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Is every chemical reaction in equilibrium?
I read that every chemical reaction is theoretically in equilibrium in an old textbook. If this is true how can a reaction be one way?

Wait, wait, what do you mean by in equilibrium? Possibility of achieving equilibrium?

Yes, every chemical reaction can theoretically be in equilibrium.  Every reaction is reversible.  See my answer to chem.SE question 43258 for more details.
This includes even precipitation reactions and reactions that release gases.  Equilibrium isn't just for liquids!  Multiphase equilibria exist.
The only thing that stops chemical reactions from being "in equilibrium" is the lack of the proper number of molecules.  For a reaction to be in equilibrium, the concentrations of reactants and products must be related by the equilibrium constant.
$$ \ce{ A <=> B} $$
$$ K = \frac{[B]}{[A]} $$
When equilibrium constants are extremely large or small, then extremely large numbers of molecules are required to satisfy this equation.  If $K = 10^{30}$, then at equilibrium there will be $10^{30}$ molecules of B for every molecule of A.  Another way to look at this is that for equilibrium to happen, there need to be at least $10^{30}$ molecules of B, i.e. more than one million moles of B, in the system for there to be "enough" B to guarantee an equilibrium, i.e. to guarantee that there will be a well-defined "equilibrium" concentration of A.
When this many molecules are not present, then there is no meaningful equilibrium.  For very large (or very small) equilibrium constants, it will be very difficult to obtain an equilibrium.  In addition to needing a megamole-sized system (or bigger), the system will have to be well-mixed, isothermal, and isobaric.  That's not easy to achieve on such large scales!

Update Commenters suggest that "irreversible" reactions do not have an equilibrium.  This is true, but tautological.  In the real world, all reactions are reversible, at least to a (perhaps vanishingly small) degree.  To say otherwise would violate microscopic reversibility.  A reaction that was 100% irreverible would have an equilibrium constant of infinity.  But if $K= \infty$, then $\Delta G^{\circ} = -RT \ln{K}$ would turn into $\Delta G^{\circ} = -\infty$. So to get infinite energy we would just have to use 100% irreversible reactions!  Hopefully the problems with the idea of "irreversible" reactions are becoming apparent.

@user109987 - Unfortunately this answer is just wrong. Irreversible reactions by definition only go in one direction and hence there is no reverse reaction. Without a forward and reverse reaction (a reversible reaction) there can be no equilibrium between the products and the reactants.

I'm confused why you commented on my answer but flagged @user109987.   Also, the answer is not wrong because no reaction is 100% irreversible.  Saying that there is no reverse reaction violates the principle of microscopic reversibility.

I flagged user109987 because that user accepted your answer which is wrong by definition of an irreversible reactions. // There is a fine difference here. There is not an equilibrium constant K between reactants and products for the reaction of barium chloride and sulfuric acid to give a barium sulfate precipitate and hydrogen chloride. If there is in fact such an equilibrium constant, then derive it. Show how the equilibrium shifts in the solution when solid barium sulfate is added after the reaction.

The point is that there is no equilibrium like:  $$\ce{K_{eq}} = \dfrac{\ce{[BaSO4][HCl]^2}}{
\ce{[BaCl2][H2SO4]}}$$

Yes of course there is an equilibrium!  If you want to know how to calculate the equilibrium constant and want to know how the composition of the system shifts in response to adding more barium sulfate, then I suggest you ask a new question right here at chem.SE.

Ok, I'll pick up the gauntlet. Question posted here http://chemistry.stackexchange.com/questions/47384/does-an-irreversible-reaction-have-an-equilbrium-between-reactants-and-products

The barium sulfate example is a poor choice. Equilibrium equations should really be defined using activities, and the activity of solid barium sulfate is by definition 1. I pushed the same bad example in the linked question above.

Equilibrium can only apply to a closed system.
Reactions which form insoluble precipitates or gases which escape do not exhibit the behavior of a closed system.  Therefore, these reactions may not be in equilibrium.  However, these claims are pragmatic rather than real.
As it turns out, in the above answer, barium sulphate has a $\ce{K_{sp}}$ of $\ce{1.1 x 10^{-10}}$, so formally there is some small equilibrium related to the amount of barium sulphate in solution $\ce{1.05 x 10^{-5}}$
As gasses are escaping in solution, they may be readsorbed, an thus there would be some small equilibrium for processes like that.
But pragmatically, these reactions are not at equilibrium.

If there were an "equilibrium" for the reaction then the equation would be something like: $$\ce{K_{eq}} = \dfrac{\ce{[BaSO4][HCl]^2}}{ \ce{[BaCl2][H2SO4]}}$$ and such an equilibrium just doesn't exist.

I cannot rationally interpret the idea of a chemical equilibrium not existing.  I can rationally interpret is it pragmatically zero.

Look at it this way - adding solid BaSO4 to the solution won't shift the "equilibrium" to the left .  So how could there be an equilibrium?!?

(1) Equilibrium can be applied to open systems in some cases.  (2) insoluble precipitates and gases can and do participate in equilibrium reactions, whether or not they are "reabsorbed".  An example is rising bicarbonate levels in the ocean due to higher CO2 levels in the atmosphere.  There is an equilibrium. (3) It is unclear what definition of "pragmatic" you are using.

@MaxW  if the amount of BaSO4 was under the incredibly small solubility limit, adding more of it would shift the equilibrium.  The effect would be so small that you could pragmatically say there is no change in the equilibrium, but that is not quite true.  It is however, a useful imprecision.

Yes every reaction is an equilibrium. A complete reaction is a equilibrium with high equlibrium constant. If you write the expression for equilibrium constant, you will find that high equilibrium constant implies that the conc. of the products is very high, i.e. the reaction has reached completion.

$$
\ce{BaCl2 (aq) + H2SO4 (aq) -> BaSO4 (s) + 2 HCl (aq)}
$$
Let's just think about what (aq) means; it means you have ions floating about in there which are in equilibrium with their solids.
If you start from thinking there is no precipitate, just ions dissolved, we have $\ce{Ba^{2+}}$, $\ce{H+}$, $\ce{Cl-}$, and $\ce{SO4^{2-}}$. Then you consider the $K_{\mathrm{s}}$ of the different salts, which are $\ce{BaCl2}$, $\ce{HCl}$, $\ce{BaSO4}$, and $\ce{H2SO4}$. They will all go to $K_{\mathrm{s}}$, so all the salts would be forming and be dissolved unless blocked, e.g. things can become supersaturated. $\ce{BaSO4}$ has an extremely low Ks so most will precipitate at the same time. $\ce{BaCl2}$ will go to $K_{\mathrm{s}}$ with the ions $\ce{Ba^{2+}}$ and $\ce{Cl-}$ which are still in solution, and $\ce{H2SO4}$ would also go to $K_{\mathrm{s}}$, which means $\ce{BaCl2}$ is being formed, and therefore there is a reverse reaction.
Note if I had $\ce{BaSO4}$ in water, which would be in equilibrium (so tiny dissolved/tiny bit of $\ce{Ba^{2+}}$ ions and $\ce{SO4^{2-}}$ ions), and I added $\ce{Cl-}$ ions, a negligible amount more $\ce{BaSO4}$ will dissolve, as $\ce{BaCl2}$ would go to equilibrium, reducing $\ce{Ba^{2+}}$ ions, leading to negligible amounts of $\ce{BaSO4}$ dissolving to remain at $K_{\mathrm{s}}$. This also shows Le Chatelier's principle.

No, every reaction isn't in equilibrium with its products. Consider the following irreversible reaction: $$\ce{BaCl2(aq) + H2SO4(aq) -> BaSO4(ppt) + 2HCl(aq)}$$.
By definition if the reaction is irreversible then there is no equilibrium for that reaction.

If there were an "equilibrium" for the reaction then the equation
would be something like: $$\ce{K_{eq}} = \dfrac{\ce{[BaSO4][HCl]^2}}{
   \ce{[BaCl2][H2SO4]}}$$ and such an equilibrium just doesn't exist
since when the barium sulfate precipitates there could be a microgram
or a kilogram as the product.
Think of it another way - adding $\ce{HCl}$ (in dilute solution) or $\ce{BaSO4}$ won't shift the reaction to the left. (Adding more HCl would shift $\ce{HSO4^{-} <-> H+ + SO4^{2-}}$ in concentrated solutions, which is besides the point I'm trying to make.)
There is a solubility product for barium sulfate, but the solubility product doesn't depend on the amount of barium sulfate precipitate, nor the concentration of HCl. So the solubility product isn't for the overall reaction but rather for part of the system:

$$\ce{[Ba][SO4^{2-}] = K_{sp}}$$
(Full disclosure - Theoretically the barium sulfate solubility product wouldn't depend on the HCl concentration, but really that isn't quite true. The barium sulfate solubility product really depends on the activity of the barium and sulfate ions, so the ionic strength of the solution matters.)

Isn't this theoretically in equilibrium, just that for all intents and purposes the equilibrium always lies far to the product side?

There is no equilibrium for the BaSO4 product. Since the ppt is another phase there could be 1 microgram or 1 kilogram.

This seems wrong.  Multiphase equilibria exist!  Liquid water at 100 °C is actually in equilibrium with gaseous water (steam) at 100 °C even though there could be 10 microgram or 100 exagrams of steam.

The OP's question was if "all reactions" are in equilibrium, not if at one equilibrium exists for every chemical system. For example in aqueous solution $\ce{[H^+][OH^-]} = 1 \times 10^{-14}$. But by definition if the reaction is irreversible then there is no equilibrium for the reaction.


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Determine formula of hydrocarbon from combustion data
Source: Oct/Nov 2015 Paper 33 IGCSE Chemistry exam (discounted during marking)
Can someone explain how to find the answer for the last part of the question?
Note: I got the answer pentane with 5 carbon atoms. Is it the correct answer?

Hydrocarbons burn in excess oxygen to form carbon dioxide and water. $\pu{20cm^3}$ of a gaseous hydrocarbon burned in an excess of oxygen ($\pu{200cm^3}$). After cooling, the volume of the residual gas at r.t.p. was $\pu{150{cm}^3}$, $\pu{50{cm}^3}$ of which was x=oxygen.
(i) Determine the volume of the oxygen used. [1]
(ii) Determine the volume of the carbon dioxide formed. [1]
(iii) If the hydrocarbon was an alkane, deduce a formula of the hydrocarbon.

