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To heat or not to heat; distillation?
One of the questions in a recent past paper I completed was:

One way to make water from the reservoir suitable for drinking is by distillation.
Describe how water is distilled.

And I answered (perhaps too succinctly) that "The water is vaporised and the condensation is collected."
I would have thought my answer correct, in that it fully addresses the question, however, the mark scheme mentions two points I did not make:

I should have mentioned heat.
I should have mentioned cool (the mark scheme's words, not mine).

But I was wondering, are these two (common) aspects of distillation actually necessary to distill water?

Surely heating and cooling are not really necessary to distill water; couldn't pressure be reduced to evaporate the water, and increased again to condense it?
Or, in the context of the question, is the heating perhaps necessary to sterilise the water (for killing microbes)?

Welcome to chemistry.se! If you ask me, and by extension you are already doing this, your explanation is quite fine. That is the general process of distillation. In the case of water, however, reducing pressure might just not be enough to evaporate it.

@Martin-マーチン According to this table (and my own - perhaps incorrect - presumptions about how vapour pressure works) pressure would need to be reduced to 0.0121 atm for water to boil at 10°C, which should be doable with even just vacuum pumps.

Of course, 10°C being an amount chosen to represent the reservoir water temperature; a higher temperature means water can boil at a higher pressure.

Why, of course you can make water boil and then condense by manipulating pressure alone. It would be slow and cumbersome, but very much possible. The problem is, as some water evaporates, the rest of it cools down, and continues to cool all the way down until it becomes ice, so you'll have to actually heat it to keep it at room temperature, and if you have the heater anyway, why bother with all that pressure stuff.

You can but the instrumentation is far more complicated. I a laboratory I would prefer do not use a noise devise. Reach a pressure low enough is not a problem. In do not know which alternative is cheaper.

You obviously already have a general idea of what is going on in a distillation, so I will spare you the picture and everything.
Your question can be broken into two parts:

Do I need heat for vapourisation?
Do I need cooling for condensation?

I am going to address them one after the other.
Vapourisation
As you probably also know, reducing pressure is one means of evaporating liquids. This is used in rotavaps which use a $40~\mathrm{^\circ C}$ bath and reduce the pressure to make sure the solvent evaporates. You could think that just reducing the pressure should be enough to vapourise any liquid.
But if you vapourise a liquid only by reducing pressure, you will reduce the temperature of the liquid. This is the evaporation cooling that your body uses when sweating: If the water in sweat evaporates it cools the surface it evaporates from. Thus, after a certain time and especially if you attempt to distill large quantities of water, it will freeze and further evaporation will become exponentially harder. Therefore, a practical setup will always require heating.
Condensation
Effectively, this is the same discussion but in reverse. Note that if you boil water in a closed pot, the lid of the pot will get successively hotter due to condensing water releasing heat. After a certain time, the temperature of the surface you want to condense on will be so hot that the liquid will no longer condense.
This is especially a problem if you decided to reduce pressure for evaporation: If you have to reduce the pressure so far that the liquid would evaporate at $10~\mathrm{^\circ C}$, then there is no way that it will condense at a cooling trap of $25~\mathrm{^\circ C}$. Therefore, the condensation area must be cooled which is typically achieved with Liebig coolers or similar.
What about vapourisation followed by condensation in the same vessel with changing pressure?
Well, that is possible. It is not practical, though, since the inside of a single vessel can only have a certain pressure (within slight margins). And vapour spreads out uniformly. Your vessel would have to have an initial ‘bowl’ and a collecting one. You reduce pressure to evaporate out of the initial bowl and then re-raise pressure. The liquid will condense uniformly and you will lose a significant amount of it because it will drip back into the initial bowl. Your yield with a typical common setup is much better.

So you're saying that heating and cooling are necessary processes for distillation?

@theonlygusti Necessary for a practical distillation setup, yeah.

but in terms of the question, is my answer technically fully correct? And so, the mark scheme wrong?

@theonlygusti The question asks ‘describe how water is distilled’. Therefore, it is reasonable to require heat and cooling since the other setup is very impractical and I am not aware of it being used.

Ah, so you're saying I should have read it as "Describe how water is distilled" instead of "Describe how water can be distilled," therefore making my evaluation afterwards incorrect?

Do you have any hard numbers to back up just how impractical my suggested set-up would be?

@theonlygusti Well, let’s put it this way. A single pot which is heated and which has a cooling device can give a constant flow of distillate. Working with pressure will require a stepwise process making it much less efficient per se because it is not constantly flowing.

The question is trying to explore whether you understand how distillation works. Yes, vapour is created and, yes, the condensate is collected. But your answer doesn't describe how that is done.
You are correct that you could use reduced pressure to create vapour, but you will still need some input of heat and some way to cool the vapour. Or you could have described the simpler process of applying heat at normal pressure and condensing the resulting vapour and the examiner would know you understood the basics.
It may be possible to do distillation purely with pressure swings, but the apparatus would be really complicated (how, for example, do you collect the low pressure vapour and then repressurize it to cause condensation?). Besides the reason why distillation used lowered pressure isn't to avoid the need for energy input but to avoid the need for heating the liquid to excessive temperatures to achieve energy input (some things will decompose at high temperatures). In a constant pressure vessel, the only way to condense the vapour is to cool it so you can't avoid the need for either cooling the vapour or heating the liquid.
Show you understand the basics and you will get full marks in the test. Don't overcomplicate.

Obviously, as I sat the test, I didn't think through the question in as much depth as I lay out in my post; if anything, I under-complicated my answer, in failing to mention how distillation can be achieved. I was simply a bit annoyed to find out that I had the "wrong" answer because I had failed to mention two processes which, in contemplation afterwards, don't even seem to be necessary for distillation.

Anyway, I don't feel that this answer answers the question I've asked definitively enough; is heating/cooling necessary for distillation?

@theonlygusti The only way you can possibly avoid heating and cooling is to have some immensely complicated apparatus that allows significant pressure swings. In practice and in real labs you will need heating or coolling or usually both. (BTW the answer does say this.)

@theonlygusti And you have speculated that pressure alone could do the job: it can't (unless you have some very clever design of device, which you have not demonstrated). Low pressure makes the temperature differential required different, not non-existent (in a constant pressure apparatus.)

You will need heat even if you play with pressure: this heat may come from a reservoir which stays at room temperature, yet heat will flow back and forth during evaporation and condensation.
And let us be honest: you gave a half-baked answer, which is half baked even if your teacher didn't write an essay about what is missing

If distilling means vaporization and then condensation of a liquid without applying work to the system, then the energy needed to transform the liquid to the gaseous state has to be supplied as heat.
This is called heat of vaporization, or latent heat of vaporization, since this does not necessarily mean that the temperature has to be raised.

I'm sorry, i don't really understand this answer; heat has to be applied, yet the temperature does not raise?

Consider water at 100°C. You have liquid at 100°C and above it you have vapor at 100°C. If you now apply heat to the liquid water it would not raise the temperature. Instead the heat would be consumed to transform more liquid water to gaseous phase. This is the latent heat of vaporization. Follow the link to get additional information.


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How to determine mobility when two mechanisms exist simultaneously

Two scattering mechanisms exist in a semiconductor. If only the first
  mechanism is present, the mobility will be $\pu{250 cm2/V s}$. If only the
  second mechanism is present, the mobility will be $\pu{500 cm2/V s}$.
  Determine the mobility when both scattering mechanisms exist at the
  same time.

I'm not sure what the last sentence is trying to tell me. I know mobility is $q\tau_c/m_n$, and I think I need to solve for mean free time here to help, but I'm not sure what I should use for $m_n$ since I don't know what the semiconductor is, and I'm not given a value.
Although, since I'm given two mobilities to start with, should I solve for $\tau_c$ twice and average those? Really not sure what direction I should be going in (if only my professor answered his emails).

I needed to add the inverses of the given mobilities and take the inverse of that to get my final result.
$1/\mu = 1/\mu_L + 1/\mu_I$

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In a condensation reaction, which of the reactants loses their OH group?
You have Propanoic Acid ($\ce{CH3CH2CO-O-H}$) so an $\ce{OH}$ on the end
You have Ethanol ($\ce{CH3CH2-OH}$) so an $\ce{OH}$ on the end
When you combine the two for a condensation reaction you have the $\ce{OH}$ group from 1 reactant and the $\ce H$ group from the other that joins to release a water molecule correct?
So which of these groups loses its $\ce{OH}$ group and which group loses its $\ce H$ group?
How can i determine this is im guessing it may have something to do with strength of bonding and shielding but can somebody please explain i cant find the answer

related http://chemistry.stackexchange.com/questions/18378/are-all-esterification-reactions-reversible-what-is-their-general-mechanism

Acid loses its OH and alcohol loses its H, that's the way it is.

Long story short, oxygen from alcohol's $\ce{-OH}$ attacks the carbon from the acid's $\ce{-COOH}$. That carbon, being linked to two oxygens, carries significant positive charge, so this is much more likely than the other way around.
Also, it has been proven rigorously with the use of isotope-labeled compounds. Say, you take some acid with "normal" oxygen ($^{16}\ce O$) and alcohol with $^{18}\ce O$, and behold, the ester comes out with one $^{16}\ce O$ and one $^{18}\ce O$, so the latter must have come from alcohol, so it is the oxygen from acid that was lost in the form of water.

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Physical reason for why delocalisation leads to stability?
Why does delocalization of electrons generally make compounds more stable (e.g. in carboxylate anions, where the lone pair on the negatively charged oxygen is delocalised into the C=O π* orbital)? Is there any proper explanation involving quantum mechanics?

Electrons are always delocalised across the entire molecule according to quantum mechanics. It is just that the Lewis depiction of localised bonds is so much simpler to teach that we use it, and then we need resonance to overcome its weaknesses.

