text:
Does the hydrolysis speed increase when more H+ ions are in the solution?
In hydrolysis of polysaccharides, does the hydrolysis speed increase when more $\ce{H+}$ ions are present in the solution?
So if I were to use $\ce{H2SO4}$ instead of $\ce{HCl}$ (same volumes) then $\ce{H2SO4}$ will be the most effective?

The glycosidic bond of a polysaccharide chain is essentially an acetal moiety, and hydrolysis of acetals proceeds under acid-catalyzed conditions. Hence, the addition of acid, within some ambit of concentration, will accelerate the hydrolysis reaction. The optimum proportion of acid, however, is probably something that needs to be determined experimentally on a case-by-case basis. It isn't necessarily true that the rate of the reaction will continuously increase, without bound, as the concentration of catalyst increases.
The mechanism proceeds through a series of proton transfers and SN1 reactions. As far as I'm aware, the rate-determining step is typically the actual cleavage of the glycosidic bond after an initial protonation of the oxygen of the leaving group. Given that, I would expect increases in acid concentration to have marginal effect beyond some threshold value. That's just a surmise on my part, however.
I did find one study that examined the kinetics of starch hydrolysis, which found the optimum pH to be 3 (with values of 2-5 being tested).

so lower pH values than 3 would have a negative effect on the reaction according to that study?

@LievenB, all I can say is that a pH of 3 was better in that study than 2-5. It may be that pH 3 is a globally optimum value under the specific conditions of that experiment, or it may be that it was simply the best out of the set of values tested, and that values lower than 2 might be even better. Personally, the latter case would seem like fairly aberrant behavior to me, but I don't have enough information to rule it out, and I would be cautious in extrapolating too much. It does, however, refute the notion that lowering pH by some arbitrary amount always accelerates the reaction.

There are acid promoted hydrolysis and base promoted hydrolysis depending on the two different mechanisms. More $\ce{H+}$ do increase the reaction rate of the former but will decrease the latter.
Give me your specific reaction so I can figure out what kind of mechanism it undergo.
In the comparison of sulfuric acid vs hydrochloric acid. Sulfuric acid is available in much higher concentration compare to hydrochloric acid (around 12M maximum). It is preferred when you need very concentrate acid. When using at same concentration, sulfuric acid may be little more better than hydrochloric acid in terms of $\ce{H+}$ amount but will not be big difference because the second deprotonation is not significant under highly acidic condition.

it is the catalysed reaction between H2SO4 and Cellulose decomposition into glucose (or other monosaccharides)

Cellulose hydrolysis can be done in both acidic or basic condition (very strong base in non-aqueous solvent). Acidic condition is easier and preferred, and it is acid catalyzed process.


text:
Is the current situation possible for recreaction in a lab?
I want to convert cellulose to another form (crystalline => amorphous). Cellulose requires a temperature of 320 °C and pressure of 25 MPa to become amorphous in water.(Is the pressure achievable in the lab?)
Is this doable in the lab and how should I begin on reproducing that situation (which tools would I need) ?
I'm guessing that a bunsen burner would produce high enough temperatures (but I'm not quite sure about the pressure)
EDIT to my question, following up the advice of bobthechemist...
Crystalline substance to amorphous - Other ways?
You can convert crystalline cellulose to the amorphous, more reactive form by heating it to a temperature of around 320°C under a pressure of 25MPa.
But are there other, less extreme ways, to achieve that amorphous form of the compound?
So because I can't achieve these high temperatures, how can I achieve it in another way?

That kind of pressure can be achieved in commercial high pressure reactors as far as I know. So if you have access to such equipment and know how to handle it safely, this should be possible.

and without being able to use high pressure reactors? would I stand a chance?

A pretty good chance at injuring yourself, I'd say. I wouldn't dare to try such pressures with any improvised equipment, you're very likely to just blow the whole thing up.

ok - so there is no possibility what so ever to do that?

That's about 250 times atmospheric pressure.  A quick back of the envelope calculation suggests that a 160 pound person standing on a block that is 0.5x0.5 cm could generate 25 MPa of pressure.  You'd need to be a very good acrobat to do that over a bunsen burner.

25MPa is just north of 3600psi for those of us that haven't converted to SI units; 320*C ~= 608*F. Your average lecture bottle is rated to about 3000psi, while compressed gas cylinders can go as high as 6000psi. Trouble is, most of them don't stand up well to being heated by an open flame. Temperature is another beast; your Bunsen burner can easily achieve a flame temperature well in excess of 600*F (and a simple variation, the Meker burner, can hit more than twice that), but your Pyrex glassware is only rated to 500*F, and that for short periods. Bunsen burners are often used for glassworking on the lab bench, creating custom tubing rigs by softening straight tubing and then bending or stretching to create bends and bottlenecks. If the glass can be molded over a Bunsen burner, it's not going to be able to sustain the temperatures you're talking about.
The short of it is that both the temperature and pressure are possible, but neither are achievable with a benchtop apparatus in a fume hood; you'll need an HTHP reactor vessel rated for your experimental conditions. They exist, but whether you can get access to one is another question only your superior can answer. If you were planning on visually studying the change in the cellulose, you'll need a sapphire viewport, and these get very expensive very quickly.
Once you have the right tool for the job, it becomes a rather academic task of calculating how much volume of reactants are required to achieve the desired pressure in the vessel when heated to temperature, then filling the vessel and evenly heating it. Only two of the vessels in the linked column hold any appreciable amount of material at the proper temp and pressure, and one of them will just barely meet your needs; if you unevenly heat any of these vessels, by the time the far end reaches the proper temperature the spot being heated will have failed.

source: Stack Exchange

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

sort: votes


text:
Why apart from when building atoms in the first place, the 3d is the lower energy orbital?
From chemguide.co.uk:

If the phosphorus is going to form $\ce{PCl5}$ it has first to generate 5 unpaired electrons. It does this by promoting one of the electrons in the $\mathrm{3s}$ orbital to the next available higher energy orbital. Which higher energy orbital? It uses one of the $\mathrm{3d}$ orbitals. You might have expected it to use the $\mathrm{4s}$ orbital because this is the orbital that fills before the $\mathrm{3s}$ when atoms are being built from scratch. Not so! Apart from when you are building the atoms in the first place, the $\mathrm{3s}$ always counts as the lower energy orbital."

Could you please explain why this is so?

This is because the commonly taught method of building electronic configurations, by assuming that all atoms have the energetic diagram for their atomic orbitals, is not a valid hypothesis any more when you read the d block. So you can learn rule of thumbs, but the reality is that energy levels of polyelectronic atoms are much harder to understand (graduate-level quantum chemistry). Here's one link that may help

@F'x That's already halfway to an answer! Would you care to turn it into a full answer on its own?

It is also from the misconception that d-orbitals are actually necessary to build up the hyper-coordinated species.

In P element, the energy of 3d is lower than 4s. Look at the link provided below. Only in K and Ca, 4s has lower energy than 3d.
http://www.cdeep.iitb.ac.in/nptel/Core%20Science/Engineering%20Chemistry%201/Slide/lect6/6_6.htm
Another reason could be because the symmetry of hybridized orbital with d orbital will be more preferred in bonding with that of s orbital, and the molecule obtain more energy compensation with sp3d over sp3s.

Ian your link is now broken. Could you somehow fix it please?

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


text:
Are aromatic amines or amides less basic?
When the lone pair of an amino group $\ce{R-NH2}$ is involved in resonance, its basicity decreases. But, between conjugation with carbonyl group $\ce{R}=\ce{R'CO}$ or resonance with benzene $\ce{R} = \ce{Ph}$, which one will reduce the basicity of amino group to a greater extent?

Measures for acid and base strength
First, you need to be clear about 'strong' and 'weak' acids.
In order to measure the strength of an acid relative to water and find out how effective a proton donor it is, you must look at the equilibrium constant for the reaction
\begin{equation}
  \ce{HA (aq) + H2O(l) <=> H3O+ (aq) + A- (aq)}
\end{equation}
The position of equilibrium is measured by the equilibrium constant for this reaction $K_{\text{eq}}$.
\begin{equation}
  K_{\text{eq}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{H2O}][\ce{HA}]}
\end{equation}
The concentration of water remains essentially constant (at $55.56 \, \text{mol} \, \text{dm}^{–3}$) with dilute solutions
of acids wherever the equilibrium may be and a new equilibrium constant, $K_{\text{a}}$, is defined and called the acidity constant.
\begin{equation}
  K_{\text{a}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}
\end{equation}
Like $\text{pH}$, this is also expressed in a logarithmic form, $\text{p}K_{\text{a}}$.
\begin{equation}
  \text{p}K_{\text{a}} = -\log (K_{\text{a}})
\end{equation}
Because of the minus sign in this definition, the lower the $\text{p}K_{\text{a}}$, the larger the
equilibrium constant, $K_{\text{a}}$, is and hence the stronger the acid. The $\text{p}K_{\text{a}}$ of the acid is the $\text{pH}$ where it is exactly half dissociated. At $\text{pH}$s above the $\text{p}K_{\text{a}}$, the acid $\ce{HA}$ exists as $\ce{A^{–}}$ in water; at $\text{pH}$s below the $\text{p}K_{\text{a}}$, it exists as undissociated $\ce{HA}$.
An acid's $\text{p}K_{\text{a}}$ depends on the stability of its conjugate base.
$\ce{HCl}$ is a much stronger acid than acetic acid: the $\text{p}K_{\text{a}}$ of $\ce{HCl}$ is around $–7$ compared to $4.76$ for acetic acid. This tells you that in solution $K_{\text{a}}$ for hydrogen chloride is $10^{7} \, \text{mol} \, \text{dm}^{–3}$ whilst for acetic acid it is only $10^{–4.76} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3}$. But, why are the equilibria so different? Why does hydrogen chloride fully dissociate but acetic acid do so only partially?
\begin{equation}
  \ce{HCl (aq) + H2O (l) <=>> H3O+ (aq) + Cl- (aq)} \quad \qquad \qquad K_{\text{a}} = 10^{7} \, \text{mol} \, \text{dm}^{–3} \\
  \ce{CH3COOH (aq) + H2O (l) <<=> H3O+ (aq) + CH3COO- (aq)} \qquad K_{\text{a}} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3} \\
\end{equation}
The answer must have something to do with the conjugate base $\ce{A^{–}}$ of each acid $\ce{HA}$, since this is the only thing that varies from one acid to another. In both the equilibria above, water acts as a base by accepting a proton from the acid. For the hydrochloric acid equilibrium in the reverse direction, the chloride ion is not a strong enough base to deprotonate the hydronium ion. Acetate, on the other hand, is easily protonated by $\ce{H3O+}$ to give neutral acetic acid, which means that acetate must be a stronger base than chloride ion.
A nice thing to remember:

