text:
Why large basis sets give better approximations to the exact solution of the Schrödinger equation?
The variational principle states that the energy of any approximate wave function will always be equal to or greater than the energy of the exact solution. Therefore the energy is minimized when performing approximations, in order to get as close as possible to the exact solution. Larger basis sets give the wave equation "more flexibility", but I am not sure I understand this fully. Here's a statistics analogy:
When performing multivariate regression analysis, we choose the regression equation of lowest order; "overfitting" the data is something to be avoided. In my head, choosing large basis sets for simple problems seems like "overfitting".
So why is it that a larger basis set leads to a better approximation? The statistics analogy may not be applicable, but it illustrates what goes on in my head.

With regression analysis, you have certain number of original data points which you can't change. In quantum chemistry there is no such limitation. Take a basis of any size, calculate all exchange integrals, and you are good to go. (It may be prohibitively expensive, but that's another story.)

The statistics analogy may not be applicable, but it illustrates what
  goes on in my head.

The overfitting analogy is indeed not applicable here. It is a totally different and unrelated phenomenon. Basically, when we are doing regression analysis we do not know the relationship between a dependent variable and independent variables. Then we can indeed have an overfitting problem when instead of fitting the actual relationship we start to describe some random noise. It generally happens when we do have too many parameters in a model, and we never know how many is too many.
But when we are doing LCAO-MO we know the relationship between a dependent variable (say, energy) and independent variables (LCAO-MO fficients). ANd we are certain that the bigger the number of AOs (and correspondingly of the coefficients), the better result could be expected with the the exact solution corresponding to using infinitely many AOs. So we should not worry about anything like overfitting. Yes, by increasing the number of AOs in a basis set we increase the number of variational parameters, but we know it can't do any harm: in the worst case1 we will get the same energy as before the just another expansion of the basis.
This is similar to a Taylor series expansion of a function. The exact result corresponds to an infinite series, and by using a finite number of terms a function can only be approximated. But the more terms we use, the better approximation is. There is even a formal proof, but I think it should be intuitively clear that if an infinite series is the exact solution, than the later you truncate it the better approximation you get.

1) The worst case would be, for instance, when you expand a basis by including new functions that can be expressed as linear combinations of basis functions from the previously used basis set. You won't improve energy by doing so, but only increase the computational time. Note, that most of the programs have a simple defense mechanism against such situations: they are smart enough to detect the linear dependence in the basis and get rid of it prior to doing the calculations.

The Taylor series expansion analogy was really helpful!

In this particular case Taylor expansion is not quite a good analogy, since it has very non-uniform error and generally behaves differently (has a radius of convergence etc.). Conceptually more similar would be Fourier series expansion. It may converge (though not at every point) even when its coefficients are unbounded — see e.g. derivative of Dirac delta.

We have the variational principle, so as long as we are not making floating point / finite precision arithmetic errors, we are golden. Even if you are given a pathologically linearly dependent basis set, you can always find a subset of functions that is complete. In this paper https://journals.aps.org/pra/abstract/10.1103/PhysRevA.101.032504 I did calculations in large Gaussian basis sets (UGBS) at sub-Å internuclear distances.

Perhaps we start with your picture first. It is indeed not applicable.
Just imagine a vector $\vec{\psi}$ in a finite $n$-dimensional space and a set of other vectors $\vec{x}_i$. Your goal is to find the linear combination of $\vec{x}_i$ that is best describing your $\vec{\psi}$:
$$
\vec{\psi} \approx \sum_i a_i \vec{x}_i
$$
If you just think of it in the usual 3D euclidian vector space with arrows... it is intuitively obvious that using more basis vectors $\vec{x}_i$ gives a better approximation. In the case of finite $n$-dimensional spaces a set of $n$ linearly independent vectors gives even the exact linear combination.
The beautiful thing about this picture is that the main task for ab initio programs in theoretical chemistry is searching for the best linear combination.
The only problem in praxis is that we have vector spaces with infinite dimensions and will never find a set that gives the exact linear combination. If we could, this would solve the Schrödinger equation.
If you look further on in text books, you are looking for the Hylleras-Undheim Theorem.
Please note that in the strict formulation this theorem is under the assumption of linear variational methods. In praxis it is even wider applicable.


text:
A "Bible" for molecular orbital theory?
Linus Pauling wrote the famous "Nature of the Chemical Bond" which is a Bible of sorts for valence bond theory.
However, VB theory was combined with crystal field theory to form molecular orbital theory, which eventually was condensed into the Hartree–Fock method.
Are there any "Bible" type reference books for these other methods?

F. A. Cotton's book "Chemical Applications of Group Theory" was the bible for crystal field theory in my college days.

Ian Fleming's "Molecular Orbitals and Organic Chemical Reactions", two editions are published: a student version, and a longer, more detailed reference version. Both are excellent texts for understanding MO theory in Organic systems and reactions.

@Dale - Yes there are forever "better" more rigorous books being published. Wikipedia in the early days used to be good for looking a basic explanation up. Now the articles have gotten so "rigorous"  in terms of being absolutely right and using all the right terms that many are nearly impossible to understand. It almost seems like three levels are needed - a dummies level, an intermediate level, and a research level.

@MaxW I have had that very thought on multiple occasions.

@MaxW There does seem to be a "Simple" Wikipedia: simple.wikipedia.org contains simplified versions of regular Wikipedia articles.

I think an excellent book for understanding molecular orbital theory from a qualitative basis is Orbital Interactions in Chemistry by Albright, Burdett, and Whangbo. It covers a wide range of topics from molecules to solids.
The book doesn't go into particular methods, but instead illustrates how various MO treatments can understand a great deal of chemistry.

I also agree with the above comment that Fleming's book is very useful, but Albright, Burdett and Whangbo cover far more ground, particularly solids and inorganic/organometallics.

I absolutely agree with you. For me it is the best book about this topic I have encountered yet - an absolute masterpiece.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
I am trying to find the solubility of benzoic acid in different solvents to prepare a liquid-liquid extraction
I need to prepare a liquid-liquid extraction using the partition coefficient (K) of benzoic acid in different solvents and water:
K ~ solubility of benzoic acid in solvent/solubility of benzoic acid in water

Several solvents are offered to me (e.g. ethyl acetate, chloroform,...), and I am trying to find the solubility of benzoic acid in each of those solvents. The problem is that I can't find all of them (e.g. benzoic acid in ethyl acetate). I can't seem to find a database of such solubilities. I would greatly appreciate if someone could please re-direct me to such database. Conversely, can solubility be predicted efficiently?
Thank you very much for your time!!

Funny you're looking at solubility of benzoic acid. A few years ago, there was an Open Notebook Science challenge to measure solubility:

The first round of this challenge calls upon people with access to materials and equipment to measure the solubility of compounds (aldehydes, amines and carboxylic acids are a priority) in organic solvents and report their findings using Open Notebook Science

So the raw data is available through Google Sheets here
There is a general web service for predicting solubility.
For example, you can predict the solubility of benzoic acid using Abraham solvation parameters
There used to be a search mechanism, but I'll have to ask the authors - it's been a few years.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How to decide between different possible substitution/elimination mechanisms under given conditions?

I'm trying to figure out the major product. At the moment I'm leaning towards an $\mathrm{S_N2}$ reaction, because there's a polar aprotic solvent, and sort of unhindered electrophile. However to me it could just as easily be an $\mathrm{S_N1}$ reaction (electrophile does have some hinderence and there's a weak nucleophile and weak base). $\mathrm{S_N1}$ would attack the tertiary carbocation and $\mathrm{S_N2}$ would go for the primary.
So how do I know if it's $\mathrm{S_N1}$, $\mathrm{S_N2}$, or even $\mathrm{E1}$/$\mathrm{E2}$?
I guess the main question is I don't know if the electrophile is hindered "enough" so we can say that an sn2 reaction is definitely not happening.

E2 requires strong base, SN2 requires good nucleophile. Water is a lousy base and a lousy nucleophile. Even if you have a polar aprotic solvent you probably wouldn't get any SN2 product imo

Generally, your substance can react in one of the following four ways that you have already identified:

$\mathrm{S_N1}$ solvolysis on the tertiary chlorine resulting in 4-chloro-2-methylbutan-2-ol;[1]
$\mathrm{S_N2}$ solvolysis on the primary chlorine resulting in 3-chloro-3-methylbutan-1-ol;
$\mathrm{E1}$ elimination from the tertiary chlorine resulting in 4-chloro-2-methylbut-1-ene or 1-chloro-3-methylbut-2-ene;[1]
$\mathrm{E2}$ elimination from the primary chlorine resulting in 3-chloro-3-methylbut-1-ene.

It is highly unlikely that one of the $\mathrm{X1}$ mechanisms happen on the primary chlorine due to the very badly stabilised intermediate primary carbocation we would need, highly unlikely for an $\mathrm{S_N2}$ reaction to proceed on the tertiary chlorine due to steric hindrance (especially since a primary one is available) and moderately unlikely (but not impossible) for $\mathrm{E2}$ to happen on the tertiary chlorine.
We thus need to consider:

How good is the leaving group (influences everything strongly except for $\mathrm{E2}$)?[2]
How strong is a base possibly present in solution (influences which elimination reaction can happen and how well)?[3]
How good is our nucleophile (influences the probability of $\mathrm{S_N2}$ most)?[4]

Starting at the base. This is a case of rule 6:[6] There is no base present in the mixture! (Yes, technically water can act as a base, but you need a strong acid for it to happen. Water’s $\mathrm{p}K_\mathrm{b}$ is $15.7$ if I recall correctly.) This definitely rules out $\mathrm{E2}$ which requires a strong base, and mildly rules out $\mathrm{E1}$ which requires a weak base.
Let’s check out potential nucleophiles next. Well, one of our nucleophiles is water, also known as very lousy, and another possible one is acetone, also known as even lousier. There is hardly any chance that either of them will nucleophilicly attack the primary chlorine site ruling out $\mathrm{S_N2}$.
Next, what about our leaving group? Well, we have chlorine which is also a pretty lousy leaving group. It means that $\mathrm{S_N2}$ will be discouraged even more and the generated chloride will probably hang around close enough to the carbocation required for the $\mathrm{X1}$ reaction paths to disallow $\mathrm{E1}$.
Finally, there are solvent effects, too. Here, the solvent is polar protic; a combination generally preferring the $\mathrm{X1}$ pathways.[5]
All things considered, this leaves $\mathrm{S_N1}$ as the only mildly probable pathway remaining. And thus I would strongly assume the main product to be 4-chloro-2-methylbutan-2-ol.

Of course, one should always include reaction times and temperatures (and yields) to any reaction-predicting exercise. Raising the temperature should increase the likelyhood of $\mathrm{E1}$ over $\mathrm{S_N1}$ because of the entropic factor (two particles become three in $\mathrm{E1}$ resulting in a rise in entropy).[7] However, in exercises, this is usually indicated by the presence or absence of a Δ above or below the reaction arrow. Since we have no Δ, we can assume room temperature.

[1] C. A. Grob, A. Waldner, Helv. Chim. Acta 1979, 62, 1854.
[2] https://en.wikipedia.org/wiki/SN2_reaction#Leaving_group
[3] https://en.wikipedia.org/wiki/Elimination_reaction#E2_mechanism
[4] https://en.wikipedia.org/wiki/SN2_reaction#Nucleophile
[5] https://en.wikipedia.org/wiki/SN1_reaction#Solvent_effects
[6] Rule 6: There is no rule six!
[7] https://en.wikipedia.org/wiki/Elimination_reaction#E2_and_E1_elimination_final_notes

I assumed that acetone was the solvent and H2O a reagent, but I agree, SN1 is most likely.