Without trying to solve the problem, it seems a bit odd to me. There is a very limited ability of water to be gaseous at room temperature. To assume all the water formed is gaseous seems wrong. You could assume liquid water, and that the volume of liquid is negligible to the final volume of 150 $\mathrm{cm}^3$. So it seems you're left flipping a coin to get an answer.

I found the no. of moles of CO2 and assumed is the amount of Carbon in the original 20cm3 of the hydrocarbon. Then I found number of moles of Carbon in one mole of the Hydrocarbon. The result is 5 Carbon atoms so it must be Pentane. I just want to know if this is correct answer because there is no answer in marking scheme

That doesn't really make sense. The initial volume of the alkane is 20 and oxygen is 150 (200-50). So the molar ratio of gaseous alkane to oxygen is 2 to 15 isn't it? (have to assume ideal gas with 22.4 liters per mole which cancels when you take the ratio.) There isn't an alkane that burns in that ratio.

n-Pentane and isopentane are also both liquids at room temp. Butane is last n-alkane that would be a gas at room temp. Butane burns in ration of 2 to 13 which is close.

The question mentions that the system was cooled after the reaction, so initially the reactants werent at rtp right?

I would assume that the starting hydrocarbon and oxygen were at rtp and that the reaction is exothermic. So the gas had to be cooled to get back to room temp. // If you assume that hydrocarbon and oxygen follow ideal gas law, and that both are at the same temperature when the reaction starts, then the starting temperature doesn't matter...

Cant we find the number of carbon atoms in the reacting alkane by calculating the amount of carbon in Carbon dioxide? (Ignoring the erratic conditions)

What happens to the water produced?!?

Could always subtract out the vapor pressure of water over water at 298 K?

@MaxW assume that the water vapour has condensed and the only gases that are left are oxygen and carbon dioxide

In order to "solve" the problem the following assumptions are necessary:

The initial 20 $\mathrm{cm}^3$ volume of the gaseous hydrocarbon and 200   $\mathrm{cm}^3$ volume of oxygen are at rtp also.
The burning of the alkane (from part iii) also creates water, but at rtp the volume of water is negligible since it condenses into the liquid phase.
All the gases follow the ideal gas law so that volumes are directly     proportional to moles.

(i) Determine the volume of oxygen used.
The starting volume of oxygen was 200 $\mathrm{cm}^3$, but 50 $\mathrm{cm}^3$ of the final 150 $\mathrm{cm}^3$ was oxygen, so $200 - 50 = 150$ $\mathrm{cm}^3$ of oxygen was used.
(ii) Determine the volume of the carbon dioxide formed.
With assumption (2), the final volume was 150 $\mathrm{cm}^3$, but 50 $\mathrm{cm}^3$ was oxygen
so $150 - 50 = 100$ $\mathrm{cm}^3$ of carbon dioxide are formed.
(iii) The hydrocarbon was an alkane. Determine the formula of the hydrocarbon.
Since each carbon atom in an alkane will burn to produce one carbon dioxide molecule, then 100/20 = 5 carbon atoms must be in the alkane. So the alkane might be $\ce{C5H12}$. (Thus the hydrocarbon could be isopentane, neopentane, or n-pentane.)
$$\ce{C5H12 + 8O2 -> 5CO2 + 6H2O}$$

Problems with $\ce{C5H12}$
(A) Given that the volumes were supposed to follow the ideal gas law, 20 $\mathrm{cm}^3$ of hydrocarbon reacts with 150 $\mathrm{cm}^3$ of oxygen. That gives a molar ration of 2 to 15. But the molar ratio of $\ce{C5H12}$ burning is 1 molecule of hydrocarbon to 8 molecules of oxygen, or 2 to 16.
(B) The initial total volume of reactants was 150+20 = 170 $\mathrm{cm}^3$. The final volume of product was 100 $\mathrm{cm}^3$. So the ration of reactants to products, neglecting water, is 17 to 10. From the chemical equation the ration is 9 to 5, or 18 to 10.

So the answer must be a cycloalkane $\ce{C5H10}$ which would react with right ratio (2 to 15) and have the right ratio of products to reactants (17 to 10).
$$\ce{2C5H10 + 15O2 -> 10CO2 + 10H2O}$$
CHECK:
Assuming 22.4 liters for a mole of the hydrocarbon, then each mole of hydrocarbon would produce 10 moles of water.
$$10 \times \dfrac{0.020}{22.4} = 0.0089 \text{ moles of water}$$
Since each mole has a mass of 18 grams and water has a density of $1 ~\mathrm{g}/\mathrm{cm}^3$, then there is:
$$ 0.0089\times 18 = 0.16 \mathrm{cm}^3$$ of water which is indeed negligible.
Also at rtp water will have a vapor pressure of about 18 torr out of 760 torr so the partial pressure of water is pretty small also in the gaseous phase.
PS: Thanks to DGS for pointing out that the alkane must be a cycloalkane.

1,1-dimethylcyclopropane. There's your 2:15 ratio, right? But I don't know where it boils, and it kind of seems like a goofy answer.

@DGS yes! - A great insight!! A cycloalkane would have right burn ratio, and right product ratio. $$\ce{2C5H10 + 15O2 -> 10CO2 + 6H2O}$$.

That and trans-1,2-dimethylcyclopropane were the only ones I found even close to being gaseous at room temperature. I'm not sure 1,1-dimethylcyclopropane has a known bp, but I would suspect it to be below the 1,2- isomers. Anyway, trans-1,2-dimethylcyclopropane has a known bp of 28.2 °C and, as you said of isopentane, that room would have to be fairly warm.

For the problem we didn't need to identify the particular alkane, I was just using that as a sanity check. I knew that n-pentane was a liquid a reasonable room temp. Cyclopentane would fit the formula but it boils at an unpleasant 49 °C (121 °F).

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Difference between biosynthesis in organism versus lab
Are there different terms for biosynthesis of a compound naturally in an organism versus the biosynthesis using chemicals in the laboratory or is it just "biosynthesis"?

More often than not, the synthesis in organisms involves the action of enzymes. In the lab, you often avoid these for various reasons. The enzymes may not be available, or you're interested in a more general procedure, that allows to synthesize a number of derivatives of the target structure or just a part of it.
When the synthesis aims to obtain a particular molecule, just as found in nature by any means, you'll often find a term like natural product synthesis.
When a procedure aims to yield a product by a reaction similar to processes in living  organism, the term biomimetic is frequently used.

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Safety regarding multiple metals in same acid solution
My son is testing the corrosion of metals in various solutions. Is it okay to place strips of multiple metals in the same container of a specific solution?
Metals being tested: $\ce{Fe}$, $\ce{Cu}$, $\ce{Zn}$, $\ce{Al}$
Solutions:

Sulfuric acid (1 part acid, 8 parts water)
Hydrochloric acid (1 part acid, 11 parts water)
Copper Sulphate
Hydrogen peroxide

Well, these things are mostly no more dangerous together than separately (which still means considerably dangerous). Anyway, I would do it in separate containers, so that the experiments wouldn't obscure the results of each other.

Ivan Neretin's comment is absolutely right! Use separate beakers to observe different reactivity, different colours, etc. Use gloves and, even more important, proper eye protection! And have fun!

Thank you for your responses. I agree with testing them separately, but still wanted to check if there were specific hazards or spoilers by combining certain metals. Looks like we will keep them separate for now.

I get nervous on a lot of inquiries like this. Proper chemical safety means being informed before some "accident" so that you know how to react. These chemicals are routinely handled and not particularly dangerous. You should use eye protection as noted. But you should also know how to flush eyes, that you'd need to strip and shower if you spill acid over a large part of your body (never take off clothes over the head - cut them off if need be...). Never work alone. Also how do you plan to dispose of the acids? So you need to plan for safety before using chemicals.


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Does the para position on benzene ring-polymers make the structure stronger?
Aramid fibers are strong synthetic fibers. Kevlar is called a para-aramid, and it's outstandingly strong. Why does the para-position give Kevlar its strength?
When the Kevlar monomers are in the para-position, will this make the chains closer (shorten the length of the bond), and make strong interactions between the chains?

This question is a bit unclear to me. What exactly would you call an "advantage"? Which aromatic molecule are you talking about?

I read that their is para, ortho, and meta position in aromatic ring and I read one of good properties of aramid relates to it's para position of molecules in aromatic ring, so does the para position relates to good properties? Is it due to cloaing the chains togeter? Thanks

This is a decent question. Had you chosen appropriate tags and given us enough context, this never would've been closed. Voted to reopen and +1.

"...will this make the chains closer (shorten the length of the bond)..." Which bond do you think will be shortened? Do you mean the length of a chemical bond or the interactions between individual chains?

I ment if the length of the bond shorter, will it be strong? and  will the para structure provide strength to material by closing the chains together, because the rings connected to chains by para position? Thanks

para isomers fits into the crystal lattice more perfectly than ortho and meta due to steric reasons and hence generally they have more melting point.
Also in certain aromatic compound ortho isomers form  intra molecular hydrogen bonding which reduces its boiling point (generally) compared to ortho and para where in these isomers would be forming inter molecular hydrogen bonding.
In the case of para aramid you can see that the two peptide linkage formed are with two other para amid molecules (always).Had it been ortho aramid then the two bonds will be more likely to get bonded to a single other ortho amid molecule.And the polymeristaion ends there (at the most polymer of para aramid will be having less length and has unsymmetrical structure).But in para amid it can polymerise to any lengh.
These are the two possible polymers for ortho.

there is no difference in the bond strength formed between two para amid molecule compared to two ortho amid molecule after all both are peptide linkages only (i don't know if the ortho aramid exists or not but what will happen is this for sure).(sorry the figure looks completely distorted.I am not really good with the paint application)
finally its the capability to polymerize to a great extent that makes it strong.
Its also worth mentioning that the functional (or substituent) groups at C1 and C4 (C1& C4-carbon numbering in benzene) can't bond themselves,as the bonds in any polyatomic group,can't bend and elongate as they want (after all the distance between C1 and C4 even if we consider diagonally,is itself quite long.And secondly thinking like this insane,say a bond bending in way they want.It will never happen with para compounds.The least may be meta compounds can.)

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Why do we assume that it is 1 mol?
A question in my book was as following:

Benzine and chloroform are considered very toxic organic solvents that dissolve in each other and in a solution of benzine and chloroform there is 0.45 mol of benzine. Find the  the percent of mass( percent concentration solutions) in this solution.