The stronger the acid $\ce{HA}$, the weaker its conjugate base, $\ce{A^{–}}$
The stronger the base $\ce{A^{–}}$, the weaker its conjugate acid $\ce{AH}$

So, the $\text{p}K_{\text{a}}$ is a good measure for the strength of an acid: the lower the $\text{p}K_{\text{a}}$ the stronger the acid.
But what about bases? To say something about the strength of a base it is convenient to take the following viewpoint: to what extent does a protonated base want to keep its proton or in other words: how big is the acid strength of its conjugate acid? For example if you want to know which is the stronger base - say, formate anion or acetylide anion - you look up the $\text{p}K_{\text{a}}$s for their conjugate acids. You find that the $\text{p}K_{\text{a}}$ for formic acid ($\ce{HCO2H}$) is $3.7$, whilst the $\text{p}K_{\text{a}}$ for acetylene is around $25$. This means that acetylene is much more reluctant to part with its proton, that is, acetylide is much more basic than formate. This is all very well for anions - you simply look up the $\text{p}K_{\text{a}}$ value for the neutral conjugate acid, but what if you want to know the basicity of ammonia? If you look up the $\text{p}K_{\text{a}}$ for ammonia you find a value around $33$ but this is the value for deprotonating neutral ammonia to give the amide ion, $\ce{NH2^{–}}$.
If you want to know the basicity of ammonia, you must look up the $\text{p}K_{\text{a}}$ of its conjugate acid, the ammonium cation, $\ce{NH4+}$, protonated ammonia. Its $\text{p}K_{\text{a}}$ is $9.24$ which means that ammonia is a weaker base than hydroxide - the $\text{p}K_{\text{a}}$ for water (the conjugate acid of hydroxide) is $15.74$.
In conclusion: the higher the $\text{p}K_{\text{a}}$ of the conjugate acid the stronger the base.
The answer to your question
The $\text{p}K_{\text{a}}$ of the conjugate acid of aniline is $4.6$. So, since this value is much lower than the $9.24$ you had for ammonia, it is much less basic than ammonia, as is expected since the lone pair of aniline is delocalized into the phenyl ring. But for amides the $\text{p}K_{\text{a}}$ values of the conjugate acid are typically smaller than $0$, so an amide is much less basic than aniline. This is to be expected because the carbonyl group contains an electronegative oxygen atom and thus has a lower lying LUMO than the phenyl ring which can interact more efficiently with the nitrogen's lone pair, thus spreading the negative charge more evenly among the conjugated groups and leading to a higher stabilization of the molecule.

Firstly, as the result, amide ($\ce{R}=\ce{R'CO}$) is much less basic than aniline ($\ce{R}=\ce{Ph}$).
The reason is carbonyl group is able to stabilize amino lone pair not only by a better resonance (negative charge on oxygen) but also by inductive effect as a strong electro-withdrawing group while phenyl group only have resonance. In a lot of cases the quality of resonance structure is more important than the amount of resonance structures.

I fear, I have to disagree with your statement "In most case inductive effect is more important than resonance effect here." The resonance is the dominant effect here.

Yes. My description is not accurate. Carbonyl beats phenyl because it have better resonance (negative charge on oxygen) and additional inductive effect.


text:
Superheating metal alloy above solidus
Is it possible to heat a material above its solidus without melting occurring?
More specifically, at a heating rate of approximately 200K/s, will a metal alloy exhibit superheating and if so by what order of magnitude i.e. fractions of a degree versus tens of degrees. What is the mechanism behind this?
There is a lot of information about supercooling but I am not certain whether this applies!

Yes, superheating can happen in some alloys. Whether it occurs at all, and its extent if it occurs, is very specific to each alloy (and heating rate dependent, obviously). It also depends on things like sample geometry (particles, solid, etc.) and size. This paper reports delays in melting by up to 100 K for some Al-Pb and Cu-Pb alloys.

That was exactly the kind of thing I was searching for - very relevant to the subject I'm interested in (friction welding process).


text:
Effect of dopant size on XRD peak shift
If we add an impurity atom (doping) to the crystalline struture of another material and take the XRD of the doped material what kind of peak shift can we expect. For the better description of my question i have added an image below

In the image black bold line ($2\theta $) is the standard XRD peak of a material; blue ($2\theta ^{L}$) and green ($2\theta ^{R}$) dotted lines are the peak shift caused due to the doping of impurity atom. Can you tell me if there is any relationship between dopent size and peak shift (right/left dotted lines).

Well, for one thing I'd say that, if the doping changes the lattice parameters then the XRD peak shifts in the opposite direction: So, if the dopant leads to an increase in the lattice constant then the diffraction angle $2 \theta$ decreases. But I fear your question might be a bit too general and it is difficult to make general predictions about what happens to the XRD shift when you dope the material. But then again, I'm no expert in this field and there might very well exist general guidelines that I'm not aware of.

Doping, not to confuse with alloying, introduces a low amount of impurities or more generally point defects in a crystal. The overall periodicity of the system is thus not changed and no shift should be expected in XRD. However doping introduces some disorder that makes XRD peaks broader.

The peak shift in the xrd is based on the dopant size. If the dopant size is smaller than the base metal it occupies the interstitial position leading to change in the lattice structure and the d-spacing between the atoms become less and there is an increase in the x-ray diffraction resulting in higher angles ($2\theta^R$). And vice versa when the dopant size is larger than the base metal. Lower angles ($2\theta^L$) is attained with increase in d spacing.

Sometimes, dopants can go to the interstitial site of the target material. We can say this makes some distortions happen.That actually change the crystal structure. Then we can see the certain peak position changes.
Also, misalignment can make peak position changes.So we have to operate the machine properly. calibration is very important.
Strain and stress of the materials can make XRD peak shift as well.
Good luck.

The OP's question was 'Can you tell me if there is any relationship between dopent size and peak shift?' This does not answer that question and so I am flagging it for deletion.

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


text:
Sulfur trioxide - vacant d-orbitals
Sulfur trioxide violates the octet rule. Upon drawing the Lewis dot structure for sulfur trioxide, we see that the central sulfur atom is bonded to three other oxygen atoms by double covalent bonds. Therefore, sulfur is in fact surrouded by 12 electrons and not 8 as per the octet rule. This is explained by the fact that the excited state of sulfur has the following electronic configuration: $1s^22s^22p^63s^13p^34s^03d^2$. As there are 3 oxygen atoms, $sp^2$ hybridised orbitals take part in single $\sigma$ covalent bonds with the unpaired $p_y$ orbitals of the oxygen atoms. In sulfur, there still exist one unpaired $p$ orbital and 2 unpaired $d$ orbitals. These remaining unpaired orbitals take part in $\pi$ covalent bonds with the unpaired $p_z$ orbitals of the oxygen atoms. My question is, why do the electrons in sulfur from the $3s$ and $3p$ orbitals in the ground state promote themselves to the $d$ orbitals rather than one electron going into the intermediate $4s$ orbital and the next electron going into the first $3d$ orbital? After all, according to the Aufbau Principle, $4s$ is less energetic when compared to $3d$.

This link might be interesting to you: http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html

Closely related question here: http://chemistry.stackexchange.com/q/6130/30

Very strongly related: http://chemistry.stackexchange.com/q/13483

I feel you are confused on atomic orbital and molecular orbital.
In the excited state of a single sulfur atom, electron do fill 4s first. You are right here. However, it is a different story in molecules.
In sulfur trioxide the bonding electrons are filling the molecular orbital, not the atomic orbital such as 4s or 3d. It is forbidden that both 3s and 4s participate any hybridization or formation of molecular orbital. So no matter how low is the energy of 4s, because 3s is already participating, 4s can not be part of the bonding.

I am repeatedly amazed at the somersaults theorists did to validate d orbital participation in main group chemistry.

@Jan  can I get the reference for this claim "It is forbidden that both 3s and 4s participate in the formation of molecular orbitals? Is it true.

@DeepakArya Why are you pinging me? I did not make the claim nor is it a claim I would put my name under.

@Jan Don't take it anything more, I tried @ Coconut and I think he/she is not here anymore. It is just that I have this doubt for over a week and searched this site and books. So I thought maybe a better knowledgeable person can at least point me to a better source.  If you disagree with my behavior, I will  delete both of my comments. (P.S. I asked a similar question but ppl took that down)

@DeepakArya in the help centre you will find that the post owner (here: Coconut) always automatically receives a ping for a comment under their question or answer. One additional user can be pinged by @-ing their username if they edited the post or commented previously.


text:
Are sulfonyl-carboxylates stable in water?
This time I am interested in a specific type of acid anhydride, sulfonyl-carboxylates, i.e. compounds with the following molecular structure:

where $\ce{R2}$ can be an organic substituents or hydrogen atoms. $\ce{R1}$ can be an organic substituent only.

In addition to the procedures described by Ian Fang, these mixed anhydrides (with $\ce{R_1, R_2 \ne H}$) have been used to prepare symmetric anhydrides of carboxylic acid in solid state reactions. If that's interesting to anyone, I can look up the reference.
As far as the effect of water is concerned, I do remember a procedure where the mixed anhydrides in question were actually prepared in a biphasic system of water and dichloromethane in the presence of a phase-transfer catalyst.

The mixed carboxylic anhydride, unless sterically hindered (the big sulfonyl helps) and electron rich, will be labile toward nucleophilic attack hydrolysis.  Acetic anhydride usefully persists in cold (acidified) water.  It goes exponentially as it exothermically hydrolyzes.  Keep it cold.  As above, use phase transfer for selective reactivity re the nylon rope trick.

No, it hydrolyzes easily to a carboxylic acid ($\ce{R2COOH}$) and a sulfonic acid ($\ce{R1SO3H}$). $\ce{R1}$ should not be a hydrogen atom by the way.
It is common to use sulfonyl choride or sulfonyl anhydride to react with carboxylic acid to make mixed anhydride which then react with amine or alcohol (or alkoxide) to generate amide or ester. This reaction can be one pot but have to be dry otherwise hydrolysis will reduce the yield a lot. Have done it many times.

