text:
False theory of how a battery works
There is an explanation of how a battery works that says that inside the battery, the electrons do a complete loop and can return to their starting point. When the electrons arrive in the positive terminal and they lost all of their potential energy, the battery does work on the electrons to  put them back in the negative terminal so they are full of energy and reenter the circuit.
This explanation is very great because it answers questions : why does putting 2 batteries in series double the voltage? (because work has to be done two times on the electrons in the battery, so the potential energy doubles). Why connecting - of battery 1 to + of battery 2 without connecting + of battery 1 to - of battery 2 doesn't create electricity (because the electrons have to do a complete loop in the circuit and return to their starting point to make electricity for a long time. There is no external wire connecting - of battery 2 to + of battery 1 so there isn't a complete loop).
However, from what I understand of batteries, there is no electron that moves from the positive terminal to the negative terminal of the battery. The electrons of the circuit all come from the negative terminal and once they reach positive terminal there is nothing else happening except making the electrolyte ions neutral. They don't return to their starting point.
So could someone prove to me that inside a battery, electrons from the positive terminal are done work on and go to the negative terminal.
Or if this theory is false, could someone answer the two questions above that are easily answered by this theory in another way?

Actually, individual electrons often do not complete a loop. The electron drift velocity is about 100 micrometers/second in copper, see http://resources.schoolscience.co.uk/CDA/16plus/copelech2pg3.html . The charge motion is more like the "Newton's cradle" demo, where each electron's EMF field pushes the next electron.

I know that electrons have a pretty slow drift velocity, but do the electrons that end up in the positive terminal of a battery return to the negative terminal? From all that I understand the answer is no, they detach themselves from the positive terminal to go in the electrolyte making some positive ions neutral but that's it.

The electrons in a battery never complete a full circuit back to the starting point. In a DC generator, they would, as the magnetic field impels them indefinitely. In a battery, as @TAR86 states, the chemicals are consumed.

Then what about the two last paragraphs of this text :  http://www.physicsclassroom.com/class/circuits/Lesson-1/Electric-Potential ?

"Positive test charge" is not "electron".

But the idea is the same, the text says that there is a field in the battery, and the electrons or "positive charge" move against this field. The work done is then converted into potential energy.  This is written in the text : "In the electrochemical cells of a battery-powered electric circuit, the chemical energy is used to do work on a positive test charge to move it from the low potential terminal to the high potential terminal. Chemical energy is transformed into electric potential energy within the internal circuit " So, is it wrong?

There are many ways of approaching the subject as has been illustrated by the answers so far.  There is nothing wrong with them, except that they do not answer your question - I think.
Kirchoff says that current must flow in a complete loop.  If the charge did not get back to the very starting point, there would be an accumulation of charge somewhere.  This would cause the voltage at that spot to increase, which would then motivate charge to move faster from that location.  So, your question, as I see it, is one of taking Kirchoff on face value and asking where and how the flow path is completed?
All batteries have a solid part and a liquid part.  The liquid part separates the two electrodes and is where the circuit is completed.  When an electron leaves one pole and goes through the external circuit one NEGATIVELY charge ion comes out of solution and attaches to that pole.  Similarly, when a negative electron arrives at the other pole through the external circuit, a negatively charged ion enters the solution.  The solution then completes the circuit, as in, one negative charge has left solution and one has been gained, leaving no net charge on any pole (except for the net that is there creating the voltage. The exact chemistry depends on the reactants involved.
This argument is simplified and very general. Positive ions also take part going in and out of solution, but in the opposite direction.
But leaving the simple model described in place, we can ask the question: Why do the ions in solution behave the way they do?  The answer is that they are motivated to do so AGAINST the voltage that exists because they "want" to chemically react at the poles.  It is the chemical energy expended that moves them against the voltage and causes them to move in that manner.
I hope this helps.

For a moment I would suggest you to forget about electrons,their flow ,batteries and stuff. Assuming that you have the basic intuitions regarding the principles of Newtonian mechanics and work done and all I would like to proceed in a different way. Suppose you are standing now on a ground and you have a ball in front of you lying on the ground. It doesn't lift itself up in the air, which is a common observation we all have made by now. But when you wish to lift it up to some height,be it h, then you expend your ATPs and chemical energies in your muscle to lift the ball slowly to that height h , and in doing so you did a work exactly equalling mgh ( neglecting air drag) . Now you take a tube of some arbitrary shape ( just to give a proper intuition) such that the ball is at its one end and the other end meets the ground. Then you leave the ball and it cleverly curves through the tube and gets back to ground level. This is exactly what we find in a battery system. It pumps electrons, against an opposing conservative field and then allows them to fall back. Here ,there's no man power but the chemicals do the trick by getting indulged into some redox reactions. Now ,if you think of those questions from this perspective hope you will be able to answer them efficiently.

I believe that it is possible to extend your theory a little bit and make it work for batteries. The crux lies in your sentence (emphasis mine):

When the electrons arrive in the positive terminal and they lost all of their potential energy, the battery does work on the electrons to put them back in the negative terminal so they are full of energy and reenter the circuit.

Remember that a battery has a finite capacity (at some point, its voltage drops to 0 because it was in use long enough or it corroded). Thus, you can think of your "negative terminal" as a reservoir for electrons on which the manufacturer of the battery has done work. Once that reservoir is empty, voltage drops. It is not necessary for the electrons to "completely" make a full loop, they merely need to move from one terminal to the other, but do not need to return. (I note here that some commenters correctly remarked that given the drift speed of electrons, one cannot very well say that a single electron travels the entire circuit. It seems more reasonable that it is like an open pearl necklace: push a pearl onto one end, on the other end a pearl will leave the string. But electrons are indistinguishable from each other anyway)
In this regard (as long as one deals with DC, not AC), a single battery can be thought of as a single charged capacitor, except that instead of an electrical potential difference, you have an electrochemical/chemical potential difference.


text:
possibility of metallic ammonium
In chemistry class, we discussed how metallic bonding occurs when metals lose electrons and become cations while the lose electrons become delocalized electrons and bind the metal cations. These electrons are then free to flow around the metal cations. But could the metal be replaced with an ammonium cation instead? (Or any other polyatomic cation, but we only learn ammonium and I'm not aware of any others minus hydronium which doesn't exist out of a solution.) Would the electrons still be free? How would you make it? And how could you prevent the following from occurring?
$$\ce{2NH4+ + 2e- -> 2NH3 + H2}$$

There are phases of solvated electrons in liquid ammonia that are metal-like. Thought these are created by dissolving metals like sodium in liquid ammonia.

Hydronium ion can be isolated in salts from very strong acids, like the perchlorate $\ce {H3OClO4}$.  But you would not want to mess with it.

A nomenclature note: conventionally, in chemistry "a metal" is an element (such as a transition metal or alkali metal). If what you're talking about is a substance that behaves as a metal, that's typically referred to as metallic, like metallic glasses.
Therefore, I would rephrase your question as "Can metallic ammonium exist?" - in fact, this has been a question for quite awhile. From what I've read, there's a theoretical basis that it should be possible at given temperatures and pressures, but it hasn't been experimentally observed. The papers I was reading were from the 50s-70s, so I'll keep looking for more recent information.
Solvated electrons are slightly related, in that there is a "free" electron, and this was in fact originally documented in liquid ammonia.

We do know that ammonium "molecules" can be incorporated into an amalgam with mercury like many metallic elements (http://www.sciencemadness.org/talk/viewthread.php?tid=12906; a technical reference is given by for example 1).  It must be chilled to retain stability, but it is a real material.  As with amalgams of actual metallic elements, ammonium amalgam is metallically bonded, so the ammonium exists as cations, with the compensating electrons intermingled with the other conducting electrons from the mercury.
Reference

Reedy, J.H. (October 1, 1929). "Lecture demonstration of ammonium amalgam". Journal of Chemical Education. 6 (10): 1767. Bibcode:1929JChEd...6.1767R. https://doi.org/10.1021/ed006p1767

source: Stack Exchange

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

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

sort: votes


text:
Chemical reaction that can produce lots of heat from 2-3 simple liquid ingredients?
Have an experiment I want to try out utilizing thermoelectric power and I want to generate the heat via a simple chemical reaction. Ideally I'd like to mix 2 liquids together (one being water or alcohol would be great) that can get to around boiling water temperature for a few minutes or longer. Byproducts need to be only other liquids that can be drained easily or gasses that aren't dangerous to humans. No solids wastes and non corrosive / carcinogen / caustic.
So far I've found one that's close to my needs but still produces a solid waste product and isn't particularly safe, which is Calcium Oxide + Water -> Calcium Hydroxide + Heat. This might work if there's another liquid I could add after the main reaction to turn the Calcium Hydroxide into a liquid for disposal.

I think the most helpful thing I can say here is that this seems unsafe.

It would only be unsafe if it didn't fit my requirements. The reaction would take place in a beaker surrounded in thermoelectric mesh and once the reaction is done it would be drained to another tank for safe disposal.

Concentrated sulfuric acid plus water would do a great job. It would readily kill your or gouge out your eyes, too.

I've worked in lab long enough to know if you are generating that big of an exotherm that you hope to make sustained, usable electric power from it, a beaker and some mesh is not even close to counting as containment.

Good point! Look at them chemists, what do they use when they have access to all those fancy reagents and want to generate some intense, but controllable heat? Electric heater, that's what.

As a chemist in a lab, ~ 100C can be a perfectly safe exothermic reaction if you are trained and take the proper safety procedures. It seems to me that @ABlankenship is actively considering the safety of his set-up. I think we should emphasize caution and safety, but that's not a reason to not answer the question. As a safety officer myself, you should really run your plan by your safety officer before implementing it. They can ensure your proposal is appropriate and conforms to the policies of your institution

You're not wrong about the electric heater but the point is to generate a bit of electricity with the reaction, not just make heat for no reason. I'm not expecting to power anything substantial, its just a proof of concept experiment.

Then go with sulfuric acid plus water, but beware. This thing can pretty easily heat itself well above the boiling point, and throw itself out of the beaker.

If it is only proof of concept, you don't need the chemical reaction for heating, do you.