In the answer, It assumed that the number of moles in the solution is 1 mole then subtracted 0.45 to get the number of moles of chloroform.
Why did he do that? Why specifically 1? It will obviously change if I assume any other number. (Unless I multiply the number of moles of benzine by that number)
Or should we just say that for every a certain amount of moles of that solution has a percentage and every other percentage is right if it follows the same instructions?

Can you update this question? "There is 0.45 mol" 0.45 mol of what? of total solution? 0.45 mol benzene? 0.45 mol chloroform? And what "percent of mass" do you want?
Without knowing any of this, my guess is that 0.45 is the mole fraction of one of the solvents. Choosing the total number of moles to be 1 just makes the calculations easier, but you could choose any n and get the same result

Sorry, but the accepted answer is wrong. The problem is indeterminate. Is there 1 liter of solution, or 5 liters, or 10 liters, or some other amount?
If the problem was to be solved using mole fraction, then the problem should have been written that way. e.g.:

Benzine and chloroform are considered very toxic organic solvents that dissolve in each other and in a solution of benzine and chloroform the mole fraction of benzine is 0.45. Find the the percent of mass( percent concentration solutions) in this solution.

There is no point in posing such sloppy problem statements.

This is the general assumption we take when we are short of information. We obviously couldn't solve the question without knowing the number of moles in the solution. There is no answer to "why specifically 1", it is generally assumed to be so. You are going to come across a lot of questions when you would require the initial number of moles. When the question maker would want you to take 1 mole,many a times he wouldn't even bother to mention that. You will be used to such assumptions as you will study more of physical chemistry.

The question is lacking some details, but in general, you can solve problems of this kind without stipulating any amount of moles. You can just call it $x$, and after doing calculations, you can find the ratio between the two substances in such a way that the $x$ cancels out.

The part where x cancel out didn't happen in my question. I kind of sure that it has to affect the concentration and it doesn't cancel out unless you have used another method.


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Why doesn't tetrachloridotitanate(II) obey the 18-electron rule?
I have worked out that the complex comes to 10 electrons, however I do not understand why it does not obey the 18-electron rule? What are the factors that account for this?

Welcome to Chemistry.SE! Find out how to beautify your posts in the help center and take the [tour] to learn more about this site. Enjoy!

My first comment is that the 18-electron rule has many more violations (e.g., typically 16-electron count for 4-coordinate complexes) than the octet rule.

But we also ask for your thoughts - why do you think the 18- or 16-electron count might be violated?

Tetra not trichlorotitanate?

There is more to complex chemistry than the 18-electron rule. It states one condition when complexes may be particularly favourable, namely when all metal d-orbitals are fully populated and the ligands favourably interact with all metal s- and p-orbitals leading to all paired spins and hopefully a large energy difference between populated and unpopulated orbitals. The list of complexes that do not obey the 18-electron rule may well be larger than that of those that obey it. In no particular preference that includes $\ce{[MnO4]-}, \ce{[CrO4]^2-}, \ce{[Fe(CN)6]^3-}, \ce{[Mn(H2O)6]^2+}$ and many more. (A few of those are not typically described as complexes but they are nonetheless.)
Further constraints that you need to be aware of include, but are not limited to:

steric stress, i.e. too many large ligands around too small a metal centre,
charge density, i.e. surrounding a weakly charged cation with too many anions,
the actual electronic configuration, which can render octahedral, Jahn–Teller-distorted octahedral or square planar complexes more favourable.

In the case of a chloridotitanate(II), we first need to realise that we are surrounding a moderately small cation (titanium(II)) with rather large anions (chloride) which may generate steric stress. We also need to observe that titanium(II) is only weakly charged and attempting to generate an octahedral configuration would give us an anion with $-4$ charge — not exactly favourable (exceptions may apply). Also, chlorido ligands are known to be weak-field ligands, so one would not expect an overly large crystal field split.
The two greatest factors are probably sterics and electronics. Consider tetrachloridocobaltate(III) $\ce{[CoCl4]-}$: while potentially six chlorides may fit around the rather small cation, even though it is $+3$ the electronic situation favours only tetracoordination. This results in a 14-electron complex rather than an 18-electron one. In analogy, I am tempted to say that the even lesser charged titanium(II) ion cannot accommodate for a total of six chlorides, allowing only four of them in its vicinity. Taking electronics into account, square planar coordination is out of the question due to lack of stabilising interactions and the complex would be tetrahedral.
In fact, the tetrahedral geometry allows for a mildly stabilised triplet state: both $\mathrm{e_g}$ orbitals are semipopulated which is generally a tad better than two of three $\mathrm{t_{2g}}$ orbitals as it would be in the octahedral case.

[TiCl4]2- is a weird one because rather than just using one effect for why its an exception it has two. It is both a pi donor ligand, and a high spin complex. I would go into more detail but its only one mark on the tutorial, so whats the point :/

Please go into more detail. It may be useful for future users and also for the OP in understanding the answer better.


text:
What is difference between T1g and T2g?
I know that the “$\mathrm{g}$” stands for symmetry of octahedral shape, but how are labels “1” and “2” given in $\mathrm{A_{1g}}$ and $\mathrm{A_{2g}}$ representations too…
Is it related to group theory? If so, then please elaborate.

Well, not quite. Here's the quick summary on subscripts in Mulliken notation for point group symmetry.
“g” stands for "gerade" and refers to the symmetry around an inversion center. (That is, the representation retains symmetry around the inversion center.) Yes, octahedral ($O_\mathrm{h}$) molecules have an inversion center, but so do others.
“u” stands for "ungerade" and refers to anti-symmetric around an inversion center.
“1” or “2” refer to symmetry (1) or anti-symmetry (2) around a secondary $C_2$ axis (e.g., in a $D_x$ point group like benzene).
The $'$ and $''$ assignments (if needed) are used for symmetry around the $\sigma_\mathrm{h}$ operation.
And, while you didn't ask, for “A” versus “B”, these are one-dimensional representations that differ with respect to the "main" rotational axis. “A” representations are symmetric “+1” and “B” are anti-symmetric “−1” for that axis.
You only need enough subscripts to uniquely identify a representation. For example, here's the character table for $D_{4h}$:
$$\begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & (xz,yz) \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & (x,y) & \\ \hline
\end{array}$$
Note that all the “g” representations have +1 or +2 for the character of the “$i$” symmetry operation. We don't need “1” or “2” to distinguish the $\mathrm{E_g},$ but we do for the different $\mathrm{A}_{x\mathrm{g}}$ representations. For those, we go to the $C'_2$ axis, which separates $\mathrm{A_{1g}}$ from $\mathrm{A_{2g}}$. These are unique and we don't need to assign $'$ or $''$ for this point group.
(If you look up the $D_\mathrm{3h}$ character table, you find $'$ and $''$ needed.)

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text:
how to prepare buffer solutions?
The buffer property of buffer solutions is most likely done by the weak acid's reversible dissociation. Can a weak base with reversible dissociation be used to prepare a buffer solution?if then what should be our choice and if not then please explain.
Note: if this has already been answered please give the url

http://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html

http://chemistry.stackexchange.com/questions/29405/why-does-a-weak-acid-and-a-salt-containing-its-conjugate-base-form-an-acidic-buf  http://chemistry.stackexchange.com/questions/9183/can-someone-please-explain-buffers-to-me


text:
Why do larger alkyl groups show a greater inductive (+I) effect?
I have been taught that an alkyl group shows a +I (positive inductive) effect. Why do they do so? I know that hyperconjugation can explain how alkyl groups could stabilize a carbocation (better than a hydrogen). But, how does a longer chain alkyl group (or a bulkier alkyl group) show more +I effect than a smaller one?

Qualitatively:
If a methyl group is more electron donating than a hydrogen, then an ethyl group would be more donating that a methyl group by the same reason.
The effect drops off with distance, and generally a pentyl group is considered no more donating than a butyl group.
As for why a methyl group is more electron donating than a hydrogen, the carbon has a higher electronegativity than hydrogen (slightly), polarizing the bond (slightly).  This increased electron density near the carbon, which increases (slightly) its electron donation (I+) ability.

How can the reason be same? A methyl group should (in my opinion) show less +I than a hydrogen as carbon's electronegativity is higher. So its greater stabilization (compared to hydrogen) of a carbocation may be explained by hyperconugation. How can we apply same reason to ethyl group's more donating capacity than a methyl group?

One thing still remains unclear (after your recent edit), in an ethyl group, one hydrogen is replaced by a methyl group, thus we replace an atom of lower electronegativity with an atom of equal electronegativity, reducing the electron density which would have otherwise been available to the central carbon of methyl group. How then is ethyl more +I than methyl group?

Electronegativity is not explicitly related to inductive effect.  The inductive effect is the ability to donate electrons, which requires an electron presence.  The methyl 'carbon' of an ethyl group owns more of the electron density from its hydrogens and is able to donate it to the ipso carbon slightly.  I appreciate the depth to which you want this resolved, but anything more than this begins to require QM math.  Hammet correlations may provide some insight, but they are empirical.

Oh! I used to naturally correlate electronegativity with +I effect... well my doubt is as good as cleared. Thank you for explaining this to me so patiently!

@Lighthart Can we say that the electron density at alpha-carbon is more in ethyl group than that in methyl group? Considering the $\ce{+I}$ effect of methyl in ethyl group.

It is because the methyl group has a positive inductive effect; this effect is due to the higher electronagetivity of carbon than hydrogen, which attracts the electron from hydrogen and becomes slightly negative and push the electrons towards other groups.


text:
Nuclear zero-point energy and quantum delocalization effects of water molecules
What do we mean by "Nuclear zero-point energy" and "quantum delocalization effects" of water molecules ? It's said that without these effects "water" becomes somewhat unbounded, which is not very well understandable to me. Can you please help me to clear my doubts.

First, I think I know what "nuclear zero-point energy" is, but I have no idea what "quantum delocalization effects" are. Secondly, usually when something is said somewhere there should be an explanation there as well. So, I want to have a look at the actual source of this knowledge you shared with us. Personally I don't like question of the "I've heard that" form here.

Chemical Physics Letters 460 (2008) 423–431

So, what was actually said in the paper1 with respect to AIMD simulations of water is the following:

Nuclear zero-point energy and quantum delocalization effects are neglected making classical water overbound.

