text:
Why is the equilibrium constant defined that way intuitively?
Suppose we have chemical equation: $$\ce{ aA + bB  <=>   cC + dD }$$
then equilibrium constant is defined: $$K=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
but why don't we define it as: $$K=\frac{cd[C][D]}{ab[A][B]}$$
or even as: $$K=\frac{c[C]+d[D]}{a[A]+b[B]}$$
As adding concentrations is additive and not multiplicative. And because: $$\ce{ aA + bB <=> cC + dD }$$
is the same as: $$\ce{ cC + dD <=> aA + bB }$$
then we should have: $$K=\frac{[A]^a[B]^b}{[C]^c[D]^d}$$
and specifically: $$\frac{[A]^a[B]^b}{[C]^c[D]^d}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
but trying some examples shows that this isn't the case.

Mostly because of $\ln K = -{\Delta G\over RT}$.

@IvanNeretin never heard of that before, could you please give more detail?

Why, https://en.wikipedia.org/wiki/Equilibrium_constant#Derivation_of_equilibrium_constant_expression

http://chemistry.stackexchange.com/questions/4262/interpreting-equilibrium-equations-correctly

The final pair (with reciprocals) is convention only.  Had we gone with reactants divided by products, the equation given by @IvanNeretin would have no minus sign, but would remain otherwise unchanged.  There's no reason that a forward and backward reaction should have the same equilibrium constant, only that their equilibria will occur under the same conditions.  The one that bugs my students more is that if you look at a rxn multiplied by a constant, K changes.  That is, K for A --> B is not the same K as for 2A --> 2B.

K (as opposed to $K_C$) can be redefined for differing reactions, though the results are likely to be meaningless.   Flipping the reaction backwards also flips time backward.   Equilibrium as time approaches negative infinity is a meaningless concept.

"Adding concentrations is additive." Yes, and multiplying them is multiplicative. But what you're probably grasping at is that to characterize a reaction, it might make sense to add and subtract energies. This translates to multiplication and division for the equilibrium constant, being defined as the log of an energy (as Ivan Neretin mentioned above).

It can be understood in simple terms with just few intuitive keys (of course it is not very rigorous).

Take in mind that $K$ is a measure of how displaced to reactive or equilibrium the reaction is. So it should be greater when products concentration is relatively high (that is why they go into the numerator) and small when reactive concentrations are relatively high.
In elemental reactions (technical term) is like a balance between the speed of the reaction in the direction of products and the reverse reaction.
A reaction only occurs when its reactives are in the same region of the space. So, if you have an elmental reaction of the form $a A + b B \rightarrow $, for the reaction takes place it is needed that $a$ molecules of $A$ and $b$ molecules of $B$ collide. This is the main point: How probably is this event? Well, if you only need one molecule of $A$ ($a=1$), the probability that an $A$ molecule be in this specific place (at some time) is proportional to the $A$ concentration. If $(a=2)$ we need another $A$ molecule, as the probability for a second molecule being in that space region (at some time, in particular at the same time) is independent, the probability of finding two $A$ molecules in the same space region (at the same time) is the product of the two individual probabilities (is like throwing a dice two times). In general, for the mixture the probability of the encounter is proportional to $[A]^a [B]^n$.
The reaction speed in that direction is proportional to the number of encounters.

That is why you need to make products. Of course equilibrium constant $could$ be defined in other ways, for example, suppose that we define them as $K_{new}=ln(K_{old})$. In such case for the logarithm properties you'll have other expressions that includes sums. But there is not advantages of defining $K$ in that way.

A background in differential calculus is necessary for the answer to be intuitive. The equilibrium constant, K, is related to the asymptotic behavior of the kinetics model used to describe the system.
For a chemical equilibrium system using linear differential equations the equilibrium constant, K, is often noted with a C subscript as follows:
$$K_C=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
The following model is one of the simpler examples.  (In this example r is defined as reaction rate.)
Chemical Equation
$$\ce{ A <=> C }$$
In the equation for $K_C$ given above the power terms (a,b,c,d) are for higher order reactions.  In this example c=1, d=0, a=1, b=0. D and B are irrelevant.
Assume the amount of A lost equals the amount of B gained.
$$ -\frac{\mathrm d[A]}{\mathrm dt} = \frac{\mathrm d[C]}{\mathrm dt} = r $$
The equilibrium constant is calculated at reaction rate equals zero.
$$ r = 0 $$
Assume the loss of reactant concentration is proportional to the concentration of reactant in the system.
$$ -\frac{\mathrm d[A]}{\mathrm dt} = k[A] $$
Boundary Conditions that fit this system:
Assume the reaction is reversible and does not to complete. At time equals infinity $(t=\infty)$ some but not all of the initial $[A]_0$ is consumed.
$$[A]_0=[A]_\infty + [C]_\infty$$
$$[C]_0=0$$
Understanding this boundary condition is the key to understanding why the equilibrium constant works only in certian situations.
Solving this system of ordinary differential equations. (This step is tough to understand without a background in differential calculus.)
$$[A] = ([A]_\infty + [C]_\infty) - [C]_\infty( 1 - e^{-kt}) $$
$$[C] = [C]_\infty ( 1 - e^{-kt}) $$
As $t \to \infty$, $e^{-kt} \to 0$, $[A] \to [A]_\infty$ and $[C] \to [C]_\infty$.  Because $[A]_\infty$ and $[C]_\infty$ are constants in the model, their ratio is constant.
$$K_C = \frac{[C]_\infty}{[A]_\infty}$$
Because of the boundary conditions, if the initial concentration of reactant is increased by some factor y the ratio of $K_C$ remains the same.
$$y[A]_0=[A]_\infty + [C]_\infty$$
$$[A]_0=(1/y)[A]_\infty + (1/y)[C]_\infty$$
$$K_C = \frac{y[C]_\infty}{y[A]_\infty}$$
$$K_C = \frac{[C]_\infty}{[A]_\infty}$$
If there is an initial concentration of the product in the system then the assumption comes into play that the reaction is reversible.   This means the reaction can be initially reversed to a point no products exists before it proceeds forward.
$$[A]_t+[C]_t = y[A]_0 $$
If one assumption in the system does not match reality, for example if a reaction is irreversible, the equilibrium constant does not work.  For a system with reaction rates defined by linear differential equations, for example reversible reactions in solutions, the equilibrium constant works.

This doesn't seem like an answer to me.

@DGS   Your background must be theoretical.

There are many, many ways to define K

I'm not sure if you didn't read the question or if you just reeeeally want to force something in that is unrelated, but you don't even give a wrong answer to this question. You just don't have any kind of answer in there at all.

The model given in the question (which is typically taught in elementary chemistry) is based on statistical mechanics.  All of the assumptions made in the original statistical mechanics equations need to be valid for the model to work.    Usually the assumptions are not valid so an empirical model is used for K.    @DGS  You would need an engineering background to fully appreciate my response.

You only talk about kinetics. You don't relate what you said back to the question on equilibrium. I've done kinetic experiments. No one coming here for the answer to the question asked is going to say "Oh, well, this answer about kinetics totally answers that question about equilibrium."

Oh...    The confusion makes sense.   The equilibrium constant is the asymptote as time approaches infinity of the kinetics model being used.

Ah. That all makes perfect sense now, actually.

This discussion reminds me of a paper addressing a similar topic on thermodynamic kinetics.  Its title is "Steps in a Minefield."   I'll try rewording the answer to avoid this confusion.

At equilibrium, the forward rate of reaction is equal to the backward rate of reaction. Let us consider the equation:
$$\ce{aA +bB<=>cC +dD}$$
The forward rate of reaction is given by
$$\text{rate}_\mathrm{f}=K_\mathrm{f} \cdot [\ce{A}]^a [\ce{B}]^b$$
The reverse rate of reaction is given by:
$$\text{rate}_\mathrm{r}=K_\mathrm{r} \cdot [\ce{C}]^c [\ce{D}]^d$$
When the two are equal:
$$K_\mathrm{f} \cdot [\ce{A}]^a [\ce{B}]^b=K_\mathrm{r} \cdot [\ce{C}]^c [\ce{D}]^d$$
$$\frac{K_\mathrm{f}}{K_\mathrm{r}}=K_\mathrm{eq}=\frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$
The equilibrium constant of the reverse reaction:
$$\ce{cC +dD<=>aA +bB}$$
Is given as the inverse of the equilibrium constant of forward reaction. They are (usually) not equivalent:
$$\frac{1}{K_\mathrm{eq}}=\frac{K_\mathrm{r}}{K_\mathrm{f}}=\frac{[\ce{A}]^a [\ce{B}]^b}{[\ce{C}]^c [\ce{D}]^d}$$

Looks good, but powers (a,b,c and d) in the rate equation are not necessarily the stoichiometric values in the balanced chemical equation.

@MaxW for elementary reactions, they are.


text:
How does fluorouracil inhibit thymidylate synthase?
I'm curious to know how just by adding a fluorine atom to uracil (to form 5-fluorouracil) permanently inhibits the enzyme thymidylate synthase when they bind together. This molecule is widely used as an antimetabolite in chemotherapy as it stops the synthesis of thymine.

What changes undergoes the enzyme-inhibitor compound so it is permanently bonded to the fluorouracil? Probably the chemistry is too complicated to be explained in this Q&A format, if that is the case, references like books or papers would be great.

Should I ask this in biology.SE?

Possibly, but I think asking it here first was warranted. Questions on biochemistry / chemical  biology have been understandably complicated. The biochemistry would typically stop at the statement, "X inhibits Y", but your question is, "What is the chemistry of the X-Y interaction that results in inhibition of Y?" It's more topical for Chem.SE, though the expertise you need might be only on Bio.SE. Wait and see what answers come, I'd say.

compare strucutures of the substrate, the product and the drug. It sees, that fluorine occupies position where hydrogen should be replaced with methyl in the reaction catalised by the ferment. To my knowledge, biogenic methylation is electrophilic (i.e. methyl is transfered as $\ce{Me+}$ in $S_N2$ -type reaction. For this to happen, a negative charge on the atom next to fluorine should occur, which would lead to dissociation of fluoride ion and active carbene generation, readily binding to the closest molecule. At least it is what I can come with out of the blue.

However, for a definitive answer you should study extensively the literatute. It is pretty possible, that there is no known agreed upon mechanism of action.

@permeakra   thanks,  you gave me a rough idea. I'll try to find a book, maybe Fluoropyrimidines in Cancer Therapy, but it seems a bit advanced for me.

I've seen many examples where substituting F for H does havoc on biology. Not sure if there is some reaction with the enzymes, or if the just bind and try in vain to do their job. Would love to see an answer on this.

I think this will solve your problem: http://www.nature.com/nrc/journal/v3/n5/fig_tab/nrc1074_F2.html

@ShankRam I agree, this does appear to answer OP's question. I recommend posting it as an answer along with a paraphrase of the key details.