A simple QM explanation I can think of would invoke Hückel theory. It would go this way:
Suppose you have a two-state system (analogous to having two orbitals) with eigenvalues of the Hamiltonian $\lambda_1$ and $\lambda_2$. These correspond to the orbital energies. So just for argument's sake, let's say that
$$\begin{equation}
\mathbf{H} = \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}
\end{equation}$$
then obviously you have $\lambda_1 = 1, \lambda_2 = 2$. Let's say this system has two electrons. They would both go into the orbital with the lower energy, i.e. $\lambda_1$. In the context of organic chemistry, this would typically mean that the first orbital would be bonding in nature (hence filled), and the second orbital would be antibonding (hence empty), although that's not always the case.
Now, you perturb the system by making the non-diagonal elements of the Hamiltonian negative (analogous to allowing the orbitals to overlap, i.e. allowing delocalisation of electron density between the two). Just for argument's sake, let's change $\mathbf{H}_{12}$ and $\mathbf{H}_{21}$ to both be $-0.1$. If you recalculate the eigenvalues you get
$$\lambda_1 = 0.99, \lambda_2 = 2.01$$
The new eigenvectors will be linear combinations of the old eigenvectors – one in-phase and one out-of-phase combination. The "bonding" or in-phase combination of the two orbitals will have decreased in energy, whereas the out-of-phase combination of the two orbitals will have an increased energy. If you put both electrons into the bonding orbital, you get a net stabilisation.

In general, interactions between filled and unfilled orbitals are stabilising in nature, since both electrons occupy a lower-energy bonding orbital.

On the other hand, interactions between two filled orbitals are destabilising, because you would have four electrons and you would have to fill both of the new orbitals. If you add up the two new eigenvalues, the total energy is greater than it was before when $\mathbf{H}$ was unperturbed.
This is not evident in my example above. The reason is that I made the assumption that $\langle i | j \rangle = \mathbf{S}_{ij} = \delta_{ij}$. In extended Huckel theory you would solve the generalised eigenvalue equation, $\mathbf{Hc} = E\mathbf{Sc}$. However with my assumption, $\mathbf{S}$ is simply the identity matrix and this reduces to $\mathbf{Hc} = E\mathbf{c}$. A theorem from linear algebra states that the sum of the eigenvalues of $\mathbf{H}$ is simply equal to the trace of $\mathbf{H}$. Therefore, as long as we assume $\mathbf{S}_{ij} = \delta_{ij}$, merely changing the off-diagonal values will not change the sum of the orbital energies.
In general, the basis vectors are not orthogonal. If you solved the full equation, $\mathbf{Hc} = E\mathbf{Sc}$ then you would find that the sum of the new eigenvalues is greater than the sum of the old, unperturbed eigenvalues.
Obviously this is ignoring other important aspects, e.g. electron-electron repulsions (that is a problem intrinsic to Huckel theory), but it's a start.

Great explanation and easily understood! Thanks!

Even simpler:  Particle in a box.  The longer the box, the lower the energy.  Delocalization equals a longer box.  However, this trite model is not quite as good as orthocresol's explanation.

Actually, is it always true that delocalisation leads to stability? Consider the case of various antiaromatic compounds. In particular, let us look at cyclobutadiene. It exists more stably in its "rectangular form" with localised double bonds, rather than with electronic conjugation across the whole molecule.

@TanYongBoon Would one describe an antiaromatic system as well delocalized? In cyclobutadiene, two bonds have constructive pi intereference, while two other bonds have destructive pi interference. I'm not an expert in it, but perhaps the instability is due in part precisely to less delocalization being available.

@Blaise I would like to disagree. In the "fully-antiaromatic" cyclobutadiene, the 4 $\ce {C-C}$ bond lengths tends towards being equivalent, favouring maximum delocalisation. The degree of delocalisation cannot be predicted by just looking at the MOs in the MO diagram. I think your description is also slightly incorrect. In the MOs of cyclobutadiene, two are nonbonding, one is bonding while one is antibonding. However, the degree of delocalisation cannot be inferred from the diagram.

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Is there wavefunction interference between alternate reaction pathways?
I'll give a specific example to clarify my question. If we look at the approach of chloromethylenecarbene's electron pair to cis-2-butene in the reaction of creating 1-chloro-2,3-dimethylcyclopropane, there's a symmetry in the choice of where the nucleophilic attack can happen.
These two alternatives can be represented by wavefunctions and so I would expect there to be an interference effect between them that would make the reaction different than if one of the spots was somehow unavailable for reaction (perhaps by the existence of some electron donating/withdrawing groups or other hindering groups attached on the periphery of the butene's double bonds).
Has reaction pathway alternatives interference been observed?

I'm not getting your point.

I've already asked the question as best as I think I could.


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Is ΔS of a system related to temperature and change in enthalpy?
I often see the equation $\Delta S_\text{universe}$ = $\Delta S_\text{surroundings}$ + $\Delta S_\text{system}$, where $\Delta S_\text{surroundings}$ can be re-expressed in terms of the enthalpy change of the system as
$$\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$$.
This leaves me wondering why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$ (Maybe as $$\Delta S_\text{system} \overset{?}{=} \frac{- \Delta H_\text{surroundings}}{T}$$ although this would make $\Delta S_\text{universe}$ equal $0$).
Instead, $\Delta S_\text{system}$ is usually given as a standard value, seemingly independent of temperature. Is there a reason why we can only express the change in entropy of the surroundings, but not the change in entropy of the system, as related to the change in enthalpy and the temperature of a reaction?

If the system experiences a change at constant pressure, the change in enthalpy of the system $\Delta H_{system}$ is equal to the heat Q transferred from the surroundings to the system.  So the heat received by the surroundings is $-\Delta H_{system}$.  Typically, the surroundings is regarded as an ideal constant temperature reservoir, and the reservoir is considered as being "internally reversible."  So, if T is the temperature of the reservoir, the change in entropy of the surroundings is $\Delta S = -\frac{\Delta H_{system}}{T}$.  The system is typically not regarded as experiencing an internally reversible change, so the same kind of approach can not be applied to the system.

Could you specify what you mean by "internally reversible?"

It's a term used in Fundamentals of Engineering Thermodynamics by Moran et al.   I particularly like their approach in this book.  "A reversible process is one for which no irreversibilities are present within the system or its surroundings.  An internally reversible process is one for which there are no irreversibilities within the system.  Irreversibiities may be located within the surroundings, however.  Using the internally reversible process concept we refine the definition of the thermal reservoir ...we assume no internal irreversibilites present within a thermal reservoir."

What would be a case where irreversibilities only occur in the surroundings (but not the system)?

Covered in section 5.3.4 of the book.  "If there were a spatial variation in temperature, say, there would be a tendency for a spontaneous energy transfer by conduction to occur within the system in the direction of decreasing temperature." Another example would be if a spontaneous chemical reaction is occurring.  Other examples are mixing and diffusion. "For reversibility (within the system), however, no spontaneous processes can be present.  From these considerations it can be concluded that the internally reversible process consists of a (continuous) series of equilibrium states.

Oh.  I didn't answer your question.  Suppose you have a process for expanding an ideal gas reversibly, but, instead of gradually removing tiny weights from a piston, you control the motion of the piston by hand.  In that case, your body is part of the surroundings, and there are a huge number of irreversibilities occurring  within your body as you gradually decrease the force on the piston.  So the gas receives the same reversible expansion in both cases, but, in the first case, the surroundings are handled reversibly and in the second case, they are not.

OK, I think I understand what internally reversible means now, but why does the system not typically experience an internally reversible process?

Let us continue this discussion in chat.

One of the purposes of defining a system and its surrounding in thermodynamics is to separate the interesting stuff from the boring stuff.

$\Delta S_\text{surroundings}$ can be re-expressed as $\dfrac{-\Delta H_\text{system}}{T}$

The best boring surrounding is a large homogeneous phase with the same temperature as the system, for example a stirred water bath. (If the temperature of the system changes, you have to quickly swap the bath to match the temperature.) In this specific scenario, the entropy change of the surrounding will be equal to the heat $-q$ transferred into the surrounding divided by the temperature. If the process in the system is happening at constant pressure, the system is closed and there is no non-PV work (no electrical, mechanical, etc. work, just expansion work), then the heat $q$ transferred out of the system will be equal to the change in enthalpy of the system, $\Delta H$.
So there are a lot of disclaimers to make before you can set $\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$.

[...] why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$

Remember we want the interesting stuff to happen inside the system. In general, there is no way of predicting the entropy change in the system from other thermodynamic values unless the system is as boring as the surrounding (for example when the system is a block of hot metal and the surrounding a block of cold metal). Of course, if you put the interesting stuff in the surrounding and keep the system boring, $$\Delta S_\text{system} = \dfrac{-\Delta H_\text{surroundings}}{T}$$ would be appropriate.

Is ΔS of a system related to temperature and change in enthalpy?

In general, it is not. Entropy and enthalpy change are not coupled. They can both be negative, both positive, one negative and the other positive and vice versa. This shows you that there is no fixed relationship between entropy and enthalpy.
There is one special case where entropy change and enthalpy change of the system are related - when the change in Gibbs energy is zero. The change in Gibbs energy is expressed in terms of entropy and enthalpy changes as (in the following, the subscript "system" is omitted because all the quantities refer to the system):
$$\Delta G = \Delta H - T \Delta S$$
When $\Delta G$ is zero (for example when a substance is at its melting temperature and both solid and liquid phase are present), the following relationship holds:
$$ \Delta S = \frac{\Delta H}{T}$$
Notice that there is no negative sign. This relationship can be used to estimate the melting point of a substance. Notice also that for chemical reactions at equilibrium, $\Delta_r G^\circ$ is not equal to zero, but $\Delta_r G$ is.
So while at equilibrium,
$$\Delta_r S = \Delta_r S^\circ + R \ln K = \frac{\Delta_r H}{T}$$
is correct, you can't calculate $\Delta_r S^\circ$ from $\Delta_r H^\circ$, i.e.
$$ \Delta_r S^\circ \neq \frac{\Delta_r H^\circ}{T}.$$
The only exception is when $K$ happens to be one, as it is by definition for melting when the substance is at the melting point.

