. If $\ce{LG-}$ weren't thermodynamically stable, then it would be inclined to attack $\ce{R-Nuc}$ and return to the status quo. (Or just grab a proton).

However, how capable is $\ce{LG-}$ of carrying out such an attack? Here we come to the kinetic part of the question. The bulkier the leaving group, the more likely it is to be sterically hindered and/or strongly solvated (trapped by solvent molecules). Trialkylamine  is a very genetic term, but when I think of amines with alkyl groups I think of quaternary ammonium compounds, which have alkyl groups that can contain upwards of 18 carbons (!). Such bulk affects the ability of the leaving group to line up properly with the substrate to perform nucleophilic attack. Nucleophilic attack often has specific steric demands; $\ce{SN2}$ requires backside attack; the approach of attacking nucleophiles with respect to flat carbonyl groups has an angle of 104.5 degrees, etc.

Here is a good summary of sterics and amine nucleophilicity.

https://books.google.com/books?id=tP4cqrYF7nwC&pg=PA231&dq=amine+nucleophilicity+steric&hl=en&sa=X&ei=nXOhVLiOD8iKgwSS5YPICg&ved=0CEUQ6AEwBA#v=onepage&q=amine%20nucleophilicity%20steric&f=false


Which orbital lobes are occupied? In the p-orbitals, there are 3 dumbbell-shaped lobes, each of which can contain 2 electrons. In the diagrams of the d- and f-orbitals, there are two or more lobes (or orbital shapes) or each value of m.  In all of these cases, which lobes/shapes do the electrons occupy for any given element?

The question implies a misunderstanding about orbitals and their representation as dumbbell-shaped lobes. These plots show the relative orientation of orbitals to each other by showing regions that represent some threshold probability (say, 90%) of finding an electron contained therein. The underlying probability distribution has the same symmetry as the orbital itself because it defines the orbital. Any (spatial) orbital can hold up to two electrons$^1$, regardless of its shape. It thus makes little sense to ask in which lobe (or, further, which half of the dumbbell, since there is a nodal plane in between the halves) an electron is.

Moreover, any single atom is spherically symmetric in the absence of external fields, meaning that the shapes of the orbitals are not actually important to understand electron distribution in a free atom. The shapes are useful tools to understand certain bonding concepts such as crystal/ligand field theory nevertheless.

$^1$ a spin orbital holds one.


Why ground-state configuration is consistent ("additive") from one element to the next? Ground-state configuration ("electron shells") is consistent throughout the periodic table, for example:

(source)

Why is it this way if the Hamiltonian of each atom is different? In other words, why Al is $\mathrm{[Ne]3s^2 3p^1}$ and not something completely different: (1) at the core, and (2) at the valence.

Well, this configurations arise in what is usually called the orbital approximation and the consistent pattern is due to the very nature of this approximation.

In the orbital approximation the overall electronic wave function describing the atom is represented as the product of the wave functions for the individual electrons known as atomic orbitals. This can be sketched as follows.
$$
\newcommand{\op}[1]{\hat{#1}}
\newcommand{\el}[1][]{_{\mathrm{e}}}
\newcommand{\nuc}[1][]{_{\mathrm{n}}}
\newcommand{\elel}[1][]{_{\mathrm{ee}}}
\newcommand{\elnuc}[1][]{_{\mathrm{en}}}
$$

Electronic Schrödinger Equation

The electronic wave function $\psi_{\mathrm{e}}$ is an eigenfunction of the electronic Hamiltonian $\op{H}\el$
$$
\op{H}\el \psi\el = E\el \psi \el \, ,
$$
$$
\op{H}\el
=
\op{T}\el + \op{V}\elnuc + \op{V}\elel
=
- \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
- \sum\limits_{i=1}^{n} \frac{Z}{r_{i}}
+ \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
which indeed is different for different atoms. Thus, the electronic wave function can also be expected to be different. However, the exact analytical solutions of the electronic Schrödinger equation are known only for one and two-electron systems, such as $\ce{H}$, $\ce{H-}$, $\ce{He}$, $\ce{Li+}$, etc., while for all other chemical systems only approximate solutions can be found.