I do. I already know that I can generate electricity from heat. What I don't know is how much heat this specific reaction generates, how much electricity can be taken from that, how long the reaction lasts and if my system can be reset to be used again autonomously (i.e. reactants and products from this reaction can be flushed out cleanly to make room for a second go)

The first thing that comes to mind, if you have access to a stock-room, is mixing two solutions: (1) NaOH; (2) HCl. This can release a lot of heat if your solutions are concentrated enough, and it forms salt-water if your NaOH and HCl are of the same molarity.
Edit: I'm on my lunch break, so I did some of the math...
The chemical reaction of interest in this case is:
$$\ce{OH- + H+ -> H2O}$$
The standard enthalpies of formation ($\Delta H_\mathrm f^\circ$) of these species are:
$$\begin{array}{lr}
\hline
\text{Species} & ΔH_\mathrm f^\circ/\pu{kJ mol-1} \\
\hline
\ce{OH-} & -229.99 \\
\ce{H+}  & 0.00       \\
\ce{H2O} & -288.83 \\
\hline
\end{array}$$
Thus, the change in enthalpy for the reaction is:
$$\Delta H^\circ = -288.83\ \mathrm{kJ/mol} - (-229.99\ \mathrm{kJ/mol}+0\ \mathrm{kJ/mol}) = -58.84\ \mathrm{kJ/mol}$$
Therefore, for $1\ \mathrm{mol}$ of $\ce{NaOH}$ + $1\ \mathrm{mol}$ of $\ce{HCl}$, you get $58.84\ \mathrm{kJ}$ of heat. Say you want to release enough heat to get the net solution up to $100\ \mathrm{^\circ C}$.
Water has a heat capacity of $4.18\ \mathrm{J/(g\ ^\circ C)}$. Say you have $1\ \mathrm L$ of $\ce{NaOH}$ + $1\ \mathrm L$ of $\ce{HCl}$, you'll need enough heat to raise the temperature of $2\ \mathrm L$ of water to $100\ \mathrm{^\circ C}$. I'll assume the water starts off at $25\ \mathrm{^\circ C}$, so you have $2\,000\ \mathrm g$ and $75\ \mathrm{^\circ C}$ to go.
$$4.18\ \mathrm{J/(g\ ^\circ C)} \cdot 2\,000\ \mathrm g \cdot 75\ \mathrm{^\circ C} = 62\,700\ \mathrm J = 62.7\ \mathrm{kJ}$$
How many moles of $\ce{NaOH}$ + $\ce{HCl}$ do you need for that much heat?
$$\frac{q}{\Delta H^\circ} = \frac{62.7\ \mathrm{kJ}}{58.84\ \mathrm{kJ/mol}} = 1.065\ \mathrm{mol}$$
That would mean that you can mix $1\ \mathrm L$ of $1.065\ \mathrm M$ $\ce{NaOH}$ + $1\ \mathrm L$ of $1.065\ \mathrm M$ of $\ce{HCl}$, and would theoretically expect to get a temperature close to $100\ \mathrm{^\circ C}$.
This might be off a bit because I've made some assumptions:

$\Delta H^\circ$ is constant with respect to temperature from $25\ \mathrm{^\circ C}$ to $100\ \mathrm{^\circ C}$. This may not be true.
the dissolved salts in water dont significantly affect its heat capacity
You have $2\,000\ \mathrm g$ of water in $1.065\ \mathrm M$ $\ce{NaOH}$ + $1.065\ \mathrm M$ $\ce{HCl}$

I think this could get you close though? I'm a bit surprised that the molarities aren't higher... It's a starting point at least.
Disclaimer of course: be careful with the $\ce{NaOH}$ + $\ce{HCl}$ solutions, they can be dangerous. Use proper chemistry hygeine protocols. The mixture should be benign, but you should confirm this with pH paper.

This is interesting, salt water would be a fine waste product though my only worry is it being damaging or cause salt build up within the tubing for disposal.

Depending on what institution you're at, some places allow you to dispose of neutralized acid/base solutions down the sink, as long as the acids and bases are benign and the pH is ~7. (Source: This is how it worked at 2 of my 3 institutions.) Your stockroom manager or assistant should be able to tell you.

Just saw your previous comment about what you meant by tubing: NaCl is incredibly water-soluble. If you're very worried, you can flush the system 1-2x with DI water between experiments, that should remove any salt left behind.

This is great, thanks so much for the expansion of the answer.

My institute even allowed pouring more concentrated acids down the drain since we had a neutraliser in the cellar.

I think this answer is off by a factor of 10. This line should be: $4.18\ \mathrm{J/(g\ ^\circ C)} \cdot 2,000\ \mathrm g \cdot 75\ \mathrm{^\circ C} = 620,700\ \mathrm J = 627\ \mathrm{kJ}$, not $62.7 \mathrm{kJ}$. So you need 10x the stated concentration of acid/base to make this work.

As Zhe says, this isn't safe. Your question mentions "draining" chemicals as if it's OK to dump chemicals down the drain or out on the ground. It's not. What possible connection is there between an exothermic chemical reaction and an experiment in thermoelectrics, I can't imagine. Use a hot plate. Here's a hint: heating water to its boiling point requires a lot of energy (per gram). Another name for a chemical reaction that produces a lot of energy is "an explosion". I can't think of any which someone who thinks "if its liquid, I can just dump it down the drain" would be safe to implement.

The chemicals don't fit my requirements if it can't go down a drain. Though by drain I simply meant its own drip system to clear the system to be reloaded. It could drip into a bucket for safe disposal later.

My assumption was @ABlankenship is doing an electricity-generation experiment, where a heat source is used to generate electricity. I think there's a very obvious connection because an exothermic chemical reaction and such an experiment.
Also, I have spent several years managing and assisting in stockrooms. Many liquids can go down the drain, but you have to be informed and careful.

Most chemicals can't be dumped down the drain. Technically, you shouldn't be able to dump large amounts of salt down the drain.

Most chemicals, but this is not a universal fact, especially for waste like aqueous sodium chloride. The rules depend on the specificities of your institutions sewer handling. State and federal regulations for most institutions I worked at allow unrestricted disposal of diluted NaCl solutions formed from the  neutralization of acid solutions. Many universities only allow the stockroom staff  or EHS to do this, to ensure it is done correctly. I reiterate that you should check with the relevant authorities at your institutions before determining your waste management plan.

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


text:
Determining the order of molar conductivity
This was a question which came in a objective type examination.

Which of the following have a $\lambda^\infty$ (molar conductivity at infinite dilution) larger than $\ce{KCl}$?
A. $\ce{CH3COOH}$
B. $\ce{HC}$
C. $\ce{NaOH}$
D. None of the above

I was not provided any experimental data to arrive at the answer.I know that $\ce{KCl}$ is a strong electrolyte and so my guess was that it must have greater molar conductivity than at least $\ce{CH_3COOH}$. But the answer key gives all A, B, and C. Is there a definite rule here?

I would agree with the answer key. The trick here is it's limiting molar conductivity, the molar conductivity at infinite dilution. For a neutral electrolyte compound, I was taught the notation $\Lambda ^0$, and $\lambda ^0$ was reserved for individual ions. (The $0$ superscript represents zero concentration, equivalent to $\infty$ dilution.)
Anyway, limiting molar conductivity is an interesting property because it does not represent a physically possibly scenario (a solute conducting electric charge when no solute is present). It can be extrapolated by the Kolhrausch Law, where it $\Lambda^0$ is the y-intercept of molar conductivity $\Lambda$ vs $\sqrt C$, concentration (note this relationship is only valid for strong electrolytes).
The degree of ionization of a weak electrolyte depends on the concentration, and tends to unity (100% dissociation) in the limit of concentration approaching zero. Limiting molar conductivity $\Lambda^0$ is defined (only) for this exact limiting scenario.
So when we consider limiting molar conductivity, we are doing so in a (abstract) condition where both strong and weak electrolyte fully dissociate. Thus for $\Lambda^0$ it makes no difference whether the electrolytes being compared are "strong" or "weak" at finite concentrations (for we are comparing them at "zero" concentration, i.e. infinite dilution). So we should have no problem accepting that a "weak" electrolyte might have a higher limiting molar conductivity value than a "strong" one, as in this case.
Why might $AcOH$ (organic chemist abbreviation), $HCl$, and $NaOH$ have higher $\Lambda^0$ values than $KCl$? Well, the $H^+$ ion is better at transporting its charge through an aqueous medium than $K^+$ because it can "water hop" via the
Grotthuss mechanism. This makes acidic compounds very effective electrolyte conductors. $OH^-$ is also an extremely effective charge carrier by an analogous "deprotonation-chain" mechanism. Because they use water itself to "tunnel"* through the bulk solution, protons and hydroxide have extremely high ion mobility in aqueous solutions and thus provide abnormally high molar conductivity.
So in general acids and bases would be expected to out-conduct a pH-neutral salt ($KCl$), and in the limit of zero concentration/infinite dilution, even "weak" acids and bases will have higher limiting molar conductivity values.
*Not in the quantum sense.

The gist here is that in water $\ce{H+}$ can often be the charge carrier. The reason that $\ce{H+}$ can move so fast through water is that a particular $\ce{H+}$ ion doesn't have to move from one point to another in the solution. The $\ce{H+}$ can attach to one side of a water molecule and "kick off" a different $\ce{H+}$ ion on the other side of the water molecule. So the $\ce{H+}$ effectively moves through solution via a chain reaction of water molecules.
In infinite dilution solutions solution acetic acid and hydrochloric acid will be completely ionized to $\ce{H+}$ and the anion. The $\ce{H+}$ moves in the chain reaction as noted above as so it moves very fast compared to $\ce{K+}$ which must actually travel the distance as the specific ion.
In infinite dilution solutions solution NaOH will be completely ionized to $\ce{OH^-}$ and the $\ce{OH^-}$ groups moves via a chain reaction like $\ce{H+}$, but a bit slower. $\ce{OH-}$ will move faster than $\ce{Cl^-}$ however.
So the answer is A, B, and C.

A good measure is to see the molar conductivities of individual ions, you will observe that:
$\ce{H+}$ >> $\ce{OH-}$ > Rest
and in general, if you have a lot of charge you come out in front.
At infinite dilution, all ions are completely dissociated.
Hence, simply by the fact that they have $\ce{OH-}$, $\ce{H+}$, you can say with high level of probability that they have more conductivity than $\ce{KCl}$. Also, all are monovalent, another nice thing for us.
The law I used was that at infinite dilution degree of dissociation = 1 for all and sum of individual ions' conductivity is equal to the compound's molar conductivity at infinite dilution.