$ΔS_\text{system}$ is usually given as a standard value, seemingly independent of temperature

Both $ΔS_\text{system}$ and $ΔH_\text{system}$ are dependent on temperature, and the heat capacity is used to describe the temperature-dependency. If you are evaluating the temperature-dependency of $ΔG_\text{system}$, however, the largest dependency comes from having the temperature directly in the defining equation:
$$\Delta G = \Delta H - T \Delta S$$
So even if entropy and enthalpy were temperature-independent (again, they are not), $\Delta G$ would still depend significantly on temperature. Oddly enough, a rigorous treatment of the temperature-dependence of $\Delta G$ gives the result that you would get from differentiating the defining equation with respect to $T$:
$$\frac{d \Delta G}{dT} = - \Delta S$$

I don't know if you have done some statistical mechanics, but maybe this is related to counting of states. First of all, we express the entropy change of the surroundings in terms of enthalpy change and temperature of the system because we simply can define our system in most cases and so can circumvent and measures quantities that defines a system (enthalpy change, for example, can be measured in a calorimetry expt.) then the entropy change of the surrounding can be inferred because the that of the universe is zero (only for reversible processes).
Second, for the counting I mentioned earlier, Boltzmann defines entropy as
$$S = k\ln W$$
$W$ represents the count of possible states of the system/surrounding. But the  number of states of the system multiplied by those of the surroundings equal the states of the universe. So if you know two quantities (here, system and universe) through knowing their entropies, you can get the remaining. Sure, this doesn't prevent you from doing what you suggested, which is expressing system entropy change as surroundings' enthalpy change and temperature, but you can't define them specifically for the surroundings and don't have a readily available tool to measure them.

We don't know the entropy change of the universe (nothing the OP said indicates a process that is at equilibrium). Even if it were at equilibrium, we still could not express system entropy via the surrounding's enthalpy change unless your system is very simple (no reactions, and enthalpy equal to heat transfer).

I probably don't have the full answer to your question, but I may have some insight. I like this topic, because it's the central idea in Chemistry, but textbooks care a lot about the math of it (which is easy), but not as much about explaining the ideas behind it in detail.
Since the reaction is assumed to take place at constant temperature, the kinetic-rotational energy of the particles of the reacting system does NOT increase of decrease following the reaction. If it were increased or decreased, the temperature would have changed.
Where does the heat released in the reaction come from (in an exothermic reaction)? The heat that is released in the reaction does NOT come from a decrease in the speed, rotation, or vibration of the particles, but from changes in electrical potential. So that's why heat can get released - bonds form with lower electrical potential energy and the difference in electrical potential is released - while the speed, vibration, rotation of the system stays the same - which is why temperature is constant.
So you see, the amount of heat released cannot be used to easily predict how the entropy of the system changed. The entropy change in the system is the result of the fact that when the system reacts, while it has the same amount of kinetic-vibrational energy, that energy is distributed in a way in which either more of it is in a form that cannot be used to do work (increase in entropy) or in a way that more of it is in a form that can be used to do work (more concentrated kinetic-vibrational energy and lower entropy).
By the way, notice that the entropy of the system is not about the electrical potential stored in the system (the source of enthalpy changes in a reaction), but is a fraction of the kinetic-vibrational energy of the particles (the higher percentage of that energy, whose amount in the system does not change when the reaction occurs in constant temperature, is distributed in a way that can NOT be used for work, the higher the entropy).

Is it possible to use the fact that since ΔHsurroundings = ΔHsystem and Tsurroundings = Tsystem, if ΔSsystem were to equal -ΔHsurroundings/T, then from the equation ΔSuniverse = ΔSsurroundings + ΔSsystem, ΔSuniverse would always be 0, in order to show that -ΔHsurroundings/T CANNOT equal ΔSsystem?

"If ΔSsystem were to equal -ΔHsurroundings/T" - I believe that you can't relate these in a straightforward way as in principle you could, for example, have a reaction which releases heat (changes in electrical potential), while experience no opposite change in the entropy of the system (no more concentrated kinetic-vibrational motion in the products compared to the reactants).

Entropy and energy have different dimensions, so the entropy can't be "a fraction of the kinetic-vibrational energy".


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Tunneling corrections in reaction rates
I've studied Transition State Theory since I asked this a while ago, but so far I have not been able to find a detailed explanation of how tunneling corrections are implemented into reaction rates computations.
I am thinking specifically in reactions involving proton transfer along a hydrogen bond. Electrons don't pose any trouble, because they are treated quantically. But for protons, which are usually beheld as classical particles, semiclassical corrections accounting for tunneling must be made, since a full quantum treatment of protons seems undoable.
First idea: If one performs a molecular dynamics simulation, one could at each step think of the proton "just for a moment" as a quantum particle, find an approximated wavefunction thanks to the WKB method and perform a Monte Carlo simulation to decide "where" will we find the proton at the next step. This seems doable, but it's not what is done, as far as I know (sometimes performing molecular dynamics simulations is not possible, or not useful).
Second idea/ problem: Imagine we have a Free Energy surface (I am concerned with two-dimensional problems, where there are two relevant reaction coordinates to be tracked), which informs us about the precise changes in free energy along infinitely many reaction pathways. For each of those, a classical reaction rate can be computed by means of Eyring equation and then we could take averages over some sets of paths via path integrals or something like that. Even better, one can employ Variational Transition State Theory to derive estimations for the reaction rate, once the Free Energy surface is known. I expect this will not be hard. But I don't have a clue as to how to make the tunneling corrections for the reaction rate in this scenario, although I suspect is in this setting where most tunneling corrections are made.
I've certainly found a lot of bibliography dealing with Quantum Transition State Theory and Tunneling Corrections, but absolutely nowhere do they explain with all the mathematical and physical details the concrete formulas they are employing. Instead, they write a lot (without much equations), give you other references, where you can find still some more references -and so on.
So, in brief, my question is: could anyone provide me with a reference containing a thorough, complete and detailed explanation, not lacking the mathematical details, of how does one compute the tunneling corrections for reaction rate constants?

Have you checked the original papers by C. Eckart, R. P. Bell, H. Shin, J. O. Hirschfelder and E. Wigner? These contain a fair degree of mathematical and physical treatment.
There is also a nice paper from S. G. Christov called "Some aspects of the theory of proton-transfer processes" published in J. Res. Inst. Catalysis 1968, 16 (1), 169-194 (pdf).
Other possible references:
Johnston, H. S.; Heicklen, J. Tunnelling Corrections for Unsymmetrical Eckart Potential Energy Barriers. J. Phys. Chem. 1962, 66 (3), 532–533. DOI: 10.1021/j100809a040.
Some papers published by D. G. Truhlar together with T. N. Truong, B. C. Garret, R. T. Skodje.


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How to prevent dripping when using a Pasteur pipette with solvents like diethyl ether?
I'm trying to use a Pasteur pipette for pipetting diethyl ether and regardless of the angle I hold the pipette at, the diethyl ether will automatically start dispensing itself (I am unsure of what this phenomenon is called). I'm sure it's not the rubber bulb since I've been using the same one for pipetting other solvents that don't do this. I've also tried using a different rubber bulb, but to no avail.
Is there any way to prevent this from occurring? If not, is there anyway to mitigate the flow rate of the drips?

Low viscocity and high vapor pressure of such solvents makes it a pain indeed. Using an automatic pippet you can reverse pippet (push plunger button all the way down, and dispense only to first check. Also work quickly, deliberately and over short distances

DCM used to be my issue. Either work really quickly if your pipette is graduated or ditch the pipette and use a polypropylene plunger syringe (not the rubber ended ones, fine for bio, rubbish for wet organic chemistry ). I also used to overfill the (plastic) pipette and let it drip to roughly the correct volume, then transfer. Depends on your application.

Usually I avoid pipetting diethyl ether because it can be a bit annoying, but recently I've realized there is a very simple technique which markedly increases ease of transfer (though admittedly I've been using it mostly for DCM). All you have to do is saturate the interior of the pipette with the vapour of the low-boiling solvent before performing the transfer.
The problem arises because the low-boiling solvent will quickly (but not instantly) evaporate into the partial vacuum inside the pipette after pulling the solvent. If you try to transfer the liquid while it's still evaporating significantly, the large volume change from the phase transition will push the liquid beneath through the pipette exit.
So how do you get around this problem in practice? Using the pipette, simply pull and dispense the solvent a few times in succession over 5-10 seconds inside the outgoing vial. Then pull what you need and transfer. In my experience this greatly mitigates dripping. If you're not too slow between transfers, you can get away with doing this only once, at the start.
Other things which can be done to help:

As mentioned in the comments, avoid dragging the filled pipette over a  large distance. Keep the outgoing and ingoing vials right next to each other, at about the same level.
Pull a little bit of air into the pipette, keeping it between the exit and the solvent. This gives you a small time buffer before it drips out.
If you have a steady hand, keep very gently pulling air while you move the pipette between vials, matching the speed at which the liquid tries to drip out. This can be somewhat fiddly, so train to get a feel for it.
Don't fill the pipette to its maximum capacity on each transfer; do more transfers, each with less volume.
There's always a drip left on the very tip of the pipette, and that can easily fall. To avoid it, contact the remaining drop onto one of the outgoing vial's walls above the surface of the liquid, then pull a little bit of air into the pipette, sucking the last droplet along and leaving the tip dry.
Practice.


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Why is astatine so radioactive?
I have heard and read about astatine's radioactivity, and it makes no sense to me. Why is it so unstable?  Uranium has a half life ranging from 70 to 4.5 billion years (natural isotopes), and it has 7 extra protons to deal with. I know that atomic nuclei can be stabler if they are spherical, implying that if you make a nucleus that isn't spherical you could destabilize it, but astatine has a maximum half life of 8 hours. Can someone explain this to me?
Information came from a few sources, including SciShow and Wikipedia.

related http://chemistry.stackexchange.com/questions/28695/why-is-tin-112-stable-but-indium-112-radioactive

That is useful, but doesn't actually answer why astatine is so short lived compared to its neighbors, as like I have mentioned: Astatine has a longest half life of around 8 hours. Using the logic from the linked question, couldn't we just shove less neutrons into the nucleus to make it less unstable? This wouldn't work, because of the way the strong nuclear force works but this si chemistry stack exchange, not physics stack exchange.

I wouldn't say there's sth unusual. After Pb with "magical" configuration, nuclei lower they stability. There's simply local minimum for this number of protons.

@Crimsonvale That’s why Mith wrote related and not duplicate ;)

Promethium sticks out much more in the PSE than astatine. Actinium, francium, protactinium are also very unstable compared to their neighbours. An odd number of protons is often not very favourable. But really this is no chemistry question.