To understand the meaning of this statement we first need to know how AIMD works. Ab Initio Molecular Dynamics (AIMD) is a molecular dynamics simulation in which potential energy surfaces for nuclear motion are obtained from first principles (ab initio) rather than represented by a force field. The quantum chemical calculation of potential energy surfaces are done in the Born–Oppenheimer approximation in which nuclei are considered to be stationary classical particles.
As its very name signifies, it is, of course, just an approximation. It ignores that nuclei move and aren't perfectly localized in space. The nuclear motion can be approximately separated into vibrational and rotational the first of which energetically dominates the second. Taking the vibrational motion on the nuclei into account results in some additional so-called zero-point vibrational energy that rises the total energy. Other quantum effects (tunneling, delocalization) can be also taking into account for the nuclei which results in some further changes in energy.
Altogether, the more and more quantum effects are taking into account, the closer are the results of the simulations to the physical reality. AIMD simulations ignore few quantum effects, and thus, do not quite match the reality. For instance, water tends to be overbound in such simulations comparing to what we have in the real world. That is, essentially, the meaning of the statement quoted above.

1) A.D. Hammerich, V. Buch, F. Mohamed, Ab initio simulations of sulfuric acid solutions, Chemical Physics Letters, 2008, 460, 423-431, http://dx.doi.org/10.1016/j.cplett.2008.06.053.


text:
Which of the following pairs show common ion effect?

Which of the following pairs show common ion effect?

$\ce{BaCl2}$,$\ce{Ba(NO3)2}$
$\ce{NaCl}$, $\ce{HCl}$
$\ce{NH4OH}$,$\ce{NH4Cl}$
$\ce{NaOH}$, $\ce{NaCl}$

Its not (1) because they are both solids.
The remaining are solute solvent pairs. So all the other pairs will show common ion effect if the solute is soluble to some extent and exists in an equilibrium with the solvent. The answer given is (3). But why cant (2) and (4) show common ion effect?

It might be related to the fact that HCl and NaOH dissociate almost completely and there's no equilibrium in their dissociation reactions to shift.

I don't understand why you rule out (1). Why can't you have aq. BaCl2 and aq. Ba(NO3)2? If you say that both of them are solids, then you should also rule out (4), because both NaOH and NaCl exist as solids as well. Furthermore you should rule out (2) because HCl is gaseous and covalently bonded so there aren't even any ions to speak of. Apart from that, @IͶΔ has the right answer.

Note that the fictitious solid compound “ammonium hydroxide” $(\ce{NH4OH})$ does not exist.


text:
hydroxide ion in water -- How does the dynamic change?
It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions. Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion. If I am correct at this point, how would the dynamics be different in these cases?

It certainly wouldn't be $\ce{H_3O_2^{3-}}$. If anything, then $\ce{H_3O_2^{-}}$.

It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions.

This is not quite right. Because of the hydrogen bonding a wide variety of $\ce{H3O+(H2O)$_n$}$ ions are in dynamic equilibrium, where $n$ can take values as large as few tens. It has been shown that $\ce{H3O+}$ can donate three hydrogen bonds, so that it's first solvation shell of 3 water molecules, and the hydrogen bonds between $\ce{H3O+}$ and $\ce{H2O}$ are quite strong. This means that the $\ce{H3O+}$ cation can actually be considered as $\ce{H9O4+}$ in solution. For more details, see here.

Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion.

Reasoning by analogy is a very dangerous road. As mentioned by sheogorath there is no possibility for a coordinate covalent bond as in the case of $\ce{H3O+}$. But the hydrogen bonding is possible and the hydroxide ion is indeed strongly hydrated as well as $\ce{H3O+}$. The details of the hydration are not well known yet, but recent studies indicate $\ce{OH- (H2O)4}$ as the prominent complex in an aqueous solution. For details, see here.

$\ce{H3O}$+ is formed by
$\ce{H2O  +  H+  = H3O+}$             where $\ce{O}$ share $2$ electrons to $\ce{H+}$ with no electron to form coordinate covalent bond .There is no possibility for oxygen to form such bond with $\ce{H2O}$ and $\ce{OH-}$


text:
Trends in the covalency of bonding across the transition metals
I am under the impression that the covalent character in the bonding of transition metal compounds increases across a group; in fact I think this is why they're called transition metals (a transition from purely ionic compound shown by the pre transition metals to covalent bonding in the non-metals). I think d orbital splitting and ligand field effects become more significant across the series (please confirm or deny) which would support this assertion. Is the reason for this that the d orbitals decrease in energy across the series, as a result of increasing effective nuclear charge, which results in a better energy match with ligand orbitals?
However, could one not argue that the covalency should decrease across the series as the d orbitals become more contracted and orbital overlap with ligands is less efficient. One trend that would seem to support this view is that the metal carbon bonds in organometallic complexes with alkyl and aryl ligands tend to decrease in strength across the series.

Seeking simple trends in the periodic table is a generally hopeless (and useless) task. Certainly, either end of the transition metal series "seems" easier to understand than the middle, but even that is not guaranteed.


text:
What do different dissociation constants mean?
What does first, second, third etc. Acid dissociation constants mean? They are written as $pK_{a_{1}}, pK_{a_2}$ etc. If any one would explain it that would be good.

Acid dissociation constants in general are the equilibrium constants that describe acid dissociation reactions. So for:
$\ce{HA + H2O <-> H3O+ + A-}$
the equilibrium constant would be
$K_a = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}$
This is enough for monoprotic acids, compounds that can donate a single proton.  However, some acids, like the common inorganic acids $\ce{H2SO4}, \ce{H3PO4},$ and many, many organic acids can donate more than one proton.  They are called polyprotic.  The issue with their protons is that they are not all donated at the same time, or under the same conditions.
So what we do not see is:
$\ce{H2A + 2H2O <-> 2H3O+ + A^{2-}}$
with an equilibrium constant of
$K_a = \frac{[\ce{H3O+}]^2[\ce{A2-}]}{[\ce{H2A}]}$
Instead there are two (or more, if there are more protons to be donated) equilibria, each of which has its own equilibrium constant.
Dissociation 1:
$\ce{H2A + H2O <-> H3O+ + HA^{-}}$
$K_{a1} = \frac{[\ce{H3O+}][\ce{HA-}]}{[\ce{H2A}]}$
and
Dissociation 2:
$\ce{HA- + H2O <-> H3O+ + A^{2-}}$
$K_{a2} = \frac{[\ce{H3O+}][\ce{A2-}]}{[\ce{HA-}]}$
Each of these acid dissociation constants can be acted on by the p operator such that:
$pK_{a1} = -log_{10}(K_{a1})$
and
$pK_{a2} = -log_{10}(K_{a2})$
Note that there are polyprotic bases as well, substances that can accept more than one proton, and they are described using $K_{b1}, K_{b2}$, etc, where $K_b$ is the equilibrium constant for the base reacting with water to form hydroxide and a conjugate acid.

The first, second, third, etc... acid dissociation constants refer to the equilibrium constant for loss of the first, second, third, and so on proton. For example, $H _2SO_4$ can lose one proton to make HSO4–,  which can then lose another proton to generate SO42–. Ka1 and Ka2 would be the equilibrium constants for these reactions. pKa1 and pKa2 are, of course, -ln(Ka1) and -ln(Ka2).


text:
Should I round an experimentally determined reaction order to an integer?
For a lab experiment I did, I'm calculating the reaction order for one of the reactants using data from two trials, both rates and the concentration of that reactant during each.
$$
\frac{\text{Rate}_1}{\text{Rate}_2} = \left(\frac{[A]_1}{[A]_2}\right)^x
$$
The number I got from that for the reaction order for reactant a was 0.967. DO I put that exactly as the reaction order, or just round and say it's a first order reaction, and the little bit of different is just a result of the number I get from my calculations not being perfect?
I'm also trying to calculate $K_\mathrm{eq}$, and need to know if I should use $[A]^1$ or $[A]^{0.967}$ in the calculation of $K_\mathrm{eq}$.

Real-world numbers are never perfect. Consider it 1.

Alright, thanks. Wanted to be absolutely sure.

It's probably 1. You can do propagation of uncertainty to make sure that your number is equivalent to 1 if you like.


text:
Oxidation state of nitrogen in HN3
The book asks me to find the oxidation state of nitrogen in the compound below (the structure is not the one given in Wikipedia for HN3):

To find the oxidation state using Lewis structure, I have to compare the electronegativity of the bonded atoms and assign the bonding electrons to the more electronegative atom. Then, the difference between the number of electrons in element state and bonded state will give the oxidation number.
Here, the bonding electrons of nitrogens bonded with a double bond will stay in between both the atoms will not be assigned to both. The $\ce{N-H}$ bonding electrons will be assigned to $\ce{N}$. What about the two $\ce{N-N}$ bonding electrons?

Throw away the book. This structure is hideously wrong. And anyway, you may assign these oxidation states more or less arbitrarily (just make sure they sum up to 0), so it does not really matter.

To be fair he doesn't say whether the book calls this hydrazoic acid, or if it's what he is calling this. Greenwood and Earnshaw does (very briefly) mention the above so I assume it is known somehow.

The given structure belongs to triazirine, not hydrazoic acid.

Ok, how is there triazirene?  It's antiaromatic if planar and highly strained.

@OscarLanzi I deleted the comment. The compounds are unstable and may have decomposed just by sitting in the comment section for 5 years. (Sorry for the terrible joke).

This answer uses electronegativities for the calculation of oxidation states as proposed in the Expanded Definition of the Oxidation State by Hans-Peter Loock in 2011.
Comparing Electronegativities
The following table shows an excerpt from Pauling electronegativities ($\chi_{\mathrm{Pauling}}$):
$$\begin{array}{cc}
\hline
\text{Element} & \chi_{\mathrm{Pauling}}\\
\hline
\ce{H} & 2.20\\
\ce{N} & 3.04\\
\hline
\end{array}$$
Looking at $\ce{N-H}$ Bond (heteroatomic)
The table shows, that the nitrogen atom is more electronegative than the hydrogen atom. It will therefore attract the electrons more than hydrogen does, and thus polarize the bond. Imagining the gedankenexperiment of an hypothetical heterolytical bond cleavage - presumed by applying pulling force to the atoms, until they get separated - it is assumed, that the electrons are allocated to the nitrogen atom during separation.

Looking at $\ce{N-N}$ and $\ce{N=N}$ Bonds (homoatomic)
Looking again at the previous gedankenexperiment, now with two identical Atoms bound to each other. Since they have, of course, the same electronegativity, there is no polarization of the bond. It is assumed, that separation would therefore result in an homolytical bond cleavage, allocating always one electron to each of the two nitrogen atoms for each bond respectively.