The key is WRONG. There is NO WAY acetic acid has a lambda° (don't know where you got ∞ from, but a rose by any other name will smell the same) that is greater than KCl's. At 0.001 m/L KCl's Λ is 146, HCl's is over 400, NaOH's is 245 and HOAc's is 49 (S cm²). For obvious (I hope!) reasons the $\ce{H+}$ and OH(-) ions' mobility in water is different from all other ions. The amount of energy "available" to an ion in solution depends mostly on the temperature of the solution, which means the kinetic energy of any two ions will be (on average) the same at the same temperature. Since K.E. = ½mv² we reach the approximation v = SqrRoot(2KE/m) for an ion of mass m. Obviously, as m increases, v decreases. So, you'd THINK that $\ce{K+}$ being larger than $\ce{Na+}$, that it would be slower, and this is correct - so far as it goes. It is more complicated, because in real solutions, ions are never naked. That is, they are always surrounded by solvation layers. This means that as charge of an ion increases, the ion will (in polar solvents) surround itself with more and more solvent molecules, and it will have to "carry" these along as it moves. This complicates the picture, since now it's not just the mass of the ion we have to consider, but also the mass of the solvation sheath - which depends on the electric field [which is greater for smaller spheres (with the same charge).] The good news is that your 'ions' are all mono-valent. So, its just their relative size that matters. The bad news is that $\ce{AcOH}$, and $\ce{HCl}$, and NaOH all dissociate to ions which can move really fast in water, so you're not just comparing Cl with, say, Br or F. For $\ce{AcOH}$ (acetic acid) we might assume that at infinite dilution it will be completely ionized, but the mass of the acetate anion still substantially slow down any charge movement. In other words, its like a team race, the speed can't be any faster than its slowest runner. Neither NaOH nor HCl have that problem, so we'd definitely expect them to move faster than KCl in aqueous solution. Chemistry has both theoretical logic, and pragmatic - it is great for people who both like to understand why and who like to get their hands dirty by DOING stuff. Unfortunately, some of the ways it is taught are all about "rules" and formulas.

"Neither NaOH nor HCl has that problem,..." How do that account for NaOH or HCl to have higher molar conductivity than KCl".

The commentary is not necessary and, in fact, discouraged.

I don't think this is a place to post rants.

Is K+ slower than Na+? Because of its higher charge density, Na+ will have a larger hydration sphere, and thus a larger effective radius. So Na+ will have a lower ion mobility - it will be slower.


text:
Hygroscopic material absorption level
I live in a high humidity location and every winter, mold forms on one of my walls due to water sticking on it during the cold times. This also ruins the paint.
I found out that certain materials absorb water from the air, so I decided to make my own dehumidifier based on sea salt. The problem is alleviated slightly, but not fully - the salt becomes moist, but I think after a certain point it just can't absorb any more humidity.
So I was wondering how much salt I would need to bring the humidity in my room to a normal level.
I looked for things like "hygroscopic materials water absorption levels" on Google, but I did not have much luck.
What I am looking for is a formula like:
X kg of sea salt will absorb Y kg of water at Z% humidity
I have no background in chemistry, unfortunately, so I am not even sure if I am searching for this formula using the right terms.

I'm looking into this problem as well. I'm considering it doing it through some sort of Gibbs energy/ equilibrium expression.

I tried to do something similar when I had a humid car.
In chemistry, we often use $\ce{CaSO4}$ or anhydrite to serve as a desiccant. You could probably get cost-effective plaster for similar purposes. You could also get gypsum and heat it to generate plaster of Paris yourself.
As you're probably aware, people also use silica gel in products (and food) to minimize water content.
Either material should be much better as a desiccant than regular salt.
One key criteria is to increase the surface area - make sure the material is a fine powder and enclose it in something porous. I've seen people use various cloth materials, paper towels, etc.
As to how much you need, you can work out how much it will eventually absorb. If I take plaster of Paris to gypsum, each mole of $\ce{CaSO4}$ will take ~1.5 moles of water. So for every 136.14 g of anhydrite or ~145.15 g of plaster, you can absorb ~27 g of water. (I use ~ here, since I'm not convinced that regular plaster of Paris will be completely $\ce{CaSO4*0.5H2O}$.. some will likely already have extra water in it.)
The trick is that humidity will continually increase in the room if there's exchange of outside air.
Personally... I'd go with an electric dehumidifier.

I have the salt covered in cloth. The cloths gets very wet, but I can't tell if the salt continues to absorb water. Hence the need for the formula.

Some of the gypsum desiccants include "indicator." For example, in a chemistry lab, the blue (dry) gypsum will change to pink when it's saturated and won't absorb further water.

Salt is tricky, since it will eventually start to dissolve in the water, so I'm not sure there's a well-defined formula. Gypsum or silica gel, it's possible to work something out.


text:
Stability of alkenes tri>di>mono, but how to explain tetra
When we tell students about the formation of alkenes (by elimination for example), we often tell them that reactions will favour the thermodynamically favourable most substituted alkene. Zaistev's rule empirically described this:

"The alkene formed in greatest amount is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents"

The stability of the alkene with increasing substitution mono>di>tri is easily explained by hyperconjugation/qualitative MO theory (Phillip wrote a concise answer about that here). As an example to clarify, if theres a choice between eliminating to form a trisubstituted double bond over a mono substituted, the tri will predominate in most situations.
What I often struggle to explain in any detailed way is why tetra-substituted alkenes don't entirely follow the trend.  They are generally quite difficult to form (metathesis aside, but that isn't an elimination in the classical sense), and if formed, are generally quite reactive (on a similar order of magnitude to di-substituted alkenes if some studies on hydrogenation and some calculations are correct).
My general answer is 'sterics', based on the fact that 4 substituents around the same alkene is quite crowded, but I'm the first to acknowledge that this 1)doesn't really address all of the deviations from expected properties and 2)isn't really an explanation.

Again hyperconjugation- it increases the nucleophlicity of the alkene. (P.S. like the photo on your profile site)

Perhaps, I'm not convinced by that argument in itself, since the 'phenomenon' actually holds even when there aren't groups capable of hyperconjugation, it also wouldnt explain why tri>di>mono but tetra isn't > tri.

Steric hinderence may cause a  destabilising effect

Well Hofmann's rule for alkene formation is due to the steric effect so steric hinderence can be an effet there.

One very important question to ask is what exactly are you comparing?

Quite a broad question @Zhe but also a fundamental one. Tetra substituted alkenes don't follow 'expected' trends relative to the less substitued counterparts. Looking to Explain using MO theory or otherwise without hand waving arguments.

That's not what I mean. You're talking about "stability," so you are comparing relative energies. Which differences are you comparing? From some cation for elimination (i.e., where Zaitsev's rule would apply), or from constituent elements (i.e., $\Delta G_{f}^{\varnothing}$)?

I think you're reading too much into the wording (do feel free to edit, however), the general point I'm trying to answer is why these tetra olefins are anomalous.

Applauds @Zhe. I also have a bit trouble understanding what you are getting  at. Are you asking why tetra-substituted double bonds are hardly formed by elimination? Or are you asking why tetra-substituted alkenes are less stable than tri-substituted alkenes for some -to be specified- set of conditions? (which I doubt) But then I don't understand where Zaitsev's rule comes into play. Stability is a relative concept, you have to state what you are comparing...

Also note that Philipp's description is in terms of Valence Bond Theory, not Molecular Orbital Theory.

I agree with the OP and have seen numerous examples where tetrasubstituted olefins are indeed quite difficult to make, both in the literature and in my own experience, leading to the formation of Anti-Zaitzev (Hoffman) elimination products.
Many question writers are reluctant to make a solution where the tetra-substituted option is the correct answer. It is not as simple as the trend in carbenium ion stability where hyperconjugation and conjugation dominate around a trigonal planar (less sterically encumbered) center.
A tetrasubstituted olefin is too sterically encumbered due to 1) two A(1,2) interactions and 2) two exo substituents (gem-dimethyl) interactions.
Such proof for this failure in the stability trend is included in the original question as the kinetic stability of such an olefin is unusually and unexpectedly high. As far as the the thermodynamic stability, torsional strain makes effective hyperconjugation less possible in the transition state which varies only marginally upon the strength of base and the lability of the bond to the leaving group.
This trend: mono is less stable than exo-di is less stable than cis-di is less stable than trans-di is less stable than tri is....that is about all you can correctly say about the trend. Consider why cis-disubstituted is less stable than a trans-disubstituted olefin. Now consider a tetra-substituted alkene.
Zaitsev's "rule" like Markownikoff's "rule" is a just fast and dirty guideline for undergraduates and which both have numerous exceptions. It does not follow that tetrasubstituted must be more thermodynamically stable than tri-substituted and product ratios indicate this repeatedly in the literature. Stereoselectivity of E over Z is a non-problem compared to the difficulty of and abysmal yields when creating a tetrasubstituted olefin.
Simply put an ellipsis ("...") after the trisubstituted alkene as that is all that can be stated to an undergraduate class without presenting a falsehood.


text:
Why amine salts are soluble in water?
Amine salts are soluble in water but insoluble in organic solvents like ether. Why is it so? I guess there is something related to polar nature of amine salts. But I am not sure. Can someone elaborate?

Amine salts are:

charged
polar
hydrogen-bonding (unless they're quaternary)

There some cross-over between those categories, but all will increase solubility in water, a polar solvent, and decrease solubility in a nonpolar solvent, like ether.
That said, if you put very long oily carbon chains on an amine salt, you will decrease the solubility in water, and increase the solubility in nonpolar organic solvents.


text:
How to determine which aqueous solution has the largest freezing point depression
Question
This is more of a conceptual question. Let's say that we are given different aqueous solutions and we are asked to choose the one with the highest freezing point depression, which do we choose? Is there a pattern? do we choose the one with salts inside of it?
Example question

Which aqueous solution exhibits the largest freezing point depression?
  $$\begin{align}(\mathrm{A})&\  1.0\ \mathrm{M}\ \ce{KBr} & (\mathrm{C})&\ 0.5\ \mathrm{M}\ \ce{MgCl2} \\
(\mathrm{B})&\ 0.75\ \mathrm{M}\ \ce{C6H12O6}& (\mathrm{D}) &\ 0.25\ \mathrm{M}\ \ce{Ga2(SO4)3}\end{align}$$

now this question is a bit different because it has concentrations, so do we choose the one with the highest concentrations or the one with salts?
BTW
The answer is A)
But my hypothesis is that we only look at the ones that contains salts/ions inside of them and then the one with the highest concentration. Please let me know if this is correct or not

Freezing point depression is a function of the amount of dissolved substance — but substance must be treated as distinct particles rather than compound. This is an important distinction because ionic compounds, when they dissolve in water, typically separate into hydrated anions and hydrated cations — two distinct species as far as freezing point depression is concerned.
Thus, for ionic compounds you first need to determine how many individual ions are generated and then multiply that by the overall concentration. In your examples:

$1~\mathrm{M}\ \ce{KCl}$. One cation, one anion. Thus, there are two moles of dissolved particles per mole of compound; giving $2~\mathrm{M}$ effective particle concentration.
$0.75~\mathrm{M}\ \ce{C6H21O6}$. This compound is not ionic. Thus, it does not dissociate and one mole dissolved gives one mole of particles. $0.75~\mathrm{M}$ effective particle concentration.
$0.5~\mathrm{M}\ \ce{MgCl2}$. This again is ionic but we get two moles of anions per mol of compound in addition to one mole of cations. Thus three moles of ions per mole of compound. Effective particle concentration $1.5~\mathrm{M}$.
$0.25~\mathrm{M}\ \ce{Ga2(SO4)3}$. Here, we have a total of two moles of cations and three of anions per mole of compound. Thus, the effective particle concentration is five times the original or $1.25~\mathrm{M}$.