What about francium? It has a half-life of 22 minutes

I don't think there is an actual answer to that. What we know about the radioactivity is based mostly on empiric observation and quantum physics is still not able to give precise answers about the process of radioactive decay.
A similar question could be asked about technetium: why is it unstable, even though it stands in the periodic table between stable elements? Probably it's a matter of their proton and neutron configuration that for some reason are particularly unstable compared to their neighbourhood, but there is still no equation capable of giving the half-life of an isotope.
This has already been discussed and probably we will have to wait until the atom nucleus is better studied before being able to answer this question.


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Software for visualization of 3D models of chemical structures with atomic orbitals
A number of programs are available that let you build, visualize, and analyze 3D models of chemical structures (I use Chem3D/ChemBio3D). Current versions usually include various possibilities for the calculation and graphical representation of molecular orbitals.

However, the resulting images can be overloaded or unclear. Introductory explanations using such images might be overdoing it. Therefore, many textbooks on organic chemistry use simple chemical structure diagrams based on atomic orbitals.

In particular for didactic purposes, I am looking for an intermediate solution. I would like to build, optimize, and visualize a 3D model of a chemical structure and show a graphical representation of the original atomic orbitals in the 3D model. Unfortunately, this option doesn’t seem to be available in Chem3D/ChemBio3D. Does any software exist that lets you build a 3D model of a chemical structure and visualize atomic orbitals?

I haven't come across something like that but I have been wanting something like it as well. I could imagine that it might be possible to achieve this with Avogadro if you are proficient enough with it and C++ to write some sort of plugin.

Does any software exist that lets you build a 3D model of a chemical structure and visualize atomic orbitals? || to my knowledge, no. However, QC packages often offer 'natural orbital analisys', that among other things prints participation of atomic orbitals in molecular orbitals. You can also visualize individual atomic orbitals in some cases and copy-paste them.

I'm not entirely sure what you want. For example, you say "can be overloaded or unclear" and "might be overdoing it." In my experience, most students can grasp molecular orbitals (or the natural orbital transformations) well. You just need to pick rendering options (e.g., partial transparency) that suit your needs.

To follow on the other comments, besides doing a natural orbital analysis (a good option - you can then use most software to view the results), I am not aware of any such software. However, there are probably ways to hack this even without coding a rendering option. If you want to pursue it, consider contacting me.

You could try JMOL as its free and has the ability to draw orbitals as well as bonds. It is fairly easy to construct structures given coordinates. See molecule-viewer.com for some examples of what you can do.

The "overloading" likely arises from default parameters for the isosurface used to display. Just increase that and the MOs will look less crazy.

As porphyrin mentions in the comments, have a look at Jmol.

It indeed does the rendering as in your original version, but with the many options (see this mo command documentation) and particularly interactivity provided by the scripting, I think you may pedagogically achieve what you are looking for. Proteopedia is a nice website where the efficiently use Jmol's scripting feature for educational purposes (try the green links).


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Why is $^3P_1$ an atomic term symbol for carbon?
I'm using these rules, $^{2S+1}L_J$ with $J=L+S$ and it seems to imply that $^3P_1$ means $S=1$, $L=1$, and $J=1$ but clearly $1+1 \ne 1$ so according to me this shouldn't be an atomic term symbol but it is. Why?

This is not the fundamental one. But it not means it is not one.

You are correct in saying that the total spin for two electrons is S = 1 as its spin multiplicity is 3 and that L =1. The J value is found in the same way as S and L values were using the Clebsch-Gordon series, $$J= |L+S|...|L-S| $$ This means find the maximum value, then the minimum make it positive, and separate any other values by 1. Sometimes there is only 1 value so be careful. We use this type of series as a short-cut for vector addition of angular momentum which is why you question '1+1 = 1'.
In your case J = 1+1 to 0 , which gives three values J = 2, 1 and 0 so three possible term symbols.  There will also be singlet terms where the total spin S = 0.
The triplet term means that electrons are in different orbitals, generally this lowers the energy, and also means that there is no Pauli principle to consider. Hund's rules are used to determine which levels and states are lowest in energy. (See for example Steinfeld, 'Molecules & radiation')

Ah ok, so it seems I was just told some quick short cut for some actual process. What's the actual process called if I want to do it "the real way"; Clebsch-Gordon addition of angular momentum or something I guess?

I'm not sure it has a name as such, but its based on the Russell Saunders coupling scheme and vector addition of angular momentum.  This is described in most undergraduate phys chem text books.

Yeah, for reference I was reading McQuarrie Physical Chemistry and this seemed to be just thrown at me without justification so I guess I need to refer to a different book for the explanation on that.

with $J=L+S$

Nope. In general, it can be shown that the permitted values of the $J$ quantum number for the total angular momentum that arise from two sources characterized by quantum numbers $J_1$ and $J_2$ are given by
$$
    J = J_1 + J_2, J_1 + J_2 - 1, \dotsc, |J_1 - J_2| \, .
$$
In this particular case, $J_1 = L$ and $J_2 = S$, so $J$ ranges from $L + S$ to $|L - S|$ in integral steps and not necessarily equal to $L + S$.


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Does "easily hydrated" imply highly soluble in water too?
Does "easily hydrated" imply highly soluble in water too?
Small ions like $\ce{Li^+}$ and $\ce{Be^{2+}}$ are said to be easily hydrated but does it also imply they are highly soluble?
I looked up some books and they say that compounds of Lithium like $\ce{Li_2O}$
and $\ce{LiF}$ are sparingly soluble.Why?

This may be a pathological example, but superabsorbent polymers hydrate very easily, swelling to many times their original volume, yet they can remain insoluble in water.

Many ceramics hydrate well, but are not soluble. The difference is the ceramic is happy to add OH groups in various places, but that does not imply that it wants to break up into constituent molecules surrounded by water. It is not a transitive property.

Note that you are considering two different things: easily hydrated cations and soluble compounds. Solubility is complicated, and compounds containing the same cations may well differ in solubility-- lithium oxide reacts violently with water to form lithium hydroxide, for example, whereas lithium fluoride is only sparingly soluble.
What causes the difference between the two compounds? For one, the $\text{O}^{2-}$ anion is an extremely strong base, and readily reacts with water to form $2\,\text{OH}^-$; likely the high enthalpy of reaction provides a thermodynamic driving force for the dissolution of lithium oxide. If we then consider lithium fluoride, no large driving force exists to favor dissolution. While $\text{F}^-$ may react with water to form $\text{HF}$ and $\text{OH}^-$, $\text{F}^-$ is only weakly basic and such a reaction would not be as exothermic as that with lithium oxide.
We must then consider smaller driving forces that were drowned out when considering lithium oxide. Even though $\text{Li}^+$ and $\text{F}^-$ are both small, highly charged species with large charge densities that will interact favorably (with regards to enthalpy) with the polar water molecules in solution, their charge densities will also cause water molecules to cluster around the ions, forming hydration shells and hence decreasing entropy; this is thermodynamically unfavorable. Furthermore, the electropositivity of $\text{Li}^+$ and the electronegativity of $\text{F}^-$, combined with their small ionic radii, lead to a large lattice energy for $\text{LiF}$, further disfavoring dissociation. These effects all contribute to the sparing solubility of $\text{LiF}$.
Related question.

Hydration is a characteristics of an ion. Solubility is dictated by a set of anions and cathions. Consider Na$^+$ in Na2CO3 (soluble) and NaHCO3 (poorly soluble) and Ca$^+$ in CaCO3 (insoluble) Ca(HCO3)2 (soluble). Anions alone or cations alone are not enough to predict solubility.
Also note the case of a proton exchange membrane. It has moieties that are hydrated well, but it is not soluble at all.


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Why is sodium bisulfite needed in syn-dihydroxylation?
On the reaction of syn-dihydroxylation with stoichiometric quantities of OsO4 (not a catalytic amount) the second step is $\ce{NaHSO3/H2O}$ or $\ce{Na2SO3/H2O}$.
According to my understanding, $\ce{H2O}$ is the only thing needed to hydrolyze the cyclic osmate ester resulting from the first step of adding $\ce{OsO4}$.
So, what is the use of $\ce{NaHSO3}$ or $\ce{Na2SO3}$ here?

You need to reduce remaining oxidizer (OsO4) and maintain pH of the solution. NaHSO3 acts both as the reducing agent and a buffer.


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Oxidation number of phosphorus in phosphine
Electronegativity of $\ce{H}$ is greater than $\ce{P}$, but according to my book in $\ce{PH3}$, $\ce{P}$ has oxidation number of -3. How is it possible? Is any of the data wrong?

Here is a way to get the answer. If you place a reagent with oxidation number of H = -1 (hydride) it would produce dihydrogen gas. For example: $\ce{KH + H2O-> KOH + H2}$.
If you place a molecule with oxidation number of $\ce{H} = +1$ dihydrogen gas is NOT formed. $\ce{NH3 + H2O -> NH4+ + OH-}$. Phosphine doesn't form dihydrogen gas on contact with water so $\ce{H}$ is in oxidation state +1 and $\ce{P}$ is in oxidation state -3. Reaction with strong acids leads to $\ce{PH4+}$ which has similar structure to the ammonium ion. For these reasons even for $\ce{As}$ in $\ce{AsH3}$ the oxidation state of $\ce{As = -3}$.
The problem with electronegativity is that we are trying to assign a single number, while we actually need a function. We take average of electronegativity of $\ce{H}$ in $\ce{HF, NaH, H2O, CH4, NH4+}$ and come up with a single number. This works well if you need to guesstimate the behavior, but it is not accurate. It doesn't even account for softness/hardness of a partner.
Analysis "will $\ce{PH3 + H2O}$ form $\ce{H2}$?" (no) vs "will $\ce{PH3 + P2O3}$ form $\ce{P + H2O}$?" (yes) is more accurate way to assign formal charges.

But H is more electronegative than P...how do you explain that?

@ZOZ please see updated version.

@ZOZ As a rule of thumb, H will pretty much always be in oxidation state +1. You can work your way through many redox reactions starting with this assumption! Only in some rarer cases like metal hydrides (the KH example above) will H be have the oxidation number -1.

Electronegativity is a mere indication of what is happening and you will find many different values for the electronegativity of H and P (and all the other elements).
If you have a look at the C-H bond, electronegativities are 2.55 for carbon and 2.2 for hydrogen, i.e. the difference is much larger than for phosphorus (2.19) and hydrogen (2.2). However, C-H bonds are generally regarded as non-polar.
What really makes phosphorus different in its reactivity is that it is extremely oxophilic, i.e. it will try to form P-O or P=O bonds on every occasion. This is why PH3 is pyrophoric, for example.