Separation of variables

The electronic Schrödinger equation is partial differential equation and one of the most frequently used analytical methods to solve such equations is technique known as the separation of variables aimed to broke one compound partial differential equations of many variables into a set of simpler partial differential equations of fewer variables. The main disadvantage of the separation of variables is that it is by no means a universal approach: for some equations a solution of the form mentioned above make the separation of variables possible, for others it is of no use.

The most common way of separation of variables for partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is writing a solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only
$$
    f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) \, ,
$$
and then substituting solution of this form into the equation.
Using algebraic manipulation one could try to break the resulting equation into a set of $n$ independent ordinary differential equations for each function $f_{i}(x_{i})$ and if succeed then one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into the product obtain the solution of the starting equation.

Orbital approximation

So, it is tempting to try to separate the electronic coordinates by representing $\psi\el$ as a product of one-electron wave functions, however, it is fairly well known that this attempt will be successful only if the potential energy can be expressed as the sum of the separate potentials for each particle. This clearly is not the case for the electronic Hamiltonian: the potential energy of the Coulomb repulsions between the electrons $\op{V}\elel$ prevents the separation of electronic coordinates
\begin{equation*}
    \psi\el(\vec{r}_{1}, \vec{r}_{2}, \dotsc, \vec{r}_{n}) \neq \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n})\, .
\end{equation*}
So, exact separation in this form is not possible, but what about approximate one? Can we somehow approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? To some extent, the answer is, yes, we can.

In the orbital approximation it is indeed assumed that the many-electron wave function can be written as a product of one-electron wave functions called orbitals. As it was already said, it will work only if the exact $\op{V}\elel$ potential is replaced by the model potential $\op{V}_{\mathrm{MF}} = \sum\nolimits_{i=1}^{n} \op{v}_{\mathrm{MF}}(r_{i})$ which allows the separation. The approximate Hamiltonian is
$$
\op{H}_{\mathrm{MF}}
=
\op{T}\el + \op{V}\elnuc + \op{V}_{\mathrm{MF}}
=
- \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
- \sum\limits_{i=1}^{n} \frac{Z}{r_{i}}
+ \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(r_{i}) \, .
$$
Physically it means that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. Hence, another name of the approximation, the mean field approximation.

Even the mean-field Hamiltonian still looks different for different atoms, however, it is now a sum of one-electron Hamiltonian operators
$$
\op{H}_{\mathrm{MF}}
=
\sum\limits_{i=1}^{n}\op{h}(r_{i}) \, ,
$$
where
$$
    \op{h}(r_{i})
    =
    - \frac{1}{2} \nabla_{i}^2
    - \frac{Z}{ r_{i} }
    + \op{v}_{\mathrm{MF}}(r_{i}) \, .
$$
And in contrast to an eigenfunction $\psi\el$ of the true electronic Hamiltonian $\op{H}\el$ an eigenfunction $\psi_{\mathrm{MF}}$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ can be written as products of one-electron wave functions (orbitals)
$$
    \psi_{\mathrm{MF}} = \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) \, ,
$$
where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equations
$$
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .
$$
Now this resulting equations that defines orbitals are not that different: there will be just a small difference in one-electron Hamiltonians due to slightly  different nuclear charges as you move along a group. Of course, there will be more and more occupied orbitals as you proceed through the group, but equations themselves and consequently orbitals will be quite simillar.


Chlorofluorocarbons rising Chlorofluorocarbons are heavier than atmospheric gases. Then how are chlorofluorocarbons produced at the surface able to reach the ozone with such a heavy mass ?

Segregation in the lower atmospheric region is prevented by various mechanisms, mainly turbulent mixing of the atmosphere by rising and falling air streams. The main driving force of the large-scale atmospheric circulation is solar heating. The most important circular air flow pattern in the troposphere is the Hadley cell circulation. (Considerable information on the movement and mixing processes in the atmosphere was gained from monitoring following numerous tests of nuclear weapons in the atmosphere mainly during 1951–1962. The radioactive substances served as tracer material.)

Winds mix the atmosphere much faster than gaseous molecules can settle according to their weight. The natural composition of the dry atmosphere remains constant in the lower 80–100 km. Only above 100 km, segregation takes place because the concentration of light gases (e.g. helium) decreases more slowly with altitude than those of the heavy gases.