@Brian It will be better if you post an answer because your level of expertise is required to explain such a complex topic properly. Here, this link is better: http://www.sciencedirect.com/science/article/pii/S0925443902000820

The inhibition comes about because:

Uracil and $5$-fluorouracil are sufficiently similar that both moieties have a strong binding affinity for the active site of thymidylate synthase.
The presence of the fluorine prevents $5$-methylation of the uracil core by a folate cofactor, which otherwise would turn it into a thymine moiety.
This inability of the enzyme to convert the $5$-fluorouracil moiety into a thymine moiety (which has a much lower affinity for the active site) means that the enzyme has no way to reject the $5$-fluorouracil or the methylating folate species, once bound.

The particulars of this interaction are nicely spelled out in Mishanina et al.$^\dagger$, conveniently available in preprint:

[5-fluorodeoxyuridylate] reacts [as normal] with the active site cysteine and undergoes condensation with [N,N-methylene-tetrahydrofolate] but prevents proton abstraction and elimination of the tetrahydrofolate [product], as a consequence stalling the reaction at the covalent ternary complex [of enzyme, inhibitor, and folate species].

Scheme 4A from the preprint linked above illustrates well the specifics of the chemistry.  (I refrain from reproducing it here due to uncertainty regarding its copyright/license status.)
As well, Figure 2 from the Mechanism Description section of the Wikipedia page on thymidylate synthase shows very nicely the enzyme interactions with the normal deoxyuridylate substrate (click image for larger version):
$\hspace{20pt}$
Image by CJerc (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0 )], via Wikimedia Commons
$^\dagger$ Mishanina et al. Bioorganic Chemistry 43: 37 (2012). doi:10.1016/j.bioorg.2011.11.005

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text:
Why is mercury's surface tension so high, when its viscosity is low?
At a basic level, both surface tension and viscosity are the result of forces between particles. Usually we'd talk about intermolecular forces, because liquids are usually molecules, but obviously that doesn't work for mercury, which isn't made of molecules at all. Stronger forces between particles give higher surface tension, and usually also higher viscosity. So water's fairly high surface tension is usually credited to hydrogen bonding, and mercury's exceptionally high surface tension is put down to metallic bonding; while the increasing viscosity of aliphatic hydrocarbons with more carbons is said to be due to increasing London dispersion forces.
It's easy to see that the two are poorly correlated, though. Ethanol has almost identical surface tension to acetone, for example, but it's more than three times as viscous. Most strikingly, mercury has really extraordinarily high surface tension, but is no more viscous than cold water.
Is it completely wrong to expect at least some kind of loose correlation between surface tension and viscosity? Is viscosity just a far more complex phenomenon, a proper explanation of which lies way beyond the hand-wavy school science that chemistry teachers use to give a vague idea about things like trends in the alkanes? Would it be better to emphasise entanglement of long molecules when trying to explain the viscosity of longer alkanes, rather than just stronger forces?
Is mercury's low viscosity mainly down to its small particle size, or is there much more to it than that?

I'd say it is due to the nature of metallic bond. See, it is undirected.

How is that significantly different from any of the intermolecular forces behind viscosity and surface tension in other liquids, @Ivan?

Sorry, I don't remember what I had in mind back then in February when writing my first comment. As it seems now, yes, it would be better to emphasize the entanglement of long molecules and all.

Related discussion on physics SE: https://physics.stackexchange.com/questions/148792/viscosity-and-surface-tension                                                                                Also, this chemical engineering article developed an empirical relationship between viscosity and surface tension: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=10&ved=0ahUKEwjglP_b-4vWAhVpyFQKHXCGCRMQFghmMAk&url=http%3A%2F%2Fwww.banglajol.info%2Findex.php%2FCERB%2Farticle%2Fdownload%2F26217%2F17599&usg=AFQjCNFTUdBYyTsAL7cg8TZncLqEHBOO-A

One important difference is that surface tension is an elastic property, another that while surface deformation requires some viscous flow (which has a similar activation energy than bulk viscous flow), it also requires a net severing (or re-forming) of intermolecular bonds, which has a different activation energy and $\Delta H \neq 0$

This is out of my area, so I'll be interested to hear from others with more expertise, but here's how I think of it:  The surface tension of a liquid under air is a thermodynamic property resulting from the difference between the free energy of bulk vs. surface liquid molecules (due mostly to liquid-liquid vs. liquid-air bond energies). Typically the former is much more favorable, hence the liquid will try to minimize its surface area:volume ratio.   By contrast, viscosity is a kinetic property determined by the activation energy involved in breaking and reforming the liquid-liquid bonds.

I think the complex phenomenon here is the nature of the metallic bond. It is commonly said there is a "sea of electrons" that are delocalized across the entire metal and that the metal ions just bash around in the sea against each other. So when they're in the metal bulk phase they feel about the same strong force on all sides (thus as Ivan said, the force is undirected). Therefore in the liquid the atoms can slip past each other OK. The problem with the vapor phase is that it's no longer metallic. An ideal gas (an excellent approximation for mercury or most other vapors at STP) has no forces between molecules, by definition. So you've got to pull the metal atom completely out of the "electron sea" (within which it is strongly-bound) to vaporize it, and that requires a lot of energy and leads to a high surface tension. Surface tension is inherently a two-phase property!
I think the contributing factors are:
Viscosity - liquid only, low energy change for motion, low viscosity
Surface tension - liquid-vapor, large energy change, high surface tension

It's beyond high school level, but I found Rowlinson and Widom's "Molecular Theory of Capillarity" gave very a good and thorough explanation of surface tension.

Ah, interesting subtlety with the uniformity of the metallic bonds. So it's not just attraction to other particles that makes a liquid viscous, it's the energy required to move from one place in a liquid to another, and in something like glycerol, even though the forces are less strong than in mercury, particles have to navigate a highly non-uniform electrical field. They can't rely on the same forces of attraction as they move from one spot to another.

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text:
Can a supersaturated solution get cooler during crystallization?
Suppose we have a supersaturated solution that is isolated from the environment. Some of the solutes condense into a crystal in the solution. Is it possible for the solution to get cooler during this process?
If so, why doesn't it violate the second law?

And why would it get cooler? Start there.

Sorry, I don't understand your comment. I want to know whether this is possible. Has it ever been observed? How does your comment help me understand that?

Compare the free energies of the two states. Since it is supersaturated, the solution would prefer (that is, have a lower free energy) if crystals formed - it is thermodynamically favored. Once crystals begin to form, where does that excess free energy go? (Hint - in supercooled liquid->solid transformations this is called recalescence).

Seriously, I'm not in need of a lesson on thermodynamics; I'm a physics PhD student. I just want to know if this ever happens. It is claimed that it does on page 584 here (http://www3.nd.edu/~powers/ame.20231/wright1970.pdf) but the reference it gives doesn't seem to corroborate it, that I can determine.

The system sodium sulfate (Na2SO4) / sodium sulfate decahydrate (Na2SO4 * 10 H2O) / Water might be an example.

Dissolving Na2SO4 in water is exothermic due to hydration.
Dissolving Na2SO4 * 10 H2O is endothermic (entropy effect).

Below 32.384°C Na2SO4 * 10 H2O crystallizes from aqueous solution, above 32.384°C water free Na2SO4.
The solubility of sodium sulphate shows the following unusual shape with a maximum at 32.384°C.

If you would now prepare a saturated solution of sodium sulphate at 32.384°C, and would then be able increase the temperature without precipitation of Na2SO4, you would get a supersaturated solution of sodium sulphate.
Putting a seed crystal of Na2SO4 into it should result in crystallization with the solution getting cooler during this process.
Note that the crystallization frees water molecules that were previously bonded through hydration. This will lead to an increase of entropy.

That's a sneaky system.  I particularly like the entropy increase via crystallization.  I hope it works.  :-)


text:
What are the roles of pyridine and DCM in the acylation of an alcohol?
What role does pyridine and $\ce{CH2Cl2}$ play in this reaction?

(Prior to this step, the corresponding carboxylic acid was converted to the acyl chloride using DMF, oxalyl chloride, and DCM.)
This reaction is taken from Bear, R. B.; Shea, K. J. Tandem Hetero Diels−Alder Reaction: Synthesis of Oxygenated Macrocycles. Org. Lett. 2001, 3 (5), 723–726. DOI:10.1021/ol007040s..

Dichloromethane (DCM) is a common organic solvent and this is its role in both reactions (and in most cases where it is used). The first reaction is indeed the conversion of a carboxylic acid to an acyl chloride. The DMF is a catalyst and the reactions proceeds through an imidoyl chloride intermediate, $\ce{Me2N+=CHCl}$.
Pyridine is a weak base, with a $\mathrm{p}K_\mathrm{aH}$ of 5.21 and it is often used to remove extra protons during the course of a reaction. It is also a good nucleophile for attacking carbonyl groups to activate them towards attack by other, less good nucleophiles (thanks to @orthocresol and @DGS for pointing this out). In this reaction it serves both these purposes.

An acid anhydride can also be used as the acylating agent. The principle is the same: pyridine acts as a nucleophilic catalyst and generates the same intermediate as above.


text:
Orbitals and Pauli Exclusion Principle without Slater determinants
I'm trying to understand how to apply the pauli exclusion principle to the Born-Oppenheimer approximation of molecular structure.
Let me preface this by saying I'm not a chemist nor physicist. I'm a mathematician by trade which has the side affect of understanding electronic structure as solutions to a PDE.
So given some set of fixed nuclei and a given number of electrons I can construct the Schroedinger equation:
$$ \left(-\frac12\sum_{i=1}^N\Delta_i^2 - \sum_{i=1}^N\sum_{A=1}^M\frac{Z_A}{|r_i - r_A|} + \sum_{i=1}^N\sum_{j>i}^N \frac{1}{|r_i - r_j|} \right) \Psi = \varepsilon \Psi $$
I've read in Szabo and Ostlund that the pauli exclusion principle manifests itself by saying the wavefunction $\Psi$ must be antisymmetric in spin+location coordinates.
The next thing most people do is to introduce Slater determinants from 1-electron orbitals. This is an approximation to the solution space of wave functions, evidenced by the fact that FCI procedures must assume a sum of slater determinants for more accurate solutions. I want to keep this theoretical, so for now I don't care much for computational considerations.
The problem becomes as such, when talking about slater determinants it makes sense to append a spin variable and to talk about spin-orbitals of single electrons. Without slater determinants, single electron orbitals don't make sense (we only have the N-electron Schroedinger equation). Appending a spin coordinate also doesn't make sense because anti-symmetry of $\Psi$ and the Schroedinger equation don't see it.
So please help me out here, assuming I can solve the electronic Schroedinger equation exactly, what conditions do I need to place on the wavefunction so that the Pauli Exclusion Principle is satisfied (giving me a physical solution)?