Could you add a reference to support this?


text:
Prediction of ionic reduction?
Here are some standard electrode potentials:-
$$\ce{SO_4^{2-} + 4H^+ + 2e^- -> SO_2 + 2H_2O}\\E^o=0.17\mathrm{V}\\
\ce{NO_3^- + 4H^+ + 3e^- -> NO + 2H_2O}\\E^o=0.97\mathrm{V}\\
\ce{NO_3^- + 2H^+ + e^- -> NO_2 + H_2O}\\E^o=0.78\mathrm{V}$$
When nitric acid reacts with a metal, either $\ce{NO_2}$ or $\ce{NO}$ is formed? How do you predict which is thermodynamically favourable and which reaction will actually take place? For example, when dilute nitric acid reacts with copper, Nitric oxide is liberated and when it is concentrated, Nitrogen dioxide is evolved? As Nitric acid formation is more favourable, shouldn't it proceed in all cases as increasing the concentration will decrease/increase the potential of both the reaction simultaneousy(Nernst equation)?
In the case of sulfate ion, although the reduction of sulfate is more favourable than that of hydrogen, it is hydrogen ions which are getting reduced when a metal reacts at dilute concentrations? why is this so? Moreover, at high concentrations, once again sulfur dioxide is evolved? How can we understand such occurrences? Is overpotential involved in these observances?
In the case of permanganate ions, there are so many possible redox pathways, but a different one is preferred in a different case based on the conditions. For example, in acidic medium, permanganate is reduced to $\ce{Mn^{2+}}$ while in neutral/slightly alkaline solutions, it reduces to $\ce{MnO_2}$. Why is this so? Here are the redox potential values for permanganate conversions (latimer diagrams)
I want to know the general way to explain all these phenomena, as to what drives the reaction to give one product under certain condition and another in different conditions. Explaining any one example would be enough to set the general method which I can extrapolate to understand the remaining Examples.I am comfortable with basic electrochemical equations and those of thermodynamics.

Fortunately, the explanation is not as complex as it might seem. It is an oft-made analysis, but one must be very careful when determining reaction spontaneity by looking at the standard free energies $\Delta G_r^o$ alone, as most reactions are not performed in standard conditions and therefore the actual reaction free energy change $\Delta G_r$ may be quite different.
You are correct that, in standard conditions, $\Delta G_r = \Delta G^o_r$ and therefore that $\ce{NO}$ is the most stable reduction product of $\ce{NO_3-}$. However, standard conditions entail, among other things, that $\ce{[H+]\ = \ 1\ M}$ (at least approximately). This corresponds to quite an acidic aqueous solution, with pH zero. If one were to examine the reaction in a more dilute acidic medium, then the value of $\Delta G_r$ for each reduction reaction would change. The reaction that produces $\ce{NO_2}$ has a $\Delta G_r$ that depends on $\ce{[H+]^2}$, while for $\ce{NO}$ the $\Delta G_r$ varies with $\ce{[H+]^4}$. It is clear then that, as one goes to a more and more dilute acidic medium, $\Delta G_r$ decreases more strongly for the reaction that produces $\ce{NO}$, and at sufficiently weak acidic solutions the relative stabilities of the products actually switches, so then $\ce{NO_2}$ becomes the most stable product. A very similar argument can be made for the switch between $\ce{H+}$ and $\ce{SO^{2-}_4}$ reduction in media with sulfuric acid. For $\ce{MnO^{-}_4}$, note that the conversion to $\ce{Mn^{2+}}$ implies the consumption of eight protons, and hence if it can happen, it will likely do so in acidic media. As the acidity of the solution decreases, the conversion to $\ce{Mn^{2+}}$ is quickly suppressed, favouring reactions for which the $\Delta G_r$ depends less strongly on $\ce{[H+]}$, and so we find that $\ce{MnO^_2}$ becomes more favourable.


text:
What is a relationship between Volumetric Analysis and Titration Theory?
I believe that Volumetric Analysis and Tetration Theory are not the same subject.
But I need to know what is a main difference between those subjects and what is a main intersection?

I think you mean titration, not “tetration”. “Volumetric analysis” is another (somewhat dated) name for titration, as explained the first paragraph of the Wikipedia page.


text:
Does it make a difference in energy if a reaction is done in many steps or in one step?
I am particularly interested in knowing about glucose breakdown. If we artificially burn glucose, will the energy released be the same if it is broken down in steps(like in a living cell)?

@Philipp I think you should put this as the answer. It answers my question exactly ! :)

Ok, done. It's only a quick and dirty answer but if you want to know more you can always ask.

nooo...acc.to hess law,enthalpy is conserved whether the rxn is done in single or many steps

You might want to take a look at Hess's law. It states that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps. But that covers only the thermodynamic aspect. In reality, you also have kinetic barriers (activation energy) which add to the energy needed for a reaction to run (or released during a reaction). That's why catalysts are so useful: They break a reaction down into steps (involving different intermediates) with lower activation barriers than the original reaction so that the reaction as a whole can proceed at a lower temperature.


text:
Common flame-retardant plastic film materials?
More questions for my DIY hot air balloon:
What are some very common (i.e. easy to purchase) plastic film materials that have very good flame-retardant properties?
How is the "degree of flame retarding" of a plastic film material rated, so that I can compare materials by their manufacturer's rating?
Are there any common "spray on" flame retardant treatments that could work? Or is a flame retardant plastic film the only way to go to really stop it from bursting into flames?

What have you found so far? In a situation like this honestly I would go ahead and research what hot air balloons are commonly made with already. Find somebody more experienced in the hobby and ask them if you can, or even an experienced fire fighter you can get in touch with.
From wikipedia: "For modern sport balloons the envelope is generally made from nylon fabric and the inlet of the balloon (closest to the burner flame) is made from fire resistant material such as Nomex"
The type of measurement you are asking for seems a bit more complicated than something like measuring mass or volume, it seems to be commonly represented with images, videos of tests, graphs, etc. You may feel more at ease by testing how useful materials are before and after flame testing them yourself, that is what I would do anyways. I hope this helps you.


text:
Which reagent can I use to remove carbon from oven?
I have a practical question: today I have to clean the oven, and it doesn't have self cleaning burn down mode, nor an official scraper I can use...
It's full of charred grease and sugar. I managed to remove 98% of the charred grease with cold grease remover. But it seems that reagent has no effect on the remaining 2%, that's black fully charred carbon or sugar. I also tried vinegar, but it doesn't have any effect on it either.
I don't want to scrape it because I fear that I would scratch the oven.
So which reagent can I use to remove those pesky carbon deposits without damaging the oven, and what will be the chemical reaction?

The charred stuff shouldn't be pure carbon: it consists of some organic "goo" mixed with carbon particles. In a labratory one might use aggressive bases like sodium butanolate mixed with sodium hydroxide but that might be too corrosive for your oven. Why don't use oven cleaner? It's active component is $\ce{NaOH}$ and so might work. But you should be careful if your oven contains aluminum parts as $\ce{NaOH}$ will destroy those.

@Philipp I used the 'Well done st moritz cold action oil and grease remover'. The bottle says it contains 25% $\mathrm{KOH}$ along with many red warning signs. I washed down 98% of the stuff with it. But it has no effect on the remaining 2%, so is either pure carbon, or something that's not originating from grease, maybe charred sugar?

Hmm, you could try going higher with the $\ce{KOH}$ concentration or use $\ce{KOBu}$ but that might be too much. The only things that I heard of that might be able to dissolve carbon are $\ce{CS2}$ or maybe $\ce{CCl4}$ but you definetly don't want them anywhere near your oven.

@Philipp I read in the Wikipedia that hot carbon can react with water: $\mathrm{C + H_2O \rightarrow CO + H_2}$. So would heating the oven to the highest possible setting, and spraying water on the carbon deposits be enough to start this reaction? (I'll need good ventilation...)

@Calmarius If I remember correctly $\ce{C + H2O -> CO + H2}$ happens at something between 800°C and 1100°C (which cannot be achieved with a conventional oven). Even if the numbers aren't correct you'd have to flow water steam over the carbon residue, which won't be easy in a conventional oven. I'd suggest you to carefully scratch it or follow Philipp's suggestion.

I recommend a brush and elbow grease.

As a disclaimer, this is for information only.  Anyone using this information assumes all associated risks.  The manufacturer's label should be followed for any cleaning products.
The exact chemical composition of the goo is not precisely known. It is likely either tar or creosote existing as some form of hydrocarbons.  These hydrocarbons are tangled up to the point it is extremely difficult to dissolve.  The temperature required to burn this material is likely too high for anything short of a ultra high temperature industrial furnace.
A good way to clean it is to use oven cleaners which take advantage of the saponification reaction. Saponification will loosen the material; however, some scrubbing and/or scraping will still be involved.
For safety when oven cleaning be sure to have good ventilation.  Wear gloves, eye and face protection, as well as not having skin exposed. Caustic cleaning compounds and vapors from the cleaning are hazardous.
Most oven cleaners have multiple components.

A caustic solution causes a saponification reaction.  The saponification reaction is commonly used to make soap from saponifiable oils.  Because the oils in the material to be cleaned are likely saponifiable,  the saponification reaction makes the material much more likely to dissolve in water.
A foaming component creates a foam. When foam bubbles pop, the popping action works to break up the tangled bonds.
A compound may be included to ensure the entire solution will stick to a non-horizontal surface.
Another compound may be included to seal the solution so as to prevent oxidation.  Oxidation may interfere with the saponification.

The trick to getting oven cleaners to be effective, in my experience, is in the timing.  Timing is important because the contents of tars are not exactly known, so the softening reaction is brief.

A warm surface is easier to clean; however, a hot surface will dry the cleaner before it has a chance to work.
The oven should not be powered when attempting any cleaning.
When any oven cleaner is first applied, it is important to leave the cleaner alone and give it time to work. Ten minutes is usually sufficient.
Once dissolution occurs there is a brief window when the tar or creosote can be easily removed with a nylon brush or sometimes even a paper towel.
When the materials dissolved, vapors which were previously trapped in the tars may be released.  These released vapors are not good to breathe.
Sometimes it is necessary to get deeper layers of tar.  The cleaner can be reapplied up to three times before the solution dries if necessary.  Each time allow it to go undisturbed for ten minutes.  (In my experience, by the fourth application, cleaning the tar is simply not worth the effort.)
If too much time passes and the solution dries, then the tar is more difficult to clean.  The caustic solution can also damage metal components.
Rinse the oven of all cleaner when done.  Caustic residues can damage the oven if left on the surface.