Oxidation numbers are generally used to determine if a molecule acts as a reducer or oxidiser in a given reaction. The same molecule can be a reducer in one reaction and oxidiser in another if it is an intermediate oxidation state. Knowing the electronegativity (however defined) does not help in this instance.
If the oxidation number of a particular atom is not known it may be deduced from the known oxidation numbers of other atoms associated with it in a molecule. For example, the oxidation state of Pu in PuO$_2^{2+}$ is not known. It is calculated as
charge on Pu in PuO$_2^{2+}$ + charges on two O's (or 4-) = charge on PuO$_2^{2+}$ or 2+
thus Pu has an oxidation number of 6+.
In some cases (S$_4$O$_6^{2-}$, Fe$_3$O$_4$) the oxidation number can be fractional. Thus oxidation numbers can be negative, positive, zero or fractional. Of course the same atom can have different oxidation numbers in different molecules. An example is shown below for nitrogen
-3 in NH$_3$, 0 in N$_2$ , +1 in N$_2$O, +2 in NO, +3 in HNO$_2$, +4 in NO$_2$, +5 in HNO$_3$
where NH$_3$ is only a reducer and HNO$_3$ only an oxidiser, but the other compounds can be reducers or oxidisers. Note that it is zero in N$_2$.
A molecule is an oxidiser of its oxidation state decreases, for example iodine oxidises the strong reducer H$_2$SO$_3$ e.g.
I$_2$ + H$_2$SO$_3$ +H$_2$O $\rightarrow$ 2HI+H$_2$SO$_4$,
the iodine oxidation state changes from 0 to -1. In a reduction the oxidation state increases.
You can see that in your compound the phosphorous' oxidation state is -3.


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What is activating effect?
In the book "Organic Chemistry" by Morrison and Boyd,It is written that

Sodium borohydride cannot reduce weaker electrophiles like esters though Lithium aluminum  hydride can reduce esters to alcohols due to stronger activating effect of Li+ ion than Na+

What is activating effect?
(Google search did not give appropriate results.)

Consider a carbonyl group. We know that, under acidic conditions, $\ce{H+}$ can protonate on carbonyl $\ce{O}$ and hence activate the carbonyl group to nucleophilic attack at $\ce{C}$, and we call such an effect an activating effect. The metal ions $\ce{Li^+}$ and $\ce{Na^+}$ serve the same purpose as $\ce{H+}$, though $\ce{Li^+}$ is a stronger activator due to its high charge density and hence stronger attraction to $\ce{O}$.


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Acetic acid reaction sequence
Can anyone tell me where I am doing it wrong for the following reaction?
In the first step, Br should be attached to the alpha carbon.
In second step, the bromide is replaced by cyanide.
I am able to get the third product that is final compound.

Heating a carboxylic acid often leads to decarboxylation. So decarboxylate your product from step 2 (well actually the product formed following acidification in step 3!)

@Leeser can you tell me the products formed

Step 1: $\alpha$-bromination
$$\ce{CH3-COOH + Br2/P -> Br-CH2-COOH}$$
Step 2: Nucleophilic Substitution by Sodium Cyanide
$$\ce{Br-CH2-COOH + NaCN -> NaBr + NC-CH2-COOH}$$
Step 3: Acid-catalyzed Conversion of Nitrile to Amide or Acid
(1) First hydration... $$\ce{NC-CH2-COOH + H2O -> H2N-CO-CH2-COOH}$$
(2) Further heating... $$\ce{H2N-CO-CH2-COOH -> HOOC-CH2-COOH + NH4+}$$
Special Notes
Be VERY CAREFUL with Step 2 as you are potentially forming $\ce{HCN}$ which could get everyone in the lab killed if you're not careful.

But @sixtytrees has made different products

@user123733 No, he only went step further with decarboxylation, without any good reason.

There are two possibilities. You will have a two step reaction
NC-CH2COOH -> HOOC-CH2-COOH This is malonic acid. It can be further decarboxylated to
HOOC-CH2-COOH + H$^+$ + heat -> CO2 + CH3COOH
the final product is acetic acid.


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Phase transition for an ideal fluid
Does a phase transition occur also for an ideal fluid (from ideal gas to ideal uncompressible liquid)?
Or ideal fluids can be modelled only as ideal gas or uncompressible liquid at all, without the possibility for a phase change?
I hope the question is well posed, thanks

Ideal gas and ideal fluid are not the same. Ideal fluid has molecular interaction that is even throughout different entities, and has no viscocity, and no compression.
Yes, ideal fluid is modeled so that it will go through phase change, but you would have to consider the gas from ideal fluid as a pseudo- real gas, since it has molecular interaction.
An ideal gas has no volume, has no interaction, will not go through any sort of phase change.
Hope this helps.

For sure any answer is helpful! So, correct if I'm wrong, if I assume that water vapour is an ideal gas, it will be always in the gas phase and hence the saturation dome will not exist. In the same way if I assume water as an (ideal) incompressible liquid?

Yes. If you assumed water vapor as an ideal gas, no saturation dome would exist. Ideal incompressible liquids will  have vapor pressure, and will saturate in a finite container given that there is sufficient amount of liquid.


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Difference between Gibbs Free Energy and Exergy
I'm wondering if there is any difference between Gibbs Free Energy and Exergy because their definition seems to me the same $\Delta H - T_\text{env}\cdot\Delta S$.
Exergy represents the maximum amount of work that can be extracted from a system going in equilibrium with its environment, while the Gibbs Free Energy is evaluated at $T,P$ of the system in equilibrium with the environment.
Practically what is the difference, if any?

https://en.wikipedia.org/wiki/Exergy

Gibbs free energy is calculated without possible PV-work, ie. without volume changes.

Exergy is equal to the Gibbs free energy if you happen to be in the isothermal--isobaric ensemble $(N, P, T)$, i.e. in a closed system if the pressure $P$ and temperature $T$ are constant. This is indicated in Eq. 7 of the Wikipedia page for exergy.

I would add a personal note: I've done thermodynamics (and taught it) for many years without ever using the concept of exergy… Although it has some historical relevance, I am not sure how useful it is in a modern presentation of statistical thermodynamics.

Thanks for your precise and clear answer. So Exergy is a way to "call" the free energies depending on the process considered (constant volume-->Helmholtz free energy, constant pressure-->Gibbs free energy)? For example if I considered a stream of hot gas, the maximum work that can be extracted, i.e. its exergy, is the Gibbs free energy if I assume to be at constant pressure?


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Nuclear spin isomerism in molecules other than H2
I was reading my teacher's material about ortho- and para- hydrogen, and I found stated that

"every homonuclear molecule, with nuclides with spin other than zero ($\ce{H2}$, $\ce{D2}$, $\ce{T2}$, $\ce{N2}$, $\ce{^{17}O2}$...) has nuclear spin isomers".

What is not clear to me is: are there any nuclides with spin zero? Is there, hence, any homonuclear molecule without spin isomers?

Nucleons of the same type tend to pair up and cancel out each others' spins, so nuclides with an even number of protons and an even number of neutrons will have zero spin. Some common nuclides with this property include $^4\ce{He}$, $^{12}\ce{C}$, and $^{16}\ce{O}$.


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Reaction mechanism of glycerol with hydrogen iodide acid
I was watching a chemistry video lecture today and I came across this:-
Basically the lecturer was illustrating the mechanism of reaction glycerol with $\ce{HI}$.See from right to left(follow the arrows :-P) in the screen shot.I have a doubt in the fourth step where it is shown by an arrow that the $\ce{I^-}$ ion shifts from terminal position to the central position therby forming a positive charge at the terminal position where another $\ce{I-}$ attacks.He says something like "The $\ce{I-}$ shifts to central position due to areal resonance(or something like that..hope I heard that word correctly) which increases stability".
Now that's my doubt!Why should the iodine ion shift its position at all?Isn't the positive charge more stabilized by hyper conjugation in the central position?And what on earth is areal resonance(I searched google too but no use)?
Can someone help me out ?Thanks in advance!

The iodine isn't changing positions, it's donating a pair of electrons to stabilize the charge by forming an iodonium ion similar to the bromonium ion formed during the halogenation of an alkene shown below. You can see that the bonded pair of electrons was not affected by the step. It looks like they confused the reaction by short-cutting the reaction step.

So the iodide ion attacks at the less hindered terminal position of the iodonium ion?Why? Is being less hindered the only reason? BTW thanks a lot for your answer!


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Why is gauche conformer of 2,3 dimethyl butane more stable than its anti conformer?
Why is gauche conformer of 2,3-dimethylbutane seen along 2,3 bond more stable than its anti conformer? It seems to be anomalous to me as in most other butane like compunds anti form is more stable than gauche form.Can someone explain me the reason why the compound shows Gauche effect?
Also does this anomaly hold in case of 2,3-diphenylbutane and 2,3-dimethylpentane?

Related: Conformations of butane and How can the gauche-effect be explained?

@Loong I saw those links.They do not explain the case of 2,3-dimethylbutane.

I am also confused by this and asked a more general question here. When it's answered, if it doesn't also answer yours, I'll be happy to come back here.

@S.Chevalier Your question is different from mine.Your one is answered in Wikipedia page I linked named Gauche Effect in my question above.

Could it be possible that the conformation where the hydrogens are anti is called gauche?

No.Even the hydrogens are in gauche position.And I found out the reason for stability  today.Will answer the question soon.


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Chromene and cumarin nomenclature
We have these nomenclatures of the same family :
4H-chromene
2H-chromene
Why do we use 2H and 4H to name them? And how ?

There are two isomers of chromene (1-benzopyran): 2H-chromene and 4H-chromene.

According to Subsection P-14.7 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name can be made specific by indicating the position of the extra hydrogen atom in the structure, i.e. the position where no multiple bond is attached. This is accomplished by adding to the front of the name an italic capital ‘H’ preceded by the appropriate numerical locant for this atom.
In 2H-chromene, the ‘indicated hydrogen’ locates one extra hydrogen atom in position ‘2’ of the chromene structure; and in 4H-chromene, the ‘indicated hydrogen’ indicates an extra hydrogen atom at position ‘4’, i.e. one hydrogen atom more than the number present if there were a double bond in the ring at that position.