Summing up
Now, having a look on the complete structure of the molecule in question, and applying the previously states rules:

Last thing to do is calculating the atoms hypothetical charge after separation, which is to be equatable with the oxidation state:
$$ \text{Oxidation state} = N_\mathrm{i}(\ce{e-}) - N_\mathrm{f}(\ce{e-}) $$
With $ N_\mathrm{i}(\ce{e-}) $ representating the number of electrons in a free atom, and $ N_\mathrm{f}(\ce{e-}) $ the one after separation (One should not forget the lone pairs).
One will end up with the following oxidation states for the different (nitrogen) atoms, with the last entry representating the mean oxidation state of all nitrogen atoms:
$$\begin{array}{cccc}
\hline
\text{Atom} & N_\mathrm{i}(\ce{e-}) & N_\mathrm{f}(\ce{e-}) &\text{Oxidation state}\\
\hline
\ce{H} & 1 & 0 & +1\\
\hline
\ce{N-1} & 5 & 5 & 0\\
\ce{N-2} & 5 & 5 & 0\\
\ce{N-3} & 5 & 6 & -1\\
\hline
\ce{N} & - & - & -\frac{1}{3}\\
\hline
\end{array}$$
And the structure with assigned oxidation states:


text:
What's the relationship between chemical polarity and vapor pressure？
A question which confuses me now.
Chemical polarity is one of microscopic properties for certain molecules.
Vapor pressure is a macro thermodynamic properties for certain materials.
But I often read about some violate organics are polar material. So, it has a  relatively small vapor pressure and  hard to evaporate.
How to connect these two concepts together? If I can access the volatile ability based on the polarity, it will help me to recognize some kind of VOCs which are eager to condense into particle phase.

The vapor pressure of a liquid is the equilibrium pressure of vapor above the liquid at a given temperature. Liquids with higher vapor pressures must have less strong forces holding the molecules together in the liquid phase, thus more of the molecules are able to escape the liquid and be in the vapor phase. This is the connection between inter-molecular forces and vapor pressure.
Polar molecules can attract each other through dipole-dipole and dipole-induced dipole interactions in addition to the induced dipole-induced dipole intermolecular interactions that all molecules experience. Thus, all else (molar mass, for example) equal, a polar molecule will generally have a lower vapor pressure than a non-polar molecule.


text:
Determining Enthalpy Change of the Combustion of Ethanol
A worked example went something like this:

Use the following experimental data to determine the enthalpy change when $\pu{1 mol}$ of ethanol ($\ce{C2H5OH}$) is burnt:
Mass of water = $\pu{150.00 g}$
Initial temperature of water = $\pu{19.5 °C}$
Maximum temperature of water = $\pu{45.7 °C}$
Initial mass of spirit burner = $\pu{121.67 g}$
Final mass of spirit burner = $\pu{120.62 g}$

Being a worked example, I saw the worked solution and understood the process completely. What's bothering me is a fundamental flaw in the procedure that has seemed to be neglected completely.
We are relying on the idea that the combustion of ethanol will produce some energy, the amount of which can be calculated with the increase in temperature of the beaker of water on top. However, if there was no ethanol, no combustion, the flame giving rise to the combustion would still manage to heat the water up; in this sense, we do not know whether it is the flame supplying the energy to the rise in water, or the exothermic reaction that takes place with ethanol. How does the existence of such a flaw make the procedure useful to any extent? Is there anything I'm missing?

I see your concern, and it's absolutely valid. Any time you perform a calorimetry experiment, you need to be certain that you're not introducing extra energy into the system. Typically, the spark used to ignite the sample has a negligible energy input, but it is still there. However, any heat released by the flame (other than the initial energy put in to generate the spark) is generated by the combustion process. Without the combustion, the flame would cease to exist. If you were running this experiment, you could use an electric sparker and measure the power input for the sparker to know how much energy was inputted. Another option would be to run a control experiment with just an ignition (but no ignitable material) and see how much, if any, the water heats up.
Of course, if you used a lighter to start the flame, and then held the lighter open throughout the experiment, you would be introducing extra energy from the combustion of lighter fluid, and this would give you false results as you feared. That's why we don't run calorimetry experiments in that way.

Change in heat=mass of water × specific heat capacity × change in temperature
But; mass of water=150•00g
Specific heat capacity=4•2
Change in temperature= final temp - initial temp
=45•7 - 19•5=26•2
Change in heat=150×4•2×26•2
=16758
But mass of ethanol burnt = initial mass of spirit bottle - final mass of spirit bottle
=121•67-120•62
=1•05
Relative molecular mass of ethanol(c²H⁵OH)=((12×2)+(5×1)+(16×1)+(1×1))
=46
Moles of ethanol burnt=mass burnt÷relative molecular mass
=1•05÷46
=0.022826087
0.022826087moles of ethanol produces 16758 joules
I mole= (16758÷0.022826087) ×1
=734,159.9986016
•°•, 1 mole produces approximately 734,160 joules per mole
734,160 joules per mole =734,160 joules per mole÷1000=734•160 Kilojoule per mole
C²H⁵OH + O²-> 2CO² + 3H²O; Change in heat = 734•160 KJ/mole

For better site experience, you may find useful these links:
notation ,  formulas/expressions ( not to be applied to titles ) and
upright vs italic

The present formatting is both outdated (e.g., the interpunct is somewhat typical for publications in English (e.g., from Britain) around the 1960 and earlier) and difficult to read.  It equally is better to carry the units from the beginning throughout the computation till the end (e.g., for dimensional analysis).


text:
Rate of Fenton Reaction
I want to determine rate of Fenton reaction with different concentration of $\ce{Fe^2+}$. These are some questions that I have encountered:

How long does it usually take for the reaction to complete?

What concentration of $\ce{Fe^2+}$ is usually used for the reaction?

I am planning to use titration of $\ce{H2O2}$ with $\ce{KMnO4}$ to determine the rate of reaction. Is it possible to quench to reaction mixture by adding an excess of EDTA solution and would it affect the titration?

NEW:
How can I determine the rate of Fenton reaction? I proposed to titrate $\ce{H2O2}$ with $\ce{KMnO4}$ to determine change in $\ce{H2O2}$ as a mean of determining rate of reaction. However, I realised that $\ce{KMnO4}$ would also react with $\ce{Fe2+}$, so I need to either find another way to measure the reaction rate or find a way to remove all $\ce{Fe2+}$. Also, I am not sure if there would be any reaction between $\ce{KMnO4}$ and the hydroxyl radicals.
I am also wondering about adding in some wastewater and determine the before and after COD to determine the rate of reaction. However, I am wondering if the water that don't really have high COD (as opposed to the industrial wastes used in other papers) would result in a very slow reaction rate. Also, it seems like the part involving oxidizing by hydroxyl radicals is very exothermic. If I don't want to add $\ce{H2O2}$ bit by bit to control rate of reaction, is it OK?
Do I have to add acid to $\ce{Fe2+}$ and $\ce{H2O2}$ for the reaction to occur?
Is it necessary to add $\ce{H2O2}$ bit by bit, if so how much should I add at each turn?

I would check this out: http://pubs.acs.org/doi/abs/10.1021/es00026a011

This is only a partial answer:
I am not sure about which step you're looking for or the substrate of your reaction, but the Fenton-based chemistry I work with is extremely fast. I use it to (supposedly) form a methyl radical from DMSO which then reacts with an α-iodoester to form another radical.
Literature suggests I add brine immediately after I've completed the hydrogen peroxide addition and work it up from there. Truth be told, I usually give it a few minutes, but the reaction is certainly quite fast. The whole process takes me less than 5 minutes, and if I wasn't using DMSO and skatole (so smelly), it would be my favorite reaction of the ones I do.
I use iron(II) sulfate heptahydrate as my catalyst, and I use 0.5 equiv. relative to the iodide. It ends up being about 0.1 g per milliliter of solvent.
As for the peroxide titration, I use a decent excess, so I've never needed to titrate it. That is why I said this was a partial answer. I will say this: don't quench large amounts of hydrogen peroxide with sodium thiosulfate at room temperature. You end up forming sulfuric acid in aqueous solution. Very exothermic (like the water in the flask can start boiling in your hand).

For quenching part, I suggest ethanol, cheap and really effective. I always used Fe2SO4.7H2O and 30%~ peroxide and a couple ml of ethanol generally helped to cease reaction less than one minute, though after vigorous part ended.


text:
Why can cloud's bottoms be flat?
I've always wondered why clouds can have flat bottoms, and have debated to myself whether it was a difference in atmospheric pressures, or some water molecule tendency? Could someone shed light?

Curiously, one might argue that clouds do not have flat bottoms, they're just invisible below a certain height! Here's a fun and very accessible video from Minute Earth which brings it up.

@NicolauSakerNeto Wow! Thanks that was really cool.

That many types of clouds have flat bottoms is a function of decreasing air temperature with increasing height, and the water vapor pressure, commonly in this context called the dew point, of the air. When some force causes a non-cloudy parcel of air to rise, moisture will condense out as cloud particles just as soon as the air temperature reaches the dew point of the air mass.  The colder air (higher up) then contains a cloud, where the warmer air (even just slightly above the dew point) below the cloud remains clear.


text:
Instability of gem dihydroxy compounds
Organic compounds containing two hydroxy groups on one carbon (gem dihydroxy) are unstable and tend to lose water molecule. However, Boron, sulfur and phosphorus with two OH groups are stable. What is the reason?

Chloral hydrate is fairly stable.

yes. Its stability is suspected  due to hydrogen bonding.

Hydrogen bonds have nothing do do with this.

The instability of gem-diols is due the fact that hydogen bonding between the 2 OH- groups increases the repulsion causing a strain on the C-O-H bonds.
In case of compounds of boron, a greater bond angle (due to sp2 hybridization of boron) prevents intramolecular hydrogen bonding.
Compounds of sulphur, phosphorous always contain a double bonded oxygen along with the OH- groups on the central atom. This oxygen atom forms hydrogen bonds with OH- groups preventing hydrogen bonding between the OH- groups, thus preventing any strain.

Hydrogen bonding should increase the overall stability of the molecule. Do you think, H-bonding makes them to first come closer and then trigger repulsion that eventually leads to instability?

Thats exactly what happens in gem-diols on sp3 hybridized atom. And, you were right about the stability of chloral hydrate. The OH- groups form H-bonds with chlorine (not exactly H-bonds since they are formed with chlorine atoms) preventing intramolecular H-bonding between OH- groups (which is the main reason for instability of other gem-diols)

You've got to be kidding me. Instability because of hydrogen bonds... Do you even know what you're talking about?