"When foam bubbles pop, the popping action works to break up the tangled bonds.", that sounds like pseudo science and I have not been able to find any information that supports it. Do you have any source for that? Sincerely.

@MadsSkjern - When I first started taking chemical engineering courses, many of the concepts sounded like pseudoscience to me until my understanding improved.  If you need literature, try googling "shearing force of a breaking bubble."


text:
Why is acetic acid more acidic than phenol?
Acetic acid (ethanoic acid, $\mathrm pK_\mathrm a \approx 5$) is more acidic than phenol ($\mathrm pK_\mathrm a \approx 10$), which is reflected in their reactivity with a weak base such as sodium carbonate (adding acetic acid leads to effervescence of carbon dioxide).
However, if we look at the structures of acetate and phenoxide ions, the delocalization of negative charge should be greater in the phenoxide ion as the charge moves inside the benzene ring. On the other hand, in the acetate ion, the negative charge is delocalized between two oxygen atoms only.
How is it possible that acetic acid is still a stronger acid, even when its conjugate base has fewer resonance structures?

There are actually several factors that affect the stabilization of negative charge. The first is what kind of atom centers the charge can be delocalized on and the second factor is how many atom centers the charge can be delocalized on. The first factor is more important than the second. Just think it like quality is more important than quantity. Symmetry is another factor but not very important in your case.
Let's look at the ethanoic acid vs phenol case. In the acetate ion, the negative charge is delocalized on two oxygen atom centers, while in the phenoxide ion, the charge is delocalized on one oxygen and three carbon atom centers. Because oxygen is much more electronegative than carbon, the delocalization of negative charge over two oxygens is better than delocalization over one oxygen and three carbons. Quality beats quantity here.

If you have 2,4,6-trichlorophenol, because the strong election withdrawing chlorine atoms is helping carbon center to stabilize negative charge, it is a much stronger acid than phenol itself. This is an example of the inductive effect at work, as opposed to resonance.

The $\text{p}K_{\text{a}}$ of ethanoic acid is $4.8$ while the $\text{p}K_{\text{a}}$ of phenol is $10$. So ethanoic acid is indead more acidic than phenol (for an explanation of $\text{p}K_{\text{a}}$ values and their connection to acid strength see my answer here). And this is only to be expected from an intuitive point of view: one might say phenol is only an alcohol (although the phenyl ring has a lot of influence of course, which can be seen at the fact that normal alcohols have $\text{p}K_{\text{a}}$ values around $16$) and one wouldn't expect an alcohol to be more acidic than a carboxylic acid. The reason for this is simple: It not only the "delocalization radius" (over how many atoms the negative charge is delocalize) that matters when determining how well a negative charge is stabilized by delocalization but the effectiveness of the delocalization. The carbonyl group contains an electronegative oxygen atom and thus has a lower lying LUMO than the phenyl ring. This LUMO can interact more efficiently with the oxygen's lone pair, thus spreading the negative charge more evenly among the conjugated groups and leading to a higher stabilization of the molecule.
You might say, in a phenolate ion most of the negative charge is still located at the oxygen and only a relatively small fraction of it is delocalized inside the phenyl ring but in an ethanoate ion the negative charge is fully delocalized over the two oxygen atoms. Furthermore the two oxygen atoms in ethanoate are more electronegative than the carbon atoms in a phenyl ring and thus are able to stabilize the electron density more efficiently.

If you look at the resonating structure of phenoxide and ethanoate ion you will find that phenoxide has more delocalistion of electron but the negative charge is delocalising on carbon but in case of ethanoate ion negative charge is delocalising on oxygen which is more electronegative than carbon and for effective resonance the negative charge should be on more electronegative element .So ethanoate ion is more stable and therefore ethanoic acid is better acid .

The acetic acid is more acidic then the phenol because in resonating structure the acetic acid forms the equivalent structure. There is rule that the acidic character of equivalent structure is more then that of non equivalent structure.
Equivalent structure are those structures in which same charge remains on same atom in resonating structure.
Non equivalent structures are those structures in which same charge resides on different atoms.
Here in acetic acid if we draw resonating structure then same negative charge resides on oxygen atom but if we draw the resonating structure of phenol the negative charge move on from oxygen to carbon and it becomes a non equivalent structure.
Hence acetic acid is more acidic than phenol.

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


text:
Transition state and free energy
We have products $\ce{A + B}$ combining to form $\ce{C + D}$ through the transition state $\ce{X}$.

Are all reactions at least virtually reversible? Is the difference between a reversible reaction and an irreversible reaction simply that for the former, $\Delta G$ is small, where as the irreversible reaction will have a difference in free energy in products vs. reactants, that the reverse reaction simply doesn't occur, but theoretically could?
Why do concentrations affect the free energy? Is it because you sum the free energy for all reactant molecules, as well as for all the product molecules?
Following is from my textbook:

[...] the transition state is too unstable to exist for long. It collapses to either substrate or product, but which of the two accumulates is determined only by the energy difference between the substrate and the product -- the $\Delta G$ of the reaction.

I would appreciate if someone could clarify this for me. Feel free to be very nitpicky, I don't want an understanding that is "approximately" correct or "functionally" good.

What I mean is that if you have all products and reactants in solution, could reactions theoretically proceed in both directions, despite the sign of ∆G? And the energetically unfavourable reaction just happens a lot less, and depending on the magnitude of ∆G, virtually not at all. Since ∆G=0 at equilibrium. And I guess at equilibrium all reactions still proceed in both directions, but at the same rate?

That's exactly right.

According to the principle of Microscopic Reversibility all elementary reactions, i.e. those that proceed in a single step, are reversible.
And you are right: For a reversible reaction $\Delta G$ is very small while for an irreversible reaction $\Delta G$ is large. This can be most easily described using a simple schematic energy curve diagram for your situation.

Each reaction is microscopically reversible. So either your reactants $\ce{A}$ and $\ce{B}$ or your products $\ce{C}$ and $\ce{D}$ can combine forming the transition state ($\ce{TS}$). As your text book says, the $\ce{TS}$, being an energy maximum, is very unstable and collapses very fast. Since the transition state represents a maximum of the energy curve the collapse may go in either the direction of the reactants or the direction of the products with equal probability, pretty much in the same way a ball placed at the top of hill may roll down on either side of it. So, once you got to the transition state you have a 50:50 situation.
The thing that determines which way your reaction goes is how easy it is to make the transition state or to "get the ball to the top". In order reach the transition state you have to overcome a thermodynamic barrier, which is the difference of the free energies of products and reactants $\Delta G$, and a kinetic barrier, which is the activation energy $E_{\mathrm{a}}$. If $\Delta G$ is a very large and negative then the reaction will proceed in the direction from $\ce{A}+\ce{B}$ to $\ce{C}+\ce{D}$ and will be irreversible because then it will take considerably more effort to "climb" the energy hill coming from the $\ce{C}+\ce{D}$ side than coming from the $\ce{A}+\ce{B}$ side. This means that in a given time a lot more $\ce{A}$ and $\ce{B}$ molecules are "elevated" to the $\ce{TS}$ than $\ce{C}$ and $\ce{D}$ molecules. So, even if $\ce{TS}$ will revert to $\ce{A} + \ce{B}$ 50 % of the time the reverted molecules will quickly be back "on top of the hill" in order to have another try while for $\ce{C}$ and $\ce{D}$ this takes much longer - the reaction rate for the forward reaction is much higher than the reaction rate for the backward reaction. So over time more and more $\ce{A} + \ce{B}$ is converted to $\ce{C}+\ce{D}$ until equilibrium is estabilished and the ratio of reactants/products is equal to the ratio of the reaction rates.
When you have a reversible reaction and $\Delta G$ is very small the "hill to climb to the transition state" is nearly the same height irrespective of whether you come from the product or from the reactant side. So, forward and backward reaction proceed nearly at the same rate.
Concerning your second question: I'm sorry, but I couldn't find a non-mathematical explanation as to why $G$ depends on the concentration. The Gibbs free energy is the thermodynamic potential of an ensemble at constant pressure and temperature. For variable particle numbers $N_{i}$ its fundamental relation is:
\begin{equation}
  dG = -S dT + pdV + \sum_{i} \mu_{i} dN_{i} \ .
\end{equation}
So, it is connected to the particle numbers (and thus to the concentrations) of the species in the system.

I'd add a tiny comment: most reactions are equilibria, so they end up with both forward and backward reaction rates equal. I write most, and not all, because there is one exception: reaction between different solid phases. There, the reaction quotient is 1 (the different solid phases all have activity equal to one), and thus the reaction proceeds in the direction of negative ∆G until all reactants are consumed.

@F'x Interesting. I've never thought of that. If you would add an answer featuring this exceptional situation I'd certainly upvote it.

@Phillip. You state that there is a thermodynamic barrier and a kinetic one, which is the activation energy. There is no barrier in thermodynamic terms as these only tell us about starting and ending states.  Any reactions barriers are that due to potential energy because along the reaction coordinate bond lengths change and so potential energy curves have  minima at different values.

@Philipp How we calculate the Gibbs free energy at the initial and final state in order to make the diagram? Do we calculate it with a specific amount of reactants and products? I mean we calculate first the Gibbs enegy for a container only with reactants (at constant $T$ and $P$) and then we do the same thing for products?

First you are correct, there is no fundamental difference in reactions being described as reversible or irreversible, unlike in thermodynamics. A reaction will be called irreversible (a)if the product is removed from reaction i.e. by precipitation or physical removal and (b) if the rate of returning from product to reactants is so slow that it cannot be measured. This is somewhat subjective, but the lifetime of the back reaction can be years, so in practice its clear cut.
Secondly, the equilibrium constant for a reaction $\ce{aA + bB <=>cC + dD} $ is defined as $$K_p=\frac{P_C^cP_D^d}{P_A^aP_B^b}$$ where $P_A^b$ etc. represents the pressure at equilibrium. As the equilibrium constant must be dimensionless so pressure is understood to be divided by unit pressure of 1 bar. If the reaction is in solution then concentrations are used instead of pressure, then these are understood to be divided by $\ce{1 mol~dm^{-3}}$ (Technically instead of pressure or concentrations activities are used but for most uses pressure and concentration are sufficient). The standard free energy is defined as$$ \Delta G^0 = -RTln(K_p)$$ and this is how the pressures/concentrations come into the equilibrium constant.
The transition state is reached only when the reactants have sufficient energy to do so. By the Boltzmann distribution the number of energetic molecules rapidly diminishes as the energy is increased. Thus most collisions do not result in reaction, unless the activation barrier is very small compared to average (thermal ) energy. The reverse reaction has a larger barrier to surmount if the reaction is exothermic, thus fewer collisions result in reaction back to products than the other way round.
The transition state is at the top of the barrier between reactants and products, in solution it may be crossed and recrossed many times before reaction is complete, it all depends upon how quickly the energy the 'molecule' has at the transition state can be removed by collisions or by energy being redistributed internally into the products as vibrations and rotations or as kinetic energy of separation.
The activation energy barrier exists because the product will have different arrangement of bonds and atoms from the reactant and so different average internuclear separation in the reaction coordinate. The  activation energy can be imagined as the point where two potential energy parabolas cross, one being slightly displaced in position and energy vs. the other. This is what the normal reaction profile drawn in text books and elsewhere represents. It is usually shown as one line(the bottom of the first parabola), a curved line to a maximum (crossing point, transition state) then onto another lower line to represent product energy at the minima of the second parabola.