@orthocresol How does that change anything? Acetone is protic (albeit slightly).

if acetone is protic, then it certainly doesn't change anything. I always thought it was considered aprotic.

This is a poorly written exercise. There is no temperature and no reaction time. Students have to be taught that these 'dimensions' are essential to describe synthetic reactions. With heating, there shouldn’t be any substitution of the primary chloride (SN2) but you would see the SN1 solvolysis product i.e. the tertiary alcohol. The carbocation intermediate can also yield the E1 alkene products (the more substituted alkene being the dominant one). This is a reference you can check: Helvetica Chimica Acta, 62(6), 1854-65; 1979. I only have access to an abstract where dioxane is the solvent in the reaction schemes but they report that acetone was also used. Since by using acetone you would achieve lower temperatures than dioxane, it is likely that you will see less of the E1 alkene products.


text:
How does a reference electrode work in a three electrode system?
This is what I understand about three electrode electrochemistry (correct me if I am wrong):

The working electrode is where the reaction of interest happens, e.g. copper ions reduced to solid copper.
The auxiliary/counter electrode supplies the current to the working electrode, i.e. it is the source of the electrons (maybe not in all cases?) e.g. zinc metal is oxidised to zinc ions.
The reference electrode lets you control the potential difference so you don't apply too much driving force to the working electrode and end up reducing water etc. No current passes through it.

I don't understand how the reference electrode lets you control the potential and how it works if no current passes through it. How is it connected to the other electrodes?

The way a three electrode potentiostat works is to

read the potential difference between the Reference electrode and the Working electrode using a very small DC current(generally pA level). (Just like MaxW mentioned)
Compare this potential to the desired voltage level as input by the user or a function generator.
Adjust the voltage of the counter electrode until the difference between the desired voltage and the measured voltage is zero.

The main advantage of a three electrode experiment over a two electrode experiment is that the electrochemical voltage that your counter reaction has stays constant. Because you are not passing significant currents through your reference electrode, it's electrochemical potential stays constant. The counter reaction that is happening on the counter electrode may change throughout the experiment without changing the measurement/control of the electrochemical potential of interest.
(In response to a comment)
When a significant current passes through the reference electrode the activity ratio of the reactants and the products are prone to changing. This change in the activity ratio ends up changing the potential based on the Nernst equation:
$E = E^0 + 59.2mV/n * log(ActivityRatio)$
The best way to make a reference electrode is to design the electrode such that the activity ratio stays constant over a long period of time and can resist high current flow. Ag/AgCl is great because of this. Both silver and silver chloride are solids, so as long as there is any Ag and AgCl in the system the potential will stay constant. The electrolyte is kept saturated by keeping solids in the solution and that assures that the electrolyte concentration doesn't change as well.

why does passing current through the reference electrode change the potential?

I edited to add more info on that. I hope that clarifies things.

The reference electrode lets you control the potential difference so you don't apply too much driving force to the working electrode and end up reducing water etc. No current passes through it.

You have the basic idea but if absolutely no current passed through the reference electrode then you couldn't measure any voltage!
The gist is that you can't measure the "voltage" of a single electrode. You need a complete cell to get a voltage. The reference electrode is used with a very very low current to measure the voltage of the electrolytic cell formed with the working electrode.


text:
Predicting the spontaneity of non-isothermal reactions
We use free energy equations -- Helmholtz or Gibbs -- to predict whether or not a reaction is spontaneous. These equations depend on constant temperature.
This forum post describes a scenario where volumes of water at different temperatures are mixed. The $\Delta G$ appears to be positive. (The poster's textbook gives $\Delta G = 99.5 \, \mathrm{Cal}$, a result I've been able to duplicate.) This seems to show that $\Delta G$ doesn't predict spontaneity for non-isothermal reactions.
So what does?

For an adiabatic spontaneous process in a closed system (like the one in your example), according to the Clausius Inequality, the change in entropy is greater than zero.  If you have a closed system in which a chemical reaction can occur, if the system is adiabatic, the same criterion applies.
The criterion that, at constant temperature, if a reaction is spontaneous, $\Delta G^0$ in going from pure reactants to pure products at 1 bar will be negative is just a rule of thumb which, over the years, has confused students to no end.  If you start out with pure reactants and mix them, spontaneous reaction will always occur to some extent.  It's just that, if $\Delta G^0$ is positive, the reaction will tend to proceed  to a lesser extent when equilibrating, and if $\Delta G^0$ is negative, it will tend to proceed to a greater extent when equilibrating.  So, in this sense, every reaction is spontaneous.


text:
Oxidation state of ions in a unit cell
Given that $\ce{Al2O3}$ is face-centered cubic (cubic-closest packed) with respect to oxide anions, I am instructed to find the oxidation state of each aluminum ion "fragment" in the unit cell.

Let's suppose that the picture above is of $\ce{Al2O3}$ and not $\ce{NaCl}$. The larger oxide ions are green; the smaller aluminum cations are purple.
There are a total of 4 oxide ions in a unit cell, and a total of 3 aluminum ions in the unit cell.
Given that 4 oxide ions impart a charge of -8, and that a unit cell must be neutral, then each of the "three" aluminum ions must have an oxidation state of +8/3 in the unit cell, correct?
I cannot find any textbook discussing the oxidation states of atoms in unit cells ... so I am at a loss at whether my work is correct or not.

Well, as you should guess, you aren't.

Are you saying I'm wrong?

Single atoms can't have fractional ox. state. There would have to be 3 at. 8+ cluster which makes no sense. If your studying sth new and get results which contradict basic knowledge then you've done sth wrong (or learning quantum mech. :))

Well first I don't believe that Al2O3 is face centered cubic. In the example you show for NaCl there are four "whole" NaCl molecules per unit cell. For sodium four 1/8 "atoms" and one 1/2 atom on top and bottom layers. The middle layer has a "whole" Na in the center plus four 1/4 atoms.

This is a really weird question.
Let's start with the general case for a FCC lattice, you'd count up the oxygen atoms per unit cell. In the $\ce{NaCl}$ example, as noted in the comment, there are four $\ce{Cl-}$ per unit cell (6 on the faces, 8 on the corners). Then you'd count up the number of Al atoms per unit cell and divide up the charge as you note above.
In the case of "fractional" oxidation state, it implies that there's a mixture of oxidation states.
Let's say, hypothetically, you had 3 cations and need a +8 charge to balance out the anions. That tells me there are two +3 cations and a +3 cation somewhere. The average is +8/3 but I need to assign formal charges as integers.
But...
Normal $\ce{Al2O3}$ is not an FCC lattice. It's triclinic.

Here, you would do the same thing - count up the number of Al and O atoms per unit cell, considering the fractional volume of the atoms inside the unit cell. It's a bit trickier, since the volumes aren't as obvious as the FCC case.
Instead
This question is about $\gamma$-alumina

Notice that the oxygen atoms actually form an FCC lattice. I have an oxygen at each corner of the unit cell, and one in the middle of each face. So that's 4 oxygen atoms per unit cell = +8 charge from the anions.
How many Al atoms are there?

There are 5 Al atoms completely in the unit cell, plus one per edge (12, each counting 1/4 = 3) so a total of 8 Al atoms!
So that's an average oxidation state of +1 per Al...

I will point out that for gamma alumina, the formula is not necessarily $\ce{Al2O3}$.

@Geoffhutchinson - is there a reason the last picture seems to have ions in and out of the cube?

@Dissenter - the oxygen atoms are on the faces and corners.

@Geoffhutchinson - thank you for the explanation! I posted the corrected question - my professor changed it from Al2O3 to an iron-based ionic lattice. http://chemistry.stackexchange.com/questions/41950/oxidation-state-of-iron-ions-in-a-unit-cell Does this change things? Or is +8/3rds wrong?


text:
What is the meaning of delta squared?
I know that $\Delta x = x_2 - x_1$. This is well known
Therefore, it follows that $\frac{\Delta a}{\Delta b}$ = $\frac{a_2 - a_1}{b_2 - b_1}$ (Unless I missed an elementary math class!)
Now, I found an equation in chemistry, while reading, that turns this principle on its head. It proposes that when considering the second in a set of values:
$\frac{\Delta^2 a}{\Delta b^2} = (a_3 - a_2) \times \frac{1}{\Delta b}$
My issue with this is that $a_3$ hasn't been measured yet, we have only $a_1$ and $a_2$ so how can this be?
Secondly, they propose this:
$\frac{\Delta \left(\frac{\Delta a}{\Delta b}\right)}{\Delta b} = \frac{\Delta^2 a}{\Delta b^2}$
Can anyone please help with the logic behind the first equation and algebra behind the second one?

I hate to break it you, but they're hiding the math from you that is really going on behind the scenes.
It is very common in situations where the students are not expected to know calculus, that wherever a derivative would be needed, to replace, for example, the $dx$ with a $\Delta x$. This is because for many simple physical systems, taking the derivative results in simply measuring the average change in a property with respect to some other property.
For instance, consider a system where you know the position of some particle and how this position changes with time. If you want to know the velocity of that particle at any given time, you would take the derivative with respect to time, $$v=\frac{dx}{dt}$$but, in many cases you might be only interested in the average velocity and could thus skip the derivative part and write that $$v=\frac{\Delta x}{\Delta t}=\frac{x_2-x_1}{t_2-t_1}$$
But, and now we're getting to your question, let's take it one step further and say we wanted to know the acceleration of that particle. This would be the derivative of velocity which is also written as the second derivative of position with respect to time. Thus, we would write $$a=\frac{d^2x}{dt^2}$$Notice how similar this looks to your question about $\Delta^2$ if you simply replace the $d$ with $\Delta$.
Again, if we decided we were only interested in the average acceleration, then we would write $$a=\frac{\Delta^2 x}{\Delta t}$$
The reason why they say that $$\frac{\Delta(\frac{\Delta a}{\Delta b})}{\Delta b}=\frac{\Delta^2 a}{\Delta b^2}$$is again because that is true of derivatives. This would be written as $$\frac{d}{dt}(\frac{dx}{dt})=\frac{d^2x}{dt^2}=a$$
So, to summarize, the symbol $\Delta$ is often used in situations where you care about a change from one state to another (as with a change in energy) or when you only care about the average of some quantity.
As to why they are able to write$$\frac{\Delta^2 a}{\Delta b^2}=\frac{a_3-a_2}{\Delta b}$$ I'm not certain about. I would have to know what this dealt with specifically. My guess is that the quantity $a_3$ is somehow related to the second derivative of $a$ with respect to $b$ and they did some algebra behind the scenes.
Hope that helps.

Thank you so much for your exceedingly easy to understand explanation. This is a bit embarrassing because I do know calculus, however, taking the $\Delta$ symbol to mean "change in " it was only natural for me to want to divide the numerator by $\Delta b^2$. If the textbook had explained the substitution or that differentiating was at the heart of this, I'd have gotten it right away. Thank you ever so kindly


text:
Which equilibrium constant is appropriate to use?
I have learnt that the standard free energy change is related to the equilibrium constant of a reaction by,
$$\Delta G^\circ = -RT \ln K$$
Here, does $K$ refer to $K_p$ or $K_c$?
Also, please give me the derivation of this formula.
On the net, I saw that we can use either $K_p$ or $K_c$. But doesn't that give two different free energy values?
Moreover, if the formula was derived for gaseous reactions (using $K_p$), how can we just extend this to other reactions saying that we can use $K_c$ as well?