I think what you are asking is "What form does $\Psi$ need to take so that the exchange interaction exists when the electronic Hamiltonian given acts upon it?". (Noting that the exchange interaction is due to the fact that electrons are indistinguishable and antisymmetric with respect to their positions).
From what I understand the Hamiltonian you have given acts upon electrons and so does not include the exchange interactions. As far as I'm aware the analytical form of the electronic Hamiltonian that includes exchange is not known. If it were, then density functional theory would not have so many approximate exchange-correlation functionals.
I will try to show the problem by presenting the form of the Schrodinger Equation in a slightly different way than is presented in Szabo and Ostlund's book.
Firstly, it is understood that $N$ in the Hamiltonian you give in the question represents electrons - not spin orbitals and we know that the electron density $\rho$ at $\textbf{r}$ is represented: \begin{equation} \label{eq1} \rho(\textbf{r}) = |\Psi(\textbf{r})|^2 = \langle \Psi|\textbf{r}\rangle\langle \textbf{r}|\Psi \rangle = \Psi^*(\textbf{r})\Psi(\textbf{r}) \end{equation}
The Hamiltonian you give acts upon the electron density by using spin orbitals to represent it.
Note, a physical electron density must therefore arise from a wavefunction that is anti-symmetric.
The Hamiltonian you have given can be seen to act upon the wavefunction as follows:
\begin{align}
E = &\int \Psi^*(\textbf{r}_1)(-\frac{1}{2}\nabla^2-\sum_{A=1}^{M}\frac{Z_A}{|\textbf{r}_1-\textbf{r}_A|}) \Psi(\textbf{r}_1) d\textbf{r}_1 \\ &+\frac{1}{2}\int\int \Psi^*(\textbf{r}_1) \Psi^*(\textbf{r}_2) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \Psi(\textbf{r}_1)\Psi(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2
\end{align}
Note that the second term - concerning two-electron integrals shows the electron density at $\textbf{r}_1$ interacting with the electron density of $\textbf{r}_2$; this is clear when rearranging the final term (knowing that the $r^{-1}_{12}$ operator is a Hermitian operator) and substituting in the first equation:
\begin{align}
\langle 12|12\rangle &=\int\int \Psi^*(\textbf{r}_1) \Psi(\textbf{r}_1) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \Psi^*(\textbf{r}_2)\Psi(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2 \\ &= \int\int \rho(\textbf{r}_1) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \rho(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2
\end{align}
Now, it is well known that electrons are indistinguishable, (i.e. they exist with a given probability density over all space and an electron cannot be labelled) such that: $$ |\Psi(1,2)|^2 = |\Psi(2,1)|^2 $$ Where $1=\textbf{r}_1$ and $2=\textbf{r}_2$.
It is also known that, as fermions (half-integal spin particles) they are anti-symmetric:
 $$ \Psi(1,2) = -\Psi(2,1) $$
Applying this antisymmetry to the two electron integrals we know that there has to be an antisymmetric electron interaction due to the interchanging of electronic positions. Therefore, the two electron integrals can be written: $$ \langle 12||12\rangle = \langle 12|12 \rangle -\langle 12|21 \rangle $$ $$ \langle 21||21\rangle = \langle 21|21 \rangle -\langle 21|12 \rangle $$ both the negative terms are the exchange interactions associated with the interchanging positions. Which are not included in the electronic Hamiltonian given in the question and their integral forms cannot be rearranged as shown before into terms of the electron density at a single given position.
This is why the Fock Operator is used instead of the electronic Hamiltonian!
I hope this all makes sense! I have probably made a mistake somewhere! So please let me know!

If you would like to know more about calculating the exchange from the electron density for known wavefunctions I recommend the paper: Pendás, A. M.; Blanco, M. A.; Francisco, E. Two-electron integrations in the quantum theory of atoms in molecules. J. Chem. Phys. 2004, 120 (10), 4581–4592. DOI: 10.1063/1.1645788.
For normailised wavefunctions, the exchange is given:
\begin{equation}
K = - \int \int r_{12}^{-1} \vert \rho_{1}(\textbf{r}_{1};\textbf{r}_{2}) \vert^{2}d\textbf{r}_{1}d\textbf{r}_{2}
\end{equation}
Where:
$$\rho_{1}(\textbf{r}_{1};\textbf{r}_{1}^{\prime}) = N_{\text{elec}} \int \Psi_{\text{elec}} (\textbf{r}_{1},...,\textbf{r}_{N_{\text{elec}}})\Psi_{\text{elec}}^{*}(\textbf{r}_{1}^{\prime},...,\textbf{r}_{N_{\text{elec}}}) d\textbf{r}_{2}...d\textbf{r}_{N_{\text{elec}}}$$

So please help me out here, assuming I can solve the electronic Schroedinger equation exactly, what conditions do I need to place on the wavefunction so that the Pauli Exclusion Principle is satisfied (giving me a physical solution)?

If the wavefunction is a totally antisymmetric tensor then Pauli Exclusion principle is satisfied. Take a look at another post I wrote where I give a reference
that explains the relation of indistinguishability, antisymmetry and Pauli Exclusion principle.
The wavefunction has also to be normalizable, with first derivative with respect to coordinates integrable.
The problem of finding the ground state (after Born-Oppenheimer) is presented as a minimization
$$ E_0 = \text{inf}\{ \langle \Psi_e,H\Psi_e \rangle, \Psi_e \in \mathcal{H}_e, ||\Psi_e||^2_{L^2} = 1 \} $$
where $\mathcal{H}_e = \wedge^N H^1(\mathbb{R}^3 \times \{1,0\})$.
Hartree-Fock variationally restricts $\mathcal{H}_e$ to the space of functions that can be written as single determinants in terms of functions of $\mathbb{R}^3$ (spin-orbitals).
Note that a determinant is just the simplest element of the alternating space $\mathcal{H}_e$. That is when exchange appears.
After projecting the problem in a finite dimensional space (Galerkin) and imposing first order optimality you get to the Fock matrix, which you don't know while you also don't know the coefficients of discretized wavefunction in a one electron basis. So your linear problem becomes a non linear problem.

Let me preface this by saying I'm not a chemist nor physicist. I'm a mathematician by trade which has the side affect of understanding electronic structure as solutions to a PDE.

Given that you are a mathematician you will have an easier time reading from a mathematician. I highly recommend you to read Computational quantum chemistry: A primer. It is only about 250 pages (shorter than Szabo) and explains many methods.

There are some points in the answer of CompChem that are not right and I would like to clarify. In comments there is not enough space.

From what I understand the Hamiltonian you have given acts upon electrons and so does not include the exchange interactions. As far as I'm aware the analytical form of the electronic Hamiltonian that includes exchange is not known. If it were, then density functional theory would not have so many approximate exchange-correlation functionals.

If the Hamiltonian acts only on electrons does not mean that it does not include exchange interactions, it only means that the nuclei have been decoupled from electrons. A more technical explanation in terms of the spectrum of the Hamiltonian is given in the reference above. In fact the Hamiltonian given includes electron-electron interactions so it has its sibblings Coulomb and exchange interaction.
It is not true that exchange and correlation expressions are not now, CompChem is mixing DFT with wavefunction methods. OP never says DFT. The problem in DFT is that the main variable is the density and the exchange and correlation energies are functionals of the pair density, not the density. So there is not known way to express it in terms of the density.

The Hamiltonian you give acts upon the electron density by using spin orbitals to represent it.

The same as above, the Hamiltonian given by OP acts on the wavefunction ($\Psi$), not the density. That would be true for DFT.


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How do I convert 4-(Methylthio)benzaldehyde to 4-(thioacetate)benzaldehyde?
Sorry if I have the incorrect name of the final product. I'm trying to acylate the methyl mercaptan substituent to a thioester (thioacetate). I appreciate any help!

The preferred prefix for the group $\ce{CH3-CO-S-{}}$ is “acetylsulfanyl”, which would lead to the product name “4-(acetylsulfanyl)benzaldehyde”. The former name “acetylthio”, which would lead to “4-(acetylthio)benzaldehyde”, may still be used in general nomenclature. However, the ester actually has seniority for citation as suffix, which leads to the preferred name “S-(4-formylphenyl) ethanethioate”.

You could react 4-(methylthio)benzaldehyde with $\ce{HBr}$ to form a thiol, and then react the thiol with acetyl chloride in triethylamine to form 4-(thioacetate)benzaldehyde.


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Chemical reactions as exponential functions
Are all graphs of molar concentration vs. time, for the reactants, exponential decay functions? (Until equilibrium is reached, if it goes to equilibrium at all.)
If so, intuitively, why is that the case?
EDIT: @ShankRam has brought it to my attention that zero-order reactions do not entail exponential decay functions. However, I would still like to know if first-order and second-order reactions do, and why they do.

No, they are not: http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Zero-Order_Reactions

Second-order reactions don't do it either: http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions

Intuitively, we can understand why only a first order reaction (at least for noncomplex reaction orders) displays exponential decay because a first order reaction implies that the rate of decay ($d[A]\over dt$) is proportional to the concentration ($[A]$). This relationship precisely describes an exponential function.


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Carboxylic Acid to Acyl Chloride
Instead of using oxalyl chloride and DMF then $\ce{AlCl3}$, could $\ce{SOCl2}$ followed by $\ce{AlCl3}$ been used to achieve the same result? If yes, is one pathway preferred over the other and why?

This reaction comes from Simmons, E. M.; Sarpong, R. Ga(III)-Catalyzed Cycloisomerization Strategy for the Synthesis of Icetexane Diterpenoids:  Total Synthesis of (±)-Salviasperanol. Org. Lett. 2006, 8 (13), 2883–2886.

They should both work. I would use oxalyl chloride with a solvent like DCM. DMF should act as a catalyst here for the acid chloride formation. (I don't have access to the paper but this is most likely what they have done) The oxalyl chloride reaction is maybe easier to work-up although I don't think there is a big difference when working in lab scale. In this case you can skip work-up (assuming you have used DCM as solvent and 1-2 equivalents of oxalyl chloride) and just add $\ce{AlCl3}$ for the cyclisation step.
EDIT: You can read the experimental in the supporting info -page S7


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How can the bulky octadecyltriethoxysilane molecule promote single-crystal growth?
Octadecyltriethoxysilane is very bulky (source),

yet it was used to promote single crystal growth (source). How can such a bulky molecule induce crystal growth? (For example epitaxial growth happens only when both materials have similar unit cell dimensions.)

The function of the octadecyltriethoxysilane monolayer in this case is not to promote epitaxial growth but to reduce the surface energy of the dielectric layer (SiO2), on top of which the semiconductor (pentacene/rubrene/etc.) is about to be deposited. The reduced surface energy allows the crystal to grow "comfortably".
Organosilanes are typical compounds for modifying oxide layers (Pernstich et al., 2004):


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Preparation of Sodium Peroxoborate

Write balanced equations for each step, and an overall equation, for the preparation of sodium peroxoborate hexahydrate from the reagents borax, $\ce{NaOH}$, and $\ce{H2O2}$.