The same atmospheric mixing processes apply to chlorofluorocarbons. For example, trichlorofluoromethane $(\ce{CCl3F})$ is nearly uniformly distributed throughout the lower 15 km. At higher altititudes, the abundance of trichlorofluoromethane decreases – not because of its density, but because it is decomposed by UV radiation. By way of comparison, the less reactive tetrafluoromethane $(\ce{CF4})$ is uniformly distributed up to at least 50 km.
(Source: Frequently Asked Questions About Ozone (1): How Can Chlorofluorocarbons (CFCs) Get to the Stratosphere If They’re Heavier than Air? in Assessment of Ozone Depletion: 1998 – World Meteorological Organization Global Ozone Research and Monitoring Project – Report No. 44 [html] [pdf])


Modelling pH as function of time and position I already posted this on the Engineering Stack Exchange under Chemical Engineering as I don't know where it's a better fit.
I have a system whose initial state consists of three different sections with different chemical composition. All sections are aqueous solutions of different $\mathrm{pH}$ that are brought into contact for $t \gt 0$. The system is isothermal in each section and is of equal temperature in all sections and is assumed to remain isothermal, as heat generation is not considered. Section 1 (the leftmost section) is basic and contains reagent 1, section 2, the middle section, contains just water at a $\textrm{pH}$ of 7 and section 3, the rightmost section, is acidic. Section 1 and 3 are of equal size, whereas section 2 is larger. The system is three-dimensional, but can be reduced to one dimension due to symmetry (it is a tube in which angular and radial coordinates are assumed to not play a role). All sections can be considered equal in terms of physical properties (viscosity, density, etc.). I would like to model the $\mathrm{pH}$ as a function of time and position as there are reactions occurring whose reaction rate is a function of $\mathrm{pH}$. There is no convection, only diffusion. I already tried setting up a PDE system using
$$
\frac{\partial c_i}{\partial t}= D\nabla^2 c_i + r_i(\mathrm{pH},T)
$$
and solving it in Matlab, while setting a very high rate constant for the recombination of $\ce{H+}$ and $\ce{OH-}$. The Matlab solver fails as the recombination kinetics are too fast, but lowering them is not an option, as this wouldn't represent reality anymore (acid-base reactions are thought of the be instantaneous). How could I rewrite my PDE system to model the evolution of the $\mathrm{pH}$ at every position as well as calculate the concentrations of products and educts of reactions that depend on $\mathrm{pH}$ (but are only catalysed by it), without using reaction kinetics for the recombination of $\ce{H+}$ and $\ce{OH-}$ to make use of the fact that this reaction is instantaneous?

To answer my own question:
Using either the diffusion limited or the literature values of the acid-base neutralisation (Baldyga et al., Non-isothermal Micromixing in Turbulent Liquids: Theory and Experiment), I was able to simulate the $\mathrm{pH}$ as a function of time by making the simplifying assumption of equal diffusion coefficients of proton and hydroxide ions. I re-expressed the proton concentration in terms of $\mathrm{pH}$:
$$
\frac{\textrm{d} c_{\textrm{H$^+$}}}{\textrm{d} \mathrm{pH}} = -\ln (10) \cdot 10^{-\mathrm{pH}}
$$
plugged this into my original equation and made everything dimensionless.


What is the proper name for the but-1-ene's isomer methylpropene? Give the name and draw the graphical formula of an alkene that is an isomer of but-1-ene and that has a different carbon skeleton.

I think it's 2-methylpropene, but I'm not sure whether the correct name is 2-methylprop-1-ene or 2-methylpropene.

Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information. Nevertheless, the practice of omitting locants when there is no ambiguity is widespread in general nomenclature. For preferred IUPAC names, however, locants are omitted only in a few exceptional cases described in Subsection P-14.3.4 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).
Ignoring the rules for omission of locants, the complete systematic name of the parent compound would be ‘prop-1-ene’. According to Rule P-14.3.4.2, however, the locant ‘1’ is omitted. (Note that no other isomer can be generated by moving the double bond from its position to another.)