As noted in this previous question, the correct definition of the equilibrium constant $K$ depends on activities. If you are interested in the derivation of the equation $\Delta G^\circ = -RT \ln K$ (which requires "proper" thermodynamics), read Philipp's answer to that question. For a reaction
$$0 \longrightarrow \sum_i \nu_i \ce{J}_i$$
(this is a fancy way of writing the reaction which makes sure that the stoichiometric coefficients $\nu_i$ of the reactants are negative), the equilibrium constant is defined as follows:
$$K = \prod_i a_i^{\nu_i}$$
where $a_i$ is the activity of the species $\ce{J}_i$ at equilibrium. (Note that since $\nu_i$ for reactants is negative, the terms for the reactants will appear in the denominator of a fraction.)

The question is, why are $K_c$ and $K_p$ taught at an introductory level despite them not actually being the equilibrium constant $K$? The answer is likely partly because it is difficult to introduce the idea of an activity, but also because $K_c$ and $K_p$ are good approximations to the actual equilibrium constant, $K$. To see why, we have to look at the mysterious quantity called the activity and see how it is related to the concentration of a solution, or the partial pressure of a gas.
If we assume certain ideality conditions, then we can write the activity of a gas as:
$$a_i = \frac{p_i}{p^\circ}$$
where $p_i$ is the partial pressure of the gas and $p^\circ$ is the standard pressure, defined by IUPAC to be equal to $1~\mathrm{bar}$. Therefore, $a_i$ is nothing but the partial pressure of the gas in bars, but without the units. If we therefore define $K_p$ as follows:
$$K_p = \prod_i p_i^{\nu_i}$$
then, as long as the pressures you use are in bars, the numerical value of $K_p$ will be the same as that of $K$, the true equilibrium constant. As such, if you ignore the fact that you are taking the natural logarithm of a quantity with units, the value of $\Delta G^\circ$ that you obtain will be "correct".
Similarly, for a solution, we can write:
$$a_i = \frac{c_i}{c^\circ}$$
where $c_i$ is the concentration of the species and $c^\circ$ is the standard concentration, again defined by IUPAC to be equal to $1~\mathrm{mol~dm^{-3}}$. So, the same idea applies. If you keep all concentrations in units of $\mathrm{mol~dm^{-3}}$, then the numerical value of $K_c$ will match that of $K$.

An example
Consider the Haber process (all species gaseous):
$$\ce{N2 + 3H2 <=> 2NH3}$$
We would define the "true" equilibrium constant as:
$$K = \frac{(a_{\ce{NH3}})^2}{(a_{\ce{N2}})(a_{\ce{H2}})^3}$$
and you would define $K_p$ as:
$$K = \frac{(p_{\ce{NH3}})^2}{(p_{\ce{N2}})(p_{\ce{H2}})^3}$$
Let's say (just as an example - the numbers are not correct!) that at a certain temperature $T$, we have the equilibrium partial pressures:
$$\begin{array}{c|c|c}
p_{\ce{NH3}} & p_{\ce{N2}} & p_{\ce{H2}} \\ \hline
20~\mathrm{bar} & 50~\mathrm{bar} & 50~\mathrm{bar}
\end{array}$$
Then you would have $K_p = 6.4 \times 10^{-5}~\mathrm{bar^{-2}}$. Now you can see why it is not correct to use $K_p$ in the equation above: you would have
$$\Delta G^\circ = -RT \ln(6.4 \times 10^{-5}~\mathrm{bar^{-2}})$$
which makes absolutely no sense since you can't take the logarithm of a unit. In a simplistic treatment you would just ignore the fact that a unit existed and just take the logarithm of the number.
However, you will not run into that problem if you use the "true" equilibrium constant, as you really should. If we assume ideality, i.e. $a_i = p_i/p^\circ$, then we find that $K = 6.4 \times 10^{-5}$. Since activities are dimensionless, you will find that $K$ is dimensionless and you can now take the logarithm. As I mentioned earlier, the numerical quantity of $K$ is exactly equivalent to $K_p$ as long as you keep your pressures in units of bars.

I have seen some questions where students are requested to determine $K_c$ for a gas-phase reaction using the ideal gas law and hence $\Delta G$:
$$\begin{align}
pV &= nRT \\
c = \frac{n}{V} &= \frac{p}{RT}
\end{align}$$
The value of $c_i$ and $p_i$ will therefore differ by a factor of $RT$. Therefore, the numerical values of $K_c$ and $K_p$ in gas-phase reactions will, in general, differ. (If you have the same number of moles of gas on both sides, then the factors of $RT$ will cancel out and numerically $K_c = K_p$, but that is not always the case.) As established earlier, because of the equation for the activity of an ideal gas, only the use of $K_p$ will lead to the correct numerical value for $\Delta G^\circ$.
If a question requires you to do so, then do it. Just bear in mind that it is not only wrong in terms of taking the natural logarithm of a quantity with units; it will also give you a wrong numerical result for $\Delta G^\circ$.

Does this mean that in any (ideal) equilibria (including heterogeneous equilibria also) the activities can be approximated - to partial pressures in case of gases, to concentrations in case of liquids and to 1 in case of solids. These 'so called activities' can give us K (equilibrium constant) using the equation mentioned in your answer and hence this K when substituted in the equation Δ G=RTln(K) gives the standard Δ G - 'standard' in the way it is defined by IUPAC. So we are infact complicating things using Kc and Kp (atleast in my opinion). Is my understanding right?

@user104014 yes that is right.

"why are Kc and Kp taught at an introductory level despite them being wrong? " They are not wrong any more than the ideal gas law is wrong! They are approximate and converge on exactness for dilute systems (eg gases at low P).

@TryHard, if you re-read my answer, I did write that they are approximations which are numerically correct for ideal gases/solutions. However, they are dimensionally inconsistent and should not be used as-is in the relation ΔG = -RT ln K. If your only objection is my choice of phrasing, then I am happy to reword it later when I am off work. I agree it was an exaggeration.

@orthocresol For a gas phase reaction or a heterogeneous solid-gas phase reaction, if $K_c$ is given, then it has not considered the standard state of reactants and products. So, by using $K_p = K_c(RT)^{\Delta n}$, we cannot get the value of $K_p$ at standard state? Then, what's the use of this expression?

@Apurvium I’m afraid I don’t understand what you’re asking. What do you mean by “considering the standard state”? What about it needs to be “considered” and how does this consideration take place? Are you talking about division by the standard concentration?

Anyway, just like how $K_c$ is not divided by $c^\circ$, $K_p$ is also not divided by $p^\circ$, so I don't see what exactly is the problem with the equation $K_p = K_c(RT)^{\Delta n}$. It's not incorrect at all (it's perfectly correct), but just be aware that it is relating a non-standard equilibrium constant to another non-standard equilibrium constant. If you are using $K_c$ or $K_p$ and trying to go between the two, then that equation is perfectly useful and valid. It's just that it has nothing (well, not much) to do with the "true" equilibrium constant $K$.

@orthocresol I meant that for gas phase reactions, for $K_c$, the concentrations are in mol/L, which is not the standard state of gases (1 bar). So, in this case, $K_{eq}=K_p$. As you mentioned "... $K$ is exactly equivalent to $K_p$...".

@orthocresol you say "just like how  is not divided by ∘,  is also not divided by ∘", I'm confused by this. I'm quite sure that  and  are dimensionless just like $K_{eq}$. So what do you mean by this?

@Boson I meant exactly what I wrote: Kc and Kp have units, and K doesn’t.

@orthocresol This answer seems to suggest otherwise. Does it depend on the reference text?

@Boson Er, well, let’s put it this way. If your equilibrium “constant” simply uses pressures, then in general it will have units and it won’t be a true constant. If your equilibrium constant uses fugacities or some dimensionless quantity, then it won’t have units. Whether you use K_p to refer to the former or the latter seems to depend, although I’d contend that the IUPAC definition of “equilibrium constant” suggests the former as it says K_x uses the product of x’s (which means K_p is a product of pressures). https://goldbook.iupac.org/terms/view/E02177

Ok, I understand now. So $K_f$ is dimensionless, not $K_p$. Thanks!

You are correct that you can use either $K_c$ or $K_p$; the trick is to carefully choose your values of $\Delta G_f$ to reflect whether you are working with gaseous compounds or compounds dissolved in solution.
For example, if you are interested in the equilibrium constant for the reaction of $\ce{CO2}$ with $\ce{H2O}$, you can imagine this happening in the gas phase or in solution:
$$\ce{CO2(g) + H2O(g) <=> H2CO3(g)}$$
vs.
$$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)}$$
In both cases, we have
$$K = \exp\left(\frac{-\Delta G_\mathrm{rxn}}{RT}\right)$$
However, in the first case, we use
$\Delta G_\mathrm{rxn} = \Delta G_\mathrm{f}\ce{[H2CO3(g)]} - \Delta G_\mathrm{f}\ce{[CO2(g)]} + \Delta G_\mathrm{f}\ce{[H2O(g)]}$
corresponding to $K_p$ and in the second case, we must use
$\Delta G_\mathrm{rxn} = \Delta G_\mathrm{f}\ce{[H2CO3(aq)]} - \Delta G_\mathrm{f}\ce{[CO2(aq)]} + \Delta G_\mathrm{f}\ce{[H2O(aq)]}$
corresponding to $K_c$.
A good table of thermodynamic values will list several different values for the same compound, corresponding to various possible phases, including (aq).  Unfortunately, the tables in most chemistry textbooks are quite truncated, so you won't always find both (g) and (aq) Gibbs energies of formation.
Contrary to the simplified definition given in most introductory texts, the equilibrium constant is defined in terms of activities. For gases, this is often approximated as the partial pressure divided by 1 atm (maybe it's 1 bar these days? I learned it as 1 atm…).  For liquids, it's approximated as the molar concentration divided by 1 M.  This is what you actually end up using in more complex systems where you have both gases and liquids (and solids) all in equilibrium - you no longer have the simple distinction between $K_p$ and $K_c$, you simply use K with the appropriate choices for activities and the appropriate choices for $\Delta G_\mathrm{f}$.

For a liquid-gas phase heterogeneous equilibrium, $K_{eq} \neq K_c \neq K_p$. For calculating $K_{eq}$, the substances in aqueous form must be in 1 M concentration and substances in gaseous form must be in 1 bar pressure?

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
S-O bond length in SO2 and SO3

Question: which molecule out of $\ce{SO2}$ and $\ce{SO3}$ has larger $\ce{S-O}$ bond length?

Both are $sp^2$ hybridised. Since $\ce{SO3}$ has more bond pair - bond pair repulsions, shouldn't $\ce{SO3}$ have more $\ce{S - O}$ bond length so it can reduce repulsion?

I looked it up and $\ce{SO3}$ does have a slightly shorter bond than $\ce{SO2}$ which is indeed unusual. You should maybe add that fact to your question and you might draw a little more attention because your reasoning seems good to me.

While the question itself may not be too related, the answer seems to answer exactly what OP is asking about.