The actual formula for freezing point depression is based on a constant factor of degrees per dissolved mole; and this analysis was used to confirm that ionic compounds separate into ions and to determine molecular weight in the past.

Depression of freezing point of a solvent(here water) is a colligative property of a solution i.e., it depends on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. Therefore, more the number of solute particles, more is the depression of freezing point of the solvent. Also ionic compounds when dissolve in water, dissociate into anions and cations which depends on dissociation constant, so the van't Hoff factor, i must also be taken into consideration.
So, applying above stated logic, option (A) is the most appropriate answer for the given question.


text:
Will the bond order of CO be greater in its existence as carbon monoxide or in metal carbonyl?
$$\text{Bond order} = \frac{N(b) - N(a)}{2}$$
Where $N(b)$ is the number of electrons in bonding orbitals and $N(a)$ is the number of electrons in antibonding orbitals.
As in metal carbonyls, there is synergic effect during formation of the π back bond from metal to $\ce{CO}$, so $\ce{CO}$ accepts electrons in its π* anti bonding orbital. As $N(a)$ increases, Bond order decreases. So, $\ce{CO}$ in its free state has greater bond order in comparison and hence is more stable.
Would the above concept be correct?

In figure 1 you can see the molecular orbital scheme of a $\ce{[ML6]}$ complex including σ and π interactions. The class of ligands is not important; however, care should be taken when correctly assigning orbitals.

Figure 1: Octahedral $\ce{[ML6]}$ complex including σ and π ligand-metal interactions.
The first thing we should be doing is assigning the original, pre-bonding orbitals. The metal orbitals on the left are obvious. The ligand group orbitals of lower energy, $\mathrm{a_{1g} + e_g + t_{1u}}$ are the six ligand orbitals that interact with the metal in a σ fashion; in carbon monoxide, these represent the lone pair on carbon (the HOMO). The twelve p type orbitals, of which nine are omitted and the remaining transform as $\mathrm{t_{2g}}$ are π symmetric with respect to the $\ce{L\bond{->}M}$ bond and represent the $\pi^*_{\ce{C=O}}$ bonds of carbon monoxide; the LUMOs.
In a metal carbonyl complex, the left-hand side is usually higher in energy due to the metal’s lower oxidation state. Thus, the lower an orbital in the centre of the scheme is, the more ligand-centred it is.
While filling in the final complex’ electrons, we start from the bottom; first the twelve ligand electrons that represent the carbon monoxide lone pairs. Once we are done with this, the next orbitals happen to be $\mathrm{t_{2g}}$ — the bonding ones which have more ligand contribution. These can not only be seen as bonding with respect to the $\ce{M\bond{<-}L}$ bond but also antibonding with respect to the $\ce{C=O}$ bond. The maximum we can fill in here is 6 electrons. These six electrons reduce the $\ce{C#O}$ bond order as can be seen in the formula:
$$\text{B.O.} = \frac{(12 + 2\times12) - 6}{2\times6} = \frac{30}{12} = 2.5$$
The additional factor 6 in the denominator is because I am analysing six $\ce{C#O}$ bonds at the same time.
The result ($2.5$) clearly shows that the bond order is reduced with respect to noncoordinating $\ce{CO}$.
The discussion I did with an octahedral complex can be done with any coordination geometry; octahedra are just a very typical case.

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


text:
Why a low conductance ion such as lithium is used in batteries?
Lithium ions have a relatively low conductance (Wikipedia) so why are they used in batteries, and not sodium, potassium, etc.?

https://en.wikipedia.org/wiki/Lithium-ion_battery There are other factors, including how much energy lithium stores when adsorbed inside the structure of a lithium ion battery and how reversible the reaction is (for recharging).

The conductivity values you cite are for ions in water. there is no water in a lithium battery otherwise, it would explode and burn before leaving the factory. Beside low conductivity can be compensated by increasing concentration of the ions or a variety of other architectures.
Further, there's a few reasons lithium is optimal in batteries.  It's not environmentally toxic (like beryllium), has a higher oxidation potential than the other alkali metals, but most importantly is its high energy density in both Joule per weight and per volume, as well as power per weight, which no other metal can beat.  The only disadvantages for lithium are its flammability and cost.
(source)

on Wikipedia it says that lithium ions do travel. Could you clarify?

I edited the answer but the reasons for lithium still stand

Lithium ions are also very small, giving it good kinetic properties when diffusion through a solid is a factor.


text:
Which chlorine species can act as an oxidizing agent but not as a reducing agent?

Which species can act as an oxidizing agent but not as a
  reducing agent?
(A) $\ce{Cl2}$
  (B) $\ce{Cl-}$
  (C) $\ce{ClO2-}$
  (D) $\ce{ClO4-}$

I thought the answer was A because if $\ce{Cl}$ is reduced, it becomes very stable with a full octet and therefore cannot be the reducing agent; however, according to the 2007 Chemistry Olympiad answer key, the answer is D. Why is that?

There is a very simple and well-known reaction in which Cl2 acts as a reducing agent. Then again, I believe perchlorate also would act that way if confronted with elemental fluorine or something. The question is badly posed. I guess the implication was that chlorine is already in its topmost oxidation state, hence...

For a species to act as a reducing agent, it should have the ability to get oxidised. Any element in its highest oxidation state, cannot oxidise anymore, and can hence not act as a reducing agent. $\ce{ClO4-}$ has $\ce{Cl}$ in $+7$ oxidation state, and cannot have a higher oxidation state, and hence cannot act as a reducing agent.


text:
Rotational-vibrational spectroscopy of a molecule
I want to ask if the rotational-vibrational spectra of a molecule is related to any one mode of frequency or if it is related to the whole molecule?
Additionally, the vibrational levels that we say are composed of lower rotational energy levels. Are these vibrational energy levels of certain vibrational normal modes or of the whole molecule?

I'm not quite sure what you mean by 'vibrational levels composed of lower rotational energy levels' so I try to explain the whole thing, briefly.
In analysing the electronic, vibrational &  rotational spectra of molecules the Born-Oppenheimer approximation has to be used. This is applicable because the vibrational period (a few femtoseconds) is far faster than rotation period (a few picoseconds) thus these motions do to interact appreciably with one another to a good approximation. This allows the energy levels of, rotation and vibrational motions to be added together.
A normal mode vibration is the motion of all atoms in the molecule in a fixed phase relationship with one another. For N atoms there are $3N-6$ normal modes ($3N-5$ for a linear molecule). How how far and in what direction each atom moves has to be determined by calculation, but the overall symmetry species ($A_1, B_g, E_u$ etc) of the total motion of all atoms  is given by the Point Group of the molecule, $C_{3v},D_{2h}, O_h$ etc. The symmetry species determines the relative phase of each atoms motion with respect to all the others. By 'phase' is meant that as some bonds contract  others extend but they always do this together.
As energy levels add each vibrational normal mode has its own stack of rotational levels. (Because the B.O. approximation is not exact a more detailed analysis in needed to add the interaction between vibration and rotation.)
The figure shows a typical spectrum (calculated)  of a (perpendicular band) of a symmetric top molecule such as $\ce{PCl3 , CHCl3} \mathrm { ~or~ } \ce{CH3Cl}$. The total spectrum is shown on the bottom line and its separation into individual PQR rotational bands above this.  There will be similar complicated features for each vibrational normal mode. Clearly the analysis of spectra such as this is non-trivial :)
(In symmetric top molecules two of its three moments of inertia are the same but different from the third. Two quantum numbers are needed J for total angular momentum and K to fix the angular momentum about the symmetry axis and this is why there are so many rotational bands for a single vibration.)

(Spectrum is taken from G. Herzberg, ' Infra-red and Raman Spectra of Polyatomic Molecules')

I want to ask if the rotational-vibrational spectra of a molecule is related to any one mode of frequency or if it is related to the whole molecule?

Both. The frequencies at which you see "lines in the spectra" are a consequence of the structure of the molecule - if you change the structure, the frequencies (and the spectra) will change as well.

Additionally, the vibrational levels that we say are composed of lower rotational energy levels.

I'm not sure that's true. There is energy transfer between rotational and vibrational modes, but vibrational and rotational levels are distinct. Rotationally resolved IR spectroscopy just happens to let you see both...


text:
Why is fullerene 60 an insulator while graphite is a conductor?
Graphite is good conductor of electricity because one carbon atom is bonded only three carbon atoms, which enables the presence of free electrons. In fullerene 60 the carbon atoms are also bonded to three atoms so why is fullerene concidered a bad conductor of electricity?

Fullerene is considered a bad conductor of electricity because it is a bad conductor of electricity. Let's distinguish facts from interpretations. Also, related: http://chemistry.stackexchange.com/questions/37297/structure-of-buckminsterfullerene/37298#37298

To reiterate Ivan's comment fullerene is a bad conductor because that's what the measured properties produce as a result. The mechanism that makes it a bad conductor is that it has shorter range continuity than graphite. In graphite the carbon is made of sheets that can be as long as the sample. These sheets have fairly low conductivity, but when the electrons must jump between sheets to continue its path, this is when the conductivity value for the bulk material lowers. Extrapolate this to fullerene. If the fullerene molecule itself had high conductivity, it would still require the electrons to jump between molecules every few nanometers. This can produce a rather large resistivity for the bulk material much more so than observed in graphite.

The sheets in graphene basically extend all through the entire material; you essentially have one macroscopic orbital going from one side of a pencil to the other. Push an electron in on one side and one will come out on the other. Great.
In fullerene, however, you have distinct little footballs that are just lying next to each other. You can easily push electrons around within one football, but it’s hard to get them moving between footballs. You would need to push them from one end of your material to the other, though; so it doesn’t help that they happily circle their football.

I think it's worthwhile mentioning that (ideal) graphene is a 2D conductor. And a single Bucky ball should be quite a good conductor... but that's just atomic scale thought experimentation.


text:
How to find the oxidation state of complexes with bridging ligands?

How can I find the oxidation states of $\ce{Fe, OH-, NH2-}$ in the given complex?

Where is your effort? Note that Chemistry.SE is not a homework solving service; do show some effort to work through the problem.

@MAFIA36790 I know, I shouldn’t have answered ^^'

Homework posts aren't unanswerable, @Jan; it depends on the context - you know better.