First question:
The answer is true from a reaction view. You are right. No matter how the reaction is favored in terms of Gibbs free energy change. The reverse process will happen in small or tiny amount.
From a thermodynamics view, they are different. Similar to the reversible and irreversible process in physics, reversible reaction has no energy loss but irreversible reaction has energy loss.
Second question:
Gibbs free energy change is a function depending on concentration for solution and partial pressure for gas. So concentration will obviously affect it.
Do not confuse it with the standard Gibbs free energy change, which assume that you have a standard condition (1 Pa pressure or 1 mol/L concentration). Since the concentration or partial pressure is a constant (precondition), the standard Gibbs free energy change is independent to concentration or partial pressure.
Both of your question is not related to transition state. Transition state is part of the chemical kinetics, not thermodynamics.

I don't understand what you mean by “Similar to the reversible and irreversible process in physics, reversible reaction has no energy loss but irreversible reaction has energy loss” — or rather: I think I understand it, and I disagree

If you understand reversible process in physics and treat reaction as a process. A restrict defined reversible reaction is an ideal process that goes so slow that at the reaction is kept very close to equilibrium until finish. In that way, all energy (free energy change) is transformed to work or other kind of energy without loss (heat). That is how the textbook "Chemistry A molecular approach" (Nivaldo J. Tro) defined "reversible" and "irreversible" reaction. On page 798 in second edition.

@ Ian Fang There is no fundamental difference in reactions being described as reversible or irreversible, unlike in thermodynamics. A reaction will be called irreversible (a)if  the product is removed from reaction i.e. by precipitation or physical removal and (b) if the rate of returning from product to reactants is so slow that it cannot be measured. This is somewhat subjective, but the lifetime of the back reaction can be years, so in practice its clear cut.


text:
How do I get MarvinSketch to use the ACS style of drawing?
I have a Wikipedia account and the people there have a tendency for getting upset over chemical structures that I created using MarvinSketch not conforming to the ACS drawing style.
Is there any way to make MarvinSketch use the ACS drawing style?

You can find some information on JACS style here.
If you are using Marvin Sketch 5.1, first clean your structure, then got to edit then format. Click the Apply changes for all the atoms / bonds and then Load Style. Pick J. Am. Chem. Soc and OK.
In Marvin Sketch 6 open File then Document Style and continue as above
For more information see the documentation here.
That should work

How do I set JACS as the default style? Whenever I set the style to JACS it reverts back


text:
Is neon a molecular element?
Does the chemical element neon consist of a molecule? For example, oxygen consist of two atoms together. But is that the same with neon?

Are you asking if an atom of Neon is classified as a "molecule"? Or are you asking if there are molecules which contain Ne atoms?

Since neon is a noble gas, it has its full share of valence electrons, which makes it unlikely to bond with other atoms.  Under certain conditions other noble gases, specifically xenon and krypton, can form compounds in extreme temperatures and pressures.
See the Wikipedia article on noble gases,

[the noble gases] are all monatomic gases under standard conditions

and, according to the article

Neutral compounds in which helium and neon are involved in chemical bonds have not been formed

I think the question is asking if a monoatomic molecule is ... well ... a molecule.

@ManishEarth Hmm, well, true, that's how it reads, but I think this is what the OP is actually asking about.  Thanks, I have an "it's" problem, doctors are checking into it :P

Then what keeps solid, frozen neon together?

@EricBrown that's a different (and interesting) question :)

@EricBrown So known van der Waals forces. They have different nature than covalent bonding, and are much, much weaker.

@permeakra It seems to me that a "van der Waals force" is a magical force that is invoked to describe bonding in situations that don't fit the empirical valence rules. Since these rules don't describe every circumstance, I conclude they are somewhat bogus, and there is little sense creating a force to shore things up.  It's electrostatics in each and every bonding circumstance.

@EricBrown They are not magical. main type of vdw forces are London dispersion forces and comes from electron correlation, so are very difficult to describe in a simple manner. wiki has a reasonable simple explanation http://en.wikipedia.org/wiki/London_dispersion_force

@permeakra The whole gamut of chemical bonding topologies can be found with variational methods using only the K.E., P.E. in its simplest form, and the antisymmetry requirement of the wfn. My point is that many of these anomalies are not really anomalies at all, if one did not have prior knowledge of the empirical valence rules (e.g. noble gases make 0 bonds)

Neon exists as single atoms in its elemental form. It is a noble gas, and as @jonsca said, that means (for various reasons) that it is extremely difficult for it to form bonds (I don't know of any neon compounds).
I think the confusion may be coming from a missing definition: what is a molecule?
A molecule is two or more atoms joined by covalent bonds.
Since neon exists as solitary atoms, it is not a molecular element.

http://goldbook.iupac.org/M04002.html


text:
Predict photochemical decomposition of organic compound?
Is it possible to predict wether an organic compound will photochemically decompose?
Also, is it possible to predict the products which it will decompose into?

Just as an experienced synthetic chemist can make an educated guess about how compound A and compound B will react, an experienced photochemist can usually do the same.  There are many, many well-known photochemical reactions that are mechanistically understood and sometimes even used in synthetic procedures, some commercially.  A few such reactions are the
Norrish Type I reaction

Norrish Type II reaction

Paterno-Buchi reaction

di-pi-methane reaction

photochemical induced halogenation, olefin dimerization and so many more.  So yes, an experienced photochemist can make reasonable predictions about which chemicals will react and what the possible products might be.


text:
How to get the freeze point of mixed liquids?
I recently bought in Italy a bottle of "Alcool puro" which is in fact 95% Alcohol and 5 % Water. So alcohol has a freezing point of -114°C and water a freezing point of 0°C. How can I calculate then the freezing point of the "Alcool puro"?

In general it is extremely hard to get the freezing point of a mixture purely by calculation. The reason is that the specific molecular interactions of the two substances you have mixed cause pretty much every mixture to need a unique formula for the calculation.
That said, there are some ways to get a rough estimate but these estimates mainly work for (linear) hydrocarbons because those are (fairly) simple to generalize. (google for "freezing point hydrocarbon mixture")
In your specific case, because it is ethanol-water your best way to find out the freezing point is to look up the experimental results of someone else. The engineering toolbox has a dataset on ethanol-water which you could use to get a reasonable estimate for your case (they don't list 95v%, but they do have the whole range from 0-100v% in steps of 10v%).

Too bad the ethanol-water eutectic point is just between 90v% and  and pure ethanol.


text:
Calculating standard enthalpy of reaction, which reactant to choose for the amount of substance
For example, this reaction: $$\ce{2NaOH + H2SO4 -> Na2SO4 + H2O}$$
Suppose both concentrations are $1\ \mathrm{mol\ l^{-1}}$ (standard condition). There is $0.02\ \mathrm{mol}$ ($20\ \mathrm{cm^3}$) of $\ce{NaOH}$, and $0.01\ \mathrm{mol}$ ($10\ \mathrm{cm^3}$) of $\ce{H2SO4}$. So neither reactant is the limiting reactant.
Well, I know one way is to find the standard enthalpy of formation of all components and subtract the ones of products from reactants'. My question is if only experiment data is given.
For temperature change of $\Delta T$,
$\Delta H = mc\Delta T = (0.03)\times(4200)\times\Delta T$,
where $m$ is the mass of the solution. Since the density should be similar to water $1000\ \mathrm{kg\ m^{-3}}$,
$20\ \mathrm{cm^3} + 10\ \mathrm{cm^3} = 30\ \mathrm{cm^3} = 0.00003\ \mathrm{m^3}$
which translates to $0.03\ \mathrm{kg}$. $c$ is just the specific heat capacity of water, it was assumed that the specific heat capacity of the mixture is more or less the same as the water's.
Now the standard one $\Delta_\mathrm{r}H^\circ$ requires division by the amount of substance, but which one – $0.02\ \mathrm{mol}$ or $0.01\ \mathrm{mol}$? What if $\ce{NaOH}$ is $0.021\ \mathrm{mol}$? Or $\ce{H2SO4}$ is $0.011\ \mathrm{mol}$?

Welcome to Chemistry.SE! Some parts of your question are a little bit puzzling to me. What do you mean by "0.02 mol (20$cm^{3}$)" - should this be a concentration? And what do you mean by "limiting reagent"? Where does "For temperature change of $\Delta T$, $\Delta H$ = $m C \Delta T$ = (0.03)(4200)$\Delta T$" come from, especially the values you inserted for $m$ and $C$?

Well. Under standard condition the solution used should be 1M, right?

$C$ is just the specific heat capacity of water, I assumed the specific heat capacity of the mixture is more or less the same as the water's.
$m$ is just the mass of the solution. Since the density should be similar to water $1000kgm^{-3}$, 20 +10 $cm^3$=30 $cm^3$ = 0.00003$m^3$ which translates to $0.03kg$

Thanks for the clarification. Could you add those to your question? It is always nice if every symbol or variable is properly declared inside the question. This makes it easier to grasp where your problem lies.

Done it. But aren't those just basic knowledge of high school chemistry?

The calculations itself, yes. But you made a lot of assumptions without mentioning them (for example: "I assumed the specific heat capacity of the mixture is more or less the same as the water's") and so those implicit calculations where not all that apparent to me. Another reason why it is good to explain how you got those numbers is, that a potential answerer can check if some of your assumptions might not be applicable or if you made another mistake there.