Borax was added to a solution of $\ce{NaOH}$, then it was treated with $\ce{H2O2}$, then cooled to form a precipitate. Please help.
(No structures are necessary, simply the chemical formulas will suffice)

Possible duplicate of Reaction of borax with NaOH

Sodium peroxoborate can be prepare using two methods.

Electrolysis of sodium borate with some sodium carbonate
Oxidation of sodium meta borate using $\ce{H2O2}$

Heating borax strongly will give a mixture of sodium meta borate and boric anhydride.
If you can remove boric anhydride from the transparent glass like mixture, you can then add $\ce{H2O2}$
$$\ce{2NaBO2 + 2H2O2 + 6H2O -> Na2[(OH)2B(O - O)2B(OH)2].6H2O}$$

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


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How do we decide which ligand is monodentate, bidentate, etc.?
I went through a question about the EAN rule. At first, it was all clear to me, but now I'm just confused on how to decide which ligand will donate how many electrons. How do we decide which ligand is monodentate, bidentate, etc.?
This question came to my mind when I saw cyclopentadienyl complex in which cyclopentadiene donates 6 electrons to the metal atom. Does that mean all compounds try to donate all the pi electrons to the metal? No, am I right? Carbonyl group donates only 2 electrons though it has 2 pi bonds i.e. 4 pi electrons. How does carbon of carbonyl group donate an electron pair, considering it has only one electron left in it's octet? Again, for $\ce{CH2=CH2}$, how many electrons will it donate? The answer is two. Which electrons are donated? Obviously the pi bond since carbon has no free electrons.
In which case electrons are donated from pi bond, in which from lone pair, and in which from unpaired electrons?

"Donate" is understood here in a peculiar sense. What you donate, still remains yours. As to the question of which electrons are donated, the donor's occupied orbitals must match the acceptor's unoccupied ones in energy, symmetry, and size.

1st, all of your examples are monodentate. 2nd, what is the EAN rule? 3rd, I don’t fully understand your carbonyl reasoning, but I think that’s a side issue.

I believe what he's referring as EAN rule is the Effective Atomic Number Rule.


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Why is phosphorus pentachloride not used in the synthesis of chlorophenols?
The best method (as I have read) for preparation of chlorobenzene is by reacting it with $\ce{Cl2}$/$\ce{FeCl3}$. Why its not the best practice to react phenol with $\ce{PCl5}$ ?

Well you see, phenol do not react with PCl5 to give chlorobenzene as it undergoes EAS mechanism.In order to make the reaction possible you will have to activate phenol with nitro group.

well how can you activate phenol by nitro group ? nitro group is deactivating in nature.

I mean to say activate it towards nucleophilic attack


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In equilibrium, if bonds are constantly being created and broken, where does the energy that does so come from?
I'm a student who recently learned about equilibrium and how it is not a state where nothing is made, but rather, a state where the product is decomposing at the same rate as the reactants are combining. If breaking and creating bonds takes energy, where does this energy come from? Does it have to be added to the system? Also, if a closed system, say, a glass jar, reaches equilibrium, and heat slowly leeches out of the jar, does the balance of equilibrium shift, or does the reaction slow down but stay in equilibrium?

The energy comes from the molecules themselves. See, they have certain energy of thermal motion. You might have heard of the ${3\over2}kT$ thing, but that's only an average value. Molecules have different energies, described by the so-called Maxwell distribution. Those with higher energy might take part in some reactions which are not possible for the entire bulk.
As for your second question, it's option 1: the equilibrium typically will shift. (Though the reaction might slow down a bit, too.)

Thank you! I would give you an upvote thing but I don't have enough reputation for it :)

Recall that the molecules have a certain total energy, split into kinetic, vibrational and rotational energy. A molecule's chemical bonds can vibrate and the whole molecule can rotate as can groups within it, such as around single bonds.  The total energy is determined by the temperature and energy can move from one form to another in collisions, e.g. vibration to translation. But as mentioned by @Ivan Neretin there is not a single energy but a distribution of these (with the average $3k_BT/2$). The higher the energy a molecule has, the fewer there will be with this energy.
Now to go from reactants to products the molecule has to 'wait around' until by random collisions with other molecules in the gas or in solution it acquires enough energy to surmount the activation barrier between reactants and products. [You realise that there must be an energy barrier here otherwise everything would react instantly and we could not exist ]. By a similar argument, products have to wait to get enough energy to return to reactants and so the equilibrium is established. The rate constants of forward and back reaction will be different.
In you second question, the equilibrium will shift, the lower energy species, could be reactant or product, will start to dominate. Conversely if you were to add lots of heat then the amounts of product & reactants will become more similar as there is more energy to cross the barrier both ways.


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Chemical test for ferrous ions
I have a reference sheet for inorganic analysis which says that the confirmatory test for ferrous ions is as follows:

Make a solution of the given salt (whose cation is to identified) and add potassium ferrocyanide [ potassium hexacyanidoferrate(II) ] solution to it.
If a dark blue precipitate is formed, then the presence of ferrous ions in the given salt is confirmed.

What is the chemical composition of this dark blue precipitate? Or is there a flaw in the test that I have specified? If it is flawed, then what is the appropriate test?

The test is better suited for $\ce{Fe^3+}$.
$$\ce{Fe^3+ + [Fe(CN)6]^4- -> Fe4[Fe(CN)6]3}$$
The compound $\ce{Fe4[Fe(CN)6]3}$ has a specific name. Its called Prussian Blue.
If you perform the test for $\ce{Fe^2+}$, the precipitate won't be that dark. First partial conversion of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ takes place. Then the above reaction takes place to give the prussian blue precipitate.
There are specific tests for $\ce{Fe^2+}$ as well. You can use Dipyridyl to get a red colouration which is due to the formation of $\ce{[Fe(dipy)]^2+}$.
Or, you can use Dimethylglyoxime to get red colouration again which is due to $\ce{[Fe(DMG)2]}$.
Reference
JD Lee, Inorganic Chemistry (IITJEE Edition), page 278.


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Explain the difference in stability of permanganate ions in acidic/ alkaline solutions?
The question is such:

Use the data to comment on the stability of $\ce{MnO4-}$ in acidic and basic solution. Suggest a reason for this difference.
$$\begin{align}
&\text{acidic solution:}&\ \ce{MnO4- + 4H+ + 3e- &-> MnO2 + 2H2O}\qquad &E^\circ&=+1.70\ \mathrm V\\[6pt]
&\text{basic solution}:&\ \ce{MnO4- + 2H2O + 3e- &-> MnO2 + 4OH-}\qquad &E^\circ&=+0.59\ \mathrm V
\end{align}$$

I know that the permanganate ion is more stable in alkaline solution than in acidic solution, as the more positive electrode potential in acidic solution means more negative Gibbs free energy, therefore the reaction is more feasible in acidic solution, therefore less stable...
However, is there a chemical reason for this?

Are the words 'Nernst equation' familar to you ?

@permeakra yes it is! Could u please elaborate what you mean?

I added an elaborated simplified answer below.  In short, you have exactly same reaction, actually, but at different pH. Nernst equation can be used to recaalculate electrode potential at different pH. Same here.

@permeakra but this is an explanation of the electrode potential difference, but does not explain WHY permanganate is more STABLE in alkaline conditions

The "stability of $\ce{MnO4-}$" seems to be the wrong focus. Shouldn't it be the "reactivity of $\ce{MnO4-}$"?!?

Actually, yes.
As we know, $\ce{H2O = H+ +OH-}$ . Chemical equation can be threated as algebraic equations as long as the final result has chemical meaning (i.e. no negative coeffecients ... unless you understand their meaning), thus we can add $\ce{4H+}$ to both sides of the second equation. This gives us equation
$\ce{MnO4- +4H+ + 2H2O +3 e- \rightarrow MnO2 + 4H+ +4OH-}$
Obviously, we can collapse four pairs of $\ce{H+}$ and $\ce{OH-}$ into $\ce{4H2O}$
$\ce{MnO4- +4H+ + 2H2O  +3 e- \rightarrow MnO2 + 4H2O}$
and now we can drop two $\ce{H2O}$ from both sides
$\ce{MnO4- +4H+   +3 e- \rightarrow MnO2 + 2H2O}$
which is exactly as your first equation. So, in fact it is exactly the same reaction. However, WTF they are written differently and the corresponding potentials are different? This is because standard potentials are implied to be measured at concentration of all ions and molecules (except water) in corresponding half-reaction  1 mol/l. However, what happend when one concentration deviates from 1 mol/l ? The corresponding potential deviates from its value as well, in accordance with Le Chatelier's principle.
When we set up concentration of $\ce{OH-}$ at 1 mol/l, we set $\ce{H+}$ at roughly $10^{-14}$, since $\ce{[H+][OH^{-}] \approx 10^{-14}}$ (water autoprotolysis constant). According to Le Chatelier's principle the equilibrium should move to the left, which is observable as a change in electrode potential.
The quantitative expression of this kind of dependence is known as Nernst equation.

but this is an explanation of the electrode potential difference, but does not explain WHY permanganate is more STABLE in alkaline conditions

@justbehappy The electrode potential is the measure of permanganate ion thermodinamical stabiliy, as it reacts almost exclusively via reducation except very exotic conditions. So it is actually one and the same in this particular case

The first sentence of this explanation should be "In this case the Nernst equation shows the dependence of the half cell's EMF is a function of pH."


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How much energy is there in 1 mol of diamond and how can you get it?
Diamond is made of carbon which is during maybe 3 billion years under pressure is formed in the earth. Imagine you found a big source of diamonds such as coal...how much is the energy of the diamond compared to the coal, and how should you get the energy out of it?

Marginally more. 2. By burning it instead of coal.

@IvanNeretin marginally more than graphite, but significantly more than coal

Depends on quality of the coal, also it's quite hard to burn diamond.

Diamond is just one of the allotropes of carbon. There is nothing unique about diamond that allows it to store significantly more energy than any other compound. It is true, however, that much energy is required to separate carbon atoms in a diamond lattice. This is due to the high bond dissociation energy of a carbon-carbon covalent bond.
Although bond dissociation energy in a C—C bond is relatively high, it is dwarfed by the energy predicted to be released by the mass-energy equivalence theorem.
$E = mc^2$
The nuclear mass of 1 mole diamond is about 12 grams. So the theoretical energy released in a fusion reaction involving one mole of diamond is around 1*$10^9$GJ.
I don't think this is the answer you were expecting. But this is the amount of energy in 1 mole of carbon. The method of releasing all this energy is a different story. Nuclear fusion would release a significant portion of this energy but not all of it.

I like this answer but feel it's incomplete without mention of the possibilities of (i) burning the diamond and the energy obtainable therefrom, and (ii) converting the diamond to graphite and the energy obtainable therefrom.