Just follow the normal rules for determining oxidation states. Any bond between elements is cleaved heterolyticly giving only the electronegative element all electrons of that bond. Then count electrons.
All oxygens in there will have $\mathrm{-II}$, all nitrogens $\mathrm{-III}$, all hydrogens (they are either connected to oxygen or to nitrogen) $\mathrm{+I}$.
You also know that your cationic complex $\ce{[Fe(H2O)4\mu{-}OH\mu{-}NH2Fe(NH3)4]^4+}$ has to have a $4+$ charge due to the two sulphate counterions. That allows us to determine the average oxidation state of the iron ions:
$$\begin{align}5\times (-2) + 5 \times (-3) + 23\times (+1) + 2\,x &= +4\\
-10 - 15 + 23 + 2\,x &= +4\\
-2 + 2\,x &= +4\\
2\,x &= + 6\\
x &= +3\end{align}$$
Thus, these are most likely $\mathrm{+III}$. You can also take it down to a per-complex level, whereby you would equally distribute the sulphate charges to the two irons giving both a formal $2+$. Then, there are two bridging ligands each as monoanions, so they require each iron cation to have another positive charge resulting in $3+$ or iron(III).
Note that while iron(III) is the most likely configuration, it does not have to be that. It may also be an iron(II,IV) complex theoretically. Be prepared for unexpected exceptions.

If this answer helped you, feel free to accept it by clicking the little checkmark below the voting arrows on the left.


text:
Is it possible to design molecules of desired qualities?
I was thinking if it would be possible to take into account every property of every atom known to man so that they can compute any type of compound they want. Is it possible to accomplish something like that with the present computation technologies or are the variables that are accounted in making of any kind of molecule and determining its properties too high?

I don't think it is even possible to enumerate all the possible combinations of every element known so doing properties as well would be a serious stretch.

But there must be a way to generate some heuristic measure..afterall there already are documents that tell us properties of atoms that make a compound under different conditions and combinations

sigh You should have some knowledge to properly formulate questions. I guess you're interested in computational design of materials, proteins etc. That's so broad topic that it would be hard to summerise.

Well...I did ask what came to my mind. I was curious. Needed an idea on where to start looking.

OK, but for future, questions about specific well described things are much more welcome here than vague ideas, IMO.

Your title and the question itself arise two different issues, and I'll try to address both.

Yes it is possible to design molecules of desired qualities, only it's not all that simple. It requires a great deal of calculations, practice, empirical laws, heuristics, experimentation, guesswork, and magic. Eventually this body of knowledge grew so wide that people started to count it a science in its own right. It became known as chemistry.
Yes we can take into account every property of every atom and compute the properties of any compound they constitute. The problem, though, quickly becomes prohibitively expensive in terms of computation time. The reverse problem (to come up with a molecule of desired properties via calculations alone) is even more costly.

So it goes.


text:
Is this the true expression for isochoric enthalpy of non-ideal gases?
$\def\d{\mathrm{d}}$
$\def\D{\Delta}$
I recall seeing an equation in a thermodynamics lecture,

$$\tag{non-ideal gas at constant $V$} \D H = n \int_{T_1}^{T_2}\! C_{P,\mathrm{m}}\ \d T$$

where $C_{P,\mathrm{m}}$ denotes molar heat capacity at constant pressure. There was not really a derivation of this, merely a statement whose mathematical equivalent is

$$\underbrace{\D H = nC_{P,\mathrm{m}}\D T}_{\text{true for ideal gases}}\to\overbrace{\D H=n\int_{T_1}^{T_2}\! C_{P,\mathrm{m}}\ \d T}^{\text{the simplest generalisation}}.$$

I have to admit that the explanation made sense at the time. However, trying to justify the result from first principles$-$the $1^{\text{st}}$ law of thermodynamics$-$has proven unfruitful thus far.
$$\tag{$\D V=0, \text{def of}\ C_{V,\mathrm{m}}$} \d H=nC_{V,\mathrm{m}}\,\d T + V\,\d P$$
$$\tag{$C_{P}-C_{V}=\frac{\alpha^2TV}{\kappa_T}$}\d H=n\left(C_{P,\mathrm{m}}-\frac{\alpha^2TV}{n\kappa_T}\right)\,\d T+V\,\d P$$
Assuming the lecture formula holds, then
$$\tag{volume constant}\frac{1}{n}\int_{T_1}^{T_2}\!\frac{\alpha^2T}{\kappa_T}\,\d T=\D P.$$
Is this the case? Why?

The equation in the lecture notes is incorrect.  It omits a term involving the partial derivative of entropy with respect to pressure at constant temperature.  See the fleshed-out version of my Answer below.

$\def\d{\mathrm{d}}$The general equation for the change in enthalpy of a non-ideal gas, a liquid, or a solid is $$\d H=C_p\d T+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]\d P$$  The derivation of this equation is in every thermo book.  So your lecture equation is not correct; it omits a term related to the partial derivative of entropy with respect to pressure at constant temperature.
In terms of the coefficient of thermal expansion, this becomes$$\d H=C_p\d T+V\left[1-T\alpha\right]\d P$$
Since, $V=V(T,P)$, we have:
$$\d V=\left(\frac{\partial V}{\partial T}\right)_P\d T+\left(\frac{\partial V}{\partial P}\right)_T\d P$$
So, for constant volume, $$\d P=-\frac{(\partial V/\partial T)_P}{(\partial V/\partial P)_T}\d T=\frac{\alpha}{\kappa}\d T$$
Therefore, at constant volume,  $$\d H=\left[C_p+V\frac{\alpha}{\kappa}\left(1-T\alpha\right)\right]\d T=\left[C_p-VT\frac{\alpha^2}{\kappa}\right]\d T+V\d P$$

Great answer! I wouldn't say that the derivation of the first equation is in every thermodynamics book, though; I've looked over four and it wasn't in any of them ;) then again during a lab test I was given seven dysfunctional pipette squeeze bulbs in a row

@LinearChristmas If you can't find the derivation, I have provided one

All the books I have in my library have the derivation, including Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness and Fundamentals of Engineering Thermodynamics by Moran et al.

The derivation for the first part of your question is shown below. I'm not sure what you are trying to do in the second part.
The heat capacity is the heat required to raise the temperature of  1 mole of substance by 1 K. Suppose  for the moment that heat capacity C is constant over a temperature range $T_1 $ to $T_2$, then the heat q required to raise 1 mole of substance  over this temperature range is
$$q=C(T_2-T_1) = C\Delta T$$
If the heat capacity is not constant, which is generally the case as real molecules have vibrational and rotational energy levels then,
$$ q=\int _{T_1} ^{T_2}C(T)dT$$
where in many cases the heat capacity can be approximated as $C(T)=a+bT+cT^{-2}$. For a gas the partition function can be calculated, if spectroscopic values are known, and $C(T)$ calculated that way.
The heat absorbed depends on the path taken to go between the two end points so that C is not well defined. To make it well defined either T or P must be constant.
At constant volume, $q_V=C_V\Delta T$ and the work done as expansion against external pressure is zero, thus from the first law with internal energy as U,
$$ \Delta U = C_V\Delta T $$
and a similar integral as described above is used when the temperature range is too large for $C_V$ to be considered a constant. (This equation shows also, in the limit of small changes,  that heat capacity is the slope of internal energy vs. temperature.)
In the case of constant pressure $q_P=C_P\Delta T$. The change in enthalpy is
$$\Delta H = C_P\Delta T$$
and so in general
$$ \Delta H=\int _{T_1} ^{T_2}C_P(T)dT$$
To demonstrate that $q_P=\Delta H$ we note that in this case the volume is not constant and so the heat transferred at constant pressure $q_P$  is related to internal energy change $\Delta U$ as
$$ \Delta U = q_P-P\Delta V$$
where $P\Delta V$ is the work done. Rearranging and expressing the $\Delta$'s directly gives
$$ q_P= (U_2+PV_2) - (U_1 +PV_1)$$
from which using the definition of enthalpy $H=U+PV$ gives $q_P=H_2-H_1$ and shows that $q_P=\Delta H$.

I have no issue with $$\Delta H = \int_{T_1}^{T_2}C_P\ \mathrm{d}T$$ when pressure is constant. The question is about constant volume. What I am attempting to do is 'prove' the same formula holds for isochoric processes, using only the first law of thermodynamics, definition of entalphy and the condition of constant volume. It is easy to show that $$\tag{constant $V$, ideal gas} \Delta H = C_P \Delta T.$$ But it is not mathematically rigorous to conclude from this the general form. In other words, induction ought to be justified.

my second equation serves both to produce $C_V$ and $C_P$. I can see where you get your penultimate and antepenultimate equations from but not the last.

$\def\d{\mathrm{d}} \def\D{\Delta}$Yes, I understand that $q$ itself is the integral you have given. But to show that $q$ is equal to entalphy when only volume (not pressure) is constant would still need demostrating. $$ $$ In the final step, $$ H=n\left(C_{P,\ m}-\frac{\alpha^2TV}{n\kappa_T}\right)\d T+V\d P,$$assume that $$\D H = \int_{T_1}^{T_2}C_P\ \d T$$ is true and substitute. Therefore, the other terms must amount to $0$, or equivalently $$\frac{V}{n}\int_{T_1}^{T_2}\frac{\alpha^2T}{\kappa_T}\d T=V\D P.$$ Divide both sides by $V$ to get my final equation.

$\def\d{\mathrm{d}} \def\D{\Delta}$ PS. It is highly suspicious that something like $$\tag{volume constant}\frac{1}{n}\int_{T_1}^{T_2}\frac{\alpha^2T}{\kappa_T}\d T=\D P$$ would hold, and this is why I am not sure that $$\D H = \int_{T_1}^{T_2}C_P\ \d T$$ is true for constant volume. Then again, record shows I might have gone horribly wrong somewhere ;).

It follows from the first law that $\Delta U = C_V\Delta T$ since V is constant and thats that; no more is needed as the work done is zero. Similarly for $\Delta H =C_P\Delta T$ when volume is allowed to change because pressure is constant. These eqns  are quite general.  The $\Delta H$ eqn above is only true  at constant pressure, which was assumed during its derivation. You can check if your  eqns in $\alpha$ etc are correct  by assuming ideal gas law and working out derivatives.