Also related: molecular orbitals of $\ce{SO3}$

Maybe we used the same book. There is a question about arranging 4 molecules in decreasing order of S-O bond length. In the hint given, the author says to apply bents rule for SO2 and SO3.


text:
Predicting pH of an acidic salt of an weak acid and weak base
Suppose we want to calculate pH of an $\ce{NH4HCO3}$ solution of known concentration.
Now, $\ce{NH4+}$ will be hydrolised to give $\ce{H+}$ ions.
Some $\ce{HCO3-}$ ions will be converted to (i) $\ce{H2CO3}$ and $\ce{OH-}$
and some will give (ii) $\ce{H+}$ and $\ce{CO3^2-}$
\begin{align}
\ce{HCO3- + H2O &<=> H2CO3 + OH− }& 1/K_\mathrm{a1}\\
\ce{HCO3- + H2O &<=> CO3^2- + H3O+ }& K_\mathrm{a2}\\
\ce{NH4+ + H2O &<=> NH3 (aq) + H3O+ }& 1/K_\mathrm{b}\\
\end{align}

I am not sure how to take account of both $\ce{H3O+}$ and $\ce{OH-}$ in the calculation of pH.
Will the third equilibium costant be $1/K_\mathrm{b}$?

Can you use the formula given in my answer and check if if it gives you the correct answer?

3rd equilibrium is written terribly - it's simply NH4+ +H2o = NH3 +H3o+ (proton exchange)

@Mithoron: That's the reaction given in my textbook. User intends hydrolysis of cation. BTW, did the formula work?

@AdityaDev You should get new textbook ;) "NH4OH" is only in old ones - there's no such molecule, there's NH3(aq) written instead now

My supervisor spent about five minutes grinning and laughing about a practical joke a TA did in an undergrad lab course to a fellow student, sending her away to get ‘solid $\ce{NH4OH}$’ while telling us the story in the post-serious drinking session during a symposium.

The main reactions which take place are
1)Ionization of $\ce{HA-}$: $\ce{HA- -> H+ +A^2-}$
2)Hydrolysis of $\ce{HA-}$: $\ce{HA- + H+ -> H2A}$
3)Hydrolysis of cation: $\ce{B+ + H2O -> BOH + H+}$
for reaction $1$, $$k_{a2}=\frac{[H^+][A^{2-}]}{[HA^-]}$$
for reaction $2$, $$k_{a1}=\frac{[HA^-][H^+]}{[H_2A]}$$
for reaction $3$, $$k_h=\frac{k_w}{k_b}=\frac{[BOH][H^+]}{[B^+]}$$
At any moment, $$\ce{[H^+]=[BOH] +[A^{2-}]-[H2A]}$$
$$[H^+]=\frac{[NH^{4+}].k_w}{[H^+]k_b}+\frac{k_{a2}.[HCO_3^-]}{[H^+]}-\frac{[HCO_3^-][H^+]}{k_{a1}}$$
Rearranging the terms to get $\ce{[H+]}$,
$$[H^+]=\sqrt{\frac{\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]}{1+\frac{[HCO_3^-]}{k_{a1}}}}$$
$$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{k_{a1}+[HCO_3^-]}}$$
Now we will have to make 2 assumptions. First lets assume that $k_{a1}<<[HCO_3^-]$ so that $k_{a1}+[HCO_3^-]=[HCO_3^-]$
$$[H^+]=\sqrt{\frac{k_{a1}\cdot \big(\frac{[NH_4^+].k_w}{k_b}+k_{a2}[HCO_3^-]\big)}{[HCO_3^-]}}$$
If you assume that neither hydrolysis nor the dissociation goes too far, you can assume the concentration of bicarbonate ion to be equal to the initial concentration (say C). Since Ammonium bicarbonate dissociates to give two products in 1:1 ratio, concentration of ammonium produced is equal to the concentration of bicarbonate ion produced.

I saw this formula for calculation pH of such salts in my textbook but
  no derivation was given:
  $$[H^+]=\sqrt{k_{a1}\cdot\big(\frac{k_w}{k_b}+k_{a2}\big)}$$ The above formula can
  be obtained if you consider the last assumption. (Clarification
  needed)

I don't understand why $\ce {[H+]=[B+]  + [A2−] − [H2A]}$ It should be $\ce [BOH]  + [A2−] − }$ [how to take acount of this? here OH- was produced]

I have edited that part. It's not B+. It's BOH. Reason is, H+ is produced by B+ and equal amount of H+ and BOH are formed. So [H+] from B+ is same as the amount of BOH formed.

And, how do you take account of the last part? (why $\ce {−[H2A]}$) I mean, how to substitute the [OH-]?

Amount of H+ used up and amount of H2A formed are same.

If you have any numerical type question, see if the formula works.

I don't have the answer. Our teacher gave this to think. But I could not figure out what to do with the OH- in the reaction 2. Thanks. Will let you know when I have the answer.


text:
Why does FeO(OH) have two CAS numbers?
Iron(III) oxide-hydroxide, $\ce{FeO(OH)}$ has two CAS numbers, 1310-14-1 and  20344-49-4. What's the difference between them?
Why does it have two CAS numbers? As I know, the CAS number should be unique to each substance.
Wikipedia reference

Why does it have two CAS numbers? As I know, the CAS number should be unique to each substance.

Yes, that’s exactly the principle and also why two CAS reference numbers are associated with $\ce{FeO(OH)}$. SciFinder gives 20344-49-4 to be the ‘boring’ laboratory iron oxyhydroxide, while 1310-14-1 is mentioned as the mineral goethite. Goethite is the α form of $\ce{FeO(OH)}$; β (akaganéite), γ (lepidocrocite) and δ (feroxyhyte) forms exist, too.
Since each of these four minerals has a well-defined structure, much better defined than simple $\ce{FeO(OH)}$, they should (and will) each get a single CAS number for unambiguous referencing. At the same time, the generic CAS number 20344-49-4 also exists for lab-grade non-mineral $\ce{FeO(OH)}$.
So what you are experiencing is not a violation of the one-substance-one-CAS-number rule but rather a confirmation of its strictness.


text:
Can I electrolyze saturated sodium hydroxide?
Can I electrolyze saturated $\ce{NaOH(aq)}$? My guess is that I can't because in the aqueous solution, the concentration of $\ce{OH-}$ is too high and thus there isn't enough $\ce{H+}$ remaining.

Why, of course you can. There are never too few $\ce{H+}$. As soon as one $\ce{H+}$ discharges, another one steps in.

At $T=25\ \mathrm{^\circ C}$, saturated sodium hydroxide solutions contain about $50\ \%$ sodium hydroxide. Hence, the concentration of $\ce{Na+}$ is about $c(\ce{Na+})=19\ \mathrm{mol\ l^{-1}}$ and the concentration of $\ce{H3O+}$ corresponds roughly to $\mathrm{pH\approx15}$, i.e. $c(\ce{H+})\approx10^{-15}\ \mathrm{mol\ l^{-1}}$ (neglecting activity coefficients).
The positive ions $\ce{Na+}$ and $\ce{H3O+}$ are attracted to the cathode.
At the cathode, the redox potentials $E$ imply that, even under the given alkaline conditions, reduction of $\ce{H+}$ to $\ce{H2}$ is still preferred over reduction of $\ce{Na+}$ to $\ce{Na}$:
$$\begin{alignat}{3}
\ce{Na+ + e- \;&<=> Na}\qquad &&E^\circ = -2.713\ \mathrm V\qquad &&E&&=E^\circ+0.05916\ \mathrm{V}\times\log_{10} [\ce{Na+}]\\
& && && &&=-2.713\ \mathrm V+0.05916\ \mathrm{V}\times\log_{10} 19\\
& && && &&=-2.64\ \mathrm V\\[6pt]
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm V\qquad &&E&&=E^\circ+0.05916\ \mathrm{V}\times\log_{10} [\ce{H+}]\\
& && && &&=0.000\ \mathrm V+0.05916\ \mathrm{V}\times\log_{10}(10^{-15})\\
& && && &&=-0.89\ \mathrm V
\end{alignat}$$
Therefore, the thermodynamically preferred reaction is the reduction of water to hydrogen according to
$$\ce{2H+ + 2e- -> H2 }$$
or in this case better (in view of the given alkaline conditions)
$$\ce{2H2O + 2e- -> H2 + 2OH- }$$
Nevertheless, under the given alkaline conditions, the concentration of $\ce{H3O+}$ is very low. The predominant positive ions are $\ce{Na+}$, which migrate to the cathode. Thus, especially when applying a high current density, the major part of the current of positive charges that reach the cathode is carried by $\ce{Na+}$. Therefore, $\ce{Na+}$ actually can be discharged at the cathode:
$$\ce{Na+ + e- -> Na}$$
The ratio of discharged $\ce{Na+}$ to $\ce{H+}$ also depends on the cathode material. For some materials, the reduction of $\ce{H+}$ to $\ce{H2}$ is kinetically inhibited, which results in a significant overpotential.
However, any generated $\ce{Na}$ immediately reacts with water (except when a mercury cathode is used):
$$\ce{2Na + 2H2O -> 2Na+ + H2 + 2OH- }$$
Therefore, the net reaction is again the above-mentioned reduction of $\ce{H2O}$ to $\ce{H2}$. The $\ce{Na+}$ ions act as electrolyte and might be considered to act as a catalyst:
$$\begin{align}
\ce{2Na+ + 2e- &-> 2Na}\\
\ce{2Na + 2H2O &-> 2Na+ + H2 + 2OH- }\\
\hline\ce{2H2O + 2e- &-> H2 + 2OH- }
\end{align}
$$

Re:"However, any generated Na immediately reacts with water" - Is that true when potential is still applied to the electrode? It certainly would be true if electrode was tuned off. // I remember doing copper plating on a Pt electrode for an anal chem class. You had to keep the potential on until after you removed the electrode from the solution to keep the Cu from dissolving back into the solution.

@MaxW Yes, the applied potential affects the remobilization of substances that have been deposited on the cathode. (I had to consider this effect when making electrodepositions of actinoids for alpha spectrometry.) But in case of sodium, I guess (considering its reactivity and standard potential), it would be difficult to completely prevent the oxidation of metallic sodium in aqueous solutions.


text:
Why is a negatively charged atom less likely to hydrogen bond than a neutral atom?
Can someone explain why a negatively charged atom let's say $\ce{CH3SH}$ is less likely to bond with an additional hydrogen than $\ce{CH3S-}$?
And can anyone relate this to nucleophilicity in a protic solvent?

Wait, I didn't quite get it: what atom is negatively charged in $\ce{CH3SH}$ and how is that different from $\ce{CH3S-}$ (which is indeed negatively charged)?

@ATP You should edit question - right now it's nonsense

I'm going to assume you mean hydrogen bond with itself.
The reason that the $\ce{CH_3S^-}$ will not hydrogen bond is because there is no longer a hydrogen which has a significant dipole. When we had just $\ce{CH_3SH}$, the hydrogen attached to the sulfur is theoretically capable of hydrogen bonding$^1$ as thiol groups are known to hydrogen bond. I am not super confident that the hydrogen bond between one $\ce{CH_3SH}$ and another $\ce{CH_3SH}$ would be very strong if it is present at all. It is possible you could see the covalent nature of the hydrogen bond, but because the dipole-dipole interaction is so weak, you would barely find an equilibrium at all, or might even find a maximum in the vibrational potential energy surface. Perhaps o interest is that there are some hydrogen bonds which are known to blue-shift as opposed to the typical red-shift seen in hydrogen bound systems$^2$.
I would point out, however, that $\ce{CH_3S^-}$ will have a significant intermolecular attraction with any other molecule that has a positive dipole. This would be an ion-dipole force which, as far as I understand it, is a purely electrostatic interaction as opposed to hydrogen bonding which has covalent characteristics.
I hope that answer was a sufficient explanation.