The trouble with the question is that is misses an important aspect of chemical energy which is the question: relative to what?
The C-C bonds in diamond are strong but this means they need a lot of energy input to break them up not that they "contain" a lot of energy. If we form diamond from gaseous carbon atoms then we release a lot of energy (though this is a theoretical idea since it isn't a practical way to make diamond).
Diamond stores slightly more energy than graphite (~2kJ/mol at normal conditions). This means that if we could convert diamond to graphite we could release a really small amount of energy. This isn't practical or desirable.
Both forms of carbon can release energy when they burn. So, in a sense, they store energy relative to carbon dioxide. This is because carbon-carbon bonds are higher in energy than carbon-oxygen double bonds so converting one to the other releases energy (this is why coal is a good energy store and fuel).
So when you talk about releasing energy from chemical, you need to specify the reaction you have in mind. Absolute energies are pretty meaningless.


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Confusion about the oxidation state of copper in copper cyanide re-dissolved with excess cyanide

$\ce{CuSO4}$ when reacts with $\ce{KCN}$ forms $\ce{CuCN}$ while it is insoluble in $\ce{H2O}$. It is soluble in excess of $\ce{KCN}$ due to the formation of the following complex:
a) $\ce{K2[Cu(CN)4]}$
b) $\ce{K3[Cu(CN)4]}$
c) $\ce{CuCN2}$
d) none

Answer given is b) but we know that $\ce{Cu^2+}$ is more stable hence anwser should be a), shouldn’t it?

Is the question title even appropriate? Where's the tautomerism?

Edited the title to be appropriate, titles should be short descriptions of problems. @Varun Edit your answer, you could have edited the title too.

True... just wanted OP's opinion. Will edit my answer shortly. Thanks for the response.

There are actually many complexing agents that stabilize copper(I) over copper(II).  For another example see here.

but we know that $\ce{Cu^2+}$ is more stable hence anwser should be a).

Ah, but do we? I hope you don’t ‘know’ that, because that ‘knowledge’ would defy years of chemical research.
The stability of an oxidation state, especially that of a transition metal, cannot be deduced easily or by a set of memorised rules. Many a time, a species will only be stable in one environment and unstable in another one. Take copper: In aquaeous solutions, copper(I) is usually unstable concerning oxidation to copper(II). However, copper(II) on the other hand is unstable in the presence of iodide and the two will react:
$$\ce{2Cu^2+ + 4I- -> I2 + 2CuI v}$$
So copper(I) is more stable in the presence of iodide while copper(II) is more stable in the presence of water. Notice a pattern? I don’t.
In your question specifically, the first step is a redox reaction between copper(II) and cyanide according to the following equation:
$$\ce{2 Cu^2+ + 4 CN- -> NCCN ^ + 2 CuCN v}$$
At least I notice a pattern here: Cyanide behaves much like a halide or the iodide introduced above.
By this reaction, we have established that copper(I) is more stable in the presence of cyanide than copper(II). So by adding more cyanide there is no way that we can reverse the reaction — not only because there is nothing that copper could reduce with its extra electron. So the redissolution and complex formation cannot be a redox reaction, it must be a simple complexation reaction which only allows for the correct solution to be:
$$\ce{CuCN (s) + 3 KCN (aq) <=> K3[Cu(CN)4] (aq)}$$

Copper (II) ions are generally more stable than Copper(I) ions in aqueous medium (however not in cases of iodides and bromides where it is reduced to copper (I)). But this does not hold good for the corresponding complexes. This is not an exception, there are many more like this, including that of cobalt.
Cu(II) is reduced when treated with cyanide to form tetra cyanide cuprate (I) complex. Simple cyano complexes of Cu(II) are unstable and are reduced to Cu(I). In fact, it is very difficult to even isolate them.
References:
http://pubs.acs.org/doi/abs/10.1021/ic50132a014 (some stable cyanide complexes of Cu(II))


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Does a weak buffer solution still maintain a mostly steady pH if the acid or base load it has to buffer is small?
I need to buffer the $\mathrm{pH}$ of a solution to $3.0$ using a citrate buffer.  I found a recipe to make a buffer solution that would use $82\%$ of $0.1\,\mathrm M$ citric acid (monohydrate) and $18\%$ $0.1\,\mathrm M$ trisodium citrate dihydrate to give a citrate buffer with a $\mathrm{pH}$ of $3.0$.  The purpose of buffering the $\mathrm{pH}$ to $3$ is that this solution contains an active ingredient that has a very short shelf-life at a more neutral $\mathrm{pH}$. The buffer solution will make up about two-thirds of the solution by weight, and isopropyl alcohol will make up about $30\%$ of the solution.  Assume all of the remaining ingredients are non-dissociable in the relevant $\mathrm{pH}$ range, except for the active ingredient which has the following properties:

Concentration of $0.03\%$ by weight
A single weakly basic moiety with a $\mathrm pK_\mathrm a$ of $11.45$
Molecular mass of approximately $420\,\mathrm{g\over mol}$

Since this will be applied to skin, I'd like to use the weakest buffer solution that would still maintain the $\mathrm{pH}$ at $3.0$.  My question is, if I use $0.01\,\mathrm M$ or $0.001\,\mathrm M$ solutions to make the buffer, would it still be able to buffer the $\mathrm{pH}$ at $3.0$?

First, a disclaimer: This answer addresses only the question of the chemical buffering capacity of the questioner's formulation, given the information provided. Neither I nor Stack Exchange / Stack Overflow can take any responsibility for the safety or efficacy of the proposed formulation, for the effectiveness of the buffer concentration indicated below for maintaining the formulation $\mathrm{pH}$, or for any other hazard, harm, or adverse event, without limitation, arising from the use of the information provided below.
To answer the general question of the title: Yes.
You haven't given enough information to answer your question in the particulars, though.
For your particular application:
Worst-case, buffer-wise, as long as the following assumptions are accurate, the maximum amount of buffering capacity you'd need can be estimated from the total number of moles of active ingredient:

None of the other ingredients have appreciable acid-base character (appears to be the case here)
Degradation of the other ingredients into acidic/alkaline species can be neglected
No acidity or alkalinity is likely to enter from the environment

At $0.03\%$ w/w, assuming the overall density of the formulation is approximately $1\,\mathrm{g\over cm^3}$ (probably reasonable, with $67\%$ w/w composed of aqueous buffer) a species with a molecular weight of $420\,\mathrm{g\over mol}$ will have a mole concentration of $0.714\,\mathrm{mM}$.  In general, I would probably want the buffering species to have a concentration about five to ten times my expected net acid/base load. So, I would choose no less than about $3.6\,\mathrm{mM}$ for the total concentration of all forms of citrate, perhaps going as high as $7.2\,\mathrm{mM}$ to be on the safe side.
Note: The $\mathrm pK_\mathrm a$ of the ionizable group on the active ingredient is sufficiently far from your target $\mathrm{pH}$ that it will be fully protonated in your formulation.  Thus, if it is possible to add this ingredient in its protonated form, the required buffer concentration is probably somewhat lower than the above.

General Principles
As long as the number of moles of acid or base to be neutralized is small compared to the total number of moles of the relevant buffering species in the solution, and as long as the target $\mathrm{pH}$ is within about one unit of the nearest $\mathrm pK_\mathrm a$, the $\mathrm{pH}$ should hold steady.
As the number of moles of acid/base to be buffered against approaches the number of moles of the buffering species, the target $\mathrm{pH}$ needs to be closer and closer to the $\mathrm pK_\mathrm a$ in order to hold the $\mathrm{pH}$ steady.
If the number of moles of acid/base to be buffered against exceeds the number of moles of buffering species, the buffer will in general fail to hold the target $\mathrm{pH}$. The main exceptions would be polyprotic species with multiple $\mathrm pK_\mathrm a$ values in close proximity (polyphosphates come to mind).

I think the general answer should work for me, but I'd appreciate it if you could guide me to give you the information that I need to give to get the specific answer.  I tried to provide that information in the question, but I'm not an expert here.  The buffer solution is about 67%(w/w) of the total product and the isopropyl alcohol is 30%, 1.5% is a non-ionic thickener, and 1% is another alcohol.  The active ingredient is .03% with a pKa of 11.45 and a molecular mass of around 420g/mol.  Is this enough information?

@Itsme2003 Are you concerned about keeping the formulation buffered against any acidity/alkalinity generated by the skin, or just buffering the formulation itself?

My concern is buffering the formulation itself in order to preserve the active ingredient.  I apologize if the following description is imprecise but it's been a long time since I took a chemistry course.  Once it's on the skin I don't care if the pH does change, and in fact I think it's probably best if the pH does change because I don't really understand the potential harm if something this acidic is applied to the skin.  My thoughts are that if it has only a very tiny amount of acidity that the acidity would be "neutralized" by the skin without causing noticeable  harm to the skin.


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Why do catalysts have limited turnover numbers in practice?
I quote the Wikipedia page definition for Turnover number:

In other chemical fields, such as organometallic catalysis, turnover number (abbreviated TON) has a different meaning: the number of moles of substrate that a mole of catalyst can convert before becoming inactivated. An ideal catalyst would have an infinite turnover number in this sense, because it wouldn't ever be consumed, but in actual practice one often sees turnover numbers which go from 100 up to 40 million for catalase.

I'm trying to find reasons for why catalysts don't function infinitely in practice. My only thoughts are that maybe they eventually degrade from slowly reacting with other species they are exposed to, but then wouldn't this depend on the environment it is in rather than how much substrate it catalyses? I also assume that all product is taken away as it is produced.

Because they are not ideal. They may get inactivated by the substrate itself (say, 1 time in a 100), or by certain impurities in the substrate (so-called catalytic poisons).

@IvanNeretin How does a catalyst become inactivated by its substrate?

@IvanNeretin means that some real catalysts can undergo more than one reaction with a substrate.  It may only occur rarely (1/100 in his example) but that 1/100 reactions is enough to inactivate a molecule of real catalyst.

@k-- Say, every once in a while our reaction takes a wrong turn, and the product (the wrong one) gets permanently stuck to the catalyst, rendering it inactive.