Only in case of gem-diols. The OH- groups bend towards each other due to H-bonding (unless it can be prevented, as in the examples above). Firstly, it causes a strain on the carbon atom because of decrease in bond angle. Also, you can see that the OH- are bent towards each due to H-bonding which results repulsions among the atoms. So preferably, the oxygen donates a lone pair to the hydrogen of other OH- group and leaves as a water molecule leaving behind a carbonyl group.

I do not think this is a valid explanation. Bending and strain in the bond angle......? Misleading...........

Strain is not the main reason (forget that I even said anything about bond angle). It is well known that attraction causes decrease in energy (in other words increases stability) and repulsion causes increase in energy (decreases stability). It is due to this fact that H-bonding is usually associated with stability. However if the attractions cause even greater repulsions, (as in some gem-diols), it results in instability. Thus elimination of water molecule is preferred.


text:
What is a 'corona phase'?
What is a 'corona phase'? I have come across this term in several papers, including this one
but all I have found concerning its definition is:

[...] synthetic heteropolymers, once constrained onto a single-walled carbon nanotube by chemical adsorption, also form a new corona phase that exhibits highly selective recognition for specific molecules...

What is the formal definition of 'corona phase'?

One of the authors has recently established a research group at my university, so I spoke to a member of their lab:
The corona phase is not exactly a phase in the thermodynamic sense, but rather the portion of the polymer in contact with the nanotube. In the below image (taken from the Nature Nanotechnology paper in question), the blue portion and the parts of the red portion in contact with the tube would be the corona phase.

A phase is region where the the chemical and physical properties of a material are uniform in a certain sense.
The corona phase is the phase formed when amphiphilic polymers are being bound to the surface of carbon nanotubes.

I understand what a phase is. But my question is whether the polymer-CNT interface is the definition of a corona phase or simply an example of a coronoa phase. If it is the latter, can you provide another example of a corona phase without a CNT-polymer interface?


text:
Why does Cl- corrode aluminum containers?
I heard that $\ce{Cl^-}$ will corrode aluminum containers. Why? Is it because it would form a galvanic cell or produce coordination compounds?

Related: http://chemistry.stackexchange.com/questions/22133/why-did-the-salt-on-the-skin-of-my-turkey-react-with-the-aluminum-foil-it-was-to

In general, it should not. Under the right conditions, i.e. acidic conditions at high temperature on a freshly exposed (e.g. scratched) $\ce{Al}$ surface, you could get some oxidation to $\ce{AlCl3}$.  One industrial process for producing $\ce{AlCl3}$ is via the reaction:
$$\ce{2 Al + 6 HCl → 2 AlCl3 + 3 H2}$$
at $\mathrm{600-800^oC}$. It would of course happen much more slowly, if at all, at lower temperatures and lower acidity.
The necessity of a freshly scratched surface is to expose the aluminum metal itself, as aluminum pretty rapidly forms a thin, inert oxide layer upon exposure to air or water.
In other words, in the absence of this perfect storm of conditions, aluminum cans should be very resistant to corrosion by $\ce{NaCl}$.


text:
Relation between Gibbs energy and net reaction rate
I have come across some papers in biochemistry that seem to claim that a reaction can be "near equilibrium" even though the flux through the reaction (net reaction rate) is large. For example, this paper measures absolute metabolite concentrations in E.Coli cells grown on two carbon sources, performs some thermodynamic calculations, and concludes that "lower glycolysis is near equilibrium on both carbon sources, with $\Delta G$ approximately 0". Yet this pathway carries net flux in these conditions according to the authors, and usually glycolytic flux is rather large.
How can this be? In my understanding, the net rate of a reaction depends on the Gibbs energy $\Delta G$ so that, near equilibrium where $\Delta G \approx 0$, there is no net flux through the reaction. And I think this dependence should be smooth, so that $\Delta G$ must be large when the net rate is large?
The authors suggest that reaction rate depends steeply on $\Delta G$, so that "Small changes in $Q$ are accordingly adequate to tip the thermodynamically favored flux direction." (From the supplementary it looks like the reaction quotient $Q$ varies between 0.5 and 6.) I have also encountered this idea in works on thermodynamic flux analysis, where only the sign of $\Delta G$ is considered important, its magnitude $|\Delta G|$ is ignored. But if this is true, the reaction direction seems unpredictable and the system should be difficult to control?
Is this all just a question of what "small" or "large" $\Delta G$ means? Can the dependence on $\Delta G$ be sharp, so that the sign of $\Delta G$ acts like a "switch"? Can a reaction really be considered "near equilibrium" if the net reaction rate is large?

Thank you for putting together an excellent, well-specified, thoroughly documented question.  The answer, as you suspected, largely comes down to what "small" and "large" mean.

I have come across some papers in biochemistry that seem to claim that a reaction can be "near equilibrium" even though the flux through the reaction (net reaction rate) is large[....] How can this be?

Consider the equilibrium approximation of Michaelis-Menten kinetics:
$$  \ce{E + S <=>[k_f][k_r] E\cdot S ->[k_3] E + P}$$
When $k_r \gg k_3$, the enzyme-substrate complex $\ce{E \cdot S}$ is assumed to be in equilibrium with $\ce{E + S}$, i.e. free enzyme and free substrate.  Making this equilibrium assumption eventually provides the rate law for overall flux through the reaction:
$$\frac{dP}{dt} = k_3 E_0\frac{S}{S+\frac{k_r}{k_f}}$$
Note that the flux is not always zero even though we have assumed that $\ce{E\cdot S}$ is in equilibrium with $\ce{E + S}$!  This is one hint -- hopefully a familiar example if you have studied biochemistry -- that assuming (quasi-) equilibrium doesn't mean zero fluxes.

In my understanding, the net rate of a reaction depends on the Gibbs energy $\Delta G$ so that, near equilibrium where $\Delta G \approx 0$, there is no net flux through the reaction.

This isn't quite right.  First, you are right that the net rate of a reaction depends on the Gibbs energy $\Delta G$.  But near equilibrium where $\Delta G \approx 0$, it would be better to say "there is little net flux through the reaction".  Then the question is little relative to what?  The answer is that "little" is relative to the exchange flux.  So mathematically,
$$\nu_{net} = \nu_f - \nu_r = k_f(E)( S) - k_r(E\cdot S) $$
$$\nu_{exchange} = \nu_r = k_r (E \cdot S) $$
Reactions that are "approximately" at equilibrium have $\nu_{exchange} \gg \nu_{net}$.  Thus, "little" is relative to the exchange flux, and not necessarily the rates of other reactions in a metabolic network.

... "Small changes in $Q$ are accordingly adequate to tip the thermodynamically favored flux direction."

Remember that $\Delta G = \Delta G^\circ + RT \ln Q$.  The question of whether a reaction is reversible is asking about the sign of $\Delta G$.  The authors' statement that small changes in $Q$ can change the sign of $\Delta G$ only make sense if $\Delta G^\circ \approx 0$.  From the context of their remark, I think they are aware of this.

But if this is true, the reaction direction seems unpredictable and the system should be difficult to control?

I wouldn't say "unpredictable" but I think you're right to say "difficult to control".  Another way of saying it would be that those reactions in the metabolic network that are approximately at equilbrium are not likely to be control points.  This 2006 paper examines that topic in more detail.

Is this all just a question of what "small" or "large" $\Delta G$ means?

Yes!

Can a reaction really be considered "near equilibrium" if the net reaction rate is large?

It depends what you mean by "large".

If you mean "large" relative to some other flux in the network, then yes, a reaction can be near equilibrium even though its flux is "large".
If you mean "large" relative to the exchange/equilibration flux, then no, a reaction near equilibrium has a net flux that is very small compared to the exchange flux.

Thank you for a great, detailed answer!  Considering the net flux relative to exchange flux really seems to be the key here.


text:
which starting material on heating with concentrated sulfuric acid generate cyclic ester

I have just learned esterification of normal reactions with aliphatic compounds, but this one is a cyclic compound. Is it different? There are both a hydroxyl group and a carboxyl group in choice B and D, but why the correct answer is B, not D?

The correct answer is glycolic acid, B.
It forms the cyclic di-ester glycolide (1,4-dioxane-2,5-dione).


text:
Changes of Internal Energy un open system
My professor says the internal energy of a system can be calculated as $U=n T c_v$. Now I always thought that we could only calculate a change in internal energy $\Delta U = n c_v \Delta T$.
Following this idea, he says that a recipient that contains gas and loses a certain number of moles at constant temperautre undergoes a change in internal energy of $\Delta U = T c_v \Delta n$.
I've never seen any of this in any book. I believe he is assuming that the internal energy at 0 Kelvin is zero, which I believe is not true. However, in the case of the recipient losing gas,I wouldn't know how to calculate $\Delta U$.
Any thoughts? thanks!

Suppose you assign an reference state for internal energy such that the internal energy per mole u of the gas is zero at $T=T_r$.  This reference state is going to be held fixed for the material both entering or leaving the open system, as well as the material within the system at any time.  Then, in the initial state of a constant-temperature system, $U_{init}=n_{init}C_v(T-T_r)$ and, in the final state, $U_{final}=n_{final}C_v(T-T_r)$.  So, $$\Delta U=\Delta nC_v(T-T_r)$$Typically, in the overall energy balance on the system, including entering and exiting streams, $T_r$ will cancel out (just as it would for a closed system).
Please note, however, that if chemical reactions are occurring, the equation for   the internal energy and the specification of the reference state has to be extended to take this into account (using the concept of "internal energy of formation").

I understand, but if there is only a exit, no entrance of gas in the recipient, still $\Delta U$ depends on $T_r$, which we don't know, so we can't calculate $\Delta U$.

No.  Not if you include the exit stream in the balance.  You have been learning about the version of the 1st law applicable to open systems, correct?  That version of the first law includes the enthalpy of the exit stream.

Yes I know the balance for an open system. Anyways I would have to calculate the point enthalpy of the leaving stream, there is no entering stream to get a $\Delta h$ term

That's not a problem.  You just use the amount that has left times its h.  Doesn't your textbook cover situations like that?  The amount entering does not have to match the amount leaving, except at steady state.

I can't find it on my book nor the internet. Could you explain me?