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Is it correct to say that H2SO4 is an acid in this reaction?
Is it correct to say that $\ce{H2SO_4}$ acts as an acid in the following reaction?
$\ce{2Ag + H2SO_4} \rightarrow \ce{Ag_2SO_4 + 2H_2O + SO_2}$
I know it acts as an oxidizing agent, but is it correct to say it shows acidic properties?

Do you get the same reaction with sodium sulfate?
I do not think so, at least under "normal" conditions.
When the sulfuric acid acts as an oxidizing agent giving off $\ce{SO_2}$, the excess oxygen comes off as oxide ions which must be somehow put into more stable form.  The protons from the sulphuric acid do that by turning them to water.
Getting the reaction to go requires the sulfuric acid to act as both an acid and an oxidizing agent.


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Degenerate orbitals in the Hydrogen atom
I came across the following link:
Which orbitals of the hydrogen atom are degenerate for $n=3$
And all the answers said that for the hydrogen atom, the energy of an orbital depends only on n. Is this really true? And also, isn't it true that for other atoms (at least according to those who answered the other question) the energy depends upon both $l$ and $n$?

Yes, that's true. In a hydrogen atom, orbitals with same $n$ and different $l$ are degenerate, that is, have the same energy; in multi-electron atoms, they are different. Also, welcome to Chem.SE.

Oh, I didn't know about this!(Hmph. Why can't they mention this exception on textbooks?!) Do u know why this is so? Also, thanks! :)

"Why" is a tricky question. What kind of explanation would you be satisfied with? We just solve the Schrödinger equation (exactly for H, approximately for other atoms), and that's it. Some people say the degeneracy is due to certain (and quite unobvious) higher symmetry.

Well, OK, although that's not much of an explanation as to "why", I don't think I'd be able to understand anyway, since I only just graduated high school. Thanks for ur help!

It is true that if you solve the Schroedinger equation the solutions depend only on the principal quantum number $n$ and the energies (often called eigenenergies or eigenstates) have the values predicted by the Bohr formula, $E=-hcR_\infty /n^2$ and energy differences as predicted by the Balmer formula. The calculated energy does not depend on $l$; this degeneracy of wavefunctions with different $l$ is a peculiar and special feature of the symmetry of the Coulomb potential.
However, nature cares little for our equations and experiments show that in fact the energy of the $l$ levels are different in energy to one another. This is called 'fine structure' and the energy shifts are very small as the name suggests. The first cause is explained as being due to the fact that the electron has an intrinsic quantum property called 'spin', which does not mean that it is literally spinning, but crudely speaking that it can be described by the same equations as angular momentum. As the electron is charged its spin produces a magnetic field  which interacts with the magnetic field of its orbital motion, which in turn depends on $l$. This interaction is called spin-orbit coupling, and technically it is a relativistic effect.The second effect is called the Darwin term, for electrons with $l=0$, and the third effect is the Lamb Shift, which needs explanation beyond relativistic effects and requires Quantum Electrodynamics.
The fine structure due to spin-orbit coupling can be observed easily in many atoms so the $l>0$ levels are split for each $n$ and so are no longer degenerate. The common example is the two lines of Sodium 'D' emission. The difference is small, the two transitions are 589.59 nm to 588.99 nm. This is the yellow colour associated with sodium easily seen if you sprinkle salt into a gas cooker's flame and in some types of street lights. The splitting can be measured with a simple spectrometer and is a common teaching lab experiment. The other effects mentions are far,far more difficult to measure and in a Chemistry course you are unlikely ever to meet them.

I would clarify that energy differences for "fine structure" are very small. Avoid this clarification may result in a misinterpretation of your nice details.

@ user1420303, thanks, have added a line to emphasise  this.

You are right, nature is more important than our equations (and moreover, Schrödinger equation is not even the "ultimate equation of everything", for it does not include relativistic things). I stand corrected. You might want to add Wikipedia links to Darwin term and Lamb shift, and maybe also to the thing I mentioned earlier which explains that the said degeneracy is, in fact, not accidental.

'accidental', I take the point and I have reworded this part of the text.

I took the liberty of formatting the links in a different way.


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Carius method for quantitative analysis of halogens

Carius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, (Fig.12.17) in a furnace

I want to know :

What is the significance of using $\ce{HNO3}$ ? Can we carry out oxidation with other acids like $\ce{H2SO4}$?
Why $\ce{AgNO3}$ is added to $\ce{HNO3}$ and not to the organic compound? Will it be wrong if I add $\ce{AgNO3}$ directly to the organic compound after it is subjected to fuming nitric acid?
Wouldn't precipitation of compounds other than that of halogens occur since we are carrying the reaction in a closed furnace?

Any help is greatly appreciated.

1) $\ce{HNO3}$ is a more oxidizing acid than $\ce{H2SO4}$.
2) The silver nitrate would not dissolve well in the organic solution.
3) Sulfide is an example of something you might worry about interfering as it forms a very insoluble compound with silver. But silver sulfide and most anything else that might interfere will decompose in the fuming nitric acid.


text:
What type of isomers are the following two compounds?
What type of isomers are the following two compounds?They do not seem to be geometrical isomers as the methyl groups at terminal position are not in same plane as such.Are they conformers?

Believe it or not, they are enantiomers. Oh, and you don't normally draw them like that, because like you said, those methyls are out of plane.

Could you explain a bit more elaborately if you have time?Thanks. @IvanNeretin

@IvanNeretin I did'nt draw them :-P!

Then these pictures are meaningless and don't represent anything. Or, if we put it another way, they are poorly drawn pictures of the same compound (like in @julien's answer), which, BTW, exists as a pair of enantiomers, but the pictures don't tell you which is which.

@IvanNeretin Oh well I think I understood why they are enantiomers!

Related: Why are allenes chiral? and How are the Cahn-Ingold-Prelog rules applied to chiral molecules without a stereogenic carbon center? i.e biphenyls and allenes

It is easier to visualise if you draw, for example,  the methyl and the hydrogen on the left  in the plane of the paper, then the methyl and hydrogen on the right, are at right angles to this and so in and out of the plane.

@Sanc Are you sure, though, that those questions are unrelated? Perhaps they're very relevant and you just stopped someone from helping you.

@TIPS I'm sorry.


text:
Number of geometrical isomers when a cyclic ring is involved along with axis of symmetry

This is an question from my Chemistry study material and it asks to find the number of Geometrical Isomers of the following compound in the picture.

Basically the method/formula given in my book to find number of Geometrical
  Isomers of a compound is:

Using this formula I put $n=3$.Morever since both ends are same I get $p=(3+1)/2=2$.So final answer according to me is $2^{3-1}+2^{2-1}=4+2=6$.
But the answer given is $2^3=8$.
Where did I go wrong?
Related: How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry?

Didn't read the formula, but 6 looks right to me. See, the ends are either both trans, or both cis, or cis and trans (no matter which is which, since they are otherwise identical), which makes 3 options, and on top of that they are either on the same side of the cycle or not, which multiplies it by 2.

Yeah right.Even I thought the same.In all probability the given answer is wrong.BTW i wonder why don't you write your comments as answers.They are elaborate enough to be an answer! :-) @IvanNeretin

Besides, I was wrong on this one, as explained by vapid. Indeed, if we consider enantiomers, we have 8 all right.

@IvanNeretin I think he is wrong.He considered enantiomers which are optical isomers and not geometrical.

Are they? I never was able to remember correctly. What are geometrical isomers, then? Everything with similar connectivity, yet different, excluding optical isomers?

Yes you are right.See this http://www.avogadro.co.uk/organic/isomer.gif @IvanNeretin

Wait, I got lost in this thread, so deleting my previous comment. The issue is that this molecule is symmetric, so you lose a factor of two. Consider the case of 2,4,6-octatetraene. There's 3 double bonds, but there are only 6 isomers because the two ends are the same. If you consider 2,4,6-nonatetraene, then you'll have the full 8 isomers.

The answer in your textbook is correct. Here are all possible isomers:

I added a symmetry axis to the right, so you can see that these compounds are mirror images and are not superimposable. The formula you used for calculations is too complicated. The way to calculate it is simple: there are four stereogenic centers so: $$2^4=16$$There is one axis of symmetry so:$$16/2=8$$Have more faith in your books:)
EDIT:
Because my answer has raised an interesting discussion, I will try to present my point of view. First of all, the cis/trans isomerism is a descriptive way to represent a stereochemistry of a molecule, but its scope is limited. It is useful for teaching organic chemistry, but not for a systematic nomenclature (especially for a ring chirality). I think that the main source of confusion here is understanding the difference between a local and global (absolute) stereochemistry. In the above picture the controversy is whether the four enantiomers in the two rightmost columns can be treated as geometrical isomers. Let's simplify the problem and consider 1,2-dimethylcyclopentane:

The situation is clear for the first two isomers (a and b) - these are geometrical isomers, no doubts about it. What about b and c? We see that in both isomers the methyl groups are in the trans orientation, so one may conclude that they are not geometrical isomers. Period. However, cis/trans notation describes only the local geometry (E/Z notation obviously cannot be used in this example). Isomers b and c have the same relative orientation of methyl groups, but in a global scope these are two different molecules, and thus two different geometrical isomers. Therefore, 1,2-dimethylcyclopentane has one cis and two trans isomers. This may be counterintuitive at first, because there are no such situations in case of double bonds, where the cis/trans notation is more frequently used, but in rings we have an additional layer of complexity. I am aware that some people may disagree with this explanation, but as far as I know there are no official (i.e. IUPAC) directives to resolve this controversy.

You considered optical isomers(enantiomers) but the question asks for geometrical isomers only.

@Sanchayan Dutta So are you saying that the two compounds in the first column are not geometrical isomers? That makes no sense.

Tell me you definition of geometrical isomers and optical isomers first.

This is not a case of different definitions of geometrical and optical isomerism - we all agree in this matter. So if you exclude optical isomerism there are only 3 geometrical isomers - don't you agree? Still, makes no sense.

No there are 6(exclude the enantiomers on the right) as the ring too shows Geometrical isomerism.

@Sanchayan Dutta Now explain me what is the difference between geometrical isomerism in the ring and optical isomerism.

No difference.However the rightmost two cannot be considered GI in any way as they have been taken before.