$^1$YUAN, K., Liu, Y., Zhu, Y., Zhang, J., & Zhang, J. (2009). Structures and Properties of Halogen Bond and Hydrogen Bond Formed between CH3SH and HOCl. ACTA CHIMICA SINICA, 67(6), 499-506.
$^2$I highly recommend this paper if you are interested in understanding improper hydrogen bonds which would be those hydrogen bonds where you get a maximum in the vibrational potential energy surface which leads to a blue-shift at that equilibrium geometry. Also, it is written as a semi-exhaustive discussion of the topic for both biologists and chemists, so it quite accessible.
Joseph, J., & Jemmis, E. D. (2007). Red-, blue-, or no-shift in hydrogen bonds: a unified explanation. Journal of the American Chemical Society, 129(15), 4620-4632.
Available for free here.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Mr of Acid getting wrong answer
I am preparing for A level chemistry exams. One of the questions I came across is

Calculate the $M_\mathrm r$ of unknown monoprotic acid. The volume is $15\,\rm cm^3$ and concentration $10.95\,\rm g\cdot dm^{-3}$.

I know how to calculate the number of moles and I know the equation to calculate $M_\mathrm r$ but could anyone give me some guidance on how to calculate the mass from this data?

@Ivan please note that units should be upright, not slanted. Thanks for the edits. :)

Is it not just $10.95 \times 15/1000$? The concentration is in $\rm g\cdot dm^{-3}$.

@surelyourejoking You would, after that, have to divide by the number of moles.

@orthocresol Indeed. Is it possible to find that given the information?

@surelyourejoking No, not the information we are given, at least. But OP mentioned that they knew "how to calculate the number of miles", so I'm guessing there's an earlier part of the question that they hid from us.

@surelyourejoking do you mean you have to divide it by the number of moles to get the mass of substance or to calculate the Mr? because you can get the number of moles this way: n = volume x concentration ,so no i'm not hiding anything

In this case, I don't think n = CV is correct. That would be correct if concentration were given, as it normally is, in moles per decimetre cubed. However, in this case it is given in grammes per dm$^3$, so in fact, concentration x volume gives the mass.

The concentration of your acid is given as a mass concentration.
Without knowing the molar mass, the molar concentration and the amount of substance in a volume of $15~\mathrm{cm}^{-3}$ are not accessible.
However, knowing that
$\mathrm{1~L = 1~dm^{-3} = 1000~cm^{-3}}$
and therefore
$15~\mathrm{cm}^{-3} = 0.015~\mathrm{dm}^{-3}$
the mass of the acid (in $15~\mathrm{cm}^{-3}$) is given as
$m = 10.95~\mathrm{g\,dm^{−3}\cdot0.015~\mathrm{dm}^{-3} = 0.16425~g}$


text:
Chemical equilibrium and free energy
I have a major confusion over the relationship between free energy change and the equilibrium constant.
I have read that $\Delta G_0= -RT\ln K$.
My first doubt is, does $\Delta G_0$ refer to free energy change ONLY at 25C?
Does $\Delta G_0$ make sense for any other temperature?
Secondly, the "T" in the equation
Does it also have to be the same temperature as the temperature for which we write $\Delta G_0$? (ie,25C)
If not so, can we write any temperature and obtain the corresponding K value?
Please explain.
If G0 can be defined for other temperatures, does that mean that S0 and H0 can also be defined for other temperatures?

It can be defined for any temperature. Second part - yes, you write the temperature for which you calculated ΔG at. I'm 99% sure this is a duplicate of something, although I will admit that the exact duplicate is not easy to find, not least because there are like 50 questions on this site with poor unspecific titles like "Gibbs free energy confusion".

Sorry about that

It's ok. That's why I chose to give you a brief answer as well. Also yes, standard entropy and enthalpy changes can also be defined for other temperatures. The thermodynamic standard state only stipulates unit activity (which can be approximated by unit concentration) and not a temperature.

But I am assuming that pressure should be 1atmosphere?

I have never come across a single book that explains all these small details of thermodynamics

Yes, its shocking.  $\Delta G^0$ refers to the change in free energy between the following two thermodynamic equilibrium states:  State 1:  Pure reactants in separate containers, each at temperature T and 1 atm. pressure.  State 2:  Pure products in separate containers each at temperature T and 1 atm pressure.

I believe the new standard is 1 bar. But yeah.


text:
Voltaic cell: why does the rate of rusting increase when NaCl is added?
There is a beaker in which there are two iron electrodes (there is no salt bridge between two beakers-there's just one beaker) and one iron electrode is being bubbled with a supply of oxygen from a pipe (causing it to rust and fall to the bottom of the beaker). There is a flow of current in an external circuit (which I think is because the oxygen prevents one electrode from dissociating into $\ce{Fe^2+}$ and $\ce{2e-}$). The one that is not being bubbled with oxygen is the anode, but I'm not sure why. I think it's because the other one can't dissociate since it's reacting with oxygen. Also, the rate of rusting increases when $\ce{NaCl}$ is added to the water. Why is that?

It is not clear what are you asking. Could you point out questions? Or the points of interest.

@JaroslavKotowski  Why is the anode electrode B and why does adding NaCl to the water increase the rate of rusting? Also I thought that electrolyte was needed for flow of electrons in the first place, so how is just water doing the trick?

If your diagram is correct (the notation of the polarity of the electrodes) then the anode should be the one that is not being pumped with oxygen, because the negative end of any electro-chemical cell is the anode.
Let's see how this happens.
First of all let us consider the ions and molecules at two electrodes the system.
We have $\ce{O2}$ at electrode A and $\ce{Fe}$ at B   .
Now since the reduction potential of $\ce{O2}$ is greater than $\ce{Fe}$ a reaction takes place at the $\ce{O2}$ pumping electrode. This is a reduction:
$$\ce{O2 + 2H2O + 4e- -> 4OH-}$$
Now at the other electrode, an oxidising reaction takes place:
$$\ce{Fe -> Fe^3+ + 3e-}$$
So you can see there's an excess of electrons at the $\ce{Fe}$ electrode. Now these electrons travel across to the $\ce{O2}$ containing electrode since the cell reaction there needs electrons. This travel of electrons give rise to an electric current. You might know from your previous lessons that current flows from negative terminal to the positive terminal inside a cell. Hence the $\ce{Fe}$ electrode becomes the negative anode.
As to your second question why does $\ce{NaCl}$ increase corrosion, for all these reactions to take place electrons should be moved about. $\ce{NaCl}$, being a strong electrolyte, increases the conductivity of water. That means electrons are moved rapidly giving a rise in the rate of corrosion.
And you have asked in the comments "How is just water doing the trick?". You should keep in mind that water exists as $\ce{H+}$ and $\ce{OH-}$ in the aqueous liquid state.

A point you didn't cover - In a basic solution you'd get iron hydroxide, not iron oxide.

but OPs question clearly states that iron(III)oxide is formed

@slhulk I thought that water only exists as H+ and OH- when it's acting as a solvent? Also at electrode A I thought that the reaction was simply between Fe and O2 (3O2 + 6e- = 6O2- at electrode A, forming rust). Why is water part of the equation to give rise to OH-? And isn't OH- already there as you say that's how water exists?

Water exists as H+ & OH- in aq medium. The reaction at A is determined by the reduction potential of elements present there. O2 reduces to OH-. Notice the change of oxidation state from 0 to -2. To balance it water is necessary.

Water exists as H+ & OH- in aq medium. The reaction at A is determined by the reduction potential of elements present there. O2 reduces to OH-. Notice the change of oxidation state from 0 to -2. To balance it water is necessary.


text:
Why does no antitoxin for botulinum type H exist yet?
Antitoxins against the botulinum toxins type A–G produced by Clostridium botulinum are known. Why does none exist against the type H botulinum toxin yet?

The premise of this question is based upon outdated information.
According to Study: Novel botulinum toxin less dangerous than thought

Researchers from the Centers for Disease Control and Prevention (CDC) and the University of Wisconsin (UW), writing in the Journal of Infectious Diseases (JID), said it turns out that the novel toxin can be blocked by available antitoxins after all. That suggests that it isn't really a novel toxin, the authors said

(Attention grammar police: The above is an exact quote from the linked reference.  If you feel the compulsion to keep editing this again and again, please do it in a way that makes clear it is no longer an exact quote.)


text:
Oil-water emulsion doesn't separate
I've steam distilled a couple of different powdered seeds/herbs to obtain the essential oils from them.
Somehow some of the extracts won't separate and stay in an emulsion.
I've seen that some fixed oils that where extracted too are separating in cool conditions, but I can't see any essential oil layer forming. Even when letting the extract (emulsion) sit for 2 weeks.
Can anyone explain why this is happening (or not) and how I could break the emulsion? Acidifying or salting it out?
Since if I would try to boil the water of off the emulsion, its vapors would carry the oils with it, so is there any way to evaporate the water but leaving the oils?
I live in a country where it's hard to obtain some chemicals, so are there any easily obtainable chemicals/solvents to use for further or better processing?
I have access to $95~\%$ bio ethanol, $95~\%$ acetone, sodium hydroxide, sodium chloride and I've managed to get some naphtha.
So my possibilities are limited as you can imagine …
I've thought about using a Soxhlet extraction with ethanol or acetone where I could simply distill the extracted and concentrated solution to recover the solvent and be left with the extract (essential oil, fixed oil, etc.).
What can I then do to separate the fixed oils from the essential oils that I want?

Hi and welcome to [chemistry.se]. Feel free to take a [tour] of the site and visit the [help] for any deeper questions remaining.

First if you going to use the oils for some sort of human use I would NOT use acteone or naptha. (Safety in this sort of question always scares me a bit...) // Not having any idea what you're doing I can only suggest splitting the sample and trying various methods. I'd start with NaCl. NaOH might chew up some compounds that form the "oil." // 95% ethanol already is  an azetrope so trying to distill with that is pointless.

Can I ask what "herbs" and what use? Or is it THAT "herb"?

First of all, no, i am not talking about mariuana plants or other psychoactive or hallucinogene plants. Some examples plant material that I've used is spices like cinamon or cloves. Or orange peels and chilis. I am not planning to consume the essential oils or use it in other forms. These experiments I'm doing are only for personal educational purpose.

If my memory doesn't deceive me, https://en.wikipedia.org/wiki/Vacuum_distillation would be the preferred way, and if not for anything else, it would protect temperature sensitive compounds from decomposing.

@Ace At least for cloves, I would recommend using steam distillation (https://en.wikipedia.org/wiki/Steam_distillation ). It is pretty straightforward.

I have encountered the same problem with plant extracts. I think the term is hydro/sol when the extract is suspended in aqueous solution. Anyway, I've had some success with heavy sodium chloride brine then chilling the solution to near freezing for an extended time (a week or so). Even then the return hasn't been that impressive. In most cases making a tincture (or gas)extraction then distilling the product (if you still need to) produces a much better yield.