@repurposer oops yeah I meant turnover number

@k-- the assumption that the products are taken away as they are produced doesn't hold for most lab reactions or commodity/speciality chem manufacturing

catalytic poisoning


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How do we compute the mole fraction of an ideal gas in a stream blended by mass flow controllers? For a non-ideal gas?
Consider a gas blending system consisting of two mass flow controllers, A and B, set to control flow to $f_\mathrm{A}$ and $f_\mathrm{B}$, respectively.  Consider that flow controller A is metering the pure gas of interest, which has an actual mole fraction of $x_\mathrm{A}$.  Flow controller B is metering a make-up gas which coincidently may contain trace amounts of gas A at a mole fraction of $x_\mathrm{B}$.
The mass flow controllers provide continuous estimates of $f_\mathrm{A}$ and $f_\mathrm{B}$; these are measured quantities.  Suppose also that we did some experiments and found the values of $x_\mathrm{A}$ (very close to unity or $\pu{1E6 ppm}$) and $x_\mathrm{B}$ (approaching zero, likely ones to tens of ppm).
To first order, the following equations provide reasonable estimates of the concentration of gas A in the blended stream at near atmospheric pressure, room temperature.  Experiments were carried out with $\ce{CO2}$ (gas A) in $\ce{N2}$ (gas B).  A $\ce{CO2}$ gas analyzer was used to estimate the mole fraction of gas A in the blend, $x_\mathrm{T}$.  We can compute the mass flow ratio of gas A by
$r = \dfrac{f_\mathrm{A}}{(f_\mathrm{A} + f_\mathrm{B})}$
or
$r = \dfrac{f_\mathrm{A}}{f_\mathrm{T}}$
where $f_\mathrm{T} = f_\mathrm{A} + f_\mathrm{B}$.  Further, we compute the mole fraction of gas A in the blend as
$x_\mathrm{T} = r \cdot \left(x_\mathrm{A} - x_\mathrm{B} \right) + x_\mathrm{B}$
My question again is: How do we compute the mole fraction of an ideal gas in a stream blended by mass flow controllers?
I'd like to do this in a rigorous manner, or correctly, possibly accounting for any non-ideal behavior of $\ce{CO2}$ in a $\ce{N2}$ make-up gas.
I came across this reference and believe the key to my problem is addressed therein.  The solution may involve the empirical rule of Lewis & Randall whereby the mole fraction of a component is equated to a ratio of fugacities of the component in a mixture to the pure component.  Certainly being able to compute the virial coefficients for the components may prove useful.  A key may also be converting a flow into a volume with an assumed time step.
Should I attempt to apply the some of the equations presented in the reference noted above?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. This appears to be a homework question, please share your thoughts and attempts towards the solution.

Thank you for the comments.  I can apply latex formatting; I wasn't aware this was supported.  This is not a homework question.  I will happily explore the solution further if deemed necessary.  Please provide feedback as to whether my most recent edits are okay, or whether I should add more content / show more work towards the solution.

What have you done in your analysis so far that says this applies only to ideal gases, and not to mixtures in general.

Can anyone make any suggestions as to further analysis?  I would be happy to add detail to the question.


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SERS Gold nanoparticles performing poorly in organic solvents?
We have been working with some organic solvents lately and noticed a sharp decrease in surface-enhanced Raman spectroscopy (SERS) performance from our gold nanoparticles deposited in an organic solvent such as methanol.
With reference to the graph attached: the top line is 5 ppm Malachite Green Oxalate dissolved in water, the bottom line is 5 ppm Malachite Green Oxalate dissolved in 50/50 water/methanol

As you can see, the SERS performance of the gold nanoparticles decrease significantly when methanol is added.
We consulted our vendor and this is the explanation given: "Since the nanoparticles are stabilized in an aqueous solutions, the addition of methanol will destabilize it. I think in this case there is too much methanol, so the nanoparticles are over-aggregating, hence the decrease in SERS performance."
I have the following questions:

Why will the nanoparticles over aggregate in organic solvents? Is this specific to my vendor's nanoparticles?
Why would that even matter? How does over aggregation lead to poor SERS performance?
Since methanol is immiscible in water, shouldn't the analyte molecules (Malachite Green) on average have the same average distance to the SERS gold nanoparticles?

have you tested your zeta potential in both cases? This is the reason for the stability of the goldnanoparticles. If you got a zeta potential between -30mV and +30mV your colloid solution is instable. You should have less ions in the methanol mixture, which can cause a lower zeta potential. Best regards!

Methanol is miscible in water in all proportions. You want lots of particles of the right size and shape, whatever those happen to be.


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Purpose of electrolyte in a galvanic cell
I am confused as to why galvanic cells have electrolytes. For example, a zinc-hydrogen galvanic cell has a zinc sulfate anode electrolyte and hydrochloric acid cathode electrolyte. The overall reaction of this may be written as:
$$ \ce{ZnSO4 + 2HCl -> ZnCl2 + H2SO4} $$
But rather, it is written as:
$$ \ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)} $$
Why is this the overall reaction? What is then the purpose of an electrolyte, if only the anodes and cathodes are written in the overall equation?

I know to write the overall cell reaction, you need to eliminate spectator ions. Is it just assumed that sulfate ions and chloride ions are spectators?

Yes they are spectators in this reaction. There should be 2HCl in the first step of the balanced equation. The electrolyte carries the charge between the electrodes.

Then why are Zn and H counted as spectator ions? They also cancel when you write ZnSO4 + 2HCl --> ZnCl2 + H2SO4 in total ionic form.

The electrons are really the subject here . What do you mean by cancel?

like the total ionic equation would be Zn2+ + S042- + 2H+ + 2Cl- --> 2Cl- + Zn2+ + 2H+ + SO42-. Wouldn't everything cancel out in this equation? So then why do they just write Zn(s) + 2H+ (aq) --> Zn2+(aq) + H2(g)?

I'm not sure what you mean by cancel out. Nothing disappears the net charge of the molecules becomes 0 but chemicals don't cancel each other out like numbers or figures in algebra. They just write the molecules that are directly involved in the electron flow and transfer which is the Zn and the H.

Sorry I am confusing you so much, but what I am trying to ask is why don't zinc and hydrogen ions count as spectators in this reaction as well?

Because they are involved in the electron transfer. The oxidation state of the Cl and SO4 remains the same therefore there not directly involved in the transfer of electrons and are considered spectator ions.

Oh okay. So then do you know how they select these electrolytes?

It depends on the cell. Variables such as gas escaping around the electrodes, loss of solvent or electrolyte due to heat or gassing , rate of corrosion at the electrodes, conductivity need to be considered , I'm sure there's a few more. For example, Electrolytes such as HCl or NaOH are highly conductive they usually corrode the electrodes to fast causing heat and gassing and therefore can be impractical as an electrolyte in an open or closed cell unless the cell is modified to contain these electrolytes.

okay and the electrolytes do not have anything to do with calculating spontaneity of cells, right? Determining if the cell is spontaneous is done through by looking up the enthalpy values of the ions (like Zn2+ and H+).

The electrolyte may still be a variable as it carries the current across the electrodes and conductivity varies significantly between electrolytes meaning this will occur at different rates resulting in different potentials. This is going off-topic from the OP. I hope I could help you out.

Yes thank you. I just wanted to know what OP means.

Original Post :)

Unless I'm badly misunderstanding your question, I think your confusion arises from an error in the balanced chemical equation you wrote.  As you have it, a salt is reacting with a strong acid to make another strong acid and a new salt. This reaction may occur spontaneously as written (I haven't figured its Gibb's free energy) , but it's not the oxidation-reduction reaction that drives a galvanic cell.  In fact, it's not an oxidation reduction at all, it's some sort of acid-base reaction, and all of the reactants and products remain in solution anyway.
The actual chemistry that occurs in this system is between the zinc anode and the acid.
$\ce{Zn + 2 HCl -> ZnCl2 + H2}$
From this it is much clearer why the net ionic equation is as you wrote.  The zinc sulfate could be replaced with zinc chloride with almost no effect on the battery's potential.  In fact, the zinc solution exists only to create standard conditions for the cell and to carry electrical current through the solution - if it were replaced by a nonreactive electrolyte solution the cell would still function, albeit at a different (higher) potential.
You might then ask what the purpose of the hydrogen cathode is.  Again, it's to create standard state.  If you did this in reality, you would be using some sort of inert electrode at the hydrogen end to carry the electrical current into the acid solution and likely would just vent the hydrogen to the surroundings.
In short, this galvanic cell does not function because of a reaction between its electrolytes, nor because of a reaction between its electrodes.  Instead it is ultimately a reaction between the anode and the cathode's electrolyte.  The anode's electrolyte and cathode are required for electrical continuity and are used in examples of this sort to put the cell in standard state.  Beyond that they don't serve much of a purpose.

zinc is the anode. So is it that in very galvanic cell, the anode and the cathode electrolyte are the parts that cause the actual chemistry?

You're right, I've switched cathode and anode throughout, thanks for the correction.  I almost wrote something like what you asked in your question, but there are sooooo many different galvanic cells that you can build that I really don't feel comfortable making that broad a statement.  Certainly in the ones that you're going to see in an introductory course this is true.

@user510 I do feel comfortable with a more general statement like: In a galvanic cells, electrons are removed from the anode, generating a substance with a higher oxidation state, and transferred to the cathode, where they generate a substance with a lower oxidation state.

The function of electrolyte is to carry the charge in the liquid phase and close the circuit in order to maintain the charge flux. In the external part of the cell (copper wires, some conductor) it is the electrons which move and carry the charge. In the solution it is the movement of ions, being the potential difference between anode and cathode the driving force for moving them (anions and cations move in opposite directions).
The 'spectator' ions may be called that way because they are not electroactive, meaning with this that they do not participate on electrode processes. As pointed out by @lcnittl, the first reaction is incomplete and does not describe the electrochemistry of reaction. For thermodynamic properties, it is mainly needed the redox reaction, reaction N°2, and potential difference associated (that can be found on Standard Redox Potentials Table). It will also depend on other variables such as valence of ions, number of electrons transfered, pH, temperature, etc.
The second reaction can be decomposed into two semi reactions, one occurring on the anode (oxidation) and the other on the cathode (reduction)
$$ \ce{2e⁻ + 2H⁺(aq) -> H_2(g)}  ---(Cathode) -- E°=+0.7618 V $$
$$ \ce{Zn(s) -> Zn^2+(aq) + 2e⁻}  --- (Anode) -- E°=0 V$$
The sum of the standard potentials of each semi-reaction gives a positive value (dE=0.7618V + 0V); a positive value means a spontaneous reaction based on the relation dG=-n.F.dE, where dG is the Gibbs free energy change, n the number of electrons transfered (n=2 in this case), F the Faraday constant and dE the potential difference between two semi-reactions.

What do you mean by 'Both reactions are OK'?

The first one is the net reaction of the cell -it does not contain practical information, only species involved and mass balance. The second one is the oxidation semi reaction

Yeah, but the first misses an HCl molecule to be balanced, and furthermore cannot be the net equation of the cell, since it's an acid-base reaction, and not a redox reaction.

@lcnittl The first of the two reactions is a reduction, which it has to be to pair with the oxidation step shown in the second reaction.  You need both for either to occur.

@JasonPatterson How can it be a reduction, if there is no change on any oxidation state of any participating atom? (Not to talk at cross purposes, we were talking about the OPs first equation)

@lcnittl I just posted a new answer to the question.  I'm pretty sure that the OP's first reaction is based on mistaken understanding of how the cell works.

and my concern is as to how they come up with the redox reactions for each half reaction (sorry, my question was not that direct).


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Where do I go to get details on setting up different instruments?
Say I want to set up a system to preform cyclic voltammetry. How do I know what instruments to get/ electrochemical cells/ etc? Is there a resource for other setups as well, like EIS?