I'll try to get at the result from first principles, let's start with $U = U(S,V,N)$. I'll just assume constant mole number, so I'll write $U(S,V)$.
We get the enthalpy by taking the "Legendre transform" of the internal energy.
$$H \equiv U[p]$$
which gives, $H = U + pV$ which is function of $S$ and $p$, i.e $H = H(S,p)$
taking the differential
$$\mathrm{d}H = \mathrm{d}U + p\mathrm{d}V + V\mathrm{d}p$$
$$\mathrm{d}U = \overbrace{\left(\frac{\partial U}{\partial S}\right)_V}^T\, \mathrm{d}S + \overbrace{\left(\frac{\partial U}{\partial V}\right)_S}^{-p}\, \mathrm{d}V$$
$$\mathrm{d}U = T\, \mathrm{d}S -p\, \mathrm{d}V$$
Using this,
$$\mathrm{d}H = T\, \mathrm{d}S + V\,\mathrm{d}p \tag{*}$$
imposing constant pressure conditions,
$$\mathrm{d}H = T\, \mathrm{d}S = \delta q_p $$
Hmm, now we that we  have identified enthalpy as the heat exchanged at constant pressure.
Now, $q_p = n\int_{T_i}^{T_f}C_{p,m}\mathrm{d}T$ and finally,
$$\Delta H = \int_{i}^{f}\mathrm{d}H = q_p = n\int_{T_i}^{T_f}C_{p,m}\mathrm{d}T $$
Your last equation seems fishy, and I don't see where it comes from but the other two seem fine.

EDIT
So after my discussion with the OP in the comments section I was informed that we are not allowed to impose constant pressure, which I assume implicitly based on the equation he quoted from his lecture notes. If that is indeed the case, then the equation provided in his notes is incomplete. I just looked at @ChesterMiller's answer and that confirms it too.
So basically, how I would proceed is start with $(*)$
Now, $H = (S,p)$, but $S = S(T,p)$ too, so let's derive and invoke the "2nd $T\mathrm{d}S$ equation.
$$\mathrm{d}S(T,p) = \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T + \left(\frac{\partial S}{\partial p}\right)_T \mathrm{d}p
$$
multipling the LHS and RHS by $T$
$$T\,\mathrm{d}S = T \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T + T \left(\frac{\partial S}{\partial p}\right)_T \mathrm{d}p
$$
Using a Maxwell relation
$$T\,\mathrm{d}S = T \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T - T \left(\frac{\partial V}{\partial T}\right)_p \mathrm{d}p
$$
Almost done,
$$T\,\mathrm{d}S = C_p \mathrm{d}T - T V \alpha \mathrm{d}p $$
Putting this in $(*)$
$$\mathrm{d}H = C_p \mathrm{d}T - T V \alpha \mathrm{d}p + V \mathrm{d}p$$
or $$\mathrm{d}H = C_p \mathrm{d}T - V[1-T \alpha] \mathrm{d}p $$
which is the equation @ChesterMiller quoted. Now if you impose constant pressure, then you recover the relation I originally derived and it resembles the one you originally quoted. If you don't wish to do that, then this is the complete expression. To get the relation at constant volume, proceed as @ChesterMiller did in his answer.

You have applied the condition of constant pressure which is not allowed. $+1$ for using \mathrm and deltas tho :)

@LinearChristmas  I thought it was implicit, the equation you quote from your notes has no pressure term ？

$\def\d{\mathrm{d}}$
$\def\D{\Delta}$No indeed. It only states that $$\tag{non-ideal gas at constant $V$} \D H = n \int_{T_1}^{T_2}! C_{P,\mathrm{m}}\ \d T$$ 'because then $$\D H = nC_{P,\mathrm{m}}\D T$$ follows for ideal gases.'

I might assume the integral form to be a good approximation for solids and liquids (at constant $V$) in most cases. Not entirely sure, though.

@LinearChristmas  hmm if that is the case then the equation you quote is incorrect. It's missing a term

Thanks for your personalised update and the derivation! For the time being, I will accept Chester Miller's answer because he was first. However, I might bounty (+50) both your and Chester's answer in the future ;)


text:
detection of phosphorus in organic compound
I searched on internet and I don't find any elaboration of " detection of phosphorus " by chemical equation please elaborate it with chemical equations

Are you asking for a chemical reaction that would detect phosphorus within an organic compound? Or would any analytical technique suffice?

I,m asking about chemical reaction (equations)  that shows complete test to detect P in a organic compound...

To check whether phosphorus is present in the organic compound or not:
The organic compound is treated with sodium peroxide($\ce{Na2O2}$). Phosphorus gets converted to sodium phosphate. Then we carry out the same procedure that we do to identify phosphate anion.
The aqueous extract is heated with concentrated nitric acid and ammonium molybdate. A canary yellow precipitate of ammonium phosphomolybdate is obtained. This yellow precipitate helps us conclude that Phosphorus was present in the organic compound.
Reactions:
$$\ce{Na3PO4 + 3HNO3 ->H3PO4 +3NaNO3}$$
$$\ce{H3PO4 + 12(NH4)2MoO4 + 21HNO3 ->(NH4)3PO4.12MoO3 (yellow ppt.) + 21NH4NO3 + 12N2O}$$


text:
How can 2-indanone be prepared?

Would adding phosgene to ortho-xylene lead to the formation of 2-indanone?

I don't think so. Also, you should avoid phosgene like the plague. It will kill you faster than that. How about just buying it?

http://orgsyn.org/demo.aspx?prep=CV5P0647.. this might help you.. if there is any confusion or want the proper answer.... plz comment. i would be writing the procudure clearly..

If you want to start from ortho-xylene 1 I would recommend dibromination using N-bromosuccinimide to obtaining ortho-di(bromomethyl)benzene 2, followed by the addition of 1,3-dithianide anion 3 (prepared from formaldehyde and 1,2-ethanedithiol followed by treatment with NaH). While formaldehyde is electrophylic, 1,3-dithiane is nucleophylic if deprotonated (eg. with NaH, forming 1,3-dithianide ion) and can attack on the bromomethyl side of the molecule to get 4. This polarity inversion is called umpolung. A second treatment with NaH gives the cyclized product 5. Deprotection of the thioacetal gives 2-indanone 6. With this reversed polarity approach no phosgene is needed for the synthesis.

Care would need to be taken in the NBS reaction. NBS is also a source of electrophilic bromine and will brominate activated aromatic rings in polar solvents (personal experience).

1,3-ditianide??

@Tetrahydrocannabinol Yes, it is the dithioacetal of formaldehyde. While formaldehyde is electrophylic, 1,3-dithiane is nucleophylic if deprotonated (eg. with NaH, forming 1,3-dithianide ion) and can attack on the bromomethyl side of the molecule. This polarity inversion is called umpolung. If you are not familiar with it I suggest you to look it up, because it is very interesting and useful.

If memory serves me correctly, the anion 3 of 1,3-dithiolane undergoes fragmentation. That is why 1,3-dithiane is used.

If you need to start with ortho-xylene, then you should use fazekazs's approach. I searched through SciFinder and found that the most commonly reported synthesis of 2-indanone involves oxidation of indene using a variety of conditions.
The following Organic Synthesis prep highlights this approach, although it certainly uses some outdated methodology (distillation by aspirator vs. rotary evaporation, for example). Newer reports use a variety of oxidations, including oxone or the the Wacker process.
In this example, indene 1 is added to a mixture of formic acid and hydrogen peroxide to afford the monoformate ester of indene 1,2-diol 2. The monoformate ester was hydrolyze with sulfuric acid to give 2-indanone 3.

However, when I first saw your question, I wanted to propose a synthesis based on a Dieckmann condensation on diethyl phenylene-1,2-diacetate 4 followed by decarboxylation of the $\beta$-ketoester 5. However, I did not find this approach in the literature, and both the diacetate ester and its parent diacid are of comparaple price to 2-indanone.

Another interesting idea that I did not find in the literature is a pericyclic approach based on photolysis or thermolysis of beznocyclobutene 5 in the presence of carbon monoxide. Following electrocyclic ring opening, intermediate 6 can react with CO in 4+1 cycloaddition. An isocyanide could be used instead of CO, and hydrolysis of the imine would give 2-indanone. Alternatively, there may be a transition metal species that could mediate this transformation.


text:
Why does NO2 react with HCl ? What is the mechanism of the reaction?
I found the following reaction given in a certain book:
 $$\ce{2NO2 + 4HCl -> 2NOCl + Cl2 + H2O}$$
I cannot think of any reason why an acidic oxide like $\ce{NO2}$ will react with an acid like $\ce{HCl}$.
So, why will they react? And, what will be the mechanism of the reaction?

It is a redox reaction. One compound is a strong oxidant; another can reduce it, albeit not very willingly. So it goes.

What do you mean acidic oxide?

@NMJD acidic oxides are oxides that on hydrolysis gives acid and on reacting with base gives salt. Basic salt on the other hand gives base on hydrolysis.

Your equation is not balanced.

@NilayGhosh - I'm aware. I wanted user39139 to answer the question; some students misunderstand "acid oxide" and assume it should act as an acid in the way of HNO3 or HCl. I wanted to see if that was at the root of this question, since otherwise the answer just lies in redox potentials (which has already been suggested). If this misunderstanding persists, that answer might not help OP. I suppose to be clearer I should have said, "Can you define what you mean by 'acid oxide'"

Yes, it is a typical redox where nitrogen dioxide is an oxidising agent and hydrochloric is a mild reducing agent as Ivan noted.
The reaction proceed in two phases:
$$
\begin{align}
\ce{NO2 + 2HCl ->[\Delta] NO + Cl2 + H2O}   & \tag{R1}\\
\end{align}
$$
Now some nitrogen monoxide react with water to form nitrosyl chloride.
$$
\begin{align}
\ce{Cl2 + 2NO -> 2NOCl}   & \tag{R2}\\
\end{align}
$$
Finally, we get our main reaction
$$
\begin{align}
\ce{2NO2 + 4HCl -> 2NO + 2Cl2 + 2H2O}   & \tag{R1*2}\\
\ce{2NO2 + 4HCl -> (2NO + Cl2) + Cl2 + 2H2O}\\
\ce{2NO2 + 4HCl -> 2NOCl + 2H2O + Cl2}  & \tag{R3}\\
\end{align}
$$


text:
How do organic chemistry mechanisms become accepted?
Now I have been learning chemistry for five years. I remember when I started organic chemistry, it was fun to draw arrows between molecules to show, as if in a mathematical demonstration, how the reactions occurred. In every lesson I had, teachers explained to us how a specific reaction (for example the Shapiro reaction) occurs step by step, explaining the chemistry of each group in each intermediate as if things were obvious (you know how teachers are).

But I've been wondering for some weeks now how does a mechanism come to be considered as accepted or still discussed?
If they use some, what kind of spectrometry techniques are used to measure the amount of each intermediate? If not how do they proceed? Do they use computational chemistry? Because for example for a reaction such like a $\mathrm{S_N2}$ it doesn't look too tricky to find how it works, whereas for Fries rearrangement (I don't know if the mechanism is considered as accepted or not) it seems to be more tricky.

(Ref.)