P-14.3.4.2 The locant ‘1’ is omitted:
(…)
(d) in unsubstituted dinuclear and trinuclear alkenes and alkynes and monounsaturated cycloalkenes and cycloalkynes; (…)

Thus, the preferred IUPAC name for the unsubstituted parent structure is propene.
Ignoring cis/trans isomerism, the substitutable hydrogen atoms of propene have three different locants (‘1’, ‘2’, and ‘3’). Therefore, the locants for substituent prefixes (e.g. ‘1-chloro’, ‘2-chloro’, or ‘3-chloro’) cannot be omitted in substituted propene.
Since, the locants for the substituent prefixes cannot be omitted, also the locant ‘1’ of ‘prop-1-ene’ can no longer be omitted for preferred IUPAC names.

P-14.3.3 Citation of locants
In preferred IUPAC names, if any locants are essential for defining the structure of the
parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. (…)

For example, (1E)-1-chloroprop-1-ene, (1Z)-1-chloroprop-1-ene, 2-chloroprop-1-ene, and 3-chloroprop-1-ene are the preferred IUPAC names for some substituteded propenes.
This principle applies to any substituent. The special situation that replacing the chloro substituent in (1E)-1-chloroprop-1-ene, (1Z)-1-chloroprop-1-ene, and 3-chloroprop-1-ene with a methyl substituent would actually lead to the names (2E)-but-2-ene, (2Z)-but-2-ene, and but-1-ene instead of ‘(1E)-1-methylprop-1-ene’, ‘(1Z)-1-methylprop-1-ene’, and ‘3-methylprop-1-ene’, respectively, does not change the preferred name for 2-methylprop-1-ene. (Note that the substitutable hydrogen atoms of propene have three different locants irrespective of the kind of the substituent.)
Therefore, the preferred IUPAC name is 2-methylprop-1-ene.


How do I evaluate the PV term when calculating a change in enthalpy? As we know enthalpy is defined by the equation $H = E+PV$.

Now to calculate this $PV$ factor for $\Delta H$, Sometime we take $P \Delta V$ and sometime $P_2V_2-P_1V_1$. Why are there so many different approaches?

I am also confused whether $P$ is external pressure or internal pressure.

The derivation for $\Delta H$ is
$$H = U + PV$$
$$\Delta H = \Delta U + \Delta (PV)$$

The $\Delta (PV)$ term expands to your $P_2 V_2 - P_1 V_1$ term. However, when the pressure is constant $P_2 = P_1 = P$, and $\Delta (PV)$ term becomes $P\Delta V$. The most common situation when the latter is true is when the system is open to the atmosphere and $P$ is the external pressure. In the former case, $P$ is more likely the internal pressure, which changes during the reaction.


Why is the notation for reactions not explicit? I hope that the title of the question is not completely wrong or misleading, however I couldn't think up of a better one.

I have a follow up question to this one. If we look at the following reaction, $$\ce{C + O2 -> CO2}$$

we are taught that gases like oxygen (looking at the reactants), when not part of a compound, always appear in molecular form.

But if to follow one of the answers in the referred question, the reaction has a little more to it. I quote:

  carbon, as normally found in nature as graphite, consists of a molecular solid with every carbon bonded to several other carbons. The reaction C + O2 is shorthand for solid carbon plus oxygen gas, so the reaction involves breaking lots of C-C bonds and O=O bonds resulting in CO2. You have to account for all the bonds broken and formed to know the energy difference.

Why is this fact not accounted for in the reaction $\ce{C + O2 -> CO2}$, in other words, why do we 'allow' carbon to be a free atom (although it's not, in the form of graphite) and 'demand' that the oxygen is in a molecular form? I do assume that chemical reactions are a form of abstraction, that is, we say that regardless of where that carbon atom comes from, it will react with the oxygen molecule into a carbon dioxide, but I still don't understand why we then apply those molecule constraints to oxygen.

To be more explicit about it: the energy spent to break the C-C bonds in the graphite crystal is not easily readable from the reaction, while the fact that the O=O bonds are split, and therefore energy is input into the system, is.

The reaction, described in the form given in the original question, i.e.

$$\ce{C + O2 -> CO2}$$

simply conveys that each formula unit of carbon reacts with one formula unit of oxygen to yield one formula unit of carbon dioxyde. As such, for example, useful to balance how many grams of carbon may be burnt with $\pu{22.7 L}$ of oxygen.  As such, yes indeed, it is a limited representation.  In an analogy, it is like book keeping; (within reason) is does not matter how you paid at a garage (cash, debit or credit) to get your car back -- but to satisfy both sides, it has to be balanced.