Brine works wonders.  I would also be cognizant of your ratio of organic-to-aqueous components. I believe that oils become more soluble in water when they make up only a small fraction of the mixture.


text:
Has a carbon compound ever been found having an ionic bond?
Though it is highly unlikely, has any carbon compound been found to make an ionic bond and to exhibit ionic properties?

Graphene oxide is not ionic at all. There are, though, certain intercalation compounds which are truly ionic with (+) on carbon sheets.

http://www.madsci.org/posts/archives/2000-09/967952226.Ch.r.html
Check this link. Hope it will cover your Question

@steveverrill I mean by the transfer of electrons, like what you see in NaF. When the ionic character is greater than the covalent one.

see this- http://chemistry.stackexchange.com/questions/9212/can-a-hydrocarbon-organic-compound-contain-an-ionic-bond

Why unlikely? Ionic compounds of carbon have been known for ages. There are ionic carbides ($\ce{Al4C3}$, $\ce{CaC2}$, etc.), graphite intercalation compounds like $\ce{KC8}$, ionic derivatives of fullerenes and more.
Come to think of it, common $\ce{CaCO3}$ is certainly ionic and at the same time a compound of carbon, but this is most likely not what you want.

There is also the classical case of the aromatic carbocations and carbanions; examples include tropylium bromide,

and sodium cyclopentadienide, which is used for preparing "sandwich compounds", among other uses.

Perhaps what you meant to ask was

Are there any known organic compounds with ionic bonds?

Organic salts would be examples of organic compounds with ionic bonds.  Here is a list of organic salts.

I think, what actually OP meant to ask is:

Are there any known organic compounds with carbon involving in an ionic bond?

The reason for my suggestion is the phrase "to make an ionic bond and to exhibit ionic properties" in OP's question, even though it did not describe the involvement of carbon. Otherwise, any ionic carbonate (e.g., $\ce{Na2CO3}$) or ionic cyanide (e.g., $\ce{NaCN}$) would qualify, none of which are not organic compounds.
Based on my suggestion tropylium bromide and sodium cyclopentadienide certainly qualify.
I think the best example is a compound containing triphenylmethyl cation (triphenylcarbenium or tritylium) and a non-nucleophilic anion (which doesn't have a strong affinity for bonding to the electrophilic carbon). Such compounds are stable at room temperature and also commercially available $($e.g., triphenylmethyl tetrafluoroborate, $\ce{(C6H5)3C+ BF4-})$:

Compounds of tritylium with other anions such as hexafluorophosphate $(\ce{PF6-})$, hexachloroantimonate $(\ce{SbCl6-})$, and perchlorate $(\ce{ClO4-})$ are also known. Further, perchlorotriphenylcarbenium Hexachloroantimonate $\left(\ce{C6Cl5)3C+SbCl6-}\right)$ has been prepared and obtained its crystal structure showing the positive charge on the carbon (Ref.1):

The title compound, $\ce{C19Cl15+.SbCl6-}$, consists of perchlorotriphenylcarbenium cations and hexachloroantimonate anions. The central $\ce{Sb}$ and $\ce{C}$ atoms occupy special position of symmetry 32 along the c axis. The cation shows a symmetrical propeller conformation. Four $\ce{Cl}$ atoms of the $\ce{SbCl6-}$ anion are disordered.

References:

E. Molins, M. Mas, W. Maniukiewicz, M. Ballester, J. Castañer, "Perchlorotriphenylcarbenium Hexachloroantimonate(V)," Acta Cryst. C 1996, C52, 2412-2414 (https://doi.org/10.1107/S0108270196007287).


text:
Which orbital in a subshell gets filled first?
Let's say we're filling electrons in subshell 2p. This subshell will have $m_l = -1$, $m_l = 0$, and $m_l = 1$ orbitals. Does the 'filling' of electrons depend on the value of $m_l$? Meaning, does the value of $m_l$ determine if it gets filled first by an electron or not?

The notation [-1,0,1] is given to address the sub shells.The energies of all these orbitals (unless you apply a magnetic field) are identical, therefore the filling is completely arbitrary. That means any of the shells can be filled first. But after 1 shell is filled the next electron goes to any one of the remaining shells. And the next goes to the other one. This is known as the Pauli's Exclusion law. Electrons with like spins are filled first and electrons with the opposite spin are filled next (note: 1 shell can share 2 electrons). This choice of spin is also complete arbitrary.

The only really important thing when filling electrons into lone atoms in vacuum is taking care to fill shells from lowest to highest principal quantum number $n$ and the azimuthal quantum number $l$, since those are the only two that affect the electron’s energy.
$n$ affects even the energies of electrons in hydrogen-like atoms while $l$ only comes into play once core electrons exist whose orbital shapes can make a difference to the overall charge distribution. This all makes macrophysical sense, since $n$ is directly part of the quantitised electron energy $E_n$ and $l$ is the quantum number determining the electron’s angular momentum with $l(l+1)\hbar$ being the eigenvalue of the squared angular momentum operator $\hat{\vec{L}^2}$.
The orientation of this (or more precisely: its $z$ component, $\hat{L_z}$ by its eigenvalue $m_l \hbar$) is determined by the magnetic quantum number your question is asking about. Stick to the image of a single atom in vacuum: There is no reason why certain directions of angular momentum (remember the value is the same, only the orientation changes) should in any way affect the energy — especially since $z$ is a completely arbitrary choice of coordinate; vacuum does not know up or down. It is not until you add surroundings, either by another atom approaching your atom in vacuum or by an external magnetic field that the orientation matters in any way.
So as long as you are adding electrons to single atoms aufbau-wise, you can choose whichever orbital you want to fill first. The energies of all three p-orbitals are identical and cannot be modified in any way. However, add another atom, and all of a sudden these become distinguishable. The system now starts disfavouring identical energy values of different orbitals if they aren’t completely populated, half-populated or unpopulated. This gives rise to effects such as Jahn-Teller distortion that you might encounter elsewhere. They all follow the basic principle of ‘reduce symmetry to distinguish directions to allow for different energy levels so my electrons can sit at lower levels’.

I hope I didn’t get any of the quantum stuff badly wrong …


text:
How does the mixing of two gases increase entropy?
I have learnt that $\mathrm{d}S = \mathrm{d}Q_\mathrm{rev}/T$, where $Q$ is the heat energy absorbed by the system. Now consider an insulated compartment with two chambers, separated by a valve. Each chamber consists of a gas at a particular temperature and volume. Now, the valve is screwed open. We know that the entropy of the isolated system will increase as the gases have more freedom.
But, starting with the assumption that the walls were insulated, where did the heat energy come from?
How did the entropy increase?

I am sure this is NOT a duplicate, as no one asked this specific question, perhaps because it is stupid.

isn't the equation you mentioned deals with  "entropy of the surrounding" ?

No, I believe this equation holds good for any system.

:) looking forward for a clear answer +1

The $Q_{rev}$ is reversible heat. Such mixing is not a reversible process. There is no way to unmix the gases. // You can heat water convert to it to steam, cool the steam to water and keep cycling as long as you want. That is a reversible process.

Sure, but we can still calculate the entropy somehow, right?

In fact, the formula is  -nR*summation(xlnx), where x is the mole fraction of the ith gas.

The entropy of mixing can be derived by considering the chemical potential of an ideal gas then finding $\Delta G$ and hence $\Delta S$, or equivalently by considering the number of different microstates in the unmixed and mixed states then using Boltzmann's formula $S = k_\mathrm{B} \ln W$.

Yes, but since entropy of the system increases, the gases should have absorbed some heat.From where?

@orthocresol Just if anyone is interested. I had written up a derivation of the entropy of mixing formula via the $\Delta G$-route in this answer of mine.

The process is not reversible, which means that $\mathrm{d}S \gt \mathrm{d}q/T = 0$. The equality only holds in the reversible case, which is why you have a subscript rev. So the whole premise of your question is wrong. The entropy change of an irreversible adiabatic process is always greater than zero.

I thought that for an irreversible process, we calculate entropy change by assuming a reversible process from the initial to the final state.Still, I don't quite understand how entropy increases from the chemistry perspective, for this situation.

Can you tell me a way of "reversibly" mixing two gases for which you can work out that $\mathrm{d}q = 0$? Because I can't think of any. As for the second part, entropy is not something that is supposed to be interpreted pictorially. It is not a function that we can observe or control. It is merely defined to be a state function with the property $\mathrm{d}S = \mathrm{d}q_{\mathrm{rev}}/T$. The best we can do is "entropy is a measure of disorder", and obviously a mixed system has greater disorder than one where the two gases are separated.

Agreed, but how can a system become more disordered without any absorption of energy?

Why does it have to absorb energy? The Second Law tells you that an isolated system (which by definition cannot absorb any energy from its surroundings) always tends to maximum entropy.

I think I get your point. The only issue is that my book has derived the formula for entropy change, considering it to be a reversible process.It has just used algebraic manipulation and other thermodynamic formulas, without mentioning where the heat came from, or commenting upon the reversibility of the process

@orthocresol I am not sure if my argument is total bs or not since my thermodynamics classes lie in a distant past, but might the following be a possible explanation: When the two gases expand to fill the larger (common) volume created by opening the compartements, they release expansion work, $-p dV$. This released expansion work is not absorbed by any surroundings but converted into heat energy, $dQ_{\mathrm{rev}}$, since the entropy of the system increases because of the expansion by $dS = \frac{dQ_{\mathrm{rev}}}{T}$.

@Philipp I think you are right. There is a theoretically reversible way to carry out this reaction, described in Levine's Physical Chemistry 6th ed., which involves separate isothermal expansion as you mentioned ($w < 0$, so $q_\mathrm{rev} > 0$ since the process is isothermal and $\Delta U = 0$) followed by a hypothetical isothermal mixing of the expanded gases using semipermeable membranes (for which $q_\mathrm{rev} = w = 0$). I've attached a pdf of the relevant pages here.

I didn't quite understand still.Can you explain a little better please?

I understood the part where we devised a way to carry the process out reversibly.Since this is the case, where DID the heat come from?

@Kaylan The process is simply a matter of energy conversion: Some kind of "volume energy" is converted into heat energy (which is linked to entropy) during the process of mixing. So, the heat comes from the gas expansion, i.e. from within the system itself.

@orthocresol Thx, that restores my faith in my lingering thermodynamics knowledge :)

Yeah, ortho Cresol has answered most of the questions that I ask on this website. Thanks for that.

Here is a reversible process to do what you are asking for (i.e., unmixing the gases reversibly).  You have a cylinder containing the two gases already mixed.  The cylinder has a piston.  Connected to this main cylinder are two other cylinders joined to it through semipermeable membranes.  Each membrane allows one of the gases to pass, but not the other.  You gradually advance the piston in the main cylinder while, at the same time moving the other pistons outward in the other two cylinders so that, at any time, the pressures of the pure gases in the attached cylinders are equal to their partial pressures in the main cylinder.  The change in entropy for this adiabatic reversible process is zero.  This is the first step.  Now, in the second step, we close off the semipermeable membranes, and compress each of the pure gases in the two cylinders isothermally and reversibly to the original total pressure that was present in the main chamber.  So, if $p_1$ was the partial pressure of one of the gases in the main chamber and $p_2$ was the partial pressure of the other gas in the main chamber, the total pressure of the pure gases in their chambers after completing step 2 will be $p_1+p_2$.  The change in entropy for the 2-step process we described is minus the entropy of mixing.  So, if the process is done in reverse, the entropy of mixing will be positive.


text:
Why does my eudiometer have this excess volume at the top?
It looks like this:

The measurements start at 2 mL and the volume remaining in the top seems to be greater than 1 mL. What is the purpose? I am trying to measure hydrogen gas.