Neglecting kinetic and potential energy effects, $$\frac{dU}{dt}=\dot{Q}-\dot{W}s+\dot{m}{in}h_{in}-\dot{m}{out}h{out}$$where $\dot{W}_s$ is the rate of doing shaft work.  I'm sure you've seen this in some form before.

Here is one of many links:  http://www.learnengineering.org/2013/03/frist-law-of-thermodynamics-open-system.html

Yes I know the balance for an open system, but I don't know how to apply it in the specific case I mentioned above in which a gas exits the recipent. No entrances, only a gas exiting at constant temperature. That would leave:
$$\frac{dU}{dt} = \dot Q - \dot m_{out} h_{out}$$.
But then how do you calculate $h_{out}$?

The value of $h{out}$ depends on what is being done.   If the tank is adiabatic, Q dot is zero, and, if  the gas exits through a valve, the value of $h{out}$ is the same as h in the tank (no enthalpy change across the valve) at the moment that the differential mass is leaving.  If the tank is somehow held isothermal, Q dot is not zero, but u and h in the tank don't change, and $h{out}$ is equal to the (constant) h in the tank throughout the change.

Sorry I truly do not understand where you are going. My question is simple, is it correct to calculate te internal energy as $U = n c_v T$? I mean the value of the Internal Energy in a particular state.

I would not calculate it that way.  I would always use a reference temperature state for the zero in U.  Furthermore, this equation would be quite inappropriate if the heat capacity were temperature dependent, such that $U(T)=\int_{T_r}^T{C_v(T')dT'}$.  I've already pointed out that using a reference temperature for U will not pose any problems in applying the 1st law because it will cancel out.  Sorry, but I don't have any ideas on how to express it any better.

Well I totally agree with you! I also believe that considering that $U$ is zero at 0 Kelvin is wrong, that we can only calculate a change in $U$ when temperature changes. But my professor uses it all the time! For example, when he calculates the $\Delta U$ of a recipent that loses gas at contstant temperature: $\Delta U = T c_v \Delta n$.

What can I say.  In some cases, such as for an ideal gas with constant heat capacity, he can get away with doing this.  But it is a bad way to teach people this stuff, and will become a problem for them in other cases when it will actually be incorrect.  I would never teach it this way.  I totally agree with you too.

Well thanks for the answer! This is actually the second time I do a course in thermodynamics. Unbelievably, the one I'm doing now is far worse than the previous one. Sorry for the misunderstanding. How would you go about calculating the $\Delta U$ in the case I stated before, in which a recipent loses $\Delta n$ moles of gas at constant $T$?

I gave that in my original answer for the case of constant Cv and constant temperature.  If the Cv were not constant but the temperature were, then $\Delta U=\Delta n\int_{T_r}^T{C_v(T')dT'}$.  If the temperature were also changing, then $\Delta U=n_fu(T_f)-n_iu(T_i)$, where $u(T)=\int_{T_r}^T{C_v(T')dT'}$.

I see, Thanks Chester! You're awesome! Sorry for insisting, I think this professor will give me a hard time during this course. Thanks a lot and hope you have a great day!

It would be preferable instead of discussing the answer in the comments section, to include improving statements into it and clear the comment section accordingly. This is beneficial for later visitors.


text:
Why does CaCO3 react with HCl, but not with H2SO4?
I have a wonderful reaction of marble chips, $\ce{CaCO3}$, with hydrochloric acid, $\ce{HCl}$, and carbon dioxide was released beautifully (fast, large volume, easy to measure and makes good visual effect too). But there is no reaction between $\ce{CaCO3}$ and $\ce{H2SO4}$. Why not?

Have a look at these picures

Who said there is no reaction between $\ce{CaCO3}$ and $\ce{H2SO4}$? - http://chemiday.com/search/?q=caco3+%2B+h2so4%3D&lang=en&m=stuff

I actually observed a reaction for maybe 2-3 seconds both times I tried to do it. So I take it back that there is no reaction. It just stops extremely soon when protective layer of CaSO4 is formed.

So, the reaction is theoretically possible but practically impossible.....

Your marble chips react on the surface.
In the case of hydrochloric acid, the resulting salt, calcium chloride, is highly soluble in the acid, dissolves and provides further attack to the (new) surface.
With sulfuric acid, the highly insoluble calcium sulfate is formed on the surface of the marble chip. With other words:
Calcium sulfate acts like a protective layer.

$\ce{CaCO3}$ reacts with $\ce{H2SO4}$ but he reaction does not go further due to the formation of the layer of $\ce{CaSO4}$ on the surface of $\ce{CaCO3}$. If we scratch the surface the reaction will proceed in the forward direction.


text:
Does the stoichiometry effect the value of the equilibrium constant of an equilibrium reaction?
Will be there any change in the $K_c$ values for same reaction, if we write stoichiometry as a fraction or as a integer? For example, assuming that we calculate $K_c$ at same temperature:
$$\begin{align}
&(1)  &&\ce{1/2A + 3/2B -> C}\\
&(2)  &&\ce{A + 3B -> 2C}\\
\end{align}$$
If we calculate $K_c$ for above two equilibrium reactions, do we get different answers?

$$K_c = \mathrm{products \over reactants}$$ So this is done by multiplying the concentrations of the products and dividing by the concentrations of the reactants. The concentrations are put to the power of their coefficient. So in your problems at equilibrium $K_c$ would be:
\begin{align}
  K_c &= \frac{[\ce{C}]^1}{[\ce{A}]^{\frac{1}{2}}[\ce{B}]^{\frac{3}{2}}}\tag1\\
  K_c &= \frac{[\ce{C}]^2}{[\ce{A}]^1[\ce{B}]^3}\tag2\\
\end{align}
Note that $[\ce{A}]$ means concentration of $\ce{A}$.

$K_c$ is a constant for a reaction at a given temperature, and these reactions ar equivalent, so $K_c$ is still constant.  However if reaction (1) were at equilibrium and the concentration of all the reactants doubled (e.g. the pressure doubles or volume halves) the reaction will no longer be in equilibrium.  The expression (2) with values multiplied by 2 will no longer equal $K_c$ and the equilibrium position will shift so that the value of the expression equals the equilibrium constant.

What perhaps needs to be added to this answer is that the law of mass action is defined for an equation int which the substances are in their lowest whol number ratios of stoichiometric coefficients.


text:
How to calculate an enthalpy change for this reaction
I've got an assignment to solve problem number 4 in this worksheet, that is to calculate change in enthalpy for reaction:
$$\ce{CHCl3 + O -> COCl2 + HCl}$$
How do I solve it? Why is oxygen $\ce{O}$, and not $\ce{O2}$?
I figured out that, it is not a combustion reaction, as combustion of any hydrocarbon will lead to formation of $\ce{CO2}$ and $\ce{H2O}$, as per this page.
Well, what I have at the beginning is not exactly a hydrocarbon, but rather, a halogenated hydrocarbon. Yet, no water and no carbon dioxide is produced. So as I mentioned above, I consider it is not a combustion. Any thoughts on this? Could it still be a combustion? Why or why not?
When I calculate $\Delta H$ of this reaction, the total outcome depends on using reaction as it is shown in the worksheet, or correcting $\ce{O}$ to $\ce{O2}$ and balancing it. Which option, do you think, is preferable, and why?

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chem.se""="), this page and this one on how to format your future posts better.

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Why should be O$_2$ ? Combustion reactions are vaguely defined, but many times refers to reactions with $O_2$, I would not call this a combustion reaction but as definition is not accurate I can give strong arguments. It could be called combustion if it a fast and very exergonic reaction (I don't know this case) but it is no common to call this way reaction with radicals. I can not view the worksheet.

To preface, the reason that this is not a combustion reaction is because this is a redox reaction that occurs at room temperature. In fact, chloroform is noncombustible.
Though at first glance, this may simply appear to be an unbalanced equation that could be rewritten:
$$\ce{CHCl3 + 1/2O2 -> COCl2 + HCl}$$
It is better to take it for what the question presents it as, an oxygen atom, so I would therefore recommend you to leave the equation as:
$$\ce{CHCl3 + O -> COCl2 + HCl}$$
Though $\ce{O2}$ is a pure element, and its $\Delta H^0_\mathrm f=0\:\mathrm{kJ\:mol^{-1}}$, for the atom $\ce{O}$, $\Delta H^0_\mathrm f=58.99\:\mathrm{kJ\:mol^{-1}}^{[1]}$ (provided thanks to user1420303).
To calculate $\Delta H$, the equation is:  $$\Delta H=\Delta H^0_\mathrm f(\ce{COCl2}) + \Delta H^0_\mathrm f(\ce{HCl})-\Delta H^0_\mathrm f(\ce{CHCl3})-\Delta H^0_\mathrm f(\ce{O})$$
Though enthalpy is a state function, this question is asking you to calculate $\Delta H$ from a specific set of reactants, and their individual energies must be considered. There is more than just one way to calculate $\Delta H$, and an analysis of the $\Delta H^0_\mathrm f$ (the enthalpy of formation of a compound from its constituent elements) for each compound accounts for which bonds are being broken and formed, and will yield (about) the same results as a bond by bond analysis. I find this method to be preferable because it bond by bond analyses are based on average bond enthalpies, while $\Delta H^0_\mathrm f$ is often painstakingly measured for each compound, and will yield the most accurate $\Delta H$ calculation.

[1] Curtiss, L. A.; Raghavachari, K.; Redfern, P. C.; Pople, J. A. Assessment Of Gaussian-2 and Density Functional Theories for the Computation of Enthalpies of Formation. The Journal of Chemical Physics J. Chem. Phys. 1997, 106, 1063.

delta H of formation and bond energy are two different things. One still needs to invest energy to break double bond between two oxygen atoms and this energy is not equal to zero.

Delta H of reaction = sum of bonds energy for bonds broken - sum of bonds energy for bonds made. There is no delat H of formation in this equation.

Regardless of what bonds are broken in a specific reaction, enthalpy is a state function. An analysis of the $\Delta H^0_f$ for each compound (which tells you the enthalpy of formation from the constituent elements of each molecule) will yield the same results as calculating which specific bonds are being broken and formed. There is more than just one way to calculate $\Delta H$.

This answer is not correct in that $O$ is not the same that $(1/2)O_2$

@user1420303 Are you trying to say half of a diatomic molecule isn't a single atom? If so I don't know what to tell you. Either way, it's only a stoichiometric concept. In the mechanism for this reaction, the $\pi$-bond in diatomic oxygen is homolytically split, and then reacts via a chain radical mechanism. Never in this reaction does there exist a true oxygen atom.