So can you explain the methods (at least the most used) to confirm a mechanism? I am aware that "confirm" does not mean that we are 100% sure, but rather that it is simply the best we have found so far.

Here's a good read: "Controversial Mechanisms in Organic Synthesis" - Baran Group Meeting, 18 Sept 2010

Great question!
When I was teaching, Anslyn and Dougherty was a decent text for this. Here are some general comments:

First, please note that you cannot be sure about a mechanism. That's the real killer. You can devise experiments that are consistent with the mechanism but because you cannot devise and run all possible experiments, you can never be sure that your mechanism is correct.

It only takes one good experiment to refute a mechanism. If it's inconsistent with your proposed mechanism, and you're unable to reconcile the differences, then your mechanism is wrong (or incomplete at best).

Writing mechanisms for new reactions is hard. Good thing we have a whole slew of existing reactions that people already have established (highly probable, but not 100% guaranteed) mechanisms for.

Computational chemistry is pretty awesome now and provides some really good insights into how a specific reaction takes place. It doesn't always capture all relevant factors so you need to be careful. Like any tool, it can be used incorrectly.

The types of reactions you run really depend heavily on the kind of reaction you're studying. Here are some typical ones:

Labeling -- very good for complex rearrangements
Kinetics (including kinetic isotope effects) -- good for figuring out rate-determining steps
Stereochemistry -- Good for figuring out if steps are concerted (see this example mechanism I wrote for a different question)
Capturing intermediates -- This can be pretty useful but some species that you capture aren't involved in the reaction, so be careful.
Substitution effects and LFER studies -- Great for determining if charge build-up is accounted for in your mechanism

For named reactions, the Kurti-Czako book generally has seminal references if you want to actually dig through the literature for experiments.
For your specific reaction, what do we think the rate-determining step is? Probably addition into the acylium? You could try to capture the acylium intermediate.
You could run the reaction with reactants that have two labelled oxygens and reactants that have no labelled oxygens. Do they mix? If not, it's fully intramolecular. Otherwise, there's an intermolecular component and the mechanism as written is incomplete.
A quick Google search suggests that the boron trichloride mediated version has been studied via proton, deuterium, and boron NMR. I didn't follow up on this, but there's clearly some depth here.
When I was T.A.ing for Greg Fu, he really liked to use an example with the von Richter reaction. I might be able to find those references...

Can a reaction mechanism consist of two different routes (happening at the same time) that lead to the same product? Like $A \rightarrow B \rightarrow C$ and at the same time $A \rightarrow D \rightarrow  C$ ?

That would be two difference mechanisms. @adosar

I meant we have A initial, which gives $B + D$ and then these give $C$ without reacting with each other. Isn't this possible?

Again, those are two different mechanisms that compete.

This doesn’t exactly concern the actual mechanism you asked for, but as part of my PhD thesis, I performed an amide alkyne coupling the mechanism of which had been researched by Arndt et al.[1] Analysing how they established an accepted mechanism may help understanding how these are accepted. The authors note five proposed mechanisms at the beginning of their paper and then proceed to state the implications of each mechanism concerning:

deuterium labelling of different reactants and how it affects the product
kinetic isotope effects when replacing protium with deuterium
whether certain ruthenium-hydride species should be observeable in NMR with either reactant
whether ruthenium-amide species should be observeable in ESI-MS
whether the intermediate species should be neutral, cationic or anionic.

After predicting the implications of each mechanism, they performed a number of experiments to prove either side of the story. Including:

Deuteration studies
Kinetic investigations via in situ IR spectroscopy; whether or not deuterium changed the picture
NMR studies of the catalyst system with one or the other reactant; and then NMR studies of all three species mixed ($\ce{^1H}$-NMR, $\ce{^31P}$-NMR, 2D-NMR, …)
MS experiments

Each of these methods has advantages and disadvantages. For example, NMR is a very slow method so one cannot expect to observe rapid transformations. However, it is a good structure determining tool, and if you ‘freeze’ the reaction (e.g. only one reactant added), you can draw good conclusions about the initial complexes by analysing the solution with NMR. Likewise, all methods have certain advantages that were used to their fullest.
At the end, the authors were able to deduce that a number of proposed mechanisms were incorrect (they did not fit the experiment) leaving one proposed mechanism that seemed plausible. Additionally, a number of MS peaks which could well represent different intermediates were discovered in the MS analyses.
The resulting most likely mechanism (mechanism D by the original numbering) was then further backed by computational studies showing the different energy differences and activation energies.
They still cannot be sure that their proposed mechanism is correct, but there is strong evidence pointing towards it.
Reference:
[1]: M. Arndt, K. S. M. Salih, A. Fromm, L. J. Goossen, F. Menges, G. Niedner-Schatteburg, J. Am. Chem. Soc. 2011, 133, 7428. DOI: 10.1021/ja111389r.

Is it true for canizzaro reaction too? The mechanism I study in A level textbook, isn't sure to be correct?

@Mockingbird Mechanisms can never be proven, only disproven. At least, until we have the possibility to videotape molecules as they are reacting.

Nowadays, the popular well-known theory for kinetic chemistry called Transition State Theory has been recognized as a tool that can be used to 'judge' or 'verify' which reaction mechanism pathway will occur truly. The computational chemistry is being able to find where Transition State (TS) is, which based on theory. Existing of TS implies how difficult chemical reactions do.


text:
Why can't a reaction go to completion?
What is the reason for why an exergonic reaction would still have some remaining reactants (or an endergonic reaction to have any products)? The explanation given on this page (see Chemical reactions and mixing) is that there is a Gibbs free energy change caused by the entropy change by mixing; therefore, it is always energetically favorable to have at least some of each.
At another page (see Dissociation of weak acids) on the site, the explanation as to why weak acids (with positive $\Delta G^{\circ}$ for reaction with water) dissociate at all seems to be that some molecules do have the energy required for the reaction, analogous to molecules overcoming activation energy in the context of kinetics.
Are these two explanations equivalent? The first one is a macroscopic view that involves entropy of the whole system; the second one involves the distribution of speeds of the molecules themselves.

You alluded to the answer when you mention activation energy. Kinetically the equilibrium constant is $K_e = k_f/k_b$ where $k_f$ and $k_b$ are the forward are reverse reaction rate constants in the reaction $\ce{A <=> B}$. The reason that there is a finite and not zero back reaction rate constant, is that the activation barrier going B to A is not infinitely high. Quite the opposite in fact and  while larger than the barrier A to B, (if the reaction is exothermic), may nevertheless be of such a magnitude that the products can return to reactants and thus an equilibrium is established. In this case molecules are continually transforming between one another. The ratio of concentrations at equilibrium is the equilibrium constant.
The reason that the activation barrier can be surmounted is that there is a Boltzmann distribution of energies in all molecules and this means that the number of molecules in either A or B with energy $E_i$ is given by $n_i=n_0\exp(-E_i/(k_BT))$. This is the Boltzmann distribution for molecules with energy $E_i$ and $k_B$ is the Boltzmann constant.
The total number of molecules of type A is given by the sum of those in each level $E_i$ from level 0 to the maximum energy level, which is
$$\frac{n_A}{n_0}=\sum_i \exp(-E^A_i/(k_BT))$$
The equation for molecules of type B with energy $E^B_i$ is
$$\frac{n_B}{n_0}=\sum_i \exp(-(E^B_i+\Delta E_0)/(k_BT))=\sum_i \exp(-E^B_i/(k_BT))\exp(-\Delta E_0/(k_BT))$$
where $\Delta E_0$ is the difference in energy between the lowest levels of A and B.
At equilibrium the constant K is the ratio of the two concentrations which is therefore the same as the ratio of the two populations and is
$$K=\frac{Z_A}{Z_B}\exp(-\Delta E_0/(k_BT))$$
where  Z represents the summations and are called the partition functions. This equation represents the equilibrium constant in terms of the molecules energy levels.
The reason that there is a Boltzmann distribution, but a fixed energy at a give temperature, is that by far the the most likely way of populating a range of energy levels is according to the Boltzmann distribution. In fact the chance of having a Boltzmann distribution is greater than all other possible ways added together.
In classical thermodynamic terms, a plot of free energy G vs extent of reaction $\zeta$ (zeta) which ranges from 0 (entirely A) to 1 (entirely B), has a minimum at some intermediate value of $\zeta$ and this value naturally depends on the particular reaction.  The minimum value is where the reaction is at equilibrium and at the minimum the slope of G vs. $\zeta$ is zero which is also when $(\partial G/\partial \zeta)_{T,P} = 0$.
The minimum in G occurs because the entropy term $-TS$ has a minimum as the mole fraction of B increases (at constant T). Initially it is the entropy of A alone in solution, but at equilibrium is a mixture of A and B. If the reaction goes entirely to B then the entropy is that of B alone. The entropy of a mixture is naturally greater than that of one species thus $-TS$ has a minimum vs. $\zeta$.
If $dn_A$ and $dn_B$ of A and B molecules react the the infinitesimal change in zeta is $d\zeta = dn_A-dn_B$. The total change in free energy is $dG=\mu_Adn_A+\mu_Bdn_B$ where $\mu$ is the chemical potential or the free energy / mole. From these two equations the change in free energy of the reaction is
$$dG=(\mu_b-\mu_A)d\zeta$$
The reaction proceeds until $(\partial G/\partial \zeta)_{T,P} = 0$ and thus to proceed further expressions for the chemical potential are needed. For a perfect gas $\mu=\mu^0+RT\ln(p) $  thus
$$(\partial G/\partial \zeta)_{T,P} = \mu^0_B-\mu^0_A+RT\ln\frac{p_B}{p_A}$$
which can be recast as
$$(\partial G/\partial \zeta)_{T,P} = \Delta G^0+RT\ln\frac{p_B}{p_A}$$
and at equilibrium as the derivative is zero
$$\Delta G^0=-RT\ln\frac{p_B}{p_A}=-RT\ln K_p$$
which connect the equilibrium constant ($K_P$ for the gas phase reaction), with the free energy.

