Understanding the bond dissociation energy of poly-atomic molecules Now I have a breaker which is containing $\ce{CH4}$ of some amount at standard conditions and then I put it on a burner.

In Thermodynamics we are taught about how to find the bond enthalpy of poly-atomic atoms. They say that such molecules (here in this case methane) all four $C-H$ bonds are identical. However, the energies required to break the individual $C-H$ bond in each successive step differs
$$\begin{alignat}{2}
\ce{CH4(g) &-> CH3(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+427\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH3(g) &-> CH2(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+439\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH4(g) &-> CH(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+452\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH(g) &-> C(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+347\ \mathrm{kJ\cdot mol^{-1}}
\end{alignat}$$

and now their bond enthalpy is the mean of all those numbers.

now going back to the diagram we heat up the beaker containing methane then the energy that is absorbed by any one of the $\ce{C-H}$ bond should be equal to the energy absorbed by the other $\ce{C-H}$ bond because all the bond in a molecule should absorb equal amount of energy at the same time because at time $t$ the temperature is same all over the beaker and  all bonds are equal. so it should be like that while calculating the bond dissociation energy of a poly-atomic molecule (take the example of methane) calculate it like $4\times 427\ \mathrm{kJ\cdot mol^{-1}}$. (is this a common sense?)

  all bonds Should broke up at once!

but we know that this does not happen, which means that all the bonds in a molecule does not absorb heat of equal amount at the same time or are they unequal? And fairly speaking it is painful to me to visualize this thing maybe because I am not able to understand this reality.

How do I comprehend this thing?

In Short

all bonds of $CH_4$ are equal
The temperature of beaker is same all over its volume
energy that is absorbed by any one of the $\ce{C-H}$ bond should be equal to the energy absorbed by the other $\ce{C-H}$ bond
bonds should absorb equal energy at time t
the should all break at once at same time t

this do not happen so how to understand this thing?

Whoa!!!

The C-H bonds in $\ce{CH4(g)}$ are not the same as the C-H bonds in $\ce{CH3(g)}$. The $\ce{CH3(g)}$ "molecule" is electrically neutral but it is a radical with an unpaired electron in one orbit. In fact the physical orientation of the molecule changes to a planar configuration.


what can 1 kJ/mol do in reality? I am using an approximation of density matrix in a high order series expansion to predict the total internal energy of chemical systems. During my research,it was found out that the precision of such algorithm is accurate up to 0.00001 Hartree by optimising the independent parameters, which is roughly 0.026kJ/mol's error fluctuation. I wonder in reality, what can 1 kJ/mol do in our daily life, what phenomena can 1 kJ/mol of energy possibly trigger? I'm quite a mathematical person and haven't really studied enough chemistry nor physics. if someone could give me some physical intuitions that would be greatly appreciated!

And I meant to leave it as a broad question. Since we are not picking a specific chemical system to chew on here, any unbiased thoughts are welcomed and shall be considered in this post! A collective answer from the various point of view is probably a better way to get one's head around rather than a single aspect answer.

For this reaction

$$\ce{A -> B}$$

a $\Delta G$ value of $1\ \mathrm{kcal\,mol^{-1}}$ will result in an equilibrium value

$$K = e^{-\frac{\Delta G}{RT}}$$

of approximately 0.18 favoring reactant. Every additional increment or decrement by this value will change the relative ratio by roughly 5.


What word is to chalcogen and pnictogen what halide is to halogen? Would "chalcide" be understood as a group 16 dianion (and something less electronegative)? Would "pnictide" be understood as a group 15 trianion (and something less electronegative)?

From what I could find in a brief search, "pnictide"(this link includes some of the origin for the name) is used, but the corresponding term used for group 16 seems to be "chalcogenide", which, while it can be used to describe all group 16 elements, typically is used in a way that excludes oxides.


Electrophilic substitution of benzene with conc. HNO₃ and HNO₂ The following question was given in my FIITJEE study material:

  $$\ce{Benzene + \text{conc.} HNO_3 + HNO_2 ->}$$ Final product of the reaction is:

,
,
,

I started with the solution by deciding the Lewis acid out of $\ce{HNO_3}$ and $\ce{HNO_2}$. I guess $\ce{HNO_3}$ should be Lewis acid as O.S. of N is higher in it. Hence, I wrote the reaction to generate electrophile:

$$\ce{HNO_3 + HNO_2 -> NO_3^- + H_2O + NO^+}$$

Now, it is a bit clear who's the electrophile. Hence, I thought that first option would be the final product. But, according to the answer key it was wrong.

What I am doing wrong?

The correct answer was:

   Second one (nitrobenzene)

Nitrosyl cation, $\ce{NO^+}$, is a better electrophile than molecular $\ce{HNO3}$ (you don't get $\ce{NO_2^+}$ in this system without sulfuric acid or other strong auxiliary acid) and so you get nitrosobenzene in the substitution reaction.  But then the nitrosyl group with its nonbonding electron pair is an attractive target for oxidation by the nitric acid and thereby is converted to $\ce{NO_2}$.  Thus nitrobenzene.  Then, the nitrobenzene, deactivated by the nitro group, will not react with the relatively mild electrophile $\ce{NO^+}$ anymore, meaning products C and D are never reached (and their nitrosyl groups would not have survived anyway).