I'd guess that because that is a cheap way to make an eudiometer. If you had makings down at the curved end then the bias in the marking and actual volume is probably hard to control. Down the tube the diameter would be more consistent. // I'd guess the problem is mass producing the markings in a machine.

authors: Đỗ Đại Nghĩa


text:
Why is pencil lead smelly?
I was writing down a chemical equation with my pencil and noticed its pungent smell. Could graphite itself be the reason for it?

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site and visit the [help] for any unanswered questions about it.

Graphite itself doesn't have smell, diamond too

If diamonds had a pungent smell then they surely wouldn't be the fad for engagement rings.

Does graphite really smell bad? o.o

I'm wondering if the smell is actually chemicals used in the treatment of the wood in your pencil or even the paint as opposed to the graphite.

Don't you find the graphite in pencil has a distinctive smell? Even the "lead" refills for re-usable pencils have the smell.

Graphite has no smell at all. If you see graphite and at the same time feel some smell, it does not follow that the smell is that of graphite.

The "lead" in a pencil isn't pure graphite of course. It is graphite mixed with clay typically and who knows what else.

Graphite definitely has a pungent smell to it, but this is probably a result of it being loosely plated, dust easily escaping into the air. Diamond probably wouldn't have such a smell.

I agree with Max W , the graphite in pencils is not pure so it would be the impurities causing the smell.

I'm voting to close this question as off-topic because it's based on false premise.

I've reworked this question. 4k views on a misleading, to put it mildly, question? Should be done long ago.

None of the carbon allotropes have a smell. If you smell something from a pencil lead (graphite), then it will be from the impurities in it.
Graphite is made from impure organic compounds that frequently contain sulphur. It is likely that you are smelling this. Mercaptans and other sulphur compounds can be very smelly. on its own

Just to add to the statement that none of carbon allotropes have a smell, this is because their molar mass is generally enormous, so they cannot form enough vapour to be detected by the nose.


text:
What is the role of sulfuric acid and mercury(II) sulfate in the oxymercuration of alkynes?
I have a homework problem where I'm given a terminal alkyne. The goal of the problem is to choose the reagent that will result in a particular ketone.
Here's a really general view of the reaction I drew out showing the reactants I chose (I don't want anyone to do my homework, just explain what's unclear to me).
$$\ce{some terminal alkyne ->[H2O][H2SO4, HgSO4] some ketone}$$
I've gone through the list of reagents given in the problem and all the others result in an aldehyde (which to me suggests anti-Markovnikov addition).
The book shows that $\ce{HgSO4}$ yields a ketone via the Markovnikov addition of water forming an enol (and undergoing tautomerization). My answer was correct, but I'm not satisfied that I really know this. The mechanism shown in the book involves only water (it doesn't show what the role of $\ce{H2SO4}$ or $\ce{HgSO4}$ is here, it only explains that it causes Markovnikov addition of water).
My only proposal is that the $\ce{HgSO4}$ adds first, in an anti-Markovnikov fashion, forcing an $\ce{H}$ from $\ce{H3O+}$ to add on the terminal carbon and thus water to add to the other side. I also am guessing that $\ce{H2SO4}$ makes the formation of $\ce{H3O+}$ more favorable.
Could someone show how this mechanism works or explain a bit about how the presence of $\ce{HgSO4}$ and $\ce{H2SO4}$ affect the reaction?

Let's first look at the acid-catalyzed hydration of an alkyne. Just like in the analogous reaction of alkenes, the new proton goes to the least substituted position, to give the most substituted carbocation. Water attacks the carbocation, and after deprotonation, an enol is produced. Under acidic conditions, the enol tautomerizes the the corresponding ketone.

The mechanism with mercury catalysis is very similar. The main difference is that the initial intermediate is a mercurinium ion, which is opened by the nucleophilic attack of water at the more substituted site. Beyond that, you can think of the mercury atom as a giant proton. The mechanism proceeds analogously, first through a mercury substituted enol, which is protonated and then demercurated.

So why is mercury typically used? We know from oxymercuration, that using mercury avoids carbocations that are prone to rearrangement. The sp2 (vinyl) carbocation shown in the first mechanism is less stable than sp3 carbocations, so we would expect rearrangement to be an issue. Avoiding the high energy carbocation also increases the rate of reaction.


text:
Which phase is more efficient for temperature changes?
In the case of attempting to heat and cool down an object of different states, if you in some case managed to achieve the same temperature of a solid, liquid and gas what state would scientifically heat or cool the object faster?
On my view point I suggested a gas to be more efficient simply due to how the particles move. However I was soon objected by someone else stating that a liquid would be far more efficient.
I was not given a reason , therefore I am asking which would be correct to assume and why.

A liquid is better than a solid for cooling an object, because it can make more complete contact with the object being cooled, plus it can undergo convection, so the layer of liquid next to the object being cooled is replaced by cooler liquid as the liquid circulates around the object.
A liquid is denser than a gas (giving better contact with the object being cooled) and also will likely have a higher heat capacity, which translates into a higher "cooling" capacity as well. But don't forget about "wind chill factor" - a moving blast of air is a lot more efficient at cooling an object than stationary air. Convection ovens take advantage of this principle too, but for heating rather than cooling. Liquids will probably be more effective than gases for most cooling applications, but there will probably be some exceptions.

The solid.
Place a warmer block of steel in contact with the same amount of water in each of its phases at the triple point. The ice will both give a lower temperature for the steel and it will cool faster since most of its heat capacity is at the freezing point. Thus there is a large temperature difference for the longest time.


text:
Nucleophiles and bases (elimination and substitution)
Is it safe to say that all elimination reactions (E1 and E2) involve bases (since in elimination reactions, a hydrogen is being removed, and thus something that accepts a $\ce{H+}$ is called a base)
Whereas, In substitution reactions, bases are not involved.
To summarise:

substitution reactions- can involve bases/nucleophiles/both
elimination reactions- must involve a base, but the base can also be a nucleophile (i.e. elimination reactions cannot occur if you only have a nucleophile)

All nucleophiles are bases and all bases are nucleophiles. It's just a question of how good they are at being a base versus a nucleophile.

The term base usually has a definition that I don't think a thiol or a phenyl group would meet, but both of which can act as nucleophiles

Show me the base in this reaction $\ce{Mg + (CH2Cl)2 = C2H4 + MgCl2}$

There are two possible definitions of base:

Substance which accepts a proton throughout the course of the reaction (catalyticly or stoichiometricly). This is basically the Brønsted definition. To accept a proton, the molecule in question needs a lone pair whose energy level and orientation is in a good range for a proton to attach (e.g: ammonia: good; phosphane: bad). We could also extend the definition Lewis-style.
Substance which generally may accept a proton in reactions. This is still basically the Brønsted definition only without the requirement of it actually happening. The molecule needs a lone pair whose energy level and orientation is in a good range for a proton to attach even if it doesn’t happen. We could also extend it to a Lewis-style definition.

Applying the first definition, nucleophilic substitution reactions usually do not require bases. However, the molecules that attack are generally bases: A nucleophile generally needs an accessable lone pair, as would a base.
Consider the case of an alcoholate. If it is methanolate, it is small and can attack nucleophilicly. However, we also know that it can act as a base to be protonated to generate methanol. Depending on the substrate, it would choose to do substitution or elimination reactions.
If the alcoholate is tert-butanolate, we have a large molecule. Technically, it still could attack as a nucleophile. However, it is usually too big to attack and rather abstracts protons to allow elimination reactions.
You cannot really draw a line. Both are bases, both could be nucleophiles.


text:
Wohler synthesis: magnesium cyanate + ammonium chloride?
Wohler famously produced urea by reacting silver cyanate with ammonium chloride,
AgNCO + NH$_4$Cl $\rightarrow$ (NH$_2$)$_2$CO + AgCl
Naive question: Could this work instead with magnesium cyanate Mg(OCN)$_2$ ?
(I ask because my undergrad chem lab has HCl, Ammonia, and magnesium strips, and I'd like to make some urea, if possible. Not sure how to make magnesium cyanate, but one step at a time...)

The reason Wohler used silver cyanate is because silver chloride precipitates out, leaving ammonium cyanate, which (in a separate step) was thermally converted to urea. If you used magnesium cyanate, there would be no precipitate since magnesium chloride is very soluble in water. However, if you evaporated the water and heated the mixture, you probably would end up making some urea. But it would not be pure.


text:
Produce dichlorine oxide by mixing sulphur dioxide and chlorine

Dichlorine oxide is used as bactericide to purify water. It is produced by the chlorination of sulfur dioxide gas.
  $$\ce{SO2(g) + 2Cl2(g) -> SOCl2(l) + Cl2O(g)}$$
  How many liters of dichlorine oxide ($\ce{Cl2O}$) can be produced by mixing 5.85L of $\ce{SO2}$ and 9.00L of $\ce{Cl2}$? How many liters of the reactant in excess are present after reaction is complete? Assume 100% yield and that all the gases are measured at the same temperature and pressure.

My attempt:
I just finished working out the problem and was hoping to get some feedback on my method of solving.

Convert both reactants from liters to moles using the ideal gas law and identify which is the limiting reactant.

n=$\frac{PV}{RT}$ = $\frac{\pu{5.85L}}{\pu{0.0821m/L}}$ = $71.2m \ce{SO2}$ (Constant T,P)
n=$\frac{PV}{RT}$ = $\frac{\pu{9.00L}}{\pu{0.0821m/L}}$ = 110m $\ce{Cl2}$ (Constant T,P)
$\ce{$71.2$m \ce{SO2 }X}\frac{1m\ce{Cl2O}}{1m\ce{SO2}} = 71.2m \ce{Cl2O}$ Excess Reactant
$110m \ce{Cl2}$ x $\frac{1m \ce{Cl2O}}{2m \ce{Cl2}}$ = $55m \ce{Cl2O}$ Limiting Reactant

Find amount of liters of $\ce{Cl2O}$ that can be made in the reaction by converting limiting reactant from moles to liters using the ideal gas law.

V=$\frac{nRT}{P}$ = $55m \ce{Cl2O}$ X 0.0821m/L = 4.5L $\ce{Cl2O}$ (Constant T,P)
4.5L of $\ce{Cl2O}$ can be produced by mixing 5.85L of $\ce{SO2}$ with 9.00L of $\ce{Cl2}$.

Find liters of excess reactant left over after reaction by finding number of moles of excess reactant that can react with limiting reactant and then subtracting number from original amount of excess.

$110m \ce{Cl2}$ X $\frac{1m \ce{SO2}}{2m \ce{Cl2}}$ = $55m \ce{SO2}$
$71.2m \ce{SO2}$ - $55m \ce{SO2}$ = $16.2m \ce{SO2}$ of excess left over after reaction.

There are 22.4 liters of gas in a mole at STP. So at STP there are  5.85/22.4 moles of SO2.