Edit:
Notes: The shape of the plot of G vs. extent of reaction can be shown by considering a general reaction of the form $\ce{A + B <=> 2C }$ . The Gibbs function is then
$$G = n_a\mu_a + n_b\mu_b + n_c\mu_c $$
for $n_i$ moles of species i with chemical potential $\mu_i$.  If we assume that the reaction is one of perfect gases then at total pressure P and with mole fraction x, the chemical potential of species i is
$$\mu _i = \mu_i^0+RT\ln(P)+RT\ln(x_i)  $$
where $\mu_i^0$ is a function of temperature only.  G can now be written as
$$G= n_a\mu_a^0 + n_b\mu_b^0 + n_c\mu_c^0+RT\ln(P)+RT(n_a\ln(x_a)     +n_b\ln(x_b)+n_c\ln(x_c) )$$
If we assume that the total pressure is 1 atm. then the $RT\ln(P)=0$ and can be removed from the equation. Using the relationship between number of moles $n_a=n_b$ and $n_c=2(1-n_a)$ gives
$$G= n_a(\mu_a^0 + \mu_b^0-2\mu_c^0) + 2\mu_c^0 + 2RT(n_a\ln(n_a/2)     +(1-n_a)\ln(1-n_a) )$$
The $\mu^0$ are the properties of the pure components so do not change during the course of the reaction which means that G is a function of $n_a$ only.
A plot of $G(n_a)-G(0)$ vs. $n_a$ is shown below.

Plotting this way gives a value of zero when $n=0$, which is the condition should all reactants be converted into product, point S. The equilibrium value is at point E. The energy T-R is that due to mixing reactants and R-E that extra energy of mixing when C is formed. The dashed diagonal line is $n_a(\mu_a^0 + \mu_b^0- 2\mu_c^0)$ which is how the reaction would behave if no entropy of mixing is included.
The important point is that there is a minimum in the free energy curve because the creation of a mixture is an irreversible process and this mixing produces a decrease in free energy.
The equilibrium constant is calculated using $-RT\ln(K_p)=2\mu_c^0-\mu_a^0-\mu_b^0$

Macroscopic view:
$$\Delta G = \Delta G^{\varnothing} + RT\ln Q$$
As the reaction progresses, $Q$ gets larger so $\Delta G$ becomes more positive. At some point, $\Delta G = 0$ but this happens before $Q \rightarrow \infty$.
Alternatively, for a reaction, you want $Q = K_{\mathrm{eq}}$. Since $K_{\mathrm{eq}}$ is finite, you won't react everything.
Microscopic view:
From the viewpoint of microcanonical ensembles, the probability of a single particle being in a state with energy $E$ is proportional to $e^{-\beta E}$, with $\beta$ defined as $\frac{1}{kT}$. This means that the relative probability between two states is $e^{-\beta \Delta E}$ if the two states are separated by energy $\Delta E$. Since the difference is not infinite, there is some nonzero probability of ending up in the less preferred state.
The fundamental issue is that we have lots of relationships to describe reactions, but they happen to deal with ratios and exponentials. In the former, setting the denominator to zero creates an infinity, which we don't have. In the latter, we require the infinity to get to 0, and so we can't.

Thanks. How are these two views related? They must be.

It's the same view. One is the individual particle (or molecule or what have you). The other concerns the combined probabilities and merges them into another number. But the concept is the same.

In very simplified macroscopic terms, you can think of every reaction having a forwards and backwards direction (being reversible). For some reactions, both directions are almost equally likely (e.g. reaction $(1)$) while for others the forward reaction greatly outweighs the reverse reaction (e.g. reaction $(2)$). But we can and should usually think of all reactions as reversible.
$$\begin{align}\ce{H2 + I2 &<=> 2 HI}\tag{1}\\[0.6em]
\ce{CH4 + 2 O2 &-> CO2 + 2 H2O}\tag{2}\end{align}$$
In again simplified terms, the more of one side you have, the more likely the reaction in the other direction will be. Thus, if you have lots of the products, the reverse reaction will be more likely giving you back some reactants. Likewise, if you have lots of reactants, the forward direction will give you a bit of product.
Thermodynamically, this is expressed with the equilibrium constant as shown in equation $(4)$ for reaction $(3)$.
$$\begin{align}\ce{a A + b B &<=> c C + d D}\tag{3}\\[0.6em]
K_1 &= \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}\tag{4}\end{align}$$
While an equilibrium constant of $0$ seems possible (no products) an equilibrium constant of infinity is not. A principle called microscopic reversibility (on a microscopic level, the forwards route is an identical but mirrored copy of the reverse route; meaning there is no reason to prefer one over the other) tells us that we should be considering any reaction as going into both directions; the equilibrium $K_2$ for the reverse reaction is the inverse of the equilibrium constant of the forwards reaction (eqation $(5)$). Thus, it is impossible for either equilibrium constant to be $0$.
$$K_2 = K_1^{-1}\tag{5}$$
All the explanations above are extremely simplified but in general they hold true.


text:
Configuration change at chiral carbon in SN2 reaction
Consider $\ce{A}=\ce{CH(CH3)(C2H5)(C3H7)}$ in R configuration. Suppose somehow in a $\mathrm{S_N2}$ reaction, $\ce{Br-}$ attacks and replaces $\ce{-H}$ (this compound we call $\ce{C}$). Now we expect change of configuration (Walden inversion), i.e. it to become S. If the structure was inverted without replacing $\ce{-H}$, we would have got the S-stereoisomer of $\ce{A}$, call this $\ce{B}$. Replace the $\ce{-H}$ in $\ce{B}$ with $\ce{Br}$, we obviously get $\ce{C}$. Now since $\ce{H}<\ce{C}<\ce{Br}$ in the Cahn-Ingold-Prelog rules, $\ce{C}$ is R. But $\ce{C}$ should have been S.
Where is my logic wrong?
EDIT: As noted by A.K. below, there is no problem in this case, and $\ce{C}$ is in-fact S. But everything depends on the atomic number of $\ce{Br}$. Is there a general proof of correctness for the scheme?

I think you need to draw out the molecule if you look at C you will see that the methyl group now has the low priority and that C is in fact S-configured.

I worked it out, and C is in-fact S-configured. But it all depends on the atomic number of the atom we use for substituting the $-H$. Can we prove this in general?

This is not $\mathrm{S}_{N}2$. There's no leaving group.

@Zhe the $S_N2$ is hypothetical and irrelevant.

C is (R)-configured unless I am seriously mistaken.

@A.K. But we have other terms for that and should avoid using a term that refers to something very specific. In this case, "substitution with inversion" would have been much better.

I am going to use ‘Pseudo-Fischer’ structures in this post, i.e. anything vertical should be considered behind the PC screen plane and anything horizontal in front of it.
Your original molecule and the hypothetic $\mathrm{S_N2}$-type inversion is here:
   Pr            Pr
   |             |
 H-C--Me ---> Me-C--Br
   |             |
   Et            Et

  (A)           (C)

If you determine both compounds’ absolute configuration, naturally we will have (R)-A but look closely at C and realise that it is also (R)-C. This is because of one of two fallacies students often have with CIP priorities (I’m presenting the other one here just for completeness):

A chemical inversion of a specific stereocentre does not have to mean an inversion of the stereodescriptor.

Just because a stereocentre does not participate in a reaction does not mean its stereodescriptor remains identical.

In our case, we can easily arrive at that conclusion by the two-step process you outlined. First, invert the original structure A to give B, then replace (under retention) the hydrogen in B to give C.
   Pr            Pr            Pr
   |             |             |
 H-C--Me ---> Me-C--H  ---> Me-C--Br
   |             |             |
   Et            Et            Et

  (A)           (B)           (C)

We already established that A is (R)-A. Going to be, this is a strict inversion and only in this case of a strict inversion will the fallacy of bullet point one actually hold true; thus, it is (S)-B. Finally, we replace the atom of lowest priority with one that will automatically become the highest priority. To explain what happens requires further elaboration.
First, assume without loss of generalisation that whichever group of B had second priority (in our case: ethyl) is pointing upwards in a Pseudo-Fischer structure while whichever group originally had third priority in B (here: methyl) is pointing downwards. This gives us depiction 1.
   Et
   |
Pr-C--H
   |
   Me

  (1)

As we have the highest priority on the left above and the lowest priority on the right above, it is easy to see that this compound has an (S) configuration. If the hydrogen and propyl groups had been switched, we would have had (R). Now, perform the replacement in a manner such that the former fourth priority becomes first. The original third priority is now fourth, second became third and first became second. This is shown in depiction 2.
   Et
   |
Pr-C--Br
   |
   Me

  (2)

Now, since our lowest priority is pointing downwards, to assign a stereodescriptor in 2 we must look from above. The first and second priorities are above the PC screen plane and on the left and right-hand sides. The third priority is behind the plane and centred. It is rather easy to see that in the case of 2 as drawn, with the right-hand group having a higher priority than the left-hand group, the stereodescriptor is (R). If they were reversed, it would be (S). We see that replacing the fourth priority with something of highest priority will invert the stereochemistry.
What happens if you replace the lowest priority with something of intermediate priority (i.e. something that will come in second or third)? The third case is easy; you are just switching two groups. Switching two groups will always give you a formal inversion. What if our replacement group takes second priority? I refer back to depiction 1 but this time replace hydrogen with $\ce{BEt} = \ce{(pin)B-C2H4}$ giving depiction 3.
   Et
   |
Pr-C--BEt
   |
   Me

  (3)

Again, we must look from above. The highest priority is on the left-hand sinde in 3, the second is on the right-hand side, the third is in the centre and behind. Our journey is counterclockwise or (S). Had the original configuration been inverted it would be identical. Thus, replacing a fourth priority with something of second priority retains the stereochemistry.

This is a one step analysis on a per-case basis. A.K. showed the usage of the switching method (two groups switch at a time) to its fullest; thereby, these independently established rules can also be derived as was shown.

I guess the two fallacies you outlined just solved my problem. A Walden inversion does not always imply a change from R to S or vv , if I got it right?

@MrigankaBasuRoyChowdhury Absolutely.

Yes you can generalize the flip in chirality but with math rather than chemistry.
Given: For a four-coordinated molecule with one chiral center, if any two of the groups switch position, then the chirality of the molecule flips.
1.)From this, we can infer that twice switching any two groups produces the original molecule.  Generalizing further we get that an odd number of two-group switches results in the enantiomer, while an even number of switches yields the starting molecule.
2.) Molecule A can have the positions ordered from low priority to high priority  you get a set ike: [A, B, C, D] with order corresponding to position (shown below, ignore the hydrogens) and letter corresponding to priority (A-low, D-High).

Molecule A (R-configuration)
3.) Your new molecule (C) can be represented as [D', A', B', C']. the question here is how many two-group switches does it take to get the molecule in the form of [A', B', C', D'] without any rotations of the molecule.

Molecule C as substituted
4.) Crunching through we get:
[D', A', B', C'] (0)
[C', A', B', D'] (1)
[A', C', B', D'] (2)
[A', B', C', D'] (3)
It took 3 two-group switches to return the molecule to an R-configuration, therefore the molecule was in the S-configuration.

I can see what you are trying to say, but can you please explain how you get [D',A',B',C'] as the new structure? I'm a new student, so I'm having difficulties. Thanks!

Because the bromine substituted the hydrogen the lowest became the highest and the rest shifted down on priority.