The difference between the Enthalpy of Reaction $\Delta_{\mathrm{r}} H$ and the Standard Enthalpy of Reaction $\Delta_{\mathrm{r}} H^{\ominus}$
The standard value is not related to a standard temperature (although Standard Enthalpies of Reaction are often given with respect to the standard temperature - they could have been given for any arbitrary temperature). The term "Standard" refers to standard pressure and the reactants and products being in their standard states. The standard states are usually those of the pure compounds. Furthermore gases are usually assumed to behave like ideal gases. This definition of standard states means that Standard Enthalpies of Reaction are often purely calculational quantities since the underlying reactions are practically not feasible. For example, the reactants usually have to be mixed in order to react with one another, meaning that they are not present in their pure form. Similarly, the products of a reaction are often formed as a mixture and not as pure substances. And real gases certainly don't show the same behavior as ideal gases.
So, enthalpically the difference between the Enthalpy of Reaction and the Standard Enthalpy of Reaction is that you would have to add the Enthalpy of Mixing of the reactants to $\Delta_{\mathrm{r}} H$ and subtract the Enthalpy of Mixing of the products from it in order to get $\Delta_{\mathrm{r}} H^{\ominus}$. Furthermore you would have to account for the enthalpic difference between real and ideal gases if there are any gases involved. Furtunately, those differences are usually quite small compared to the often very big Enthalpies of Reaction so that they can be ignored without introducing a large error. But you have to be mindful of the difference between $\Delta_{\mathrm{r}} H^{\ominus}$ and $\Delta_{\mathrm{r}} H$: for example, if $\Delta_{\mathrm{r}} H^{\ominus}$ is small and the reaction is conducted under very non-ideal conditions with respect to the behavior of gases and/or including liquid mixtures with a very large Enthalpy of Mixing, then you have to expect big differences between the (tabulated) standard values and the measured values.
So, in this sense you would have quite a hard time to calculate the exact Standard Enthalpy of Reaction from your measured $\Delta_{\mathrm{R}} H$. But, under the appropriate conditions your measured value is a good approximation.
How to calculate $\Delta_{\mathrm{r}} H$?
The key to solve your calculational problem is to look at the definition of $\Delta_{\mathrm{r}} H$:
\begin{equation}
  \Delta_{\mathrm{r}} H = \biggl(\frac{\partial H}{\partial \xi} \biggl)_{p, T} \ ,
\end{equation}
whereby the extent of reaction $\xi$ is defined as
\begin{equation}
  \mathrm{d} \xi = \frac{\mathrm{d}n_{i}}{\nu_{i}}
\end{equation}
with $n_{i}$ and $\nu_{i}$ being the amount and the stoichiometric coefficient of substance $i$, respectively.
Consider the reaction:
\begin{equation}
  \ce{\nu_{A} A + \nu_{B} B <=> \nu_{C} C + \nu_{D} D}
\end{equation}
From its definition you see that $\Delta_{\mathrm{r}} H$ is the amount of heat that is exchanged with the surroundings in a isothermal and isobaric process when $\ce{\nu_{A}}$ moles of A and $\ce{\nu_{B}}$ moles of B are converted to $\ce{\nu_{C}}$ moles of C and $\ce{\nu_{D}}$ D (= "conversion of one formula unit").
So, using the differential form of your calorimetric formula (under the assumptions you made) and "dividing" it by $\mathrm{d} \xi$ you get
\begin{equation}
  \mathrm{d} H = m C \mathrm{d}T \\
  \frac{\partial H}{\partial \xi} = m C \frac{\partial T}{\partial \xi} \\
  \Delta_{\mathrm{r}} H = m C \frac{\partial T}{\frac{\partial n_{i}}{\nu_{i}}} = m C \nu_{i} \frac{\partial T}{\partial n_{i}} \ ,
\end{equation}
where $m$ is the mass of the stuff you heat in the calorimeter (assumed to be constant) and $C$ its heat capacity.
Considering finite instead of infinitesimal changes of $n_{i}$ leads to
\begin{equation}
  \Delta_{\mathrm{r}} H = m C \nu_{i} \frac{\Delta T}{\Delta n_{i}} \ .
\end{equation}
So, in order to get your Enthalpy of Reaction you simply have to pick one substance from your reaction (it doesn't matter which) and substitute its stoichometric coefficient and the amount that was consumed/produced during the reaction into the equation above. Note that since $\xi$ is always positive the sign of $\Delta n_{i}$ is positive for products and negative for reactants (the same is true for $\nu_{i}$).
Side Note: Standard Enthalpy of Formation $\Delta_{\mathrm{f}} H^{\ominus}$
The Standard Enthalpy of Formation is a special Standard Enthalpy of Reaction for which further conditions are in effect:

the target compound is to be formed by reacting the (pure) elements
it consists of, whereby each element is expected to be in its most
stable modification for the given temperature.
The reaction equation is to be formulated in such a way that $1 \,
   \text{mol}$ of the compound in question is formed.

Thanks for the explanation. Is there any material it is referencing?
BTW, on this part "...in a isothermal and isobaric process when $ν_A$ moles of A and $ν_B$ moles of B are converted to ...", shouldn't the $ν_A$, $v_b$ be just the stoichiometric coefficients, while $n_A$, $n_B$ be instead the number of mole?
Referring back to my original example, if I pick $\ce{NaOH}$, $\Delta_rH$ = $(0.0342002*\Delta T\over 0.02)$ = $(0.034200\Delta T)\over 0.01$. While if I pick $\ce{H2SO4}$, $\Delta_rH$ = $(0.0342001*\Delta T)\over 0.01$ = $(0.034200\Delta T)\over 0.01$. So both will be the same

@Ijriims I'm sorry but I don't know what you mean by "Is there any material it is referencing?" - do you mean where I have got the information? And for the sentence "...$\ce{\nu_{A}}$ moles of A and $\ce{\nu_{B}}$ moles of B...": no, I mean exactly what is written there. The point about $\Delta_{\mathrm{r}} H$ is that it is the heat you get from the "conversion of one reaction formula unit", so you could say when $\Delta n_{\mathrm{A}} = \nu_{\mathrm{A}} , \text{mol}$ and $\Delta n_{\mathrm{B}} = \nu_{\mathrm{B}} , \text{mol}$ etc.

@Ijriims And for the last part of your comment - "So both will be the same" - yes, that's exactly what you should expect as $\Delta_{\mathrm{r}} H$ describes the heat you get per formula unit of the reaction it should give the same result irrespective of the component as long as the reaction is complete and there are no excess of any one component.

Yes. I mean the source.
And I think the formula still works even if there is excess reactant, since it uses $\Delta n_i$, not $n_i$. You calculates how much number of mole of the excess reactant is actually being reacted, then that amount is $\Delta n_i$.

@Ijriims Well, as for the source: Most of it comes from memory :) Apart from that you should find most of the stuff in any undergrad text book on physical chemistry. I often use a German one by Gerd Wedler. And, of course you are right about that it still works if there is excess reactant because you only need $\Delta n$. It's nearly 5 am here in Germany and I'm a bit tired so this one slipped into my comment. I guess I thought of $\Delta n$ being easier to measure when there is no excess of the component.

"$\Delta_{\mathrm{r}} H$ is the amount of heat that is exchanged with the surroundings in a isothermal and isobaric process..." I am wondering: if the reaction takes place in an adiabatic container, the energy released by the reactants converting to product can heat the solvent and so overall make the temperature increase, but still the $\Delta_{\mathrm{r}} H$ (as applied to the reactant and product molecules) is the same ($dH = \delta Q_{P}$, right?), so for $\Delta_{\mathrm{r}} H$ to be equal to the heat released, it suffices for the process to be isobaric. Is that a correct reasoning?


text:
Use of osmosis to measure molecular weight of macromolecules
I found a question in a biochemistry book.

How can osmosis be used to measure molecular weight of macromolecules? Is it applied practically?

A general answer without the usage of many technical terms would be most suitable for me.

Maybe this helps: http://www.brooklyn.cuny.edu/bc/ahp/ChemInvest/CI.MMOsmosis.html

Out of curiosity... are you expecting a General Chemistry level answer or a Physical Chemistry level answer?

I am a "lay man" in terms of chemistry.  I found the question in a biochemistry book. I think a general chemistry answer using not many technical terms would be best :)

It has to do with the fact, that osmotic pressure of any solution is dependent only on the relative number of particles in the solution (in case of dilute solutions), irrespective of their nature. Hence osmotic pressure is called a colligative property.
Therefore, you take a certain known weight of the substance you want to calculate the molecular mass of, prepare a solution and measure the osmotic pressure. Then you compare this to the known osmotic pressure of standard solution which tells you the relative number of particles required to  create such a pressure. From this and the knowledge of the weight you used to prepare the solution, you can get the required molecular weight (weight of Avogadro's number of particles). In practice, certain complications, like dissociation and association might complicate the matter a bit.
This method is widely used to calculate the molecular weight (especially) of polymers. See Berkeley and Hartley method.

The relationship of osmotic pressure and molecular weight is given by the equation
$$\dfrac{\pi}{C}=\dfrac{\mathrm{R}T}{M_r}$$
In a graph of $\frac{\pi}{C}$ against $C$ the intercept is
$$\mathrm{intercept}=\dfrac{\mathrm{R}T}{M_r}$$
where

$C=$ concentration
$\mathrm{R}=$ gas constant
$T=$ absolute temperature
$M_r=$ molecular weight

We have MathJax enabled for formatting chemistry and maths. For more information, see here: http://chemistry.stackexchange.com/help/notation. I have edited this post for you.

Osmotic pressure is measured at room temperature while other colligative properties are measured at high temperatures.
OP uses molarity while other CP uses molality.
OP is used at standard pressure i.e. 1atm.
it can also measure molecular mass for a diluted solution.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.


text:
Why is it not dangerous to dissolve NaCl?
$$\ce{H2O~(l) + NaCl~(s) ->[\Delta] Na+~(aq) + Cl^{-}~(aq)}$$
When table salt is placed in water, it dissolves due to the polarity of water molecules. When solvation takes place, negatively polar sides of water molecules attach to $\ce{Na+}$ ions, and positively polar sides of water molecules attach to $\ce{Cl-}$ ions. The ions are then carried away by water molecules through diffusion. Since the sodium is now isolated from the chlorine, what keeps it from violently exploding with water in the following chemical reaction?
$$\ce{2Na + 2H2O -> 2 NaOH + H2}$$
Also, since chlorine is poisonous, wouldn't that make drinking salted water extremely lethal?

The first reaction involves Na+ but the second involves Na. They're different species.