I'm lazy. Diving 9L by 2 gives you the ratio. So 4.5 is the product and 1.85L of SO2 left over.


text:
Would graphite stick to basalt cooled from its molten state in a graphite mold?
Molten metal that is poured into a graphite mold pops out easily once it has cooled. Would basalt poured the same way pop out just as easily?
I am particularly interested in lunar basalt, if that makes any difference. The image shows the composition used to make a lunar regolith simulant, taken from chapter 13 of Resources of Near Earth Space, by B. J. Pletka.

What are you going to melt that mixture in? It would seem that it would dissolve a ceramic or clay crucible.

@MaxW i was thinking graphite, with some tungsten reinforcement.

graphite burns in an oxidizing atmosphere. In a reducing atmosphere many of the metal oxides could react to form the metals. I'm sure that making such mixtures has been done before. I'm just curious...

@MaxW i'm figuring this out as i go along. I don't know anything about it, it's for a simulation of a lunar colony. Could the furnace be tungsten? The molten mixture would be poured in the open, so it is a vacuum environment. Oh, also, maybe if the temperature isn't higher than what is needed to melt the feedstock, say 1200 C, maybe there wouldn't be enough oxygen liberated for it to be an issue?

If the idea is to make bricks from basalt then the lunar sand would be sintered not melted.

@MaxW that isn't the idea. http://moonwards.com/index.html - click #5 of the second image for a very brief initial write-up. The concept needs lava in high volume. Sintering methods are a bit finicky, they need to be finely controlled, not very workable for a tele-operated approach.

If you put tungsten in contact with graphite it will form tungsten carbide

@A.K. a quick look at its properties doesn't make that seem like a problem.

No, it would not stick. In our lab we commonly use graphite dishes/cups/etc to contain small amounts of liquid rock. All you need to do then is to hold it upside down and the solidified rock (now glass mostly) falls out. Sometimes there's a bit of graphite dust on it, but it can be easily scraped off.
Doing it in a reducing (or oxygen-free) environment will make it even easier, because graphite slowly burns to carbon dioxide with time, on the scale of hours, from my own personal experience.


text:
Hybrid orbitals forming molecular orbitals

My teacher showed me this diagram on how the hybrid orbitals of two atoms combine to form molecular orbitals. I was confused by this because I thought that VB and MO theories were two separate theories. In VB theory, atoms form hybrid orbitals that overlap, and the electrons are located in the overlap.

In MO theory, molecular orbitals are formed from the superposition of regular atomic orbitals (not hybrid orbitals), so most MO diagrams I see are drawn like this:

So are molecular orbitals formed from regular (s,p,d, etc.) atomic orbitals, or are they also formed from the hybrid orbitals? And if hybrid orbitals do combine to form molecular orbitals, are the electrons still in between the two bonded atoms or are they delocalized like in the MO theory?

re: "and the electrons are located in the overlap." - The electrons have a high probability of being in the overlap, but they aren't "located" there alone. An electron in a bond "travels" around both nuclei of the atoms in the bond.

It is still LCAO-MO theory, but just dumbed down a lot. The difference is that, instead of feeding the "pure" atomic orbitals into the LCAO mechanism, you carry out an additional mathematical step in order to get orbitals that have nice directional properties, and you feed those into the LCAO mechanism instead.
Let's talk about a simple example, methane. You could hybridise the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals on carbon to obtain a set of four $\mathrm{sp^3}$ hybrids. If you now try to derive the symmetry of these $\mathrm{sp^3}$ hybrids, you will realise they transform as $A_1 + T_2$. And if you didn't bother doing the hybridisation step, guess what are the symmetries of the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals? The $\mathrm{2s}$ orbital transforms as $A_1$ and the $\mathrm{2p}$ orbitals transform as $T_2$. In the end, you get roughly the same picture through MOT regardless of whether you use hybridised orbitals or not. (Note that the $4\mathrm{sp^3}$ in the second diagram does not indicate a principal quantum number of $4$, but rather that there are four such orbitals. I should have changed it, but I was lazy.)

There is a difference between this and conventional VB theory. Conventional VB theory tells us that each of the $\mathrm{sp^3}$ orbitals overlaps with one of the hydrogen $\mathrm{1s}$ orbitals to form four identical bonds. Once you draw a MO diagram, complete with symmetries, you are entering the realm of MO theory.

However, part of the reason why that example worked so well is that methane has a central atom. When you come to larger molecules, and have to piece together different fragments of the molecule, there is a difference between LCAO using the hybridised and unhybridised orbitals. The best example is the very first MO diagram that you drew - I cannot tell what it is, but I think it is that of an alkene. So let's consider ethene, $\ce{C2H4}$.
In fact, the diagram is "wrong". There are 8 electrons in total in the constituent AOs, but the MOs only have 4 electrons. This is not because of any elementary mistake by your teacher. It is because two of the $\mathrm{sp^2}$ orbitals on each carbon do not point towards each other; they point towards the hydrogens instead. Therefore, they are assumed to be involved only in C-H bonding and not in C-C bonding. Strictly speaking, they should therefore not be shown on the MO diagram. Nevertheless, it is this assumption that gives us some insight into the difference between this "hybridisation-LCAO" treatment and normal LCAO treatment using the pure AOs.
We'll first look at the "real" MOs of ethene (image taken from Fleming, Molecular Orbitals and Organic Chemical Reactions):
$\hspace{15 mm}$
We can try and draw some parallels between your "simplistic" MO diagram and this "proper" MO diagram here. In the simplistic diagram, the C-C $\sigma$ bonding arises from overlap of the two $\mathrm{sp^2}$ hybrid orbitals. So, this is likely to somewhat correspond to the $\pi_x$ MO above, which kind of looks like what you would expect from constructive overlap of $\mathrm{p}$ orbitals on carbon pointing towards each other. However, there's likely to be a bit of the $\sigma_s$ MO mixed into it, because you are using $\mathrm{sp^2}$ hybrid orbitals which also have the $\mathrm{s}$ orbital mixed in. The other filled MOs are not likely to contribute since they are antibonding with respect to the two carbons. In fact, they are mostly there to provide the additional C-H bonding required to make the molecule stable, which is exactly what is neglected in the simplistic diagram.
However, the $\pi$ bonding orbital in your simplistic diagram is a resounding success. It's exactly the same as the $\pi_z$ MO in the real diagram! And the $\pi^*$ antibonding orbital in the simplistic diagram is also exactly the same as the $\pi_z^*$ orbital in the real diagram! This leads us to the main point: we can do this much simpler MO treatment using hybrid orbitals and get a sufficiently accurate MO picture, without having to go through a bunch of symmetry arguments to determine which AOs overlap and which AOs don't. Furthermore, guess which are the most important MOs that control the reactivity of ethene? Exactly; it's the two $\pi$ orbitals.

Now that I've gotten that out of the way, I can proceed to answer your specific questions:

I was confused by this because I thought that VB and MO theories were two separate theories.

What I wrote above mostly applies to this confusion. The crux of the matter is, hybridisation can be applied to VB theory and it is actually in this context that you first meet hybridisation, but it is not exclusive to VB theory. You can do a little mix-and-match in order to simplify your life, without sacrificing too much accuracy.

Are molecular orbitals formed from regular (s,p,d, etc.) atomic orbitals, or are they also formed from the hybrid orbitals?

It is your choice. One leads to a more accurate picture, but it becomes extremely tedious as the molecule becomes more complex.

And if hybrid orbitals do combine to form molecular orbitals, are the electrons still in between the two bonded atoms or are they delocalized like in the MO theory?

The electrons are localised. This is most obvious from the difference in the $\sigma$ orbitals shown above. Full delocalisation a la "proper" MO theory would imply that there are MOs which lead to bonding over the whole molecule, i.e. the $\sigma_s$ orbital.

Then why do you still call it MO theory if the electrons are localised?!??!!??!

You didn't ask this question, but I anticipated you might wonder about it, since I've been going on and on about MOT only to end up saying that the electrons are localised. It's still LCAO-MO theory, just that the basis set of orbitals chosen is different, which leads to a different form of the wavefunction which happens to be localised between two nuclei. In terms of the electron density distribution, it is the same qualitative picture that is painted by VB theory. But, remember that VB theory has no idea what an antibonding orbital is, or that it even exists. So, what you have done is: you have actually arrived at the same result (localised electron density) by another means (MOT).

Disclaimer: There are people who will debate exactly what "VB" and "MO" theories are. Wavefunctions exist in VB theory as well, except that they are constructed differently, and one might very well say that the simplistic MO theory above is, in fact, VB theory. Nevertheless, I am sticking to the introductory, qualitative notions of VB and MO theory here.

Thanks for the detailed answer, although I couldn't understand all of it because I haven't learned how to apply MOT to molecules that are not diatomic. So what I think you're trying to say is that in MOT, sometimes the MOs are localized in the places VBT predicts, and other times the MOs are delocalized over the while molecule. Is this correct? And thus combining VBT with MOT ends up giving you the same result as using MOT by itself (with AOs), just that it's simpler for larger molecules?

"in MOT, sometimes the MOs are localized in the places VBT predicts, and other times the MOs are delocalized over the while molecule" - Yes. It depends on the basis set of orbitals that you use to generate your MOs, which is a fancy way of saying which atomic orbitals (or hybrid orbitals) you are going to combine to form MOs.

"thus combining VBT with MOT ends up giving you the same result as using MOT by itself (with AOs), just that it's simpler for larger molecules" - You don't combine VBT with MOT, you combine hybridisation with MOT. Hybridisation ≠ VBT. The result is not exactly the same. However, the differences are not large enough to make the simpler approach invalid.

@orthocresol: Could you tell what are the advantages of using hybrid orbitals in MOT? There must be a reason to use hybrid orbitals in place of AO. Could  you please tell?

@user36790 Sorry, I forgot to reply. Yes, there is a huge advantage, and that is simplicity. Try "rigorously" constructing the LCAO-MOs for 2-fluoroethanol, and explaining why you observe the gauche effect - vs. using the hybrid-orbital approach as described in Jan's answer. One is much easier, and it's not the rigorous approach!

In simple molecules (H-H, H-F etc.), simple molecular orbitals can be constructed on a piece of paper by combining atomic orbitals, to give the molecular orbitals. It is, of course, worth remembering that these hand-drawn MO diagrams are just qualitative representations– true molecular orbitals are generated computationally using quantum mechanical calculations to approximate/calculate the various wave functions representing the MO's.
For more complex molecules (even something like ethene), it begins to get difficult to draw simple qualitative MO diagrams as there are so many orbitals to combine, and without  computation, it can be difficult to know which orbitals to combine etc.
Hybrid orbitals (also sometimes referred to as group orbitals) are a way of simplifying the idea, by allowing us to combine an sp3 carbon, or an sp2 carbon, without having to always derive from first principles.
In this sense, the MO theory provides a rationalisation for why VB theory works when using hybrid orbitals.


text:
Preparing polyurethane with IPDI (Isophorone Diisocyanate)
I am trying to prepare a polyurethane using hydroxy-terminated polybutadiene (a polyol) and IPDI (a diisocyanate) as the curing agent. I intend to use DBTDL as a catalyst. I read that both IPDI and DBTDL are very toxic and dangerous to use. Being a non-chemistry student, I do not have a lot of experience dealing with chemicals. Could someone point to finer details involved in the process (things like the PPE used, whether the reaction needs to be done in a fumehood, etc.)?
Thanks.

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site and visit the help center for any unanswered questions about it.