@SanchayanDutta I disagree. The author of your textbook also seems to disagree. I am stopping this discussion and letting others voice their opinion.

You are both right and both wrong in the sense that there is no accepted definition of geometric isomers. It seems that IUPAC discourages its usage . @vapid maybe you can edit this in your answer or make more research on it first

@K_P I agree with you that the problem is controversial. I will edit the question in the near future to show my point of view, today I am a bit busy.

@K_P The link is broken


text:
Why are alkyl halides more soluble in alcohol water solution rather than pure water?
Why are alkyl halides more soluble in alcohol water solution rather than pure water?
I could not find any satisfactory explanation using Google.

a brief answer on solubility in general  is given at https://socratic.org/questions/how-is-molecular-polarity-related-to-solubility.

As others have mentioned, it generally comes down to "like dissolves like".  Water is highly polar, low molecular weight alcohols like ethanol a bit less so.  By increasing the ratio of alcohol to water in a solution you are effectively making the solution less polar as a whole.  There are effectively fewer hydrogen-bonding opportunities in an alcohol water solution than in pure water.  In general, alkyl halides tend to fall in the low-to-intermediate polarity range.  Methylene chloride for example is only very slightly soluble in water, which makes it an excellent candidate for performing extractions.  By adding alcohol to water, you are giving the solution as a whole a more non-polar character, and increasing the potential for favorable interactions between the alkyl halide and the solvent.


text:
Electronic configuration and the periodic table
I came across the following question:

If each orbital can hold a max. of 3 e–  what is the number of elements in the 4th period of the periodic table?

I was unable to even start the thought process of answering the question because I'm rather horrible at discerning the period number using the electronic configuration, the number of elements that can be accommodated in a period etc. If someone could please explain it to me, that would be so great, especially since I've googled this too many times already and haven't been able to wrap my head around it yet!
Thanks in advance :) Cheers!

In the 4th period of the periodic table there is 1 s orbital, 3 p orbitals, and 5 d orbitals. That's a total of 9 orbitals. $9 \times 3=27$
That's 27 elements.

Since when there's supposed to be 3 electrons per orbital?

It's a hypothetical question.

It's not hypothetical, but sarcastic - there can be only two...

Yes, and in the "there can only..." stream of thought, the double slit experiment, GR, and the quantum model of the atom were attempts at sarcasm.

https://en.wikipedia.org/wiki/Spin_quantum_number - seems you're somehow not getting my point. It's 9*2 = 18 as spin quantum number can have only two different values, orbitals can be occupied only by 2 electrons. Geez you can count elements in periodic table to see there are 18 elements in this period.

Everyone knows that there are two electrons in an orbital. The question is testing your understanding of how the number of electrons in an orbital affects the number of elements in a period. It isn't a real-world problem. When a physics problem says, "a ball rolls on a frictionless surface", you don't tear up the question paper and declare,"it's impossible to have zero friction". It's a pretty basic intro chem problem: its purpose is to test understanding, not to be realistic.

Oh, my I get it... Sorry!


text:
Convergence of spectra calculated from molecular dynamics trajectories
How can we check whether or not the power spectra (vibrational density of states) calculated by taking the Fourier transform of the velocity auto-correlation function (VACF) from a molecular dynamics trajectory is converged? How to estimate the statistical error in it?
What is the meaning of the word "convergence" in this context?

There are multiple ways in which the VACF, along with other properties derived from molecular dynamics trajectories like the dipole ACF to give IR spectra, can be considered "converged".

The molecular dynamics trajectory itself must be converged by some time-dependent statistical metric, such as measuring fluctuations in the energy or temperature over a number of steps and ensuring the average is below some threshold. This will be related to the ensemble chosen, the thermostat and barostat and associated parameters (if relevant), the integrator used for time evolution, and the size of the simulation, among other things.
Molecular simulations are layers of approximations piled on top of each other. In the case of purely classical dynamics, this usually refers to the force field parameters for the solute and solvent. Force fields are designed for different systems and may or may not be transferable, but the goal is usually to accurately predict dynamics. In the case of the VACF, if the dynamics of a system are described correctly, then the resulting power spectrum should be as well. However, in the case of the dipole ACF, the force field parameters enter directly as the nuclear charges (rather than just indirectly through nuclear velocities), which places stricter requirements on the quality of the force field. One hopes that newer force fields are better, but improvement over time isn't necessarily systematic and there is no quality hierarchy.
As an extension of 2, if the simulation is quantum mechanical in any way (such as in QM/MM, Born-Oppenheimer MD, or Car–Parrinello MD), then there are approximations in how the electronic structure is treated. For spectroscopic properties, it is more clear how to systematically improve results with regard to electronic structure than force fields due to well-established quantum chemical method hierarchies. Benchmarks of this hierarchy have been performed on many properties, including geometric parameters, dipole moments, and vibrational frequencies. In theory, the error in both dynamics and spectra should decrease up going from Hartree-Fock to MP2 to CCSD to CCSD(T), or by using more advanced density functionals (say, BP86 to PBE0 to M06-2X to B2PLYP). In practice, this would require benchmarking to draw any conclusions.
As an extension of 3, there are approximations in QM/MM of how the classically-treated and QM-treated regions interact with each other. Rather than go into more detail, here is a review of different QM/MM methods covering different definitions of the boundary and how one region may interact with the other.

Regarding estimation of the statistical error in resulting spectra, I am only aware of direct comparison (absolute error) between experimental and computed spectra, because it is assumed that the MD trajectory is converged (point 1). Performing multiple independent MD simulations and calculating the VACF from each should give an error measure closer to true statistical error, though I am unsure of how this would be computed.
Here is good almost-review (with good examples) of using Born-Oppenheimer molecular dynamics to calculate power spectra, IR spectra, and Raman spectra. I am not aware of any comparisons between classical MD and BOMD for computing spectra.
On the surface, it may appear as though the introduction of quantum mechanics makes things more difficult; certainly things become more time-consuming, however it is much clearer how results will change by treating electronic structure explicitly and with less severe approximations, rather than with a force field.

whatever you are suggesting is related to the quality of MD trajectory, the way to check convergence is still not clear to me! As far as my understanding, we can check  whether VACF is converging to zero or not within the simulated time, since if the simulation time is small, it may not get converged! And, if we have long enough trajectory, we might consider spectra calculated from different portion of it, ie taking different initial starting point.

To your first comment, "whatever you are suggesting is related to the quality of MD trajectory, the way to check convergence is still not clear to me", you should read the peer-reviewed literature for systems related to yours in order to understand this. This is both fundamental and critical to running any kind of MD simulation.

To your second point, "we can check whether VACF is converging to zero or not within the simulated time", that might be valid, but what does the literature say? It sounds like you have some knowledge of this already.

To your third point, "if we have long enough trajectory, we might consider spectra calculated from different portion of it, ie taking different initial starting point", the simplest case would be to take the entirety of your production run and none of your equilibration run, assuming that your system is properly equilibrated. This would ensure your VACF captures all the points (hopefully unbiased) that your trajectory samples.


text:
Mechanism for the decomposition of HI
I've been trying to find the mechanism for the gas phase decomposition of hydrogen iodide:
$$\ce{ 2 HI (g) <=> H2 (g) + I2 (g)} $$
Anyone know of sources where I can get information on this? I found a 1927 paper in PNAS but wondered if there is anything more recent. This is the 1927 paper:
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1085202/
Proc Natl Acad Sci U S A. 1927 Oct; 13(10): 720–726.
The Photochemical Decomposition of Hydrogen-Iodide; the Mode of Optical Dissociation

Could you give us a link to that original 1927 paper?

This is a 2006 paper-http://www.cder.dz/A2H2/Medias/Download/Proc%20PDF/posters/%5BGII%5D%20Production%20from%20electrolysis%20&%20thermochemical%20cycles/336.pdf

Related:http://chemistry.stackexchange.com/questions/31225/mechanism-of-photolysis-of-hx

Are we talking about non-catalytic decomposition?

Yes, non-catalytic.


text:
Are the following two compounds isomers or not?
Are the following two compounds isomers or not?
I feel they are enantiomers(non-superimposable mirror images) however my book states that they are not isomers?
I cannot understand why?

Probably a typo. You are correct that they are enantiomers.

This is an example of a compound with no chiral center, but it is chiral.  Compare 1,1-diethyl 3-methyl propa-1,2-diene (did I get the substituents in the right order?).

http://chemistry.stackexchange.com/q/42999/16683 #3

You are right, these two molecules are enantiomers. If you rotate the bottom molecule 180 degrees, you can see that it is the mirror image of the top molecule. Enantiomers are stereoisomers, so the only reason your book would say they are not isomers is that it (unfortunately) contains a typo.

Or possibly they missed the concept that a molecule can have not a (single atom) chiral center and yet be chiral.


text:
Van der Waals real gas equation
Van der Waals corrected the ideal gas equation for real gases and gave the following equation:
$$\left(p+\frac{an²}{V^2}\right)\left(V-nb\right)=nRT$$
I’m very confused about one aspect of this equation, which is this: why add the pressure term but subtract the volume term?
Edit: If somebody could please just explain the physical significance of the corrected equation and also, exactly what Van der Waals was trying to do by correcting this equation, it would be tremendously helpful for me.

Where do you see that you substract the volume?

Ideal gas molecules are mere dots with no volume, and don't interact with each other. Real gas molecules have certain own volume which eats up some of the space, hence $-nb$. Also, they repel each other and thus push outwards harder than they would otherwise, hence $+an^2/V$.

@GM: Clearly, we are subtracting $nb$ from $V$.

@Ivan Neretin: Are you sure about this? In most texts, the situation is described quite differently; the molecules close to the walls of the container experience a force of intermolecular attraction from the neighboring molecules and hence, the net force with which they collide with the walls decreases, hence decreasing the pressure.

The equation does not have any wall-related parameters in it. It describes an idealized situation without walls at all. Otherwise things might have been different.

Without a boundary surface(walls), how is the pressure of the gas defined?!

Think of an infinite expanse of gas at certain pressure and temperature, where such-and-such volume contains the prescribed amount of gas... No, that's too far-fetched. OK, think of a gas in a container with idealized walls which do not interact with anything.

@IvanNeretin I respectfully disagree with your second comment, the repulsions act over relatively short distances, and the potential on average is attractive at intermediate distances.