[A really-good answer to this question should include quantitative evidence, but here is my hand-waving version...]
Solvated sodium ions ($\ce{Na+}$) aren't all that reactive.  Surrounded by a lot of polar water molecules, they're actually quite stable.  What you're thinking of is solid sodium (Na) which will "react violently with water" until it's all dissolved - dissolved that is, into stable sodium ions.
Same kind of logic applies to chlorine.  Solvated chloride ions ($\ce{Cl-}$) are stable and not nearly as poisonous as chlorine gas. (Note:  everything is poisonous at some level, i.e. "the poison is in the dose".  In saying chlorine gas is more poison, that really just means it is a more 'potent' poison:  it will kill at lower doses compared to chloride ions.)

Another way to think of it:
Solid sodium and chlorine gas are at high energy states.  Since they can reach much lower energy states by ionzing in the presence  with water, the WILL break down and ionize in the presence of water

Note: chlorine gas actually forms hypochlorous acid with water, but the principle is sound.

The answer is that the universe is lazy. And that means that the universe is always seeking the lowest energy states it can find.
And so $\ce{Na}$ is less lazy than $\ce{Na+}$, and similarly for $\ce{Cl}$ and $\ce{Cl-}$, because both these transitions release energy and thus are lazy. $\ce{Na}$ is dangerous, and $\ce{Cl}$ is poisonous, precisely because they have not achieved their most lazy state and thus want to react more, $\ce{Na}$ donating an electron and $\ce{Cl}$ accepting one, both releasing energy.
The "universe is lazy" hypothesis is further supported by the 'most reactions are random" hypothesis. Most molecules are randomly hitting each other, and those that release energy are going to be favored.
And finally, $\ce{Na+}$ and $\ce{Cl-}$ are not dangerous because they are in their lazy states and thus don't need to react with much of anything.

Sodium metal reacts violently with water because an electron is transferred from sodium to water (or protons in water, depending on pH) to form H2 gas. However, Na+ has already lost its electron, and Na+ does not transfer an additional electron to water.

Sodium ion Na+ should not be confused with metallic sodium Na. They are actually two different chemical species (atoms vs ions), so it is expected that they chemically behave in a different way. The same applies to chlorine Cl and chloride ion Cl-.

Na can "switch" to Na+ when it is oxidated; this process of oxidation requires a first ionization energy: Na is supplied with energy in order to become Na+, so it is an endoergonic process; a strongly energetic one. When NaCl dissolves in water, the partially negative side of water molecule bonds weakly with Na+ (intermolecular bond, which has electrostatic character); in a qualitative view, forming a bond is an exoergonic process; since this type of bond is weak, the energy lost by Na+ isn't enough to bring Na+ back to Na.

Cl- has a lower energy than Cl has, so when Cl- forms a weak bond with the partially positive side of water molecule, its energy lowers and it's thermodinamically impossible for it to become Cl.

What you asked about poisoning power of Cl is a Biochemistry topic, but it's reasonable thinking that in a healty person's body at room conditions (e.g. not a scuba diver underwater) the total amount of Cl (Cl• radicals or Cl2 molecules) isn't large enough to cause health issues and it won't appreciably raise by drinking salted water. On the other hand, Cl- is part of our daily consumption of mineral salts.


text:
Does a sulfuric acid/nitric acid mixture dissolve PTFE (ie. teflon)?
In our lab, we commonly use a mixture of sulfuric acid and nitric acid to clean borosilicate glassware.  I'd like to clean some PTFE pieces in there, too, but not sure if they will dissolve or not.
Does anyone have any first-hand experience with this, and know whether or not a sulfuric acid/nitric acid mixture dissolve PTFE?

Just out of curiosity, what is on the PTFE and why do you think you need such harsh cleaning conditions?

PTFE = Polytetrafluoroethylene, a polymer commonly known as Teflon.   http://en.wikipedia.org/wiki/Polytetrafluoroethylene

Be aware that some PTFE items (more often mechanical mechanical components)  are made from a sintered product (has a faintly dimpled texture like expanded polystyrene scaled down) that may have voids that trap the acid, I would keep these out or take care of how they will be used.  PTFE intended for chemical handling should be the virgin material and void free, I would risk some items to test.

For having used "sulfonitric acid" (sulfuric and nitric acids mixture) for at least 2 decades, PTFE is definitely stable to such a mixture.

Safety on the use of "sulfonitric acid"
I know it is off-topic but I think that safety should be learned as light touches, not as a-difficult-to-swallow course. (Please tell me if it is not in accordance with the site's rules).
You can safely use "sulfonitric acid" (i.e. usually a mixture of one volume of conc. sulfuric acid poured slowly into one volume of conc. nitric acid) to clean glassware and PTFE-ware if you follow some common sense rules. Explosions (or at least too vivid reactions) reported were the direct consequences of a poor understanding of what "sulfonitric acid" does.

Handle it under a fume hood (I told you it was common sense!!)
NEVER pour sulfonitric acid on organic material as the latter will be oxidized in an explosive way! Every piece of glassware should first:

Be cleaned as much as possible with any solvent or manually removing solid substance which is stuck where it should not be,
Be thoroughly rinced with a solvent miscible with water (acetone, methanol, ethanol...),
Be thoroughly rinced with tap water, in order to eliminate most of any organic solvent,
Only then can you pour the sulfonitric acid onto your glassware and let it do its magick.

Do not think that a plastic which can seemingly resist any solvent will actually resist exposure to sulfonitric acid! PTFE will resist but not common plastic materials.

If you follow these rules, there is no reason why you should experience any dangerous reaction in your lifetime. Be especially careful with sintered funnels.

I can't claim to have any personal experience with that sulfuric acid/nitric acid cleaning mixture specifically, but, as far as I'm aware, PTFE is generally inert towards acids. As a testament to that, it is apparently the material used for storing fluoroantimonic acid. I found a chemical resistance chart confirming its stability against strong acids. It is evidently resistant to concentrated sulfuric and nitric acids, as well as aqua regia (a concentrated mixture of nitric acid and hydrochloric acid). I have used the last of those for cleaning PTFE lab equipment before with no problems.
Having said all that, it would be wise to consult the handling and safety data for your particular labware to be certain.


text:
Transesterification of propylene carbonate and methanol
I'm going to try out the transesterification of propylene carbonate and methanol. I know that the mechanism is as follows :

-What are the products yielded after the reaction?
-And which catalyst should I use? (reaction is happening at normal air pressure and at room temperature)

What you have listed is not a mechanism. Also, those structures are not propylene carbonate nor methanol.  Maybe you should look at the actual structures and try to figure out yourself.

the mechanism is an example mechanism (See the wikipedia page about transeserification)

I know it is used to obtain  dimethyl carbonate (DMC) I think the others products should vary from which catalyst you choose. Have a look to Transesterification of propylene carbonate with methanol using $\ce{MgAlCO3}$
hydrotalcite as solid base catalyst: Indian Journal of Chemistry
Vol. 49A, September 2010, pp. 1182-1188 to choose the catalyst. I think there are two little typos in Scheme 2 where the mechanism is shown: miss $\ce{^-}$ from $\ce{CH_3O^-}$ (first step) and miss $\ce{C}$ from one part ($\ce{CH_3CO}$) of DMC formula (second step) but is a good article.


text:
What does the commercial chemical composition notation "Percent dry, wet with water"mean?
I am not a chemist, I develop software. I'm hoping for a general explanation I can understand in order to assist me with a project I'm working on.
Customers viewing a product online such as peroxide see 97% (dry wt.), wet with 25% water.
What does this mean? What would you expect the product to be in the form of? A powder, liquid/gel, solid in water? Any specific type of container?

I believe there is no general rule.
In your case (assuming you mean this) it is a solid. materials can have 25% and even more water and still be solids.

Yes, that is the type I was referring to. What type of packaging would this be shipped in? What I hope to do is create a web graphic that would incorporate the information, as well as any other useful shipping information.

In this particular case it means that the product (a peroxide, explosive!) has been phlegmatized with 25 % water.  The product is a solid, of which 25 % are water.  The dry solid has been analyzed for peroxide content, and 97 % of the dry solid is benzoyl peroxide.
You would think it came in a plastic container because glass is basic and may leach ions that could catalyze decomposition of the product over time.  But that is just a hunch.


text:
Which reaction between Lugol and Hydrogen peroxide?
The Lugol solution is a mixture of $\ce{(K+, I- )}$ and $\ce{I2}$ such that $\ce{I3-}$ are formed by:
$$\ce{I- + I2 -> I3-}$$
I know that the brown color comes from $\ce{I2}$, but in this solution one have $\ce{I3-}$ not $\ce{I2}$, why is it brown?
When I mix Lugol with $\ce{H2O2}$ without the presence of $\ce{H3O+}$ there is a variation in the conductivity of the mixture but the brown color doesn't disappear. If I add $\ce{H3O+}$ the solution becomes transparent after few minutes.
My guess is that either $\ce{I2}$ is reduced by $\ce{H2O2}$ (hence the disappearing of the brown color) or a redox reaction between $\ce{H2O2}$ and $\ce{I-}$, but in the later case I don't understant why the brown color disapear instead of appearing.
Can you enlighten me please, thanks.

Hi I can't comment your question so I ask you here, what acid did you add? I will improve my question when I will have more infos!

What exactly do you mean by "becomes transparent"? Do you mean colorless or does the solution become only yellowish-brown?

I mean colorless

I'll try to figure out the color of every compound separately, but there is a strong interaction between them and the different parameters so this could be quite tricky:

$\ce{I+ , IO3-}$ don't absorb in the visible region
$\ce{I3-}$  I think should be red
$\ce{I2}$ is blue if solid but has different colors in solution depending from the type of solvent and $\ce{H+}$ concentration see Hildebrand (1909). In polar solvent is brown, however I don't exclude this is due to the interaction with $\ce{I3-}$.