Your question is very broad and hard to answer without pointing out multiple books on the topic. Here is a good one to start: Lab Techniques in Electroanalytical Chemistry, Edited By P. Kissinger and W. Heineman
However, your specific experiment will depend on the type of your sample and what information you are looking for. The best place to start is to look for papers in your field and read their experimental sections.
Instrument manufacturers also write application notes specifically for this reason. You may find useful information on their websites as well. Here is an example : Application Notes Regarding EIS from Gamry
Full disclosure: I used to work for Gamry Instruments.


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Volume occupied by real gases

Molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller . As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.

This is what I read in a website. In the above statements, what is the volume predicted by the ideal gas law? The volume predicted by the gas law, should be the volume of container. Now if volume occupied by real gas is less than the volume of container, how can the volume occupied by it be greater than that predicted by ideal gas law ?

Also, compressibility is given by $Z = \frac{PV}{nRT}$.  In this equation is P the measured pressure or the ideal pressure?

When you are "containing" a gas in a container then you are by definition restricting its volume. Lets assume you heat the gas. If the volume is confined then the pressure will increase. Similarly, two different gases will have different pressures if confined to the same volume.
If you somehow control the pressure (there is a movable piston) then a real gas will occupy less volume because of its intermolecular interactions.
Also see this: Volume of different gas in a container

This site may help you: http://www.chemguide.co.uk/physical/kt/realgases.html
I've changed my answer to hopefully answer your question better.
At higher pressures, ideal gas laws consider particles as just points. That means they have more space to fly around, and so for a certain pressure, they need a larger volume. In reality, the particles do have a size. That means they bump into each other/their container more frequently than the ideal gas law would indicate. This bumping means a higher pressure occurs, at a lower volume than predicted.

Then how to explain the graph I just added above in the  question ?

I realized I wasn't answering your question properly at first. Hopefully this answer makes more sense :)

For low pressures, a real gas would be predicted to occupy a slightly smaller volume than an ideal gas since the molecules themselves occupy space for a real gas. However, as we get to higher pressures, the volume of the particles becomes a significant portion of the volume of the container, which makes it much more difficult to compress the gas. We could think of hypothetically compressing an ideal gas to zero volume since its particles take up no space, but we couldn't do the same for a real gas.


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Preparing Saturated Carbon Dioxide
I'm doing a chemistry lab where I'm collecting data to create a solubility curve for carbon dioxide at different temperatures. I'm using carbonated water and my concern is that, by heating or cooling the solution, will it be saturated with carbon dioxide after exposing it to the air for a small period of time? I'm guessing that this will not be the case.
I know I can add carbon dioxide to water samples by exhaling into them but, is there a way to determine that it has been saturated without using a pre-existing solubility curve?


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Is the freezing point of solution always lower?
If the freezing point of a solution always lower than that of the pure solvent or is it possible for it to end up being higher?

Depends on your definition of solution. Is $^2\ce{H2O}$ in $^1\ce{H2O}$ close enough? Because that's the only way I know to make a "solution" and have the FP go down.

Of course not. Whether you're talking only about aqueous solutions or other ones.

Thanks. There's no way the freezing point could not change either is there?

Whoops. Go up* sorry.

Consider the enthalpic and entropic contributions in the small solute limit. Entropy will win, lowering the freezing point.


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Is there a convention as to what order to write reactants and products in a reaction?
When doing homework, it can be time-consuming to check the answers in a redox reaction due to different order of the reactants and products.
For example, while $$\ce{2MnO4- + 5Fe^2+ + 16H+ -> 2Mn^2+ + 5Fe^3+ + 8H2O}$$ and $$\ce{5Fe^2+ + 16H+ + 2MnO4- -> 5Fe^3+ + 8H2O + 2Mn^2+ }$$ are equivalent, they are not visually equivalent and to mark them as correct/incorrect takes a lot of time.
Is there a convention of what order to write these reactants/products in? If so, what is this convention?

There is no convention.
There might be some cases where one choice simply makes more sense to people. For example I have yet to see anybody write
$$\ce{NaCl(s) <=>[H2O] Cl-(aq) + Na+(aq)},$$
but there is no convention dictating that you have to write $\ce{Na+}$ before $\ce{Cl-}$.


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How to determine which gas liquifies first when cooled from high T, based on formula and critical temperature?

The gas that liquifies first, when cooled from 500 K to its critical temperature given in parenthesis is
a) $\ce{CO2}$ (304.1 K)
b) $\ce{NH3}$ (405.5 K)
c) $\ce{O2}$ (154.3 K)
d) $\ce{N2}$ (126.0 K)

Answer: b
I know that the critical temperature is the highest temperature at which the gas can be liquified and above which the gas can not be liquified whatever high pressures are used.
Also,
$$\frac{p_\mathrm c V_\mathrm c}{T_\mathrm c} =nR =\frac{m}{M}R$$
i.e.,  $T_\mathrm c$ is directly proportional to $M$, when all the other quantities are constant.
So, based on $M$ $\ce{NH3}\lt\ce{N2}\lt\ce{O2}\lt\ce{CO2}$, so b should be the answer.
This is how I have solved this problem. Please tell me if I am wrong and if so, then please tell me the correct way to answer this.

The method you have used doesn't work for all cases. The equation is valid only for ideal gases. When some form of inter-molecular forces come into picture, the critical temperature doesn't depend on the mass alone. For example, $\ce{H2O}$ and $\ce{NH3}$ have similar molecular masses, but the critical temperature of $\ce{H2O}$ is higher because of stronger hydrogen bonding (~650 K). However, a better approximation of critical temperature would be
$T_c=\frac {8a}{27bR}$ which accounts for the inter-molecular forces and the size of the gas molecules.

Good question.  I think you got the right answer but by using some pretty suspect logic.
First, there isn't enough information to solve the problem in general because pressure isn't specified.  Is the cooling of the gas isobaric?  Or adiabatic?  Or ... ?  Let's assume its isobaric, i.e., it happens at constant pressure.  If so, what's the pressure?  If the pressure is below 5.1 atm, then $\ce{CO2}$ will never condense into a liquid, instead it would transition directly to a solid phase (dry ice).  Ammonia's triple point is roughly 0.06 atm.  But your question says "cooled to the critical temperature", which implies not below the critical temperature, which means the only way any of them will liquefy is if the pressure is at or above the critical pressure.  So let's assume the pressure is isobaric and is at least 11.8 MPa, the critical pressure of ammonia, the gas with the highest critical pressure among the possibilities.  But if we do that, the question is easy.  The gas that liquefies first is the gas with the highest critical temperature, which would be ammonia.
Relaxing this assumption makes the question much harder.  Complex calculations would be required.  The theory of corresponding states would likely be useful.
Your logic is OK, but in your equation, $w$, the weight, and $V_c$, the critical volume, also vary.   So the statement that

$T_c$  is directly proportional to $M$

is not true, because the other factors are not constant.$%edit$

By definition, critical temperature is the highest temp at which gas can be liquefied, and its proportional to ratio of a and b so, higher the a , more intermolecular forces strength, easier it is to liquify, since sizes here are pretty much same {b} and can't cause difference of 100K, it must be big difference in intermolecular forces, again, Tc being proportional to M is wrong like Curt specified, if it was a different case of sizes and pressure not being enough for liquification, lots of complications would be there

The Q is flawed and is being considered for deletion. Do you have objections to deletion, and/or can you see a way to rescue the OPs post through edits?

Why is it being considered for deletion?  What exactly is the flaw?

You have assumed incorrectly that all other quantities like critical pressure and critical volume are the same for the gases. In reality it is possible to solve for the critical temperature in terms of the van der Waals constants $a$ and $b$ only so that
$$T_c=\frac {8a}{27Rb}$$
Now qualitatively you can deduce that since for $\ce{NH_3}$ intermolecular forces are strong, so it's $a$ is higher than the others, and it is difficult to guess this in case of a generic problem.

You should check the type of bonding whether it is hydrogen bonding:London dispersion....
Simply check out if all are organic go for...
mass  directly proportional to. Tc
And if other then this prefer those which possess strong hydrogen bonding.
Like water possess more strong hydrogen bond than NH3....
So Tc for water is more it is easy to liquefy!


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Why methanoic acid is more acidic than phenol?
If you draw the resonatic structure, obviously, the conjugate base of phenol seems to be more stabilized than methanoic acid's. Then why this anomaly?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about how it works.

Can someone find a duplicate which says that it's not the number of resonance structures that matters, but rather the extent of their contribution? This question has been asked too many times.

obviously, the conjugate base of phenol seems to be more stabilized than methanoic acid's No, I do not think this is obvious at all. As @orthocresol pointed out, the number of resonance structures is of secondary importance compared to their individual stability and contribution.


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Lithium chloride is soluble in pyridine while other lithium halides are not. Why?

Lithium chloride is soluble in pyridine while other lithium halides are not. Why?

I know that lithium chloride is deliquescent like magnesium chloride. Lithium halides are covalent and thus are soluble in organic solvents. Then, why is it that only lithium chloride is soluble in pyridine?

I was told that, since the solvation energy of LiCl is more, so its soluble. But this doesn't clear my doubt.

@Mithoron, i know that alkali metal halides are mostly ionic solids, but, Li halides are more covalent because 1)high polarisation capabilities of Li ion. 2)small size 3) high tendency to distort electron cloud around negative halide ion.

Related: https://chemistry.stackexchange.com/questions/115112/why-do-we-use-licl-as-a-source-of-chloride-ions-as-nucleophile-but-not-other-alk/115149?r=SearchResults#115149.

I'm pretty sure the premise of question is false - probably LiI is best soluble in org. solvents, Wikipedia mentions 3430 g/L in methanol! (Only about 40 for LiCl)


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Why is Iron (Fe) difficult to reduce?

$\ce{Fe^3+ + e- -> Fe^2+} \quad +0.77$
$\ce{Fe^3+ + 3e- -> Fe} \quad -0.04$
$\ce{Fe^2+ + 2e- -> Fe} \quad -0.41$
$\ce{Zn^2+ + 2e- -> Zn} \quad -0.76$
$\ce{Co^2+ + 2e- -> Co} \quad -0.29$
$\ce{Ni^2+ + 2e- -> Ni} \quad -0.25$

I am having a hard time to understand if there is a trend in reduction potentials so predictions could be made on whether a certain 'strong' (e.g. $\ce{NaBH4}$) or 'weak' (e.g. hydrogen) reductant  reduces an iron salt. I also wonder if there is a trend in the periodic table between first row d metals.
My question is possibly best split into two: Is there a trend in the periodic table regarding reduction potentials? Why is iron (our of the first row elements) so hard to reduce compared to others?

Fe is not all that hard to reduce compared to others from that row. Zn is harder to reduce, and Ti even worse.

That's true. So is there no trend in the first row as to why that occurs?

There is no "that" in the first place. As to the trend... well, it is kinda possible to derive certain qualitative trends, but $\ce{Fe^2+}$ is in no way exceptional.