@getafix OK, then why $+an^2/V$?

@IvanNeretin I have edited my answer to include a discussion on this very issue. Correct me if I'm wrong :)

Got it. You were right, and I was not. Of course it is attraction that matters. It is attraction that is responsible for the critical point and the emergence of liquid state after that, even though the liquid itself is not described.

Let me begin my discussion by first introducing the notion, using some intuitive model building, and will try to derive, or perhaps guess the van der Waals equation of state.
The perfect gas equation, $ PV = nRT $ doesn’t allow for any interactions between molecules. The van der Waals equation is a refinement of this model, in the sense that it introduces the possibility of both repulsive and attractive interactions.
Imagine the gas molecules are small, hard, impenetrable spheres. This accounts for repulsive interactions, as now they can’t get arbitrarily close to each other. Consequently, they are no longer free to move about in a volume, $V$, and only have access to a much smaller volume $V-nb$ where $nb$ is roughly the volume taken up by the molecules.
The modified equation now becomes:
$$P = \frac{nRT}{V-nb}$$
Now to calculate the excluded volume:
$$V_\text{molecule} = \frac{4}{3}\pi r^3$$
where $r$ is obviously the radius.
Now note, a particle is surrounded by a sphere of radius $2r$ (two times the original radius) that is forbidden for the centres of the other particles. If the distance between two particle centres were to be smaller than $2r$, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do.
The excluded volume for the two particles (of average diameter $d$ or radius $r$) is: $\frac{4}{3}\pi (2r)^3$ or, $8V_\text{molecule}$. Divide this by two to get the volume excluded per molecule and we get $b \approx 4V_\text{molecule}$
Note: Of course, molecules are not infinitely hard, as van der Waals thought; there exists some “softness”. Thus, the factor four yields an upper bound; empirical values for b maybe lower.
Now turn on an attractive force between the particles.
Assume, notwithstanding the existence of this force, the density of the fluid is homogeneous.
And assume that a large majority of the particles are always surrounded by other particles, which attract each other (unless they get too close, when they repel; thus, the bulk of the particles do not experience a net force pulling them to the right or to the left. They feel that the container is infinitely large without any walls.
Now, the surface particles feel a net force from the bulk particles pulling them into the container, because there are no particles on the side where the wall is (and we assume here that there is no interaction between walls and particles, which is not true in reality).
This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density:
$$F = \frac{N_\mathrm a}{V_\mathrm m}$$
Assuming homogeneity, the number of surface particles is also proportional to the number density. So the total force on the walls is decreased by a factor proportional to the square of density.
Since pressure is force per unit area, it's decrease proportional to:
$$a'C^2 = a'\left(\frac{N_\mathrm a}{V_\mathrm m}\right)^2 = \frac{a}{V_\mathrm m^2}$$
Now, the modified equation becomes:
$$ P = \frac{RT}{V_\mathrm m-b}- \frac{a}{V_\mathrm m^2}$$
Derivation using Statistical Mechanics
For a system of $N$ particles, in 3 dimensions
The partition function:
$$ Z = \frac{1}{N!} \int \mathrm dp^{3N}\mathrm dx^{3N}\mathrm e^{-\frac{\beta}{2M} \Sigma p^2_n}\mathrm e^{-\beta u(x)} $$
This integral factorises into two, an integral over $p$, and an integral over $x$, i.e. a kinetic energy contribution and a potential energy contribution.
I write the result as
$$Z = \frac{z^N}{N!}$$
for an ideal gas, this result is
$$\frac{V}{\Lambda^{3N}}$$
where
$$ \Lambda = \frac{h}{(2\pi mk_\mathrm bT)^{1/2}}$$
(thermal wavelength)
For a van der Waals gas,
$$
U(r) =
\begin{cases}
\infty, & \text{when }r<d \\
-\epsilon\left(\frac{d}{r}\right)^6, & \text{when }r \ge d
\end{cases}
$$
basically attractive up to a certain distance, and repulsive when they come too close. r is the distance between the spheres and d is the distance where they touch. Also, each particle moves independently in an average potential field offered by the other particles
Also, for an ideal gas we integrate over all space i.e. the entire volume of the container, however, we cannot do that anymore. Instead our volume of integration is $V - Nb'$ (or $V – nb$, same as what we did before, $n$ is number of moles instead of number of particles).
$$b'= \frac{2\pi d^3}{3}$$
Now, assume that in a shell of thickness $\mathrm dr$ there are $ \frac{N}{V} 4 \pi r^2\mathrm dr$ particles, i.e. the position of the particles is averaged (reality, like always is different)
So for a single particle, the attraction felt is
$$\phi = \int_{d}^{\infty} u(r)\frac{N}{V} 4 \pi r^2\mathrm dr$$
and
$$z = \frac{(V-Nb')\mathrm e^{-\phi/2k_\mathrm bT}}{\Lambda^3}$$
I will omit the computations and quote the results,
$$\phi = -2a' \frac{N}{V}$$
where
$$a' = \epsilon (2/3) \pi d^3$$
going back and putting things together,
$$\ln Z = N\ln(V-Nb') + \frac{N^2 a'}{Vk_\mathrm bT} - N\ln(\Lambda^3) - \ln(N!)$$
Now,
$$ P = k_\mathrm bT \frac{\partial \ln Z}{\partial V}$$
which turns out to be,
$$ P = \frac{Nk_\mathrm bT}{V-Nb'}- \frac{N^2 a'}{V^2}$$
UPDATE: Physical Interpretation
I am writing this section because I get the impression (from the discussion the in the comments) that my answer hasn't adequately addressed this.
In the limit $P \to 0$ (or $V_m \to \infty $) we can expect perfect gas behaviour, i.e $PV = \text{constant}$
However, say at high pressures, we observe that real gases would have larger molar volumes than expected, so we remove a term from the volume.
Similarly, the pressure is lower than ideal gas pressure, so we add a term to it to compensate.
So, instead of saying $PV_m = \text{constant}$, we say $ (P +\text{"something"})(V_m - \text{"something"}) = \text{constant}$
It also accounts for critical behaviour, simply solve the following system to obtain critical constants (since the critical point, is a point of inflection on the PV diagram)
$$ \frac{\partial P}{\partial V_m} = 0 $$
$$ \frac{\partial^2 P}{\partial V_m^2} = 0 $$
Of course, it would be too optimistic to expect a simple, elegant model such as this to truly determine all properties of gases with complete precision; the van der waals equation of state does have its failings, but if nothing else, it improves at least the qualitative discussion of properties/behaviours of real gases

Comment have been moved to chat: http://chat.stackexchange.com/rooms/44503/discussion-on-answer-by-getafix-van-der-waals-real-gas-equation.  Please do not hold extended discussions in comments.  Thanks!

When you see the ideal gas equation, derieved from the kinetic theorey of gases, there are many assumptions taken for the ideal gases, a few like their Potential energy is 0, the molecules are spherical, and so on. What different real gas equations attempt to do is to reduce the number of assumptions taken by providing "corrections". Van der Waal attempted to correct the pressure and volume terms.
According to Van der Waal, his equation of real gas should be of this format:
{ Pressure on walls of container without any intermolecular attraction }. { Volume available for motion of molecules }  = nRT
In ideal gases, we assume that there are no intermolecular attractions between any of the molecules, thus the pressure of the gas is used in the equation. Also, we assume that the volume available for motion of the molecules is the volume of the container itself.
What exactly is the volume available for motion of molecules?
It is the space in which the molecule can move in. If two molecules are near each other, then neither can travel into the space occupied by the other molecule. Thus the volume available for motion is actually less than the volume of container.
Thus, Van der Waal introduced two correction terms, $\frac{n^2a}{V^2}$ and $nb$, which are the pressure correction term and the volume correction term respectively.
If we account for intermolecular attractions, the pressure of real gases increases. If we assume molecules to occupy space in real gases, the volume available for motion decreases.

But are you absolutely sure about the fact that pressure increases? Most texts describe the situation quite differently. Please refer to my response(comment)to Ivan Neretin's comment.

@Kaumudi, I'm certain that pressure increases.

But most of the texts seem to point to the contrary. Even Khan Academy!

@Kaumudi, the corrected pressure we plug in the real gas equation increases. Real gases' molecules have intermolecular attractions, which decrease the overall pressure of the real gas. If we were to treat it as an ideal gas, we would only put the overall pressure. But since it is a real gas and we have to exclude the intermolecular attractions, pressure increases.

Since it is a real gas, we are required to  take the intermolecular forces into consideration! You seem to have got it all wrong, I'm afraid. Read getafix's answer.

Suit yourself sir. And please avoid the insulting language.

Please refer to the answers given above. You must be prepared to accept that you may be wrong at all times and this time, you were wrong. I am not attempting to win this argument; I am simply begging you to remove your prejudices and accept my viewpoint with an open mind.

It is difficult to express my viewpoint over these restrictions between us. I am a 11th standard student studying for IITJEE with not a lot of time to answer completely. I believe it is understandable one will get frustated by this incomplete communication.

No, it is not understandable that one gets frustrated and calls somebody else an "idiot" merely because you didn't agree with what they had to say.

Im sorry I'm now in my senses please leave me alone I'm sorry for trying to answer.


text:
Dehalogenation of vicinal dihalides
Why, among different metals, only zinc is favourable for the dehalogenation of vicinal dihalides?

Magnesium + ether will also reduce a vicinal dihalide to an alkene.

I guess, mostly for safety and simplicity reasons. Magnesium tends to react vigorously with alkylhalogenides and aluminium forms oxide coating. Let's not even talk about such cases as sodium.

Zinc is used effectively in the dehalogention of vicinal dihalides because a carbon zinc complex is formed in the mechanism which is probably more stable. This reaction is also known as Simmon' Smith reaction.

The stability and usefulness of the Simmons-Smith reagent (zinc metal)
may be attributed in part to the higher covalency of the carbon-zinc
bond together with solvation and internal coordination of the zinc (which makes the zinc complex stable). (Source)
Reduction of vicinal dihalides by zinc metal proceeds by way of
oxidative insertion of Zn(0) into a C-X bond, producing an organozinc
species (stable) which undergoes dehalogenation:

(source)

The Simmons-Smith reaction is the dehalogenation of a geminal dihalide, and thus shouldn’t be the main point of interest here.