Because there is an equilibrium:
$$ \ce{I_2 + I^- <=> I_3^-}$$
So there is always a little bit of iodine in the solution and this could cause the brown color (however keep in mind that is $I$ is not so soluble in water).
Adding $H_2O_2$ should oxidize the iodide in this way:
$$ \ce{2I^- + 2H^+ + H_2O_2 -> I_2 + 2H_2O}$$
It occurs the reverse reaction too but should be negligible. In this case is probably a cinetic matter,  you can make the solution transparent reducing the colored compounds. If you have a look to the iodine clock reactions iodine can be reduced by bisulfite:
$$ \ce{I_2 + HSO_3^- + H2O -> 2I^- + HSO4^- +2H^+}$$
But I exclude that a reduction could occur in an oxidizing ambient like your. So my hypothesis are two:
FIRST: You are oxidizing the colored species. Adding $\ce{H2SO4}$ and $\ce{H2O2}$ you could obtain one of the most oxidant agent, Peroxymonosulfuric acid ($\ce{H2SO5}$),  see Piranha solution (I think this could be very dangerous please take all the precautions!). But now I don't think that the equation that if I wrote before (see J. Am. Chem. Soc., 1931, 53 (1), pp 38–44 and
J. Am. Chem. Soc., 1978, 100 (1), pp 87–91):
$$ \ce{I_2 + 5H2O2 -> 2H^+ + 2IO3^- + 4H2O}$$
could occur. I know iodic acid $\ce{HIO3}$can be produced oxidating $\ce{I2}$ but I never found a reaction involving $\ce{H2SO4}$ and hydrogen peroxide. It could be a more complex reaction maybe involving the others ions. I was thinking about potassium hydrogen iodate ($\ce{KH(IO3)2}$). I try to figure out a stechiometric reaction but is quite an hell!
SECOND: simply you are  modifying the interaction between the solvent and $\ce{I2}$ and $\ce{I3-}$ avoiding the  light absorption to take place.
I'm sorry this is now quite not an answer, but I hope it helps!
Iodine color reaction are wonderful (but very complex)!

Hi I can't comment your question so I ask you here, what acid did you add? I will improve my question when I will have more infos!

(see comments above: he used H2SO4)

I was told in a french forum that the brown color comes from $I_{3}^{-}$ and that when we put $I_2$ in water there is an equilibrium $2I_2 \leftrightarrow I^+ + I_{3}^{-}$ which gives a slight brown color, slight because the  equation is much more shifted to the left then to the right.
Things change when we put $I_2$ in $(K^+, I^-)$, the brown color increases because of $I_2 + I^- \leftrightarrow I_{3}^{-}$ equilibrium.
NOW I really still don't understand which reaction occures and why brown color disappear, is that because of $I_2$ disappearing or $I^-$ disappearing?

@pedro I never seen $2I_2↔I^+ +I_3^−$ equilibrium. I've seen $I^+$ in $I_2↔I^+ +I^- $ (J. Am. Chem. Soc., 1978, 100 (1), pp 87–91)) but this doesn't matter. I don't think however is more to the left because  $I_2$ is not so soluble in water. I think $I^-$ could act only indirectly because is color less, I think something happen to $I_2$ or $I_3^-$.

I appreciate all these details, since I'm not chemist (but theoretical physicist so far from chemestry) I think that this experiment is far from being well suited for my classroom :-) I'll try to find another (less complex) slow reaction to study with my students.

@pedro yes experimental chemistry is not the best choice for a classroom especially with strong acid and peroxide... :-) Try a question about this topic many people here can suggest you a safe and well know experiment! Best regards!


text:
Law of Definite Proportion: Do the numbers indicating moles count?
I need to find the ratio of a substance as it converts from a bicarbonate to a carbonate:
ex:

$\ce{Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$
$\ce{Fe(HCO3)2} / \ce{FeCO3}$ = $3/2$ *using molar masses

But what if the substances aren't  the same number of moles?
ex:

$\ce{2Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$

Would the ratio be:

$\ce{2Fe(HCO3)2 / FeCO3} = x$

or still:

$\ce{Fe(HCO3)2} / \ce{FeCO3} = 3/2$

How do moles factor into this equation and why?

Balance your reactions before you get the ratio.

The stoichiometric coefficients are important. They make sure you have a valid relationship between reactants and products that obeys the Law of Conservation of Mass.

As Jerry hints at in his comment, there is another fundamental law that you need to consider: the Law of Conservation of Mass. Matter cannot be created or destroyed (outside of nuclear reactions and relativist physics). In chemistry this means you need the same number of atoms of each element on both sides of the equation. You cannot create additional atoms from nothing, nor can extra atoms of something vanish.
Consider your first equation:
$$\ce{Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$$
This equation is balanced - it obeys the law of conservation of mass. Each side of the equation has one $\ce{Fe}$, two $\ce{H}$, two $\ce{C}$, and six $\ce{O}$ atoms. This is a valid chemical equation representing a real reaction relationship.
Your second equation is not balanced.
$$\ce{2Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$$
There are now twice as many atoms of each element on the reactant side of the equation as on the product side. What happens to this extra atoms? They cannot vanish! This is not a valid chemical relationship. Balancing this equation generates a scalar multiple of your first equation, which will yield the same ratio of iron (ii) bicarbonate to iron (ii) carbonate.
$$\ce{2Fe(HCO3)2 -> 2FeCO3 + 2CO2 + 2H2O}$$
The Law of Definite Proportions tells you that chemical compounds always contain the same elemental compositions by mass. I am not sure how that is relevant to the ratio of product to reactant in your stoichiometry problem. In other words, it tells you that iron (ii) carbonate is always $\ce{FeCO3}$ and never $\ce{Fe3(CO3)4}$ or $\ce{Fe2(CO3)3}$ or $\ce{FeCO4}$ or anything else. Those other formulas may refer to real compounds (the first two are real), but they are not iron (ii) carbonate.


text:
What happens in the attempted diazotisation of 1,2-diaminobenzene?

Usually, $\ce{NaNO2}$ and $\ce{HCl}$ are the reagents used to generate an aryldiazonium chloride from an aniline. However, in this compound there are two amine groups adjacent to each other, so it seems unlikely that a diazonium chloride would form. Is it possible for the two amine groups to react during the reaction and form some cyclic product?

Your suspicion that a ring forms is a good one. You can convert one of the amine groups to the diazonium chloride:

However, in addition to substitution reactions by loss of $\ce{N2}$, diazonium compounds will react with nucleophiles at the terminal nitrogen atom. The best known examples of this behavior are diazo couplings, where the nucleophile is an activated arene. However, other nucleophiles, especially intramolecular nucleophiles, can attack diazonium cations. The other amine is poised to do this:

I have a doubt sir. Why does the nucleophilic attack happen on the diazonium chloride by the other amine group rather than conversion of both amine group to diazonium chloride?

Certainly that could happen, but it is not likely. The formation of the first diazonium is probably much faster than the formation of the second. Consider that the process involves incremental increase in formal charge $\ce{0->+1->+2}$. Additionally, intramolecular reactions are generally faster (no diffusion or mass transport required). Formation of the second diazonium is an intermolecular nucleophilic attack of the second amine on a $\ce{NO2^+}$ ion. Formation of a ring is an intramolecular nuclephilic attack.

@BenNorris But isn't the geometry a problem: The $\ce{N2^{+}}$ group and the $\ce{NH2}$ group are help away from one another by the ring without much freedom to bend. Wouldn't this hinder the intramolecular reaction or are the $\pi^{*}$ orbital of $\ce{N2^{+}}$ and the lone pair on $\ce{NH2}$ just pointing well enough towards each other that there is not much need for freedom of movement?

@Philipp, I agree the geometry appears to be unhelpful, but the reaction works

I did a recent Chem Abst search on the cyclization of the diazonium salt to the benzotriazole. No hits but that doesn't in itself prove anything. There were hits on conversion of the diamine to the benzotriazole by diazotization. The former diazotization used mineral acid while the latter used acetic acid or montmorillonite clay. Is there a change in mechanism? Might the cyclization occur earlier than formation of the diazonium salt, perhaps at the -NH-N=O stage? This would avoid the strain issue for cyclization. The formulation of the question is suspect IMHO.

Yeah, ring closure probably happens even before diazo group would be created.


text:
Organic compounds that react with carbonate ions
I want to remove carbonate ions from dissolved potassium carbonate (potash) by letting an organic compound react with the carbonate ions. I don't have a very good knowledge of organic chemistry, but are there certain types of organic molecules that can react with these carbonate ions to form new compounds?
Additionally, are there reactions in which the carbon atom from the carbonate ion can be inserted into the main carbon chain of another organic compound (making, for example, propane from ethane)?

I just rediscovered this question, are you still looking for answers or was the one provided by Al enough? Or do you want to bind carbonates in this form?

Cook it up with ethylenediamine or diethylenetriamine- but don't bet on it.  What do you plan to have as the counterion for $\ce{K^+}$?  Check your carbon oxidation states for the second half of your question.
No, wait!  Add citric acid.  Yeah, that's the ticket.  Plop, plop, fizz, fizz.


text:
Why do buffers need to be composed of equal amounts of the acid and salt?
Say you have $\ce{NaOH}$ and $\ce{CH3COOH}$ and want to make a buffer.
Theory teaches 6 mols of $\ce{CH3COOH}$ and half, 3 mols, of $\ce{NaOH}$ would be needed.
Then you'd be left with 3 mols of $\ce{H2O}$, 3 mols of $\ce{NaCH3COO}$ and 3 mols of $\ce{CH3COOH}$.
However, what if you mixed 3 mols of $\ce{NaOH}$ with 12 mols of $\ce{CH3COOH}$? According to my calculations, you'd be left with 3 mols of $\ce{H2O}$, 3 mols of $\ce{NaCH3COO}$, and 9 mols of $\ce{CH3COOH}$.
In the latter example there is a clear imbalance in the salt:acid ratio. However, won't this still act as a buffer, while not as strong as an equally concentrated 1:1 buffer? So it will work, it's just not ideal?

They don't need to be comprised of equal amounts.  However, when you are near the 50:50 point, the buffer is "unbiased" (a contrived term), and it can equally accommodate assaults to its pH going in either direction.
If, as in your first example, you're sitting at equal concentrations of sodium acetate and acetic acid, dropping in a 1 mole of either NaOH or HCl won't overwhelm the buffer capacity (pH will change, but not as quickly as it would when you exceed capacity).
Conversely, in your second example, if you add 4 moles of NaOH you'll still be buffered, (acetate will rise to 7 moles, acid fall to 5), but if you add 4 moles of HCl, you'll protonate all the acetate back to acetic acid and the buffering effect will collapse.
So while they don't need to be at the 50:50 point, it's generally a good idea to stay in that order of magnitude (which is why recommended buffering ranges are roughly 1 pH unit); 10:90 is iffy (used appropriately, it works fine), but 1:99 is nearly pointless.  The pKa is the theoretical pH when at a 50:50 mix, so different chemicals are used for different pH buffers (Good's buffers shown below).