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How can I isolate glycolic acid from sugarcane juice?
This beginning of a bachelor thesis uses ethylene glycol as a catalyst in the isolation of glycolic acid. I can't find the rest of the article (it ends in the middle of the description) to find out how they did it.
I'm wondering if someone could please help me I want to learn how to do this as an amateur dermatologist. Currently studying pre-med, small knowledge of chemistry.

Ethylene Glycol was not used as a catalyst! Kataoka described in 2001 the production of Glycolic acid (GA) from ethylene glycol using glycol-oxidizing microorganisms isolated from soil, which is what the author of this thesis is trying to say (the writing is unclear though). It will be difficult to reproduce this in your home since you would need to cultivate those microorganisms inside your own house, chemical oxidation would also be a problem if you mean to use it for medical purposes (the oxidation of ethylene glycol releases formaldehyde, which is toxic and probably carcinogenic), besides the reaction needs potassium dichromate to happen which is also hazardous.
Fruits produce a small amount of glycolic acid and you can try to extract it from banana or sugarcane pulp by an acid-base workup, though I believe you won't get much pure GA using this method (specially if you don't have a deep knowledge of organic chemistry).
Glycolic acid is very easy to find online and it will be a lot cheaper to buy it from ebay or other source then to synthesize it if your goal is to use it for its skin effects.
Source: Glycolic Acid Production Using Ethylene Glycol-Oxidizing Microorganisms
Michihiko KATAOKA, Mié SASAKI, Aklani-Rose G.D. HIDALGO, Michiko NAKANO & Sakayu SHIMIZU
Bioscience, Biotechnology, and Biochemistry Vol. 65 , Iss. 10,2001


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How difficult is it to produce diatomic hydrogen?
Just to preface this, I have a very basic understanding of chemistry, so forgive me if what I'm asking is super basic.
How difficult is it to produce diatomic hydrogen? I've been doing some reading on the internet about diatomic hydrogen and a fair amount of the stuff I've come across are health articles relating to the benefits of drinking hydrogen rich water, so I assume that it's not terribly difficult to come across.

There are several easy ways to produce pure hydrogen, e.g. electrolysis of water. However, hydrogen is a gas at room temperature, so I would seriously doubt that there actually is something like hydrogen enriched water. I would further doubt that if such a thing existed, it would have no health benefit.

There is a NIH study on hydrogen water so it isn't totally poppycock. http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2831093/

As far as I knew, Hydrogen IS diatomic, in its natural form at least.
http://chemistry.about.com/od/elementfacts/f/What-Are-The-Seven-Diatomic-Elements.htm
That means to produce it, you really just have to create pure Hydrogen, and it will become diatomic on its own.
Edit: I'm not an expert on this kind of stuff, but I would say that claims like 'Hydrogen-Rich water is healthier' are probably bs. Just saying.

Diatomic hydrogen is hydrogen’s stable elemental modification under standard temperature and pressure — so if you generate hydrogen you will be generating diatomic hydrogen.
The simplest way to do this with household means is probably to use acetic acid concentrate (the stronger the better) and zinc powder. (Solid zinc works but much slower.) The reaction may be sluggish; if that is the case, very carefully and slowly heat the acetic acid. Capture the hydrogen in some glass, cup or test tube held upside down (hydrogen is the least dense gas with each molecule only weighing $2~\mathrm{u}$ (atomic mass units) so it will always rise).
If you have collected some amounts of hydrogen gas you can proove its presence. This was a common experiment in my high school chemistry courses so I’m going to tell you how to do it. However, exercise caution! If you didn’t do this at school yourself or didn’t watch someone do it before, I would advise you to not attempt it. Only do this with very small amounts of hydrogen, no more than a test tube of $2~\mathrm{cm}$ diameter. Make sure the test tube is full (by collecting enough hydrogen). Carefully bring the opening towards a candle flame. Do not use a pocket lighter where your thumb is right next to the flame for safety reasons. The hydrogen should burn with a faint ‘pop’ creating water. Considering how slurrish the reaction of zinc and acetic acid probably is, it may be too slow to actually collect enough hydrogen. Don’t attempt the burning if that is the case.
The reason for being cautious is the following: $2\,:\,1$ mixtures of hydrogen and oxygen are explosive. And with explosive I mean that they give a very loud bang. Luckily, air is mainly nitrogen so the probability of accidentally creating an explosive mixture is rather low. However, if you did, you will regret it.
Finally, on the topic of ‘hydrogen-enriched water’: Hydrogen, like all other gases, is not well soluble in water. However, it is hardly present in standard water since it is almost non-abundant in the atmosphere so it cannot dissolve. (Carbon dioxide, nitrogen and oxygen are far more abundant and therefore spontaneously dissolve in water to a certain extent — especially carbon dioxide which, in fact, reacts.) It is possible to create a hydrogen-saturated aquaeous solution or something less hydrogen-containing and sell it as ‘hydrogen-enriched water’. It would have no positive or negative effects on you or your surrounding. (Unless maybe your voice got slightly lighter in the same way as when you breathe helium — however, the amount of hydrogen is very likely too low for this to happen.) So it sounds like my dream of life: Making money with other people’s stupidity. It’s a shame that I’m not ruthless enough to do such a thing …


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How is hybridisation of a carbon atom determined by looking at the molecule's structure and observing the lone pair and/or sigma bond present?
How can I judge the hybridization of a carbon atom of any carbon containing compound just by looking at its structure? How can I say this by the help of lone pairs and sigma bonds present?

The answer you are looking for is probably this:
Count the number of $\sigma$ bonds made by that atom, and add to that the number of lone pairs present on that atom. This number is called the steric number.
If the steric number is $2$, hybridisation is $sp$.
Similarly, $3 \Rightarrow sp^{2}$, and $4 \Rightarrow sp^{3}$ hybridization.


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How to calculate Madelung's constant for NaCl?
Madelung's constant for $\ce{NaCl}$ is $1.748$.
What I tried to so is sum up all the individual interactions into $1$ component where the $X$s are the nucleus and the electron if it was a $2$ atom system. Is this the correct way to go about it? Thanks!

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

As hinted at in the question, the series as you have written it out 6 - 12/rt(2) + 8/rt(3) - ... does not converge rapidly. I am not exactly sure how the value is obtained though.

Is this the correct way to add up the interactions though?

The series $\displaystyle \frac{6}{1}-\frac{12}{\sqrt{2}}+\frac{8}{\sqrt{3}}-\frac{6}{2}+\frac{24}{\sqrt{5}}\pm\ldots$ diverges as has been proven in 1951.
The correct value can be obtained by summation over lattice cubes with edge length $2n+1$ and getting the limit for $n \to \infty$. This has been shown in 1985.
Wikipedia provides more information and further reading on this.


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What is the size of the rate constant dependent on?
In the rate equation, consider k:

I know that $k$ is fixed unless temperature changes, but what else besides temperature is $k$ dependent on?

The rate constant $k$ includes all the factors other than the dependence on concentration.  A significant factor in the rate is collision frequency and energy, so temperature has a significant influence on the value of $k$.  Other factors like the range of angles of collision necessary for a successful reaction, the strength of dipoles, solvent effects, quantum effects, and whatever factor corrects the units of the equation are not determined.
However the rate constant can be expressed by the Arrhenius equation
$$k=Ae^{-{E_a/RT}}$$
Doing this decomposes the constant into an exponential relationship including the activation energy, $E_a$, the molar gas constant, $R$, and of course temperature, $T$, which represents the fraction of total molecules that have sufficient energy to overcome the activation energy.
This equation comes with it's own new constant, the pre-exponential factor, $A$, which now includes all the factors in the rate constant that aren't the exponential relationship with $E_a$, $R$, or $T$. There may still also have some temperature dependence hidden within $A$, but usually it can be assumed constant over small temperature ranges.  Factors like collision frequency and angle, or action of a catalyst, are still obscured within this constant.
Even if these obscured effects were determined there would still be some new constant, $C$, which would include subtle effects such as interactions with surfaces in a container or non-reacting particles, temperature and concentration gradients, ionic strength, entropy changes in the solution due to changing composition, and any other factor that influences the reaction between the particles that hasn't been explicitly considered.

Nice answer!  It might be worth explicitly mentioning that rate constants can depend on pressure, ionic strength (if the reaction is in liquid), and the concentration of non-reacting species (a.k.a. catalysts) in addition to all the factors that you listed.

k is a characteristic of the reaction being considered.
It depends on the stability of the reactants with respect to intermediate and intermediate with respect to products and the mechanism of the reaction.


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Internal Cannizzaro reaction of phenylglyoxal
Phenylglyoxal $(\ce{PhCOCHO})$ can undergo both internal Cannizzaro reaction and between two molecules.
The internal Cannizzaro reaction is like this:

Which reaction will be preferred? Internal Cannizzaro or Cannizzaro reaction between two molecules?

Just a guess. The internal mechanism should be faster because there is a hydride transfer and this transfer should be faster if it happens in the same molecule.

I agree with the other answers that in general intramolecular reactions are faster than intermolecular reactions. Let's look at why this intramolecular reaction might be fast.
We can describe the Cannizzaro reaction as the delivery of a hydride from an anionic hydrate to a carbonyl group. The reaction shown certainly meets those criteria. I see this reaction as a carbocation rearrangement or a (semi-)pinacol rearrangement. In the latter class of reactions, a group undergoes a 1,2-migration to quench a carbocation and generate a carbonyl.
In the phenylglyoxyal rearrangement, if the ketone is considered as its ionic resonance structure (2), then the ketone carbon is a carbocation. We know that hydrogens can and will undergo 1,2-migration if the resulting carbocation is more stable. In this case (3->4), the resulting carbocation is very stable, because it's really a carboxylic acid (5). Furthermore, the carboxylic acid will be deprotonated rapidly under basic conditions (5->6).

As a rule, intermolecular reactions are slower than intramolecular reactions because you need two or more different molecules to collide with the right energy and the correct orientation. In an intramolecular reaction, you only need the right energy as (1) there is no need for collision and (2) the right orientation is often easier to achieve.
The reaction you describe is also documented in the paper A Highly Efficient and Enantioselective Intramolecular Cannizzaro Reaction under TOX/Cu(II) Catalysis but you'll note that the aldehyde exists as an acetal.

If you could provide a short summary of the linked information then this would be a very satisfactory answer. Links can become obsolete which leads to answers becoming useless.

Intramolecular reactions are generally always faster than intermolecular ones. This can be seen very simply from kinetics.
An intramolecular reaction has only one reactant and hence is a unimolecular reaction. In this case there would be two participating molecules in the intermolecular reaction. Hence this becomes a bimolecular reaction (the intermolecular one).
Since the feasibility of a reaction depends on molecules colliding and reacting, the probability of molecules colliding in the correct orientation and with appropriate energy reduces when there are more reacting species.
Hence intramolecular reactions are faster.