Yes @ringo, the reaction was written in the question (exist it or not) has associated a change in standard reaction enthalpy. The standard molar formation enthalpy of $O$ is close to 60kcal, while the standard molar formation enthalpy of O$_2$  is null by def.. So it is important use the chemical species involved in the reaction. I think that you can give a perfect answer if you consider this fact. You can take a look at the table in http://www.gaussian.com/g_whitepap/thermo.htm for the experimental value. I hope that the comment helps. Excuse me if it sounded rude, english is not my language.

@user1420303 I'm sorry as well, I may have been a bit too quick to reject your criticisms. I've made changes according to your comments. Thank you very much.

Let it remain as $\ce{O}$. Using $\ce{O}$, the bonds that are formed are:

1 $\ce{C=O}$ bond
1 $\ce{H-Cl}$ bond

The bonds that are broken are,

1 $\ce{C-H}$ bond
1 $\ce{C-Cl}$ bond

Using the above information and the formula given in your assignment,
$$\Delta_\mathrm{R} H=[\text{energy used for breaking bonds}]-[\text{energy used for forming bonds}]$$
you can find the enthalpy change.
Yes. The reaction given is not a combustion reaction. However, it is not necessary that the products are carbon dioxide and water vapour.
One interesting thing that you can note is that, chloroform stored in presence of oxygen will slowly convert to phosgene and this can take place at room temperature.
Anyways, the main intention of the question is simple and only asks you to calculate the enthalpy change.

I don't think $\ce{C=O} \approx 2(\ce{C-O})$.

Eh... They're not too far off.

@DGS my mistake. I thought $\ce{C = O}$ bond energy was not given. So I thought I could make an approximation :)

How do I solve it? Why is oxygen O, and not O2?

Because $\ce{O2}$ does not react.$\ce{O2}$ gets dissociated into $\ce{2O}$ and then this oxygen atoms reacts.Remember about bond dissociation enthalpy?

Diatomic molecule is one that only contains two atoms. They could be the same (for example, Cl2) or different (for example, HCl). The bond dissociation enthalpy is the energy needed to break one mole of the bond to give separated atoms - everything being in the gas state.

That is exactly my point. If I use O2, I have to consider energy I have to use for bond breaking in O2, Outcome (in terms of endothermic or exothermic) is different, if I use it with O, and if I use it with O2, and then balance equation.


text:
Comparing the reactivity towards aromatic electrophilic substitution reaction
In the following pairs of molecules, which is more reactive towards electrophilic substitution reaction?
1)  1,4-dinitrobenzene or 1,3-dinitrobenzene (don't consider the ortho isomer)

2)  benzene-1,3-diol or benzene-1,4-diol (don't consider the ortho isomer)

3)  4-methylphenol or 3-methylphenol

4)  4-nitrotoluene or 3-nitrotoluene

I am having a conflict of concept.
In the first pair of molecules, according to resonance the presence of one nitro will make C3 negative and C4 positive. So if another nitro comes at C4 then the positive charge at C4 will intensify as nitro draws more electron density. If the second nitro comes at C3 then the negative charge at C3 will be absorbed by nitro. So which of these will reduce the electron density on benzene more.
In my opinion 1,3-dinitrobenzene will have less electron density on the benzene ring as (C3 is already negative due to nitro at C1) much of the electron density at C3 is absorbed. But in 1,4-dinitrobenzene the second nitro group is absorbing electron density from C4 (in which electron density is very less already due to presence of nitro at C1). So since the electron density is less at C4 then it won't be able to take much electron from ring (as its already positive and deficient of electron density).
If my concept is wrong with respect to above context, please correct me.
All of the remaining isomer pairs have same problem, but each is a bit different. I would like an explanation for each answer and especially the 1st one in detail. And try not to provide an answer by directly copying from a source (especially large) thanks.

Is "more reactive" equivalent to a lower activation energy? If so, you might want to draw the arenium ions (Wheland intermediates) for the the addition of an electrophile to the different isomers.

@KlausWarzecha actually i mean which of the two isomers makes the benzene ring more activated.For example hydroxy group activates benzene   lesser as compared to amino group.Similarly a methoxy goup is less activating as compared to hydroxy group.(actually i have read about activation energy in nucleophilic substitution reactions only).But i don't really like thermodynamics and all the heat and entopy.If you would please explain with respect to concentration of charges,resonance,inductive effect and destabilisation caused by charge concentration it would be great and easy enough to understand

A nitro group doesn't lead to a resonance structure where there are negative charges on at meta carbons. They are simply more active than the ortho and para positions because they don't have a positive charge in any resonance structures.

@DGS only if there is a negative charge at the carbon to which nitro group  is attached then only it can show  -R effect (if i am right) as in the case of para nitro phenol (where negative charge is at C4 and nitro group can take the electron pair in conjugation with carbon and nitrogen).how is then "A nitro group doesn't lead to a resonance structure where there are negative charges on at meta carbons" your statement correct.Sorry I didn't get what you explained after that.And why is it that you specifically mentioned meta.Thanks.

I'm not sure what you're asking (if you're asking anything).

Draw out the resonance structures. A nitro group is a strong electron withdrawing group. The nitrogen has no lone pairs to donate. A negative charge will only end up within the ring with electron donating groups.

@DGS that was one point which i was pondering about.if you consider nitro benzene then in its resonance structure we will be having a comparitive negative charge at meta position (actually not due to the presence of electron pair on carbon but on the fact that since ortho and para are having positive charge then meta becomes comparitively negative).So according to you,this comparitive negative charge can't be resonated as its not an electron pair.right?

We can answer your question by first looking at each individual, unsubstituted position in the benzene ring and determining the relative reactivity at each of these positions. Next we can compare the molecules and see which one has the most activated positions.
I've carried out the first step of this exercise and show the results in the following figure.

Looking at the top molecule, m-dinitrobenzene, we see that 3 positions are deactivated by the resonance effects of both nitro groups (that is, these positions are either ortho or para to each nitro group).  One position is not deactivated since it is meta to both nitro groups. We can analyze the para-dinitro isomer in a similar manner.  In this case we find that each position is only deactivated by one of the nitro groups.  Clearly, a position that is doubly deactivated will react much slower than a position that is singly deactivated.  Now, if we try to estimate the relative molecular reactivities by summing up and averaging the reactivities at each position, we see that the meta isomer is roughly deactivated by an average of (0+2+2+2/4) 1.5 at each position, whereas the para isomer is roughly deactivated by an average of (1+1+1+1/4) 1.0 at each position.  Therefore our analysis suggests that the para isomer will be the least deactivated and consequently react faster.
In the dihydroxybenzene case, hydroxyl groups are ortho-para activating.  Applying the same type of analysis leads us to conclude that the various positions in the meta isomer will, on average, be more reactive than the positions in the para isomer.  Hence the meta isomer will react faster in an EAS reaction.
Similarly, in the cresol case both substituents are again ortho-para activating but with the hydroxyl group being more activating than the methyl substituent. In the meta isomer, 3 positions are doubly activated and one position is not activated (or deactivated).  In the para isomer, each position is singly activated, and only 2 of these positions are activated by the strongly activating hydroxyl group.  This would lead us to suspect that the meta isomer will react the fastest.

For the answer to the third question why is the meta carbon (with respect to methyl group) is not activated by the +I effect of methyl group? –

Would you please provide the answer for fourth question.I will try to make explanation  according to your answer.

Inductive effects are transmitted over a fairly short range, the "meta" carbon is just to far away for the +I effect to be significant.

For the fourth set of molecules the para isomer will react the fastest.  It is the only isomer that has (2) carbons activated by the methyl group and not deactivated by the nitro group. Try to apply the above concepts and see if you can arrive at this answer. Let me know if you encounter a problem.

but the figure shows para carbon is activated in the third question.para carbon is at a greater distance than meta then how can para carbon get activated by inductive effect if meta carbon doesn't get activated?

The effects I discussed are not inductive effects, rather they are resonance effects.  Resonance effects can occur over much greater differences.  Do you understand the difference between resonance and inductive effects?

yup i hope so.But methyl group shows only inductive effect (if i am right) as it doesn't have a lone pair on carbon to resonate.(is it no bond resonance responsible for methyl group's resonance effect.)

Hyperconjugation is the term I'm familiar with - see this earlier answer.

Well in that case its right.Thanks for the excellent explanation.

Your welcome, glad to help.

Usually substitutents that activate electrophilic aromatic substitution like OH and CH3 are most effective at positions ortho and para to the substituent, so they direct the reaction to that site.  You should be able to see that by drawing resonance structures.  Deactivating substituents like NO2 are most deactivating at those same ortho and para sites, so they tend to favor the meta position for whatever reaction does occur.
Now you should be able to tell.  Look for the molecule that has as many sites as possible in favorable positions, ortho/para to the activating groups but meta to the deactivating ones.  For instance, in the first case the 1,3 di-nitro compound has a site meta to both deactivating NO2 group versus the 1,4 isomer having no such site.

so you are saying 1,3 dinitro benzene is less reactive towards electrophilic substitution than 1,4 dinitro benzene.Right?In 1,3dinitro benzene does the nitro group at C3 lead to resonance stabilisation (as C3 appears to have  negative charge if draw the resonance structure of nitro benzene )

Look for a compound with more sites that are favorably licsted, for highly deactivating groups like NO2 the meta position is favored for any reaction that could occur.  Which isomer has a site meta to both deactivating groups?

it is 1,3 dinitro benzene.But if i am right at meta there is no resonance stabilization but rather only inductive stabilization (because,in resonance structure of nitro benzene we are not really having a pair of electron on meta carbon.right?Its just a comparitive negative charge,that is,we assigned negative charge for meta carbon on the fact that ortho and para carbon had a real positive charge i.e by forming a carbocation.But a meta,no carbanion is being formed)

I am not sure where you get a carbenium from.  I get carbocaions.  In the nitro group you have two highly electronegative oxygen atoms and you can consider resonance structures with both oxygens negatively charged.  Now you need two positive charges.  One is on the nitrogen atom, the other ends up on the ring and goes to the ortho and para positions.  Which deactivates them stringly and makes the nitro group meta directing.

the carbanion which i mentioned is in the case of resonance structure by an electron donating group as in the case of phenol (at ortho and para).And what i meant is,such sort of a carbanion is not formed (at meta ) when an electron withdrawing group is present on a benzene ring .(the negative charge is not due to the formation/presence of carbanion,rather a comparitive negative charge)