This view contrasts with many organic reaction equations as reaction mechanisms which are intentionally drawn with further detail like this:

(source)

Here, the how is important, and consequently outlined with more attention.  As if you would need to explain an apprentice how the teeth of a gear have to be adjusted in face, pitch, flank, etc. to transmit the force from the engine to the wheels.


How do you determine the amount of acrylamide in foods such as baked fries? There are many food studies which determine the concentration of a substance in a food item.

For example, this study determines the concentration of acrylamide in solid foods like fries.

By what experimental technique is this done?

Check out the methodology page for the FDA section on acrylamide quantitation, which goes into considerable detail.

In short, they start by mashing up some fries, and then extract and purify a fraction containing the acrylamide through a number of steps leading up to liquid chromatography. They then use quantitative mass spectrometry to determine the amount of acrylamide present.

They add a small known concentration of isotopically labelled acrylamide (i.e. acrylamide with all the carbons as carbon-13, which has very low natural abundance and results in a molecule that is approximately 3 g/mol heavier). This shows up in the mass spectrum as an offset peak. Then they can integrate the normal acrylamide mass spectrum peak and the labelled peak and obtain the concentration from the relative peak areas.


Thermodynamics behind Deliquescence? The phenomenon of deliquescence wherein a water soluble substance absorbs water from atmosphere and dissolve in it forming a saturated solution, has thermodynamic roots. I have read the fact that the vapour pressure of the saturated solution should be lower than the the vapour pressure of water at that temperature. Why is this so?

Moreover, since deliquescence is a spontaneous process and therefore thermodynamically favourable, there must be a negative $\Delta G$ for the absorption of water vapour and conversion of solid substance into its saturated solution? How can we derive, understand or quantify various thermodynamic parameters ($\Delta G, \Delta S, \Delta H$ etc) for the process of deliquescence?
I am comfortable with basic thermodynamic manipulations, binary phase diagrams, multiphase systems and equilibria.

The vapour pressure lowering is a consequence of Raoult's law with a non-volatile solute, i.e. a solvent that has zero vapour pressure (or at least, lower vapour pressure than the solvent).

Regarding the second part (“How can we derive, understand or quantify various thermodynamic parameters (ΔG,ΔS,ΔH etc) for the process of deliquescence?”), I am not sure exactly what you are looking for. This is a complex phenomenon, and there is a lot of research literature of the topic. To give only one example, see

  “Water-Solids Interactions: Deliquescence”, Lisa J. Mauer and Lynne S. Taylor, Ann. Rev. of Food Sci. Technol., 2010, 1, 41-63.


Why does soda explode if artificial sweetener or other powders are added? I opened a can of soda and spilled some generic artificial sweetener into the soda and it exploded. Why did this happen? I am not talking about mentos, I mean artificial sweeteners like Splenda.

Artificial sweetener powder causes nucleation of the carbonation in the soda/pop/soda-pop. The carbon dioxide concentration in the soda is higher than the equilibrium concentration for the partial pressure of $\ce{CO2}$ in the atmosphere thus over time the soda will off-gas $\ce{CO2}$ and if allowed to sit long enough will become "flat".

The off-gassing is spontaneous but not instantaneous, thus it is subject to kinetic rates. To form a bubble, enough $\ce{CO2}$ molecules need to coalesce at one time to form a large enough bubble to be stable and not redissolved in the soda (nucleating).

When a rough surface is introduced, the number of $\ce{CO2}$ molecules that must come together to form a stable bubble is far less that the pure drink, so the off-gassing occurs much faster causing the observed "explosion".

You can observe this when certain glasses (ussually beer glasses) etch a small ring in the glass, creating a rough surface and the formation of bubbles at the ring.  you can also observe this if you put a larger rough object in the soda. This could be a rock, a piece of wood (like a tooth pick), or even a Mentos.

To solve this you might be able to dissolve it in a small amount of water first. This may not completely solve issue though, as most sweeteners use silicon oxide or titanium oxide powder to prevent clumping, neither of which are soluble and will still nucleate bubbles.