How can we replace volume in ideal gas equation with V - nb? We know that ideal gas equation is $PV= nRT$ where P is the pressure of the ideal gas and V is the volume of the ideal gas. Thus, we can write :
$$ P_{\textrm{ideal}}V_{\textrm{ideal}} = nRT$$
Now from volume correction term of van der Waals' equation we know that
$$ V_{\textrm{real}} = V_{\textrm{ideal}} - nb $$ where $ V_{\textrm{real}} $ is the volume of the real gas and $ V_{\textrm{ideal}} $ is the volume of the ideal gas. Clearly, $ V_{\textrm{ideal}}$ and $ V_{\textrm{real}} $ are different.
Then how can we write :
$$P_{\textrm{ideal}} (V_{\textrm{ideal}} - nb) = nRT$$
How can we replace $ V_{\textrm{ideal}}$ with $ V_{\textrm{real}}$?

The actual volume occupied by the gas is less than the measured volume(volume of the container).
In reality gaseous particles are not point objects and don't occupy zero space, which is against one of the postulates of the Kinetic Theory of Gases.
Thus while dealing with real gases we have to account for the volume which has been occupied by the gaseous particles. We do so by assuming the volume of gaseous particles to be $nb$.
Now, the ideal volume(actual volume) occupied by the gas is the measured volume(volume of container) minus the volume occupied by the gaseous particles.
Thus, $V_{ideal} = V_{real} - nb$ and not the other way around.
Note, $V_{ideal} =V_{actual}$ and $V_{real} =V_{measured}$.
The mistake you are making is formula you are using and thus
$P_{ideal}(V_{ideal}-nb) ≠ nRT$, but
$P_{ideal}(V_{real}-nb) = nRT$


How do copper sulphate and the like change crystal structure on adding water of hydration? Copper sulphate, in its hydrated form, is crystalline, whereas the anhydrous form is amorphous.

Gypsum has a similar story-- on heating the crystalline dihydrate we get an amorphous hemihydrate. (Gypsum in fact has two different crystalline forms of the dihydrate, and has an anhydrous form as well).

I've never really grasped how the water makes it crystalline. I suspect it has to do with the relative sizes of the compounds being re-compensated by ligands, but I'm not sure. How does the water affect/create the crystalline properties?

[Manishearth, I'm not sure I'm even answering the right question below. After reading your question more carefully, I think you were more interested in how water molecules help crystals form, rather than whether or not the anhydrous forms can form crystals also (versus being strictly amorphous). So, oops. Maybe some one else has a good answer on the water molecules part?]

I suspect the answer is an issue of process rather than of fundamentals.

That is, if you could find a non-aqueous medium that readily dissolves the anhydrous form of some salt, my very strong suspicion is that you could grow excellent clear crystals of it that would look almost nothing like the hydrous crystalline forms.

The problem for metallic-anion salts in particular is that finding a liquid that dissolves the salt without simultaneously forming strong coordination complexes with it is going to be tricky, since I'm guessing it's usually the coordination effect that help ionize the units of the material to make it soluble!

Here's a wild guess: Someone, somewhere, has likely done research on how to grow crystals of substances such as anhydrous $\ce{CaSO4}$ (hmm, why does this phrase "dead burnt plaster" come to mind? :) or anhydrous $\ce{CuSO4}$ by using low-temperature molten solutions of... something or other? Not sure what!

And what a delightful crystal that would be to have! It would be way out of the ordinary despite being so very... ordinary?... in one sense in its composition.

Update: Apparently, clear crystals of anhydrous $\ce{CuSO4}$ do indeed exist, and are easier to create than I thought. You just add concentrated sulfuric acid [a] to ordinary copper sulphate pentahydrate solution and let it evaporate slowly. Since $\ce{H2SO4}$ is one of the most aggressive drying agents in existence, my guess is that the $\ce{CuSO4}$ just can't compete for the remaining water molecules and is forced to crystallize without hydration.

[a] Please, never ever attempt to use concentrated $\ce{H2SO4}$anywhere except in a real lab with protections and procedures fully in place. This is a compound that wants water so badly that it willy synthesize water right out of the hydrogen and oxygen bonded into proteins and carbohydrates, leaving only charred carbon. Ironically, the actual acidity of concentrated (vs hydrated) $\ce{H2SO4}$ is quite moderate, comparable to that of vinegar.

Addenda:

-- Another common-exotic crystal may be lithium fluoride, $\ce{LiF}$, although the accuracy of that seems to depend on who you read. Unique how, you say? Even though lithium was an early result of the Big Bang, lithium and fluorine are both consumed rapidly by large hot stars, especially fluorine. Their existence on earth thus appears to depend on processes such as proton irradiation and possibly even intensive neutrino irradiation tht only occur during the explosion of a supernova star, rather than being inherited from before the cataclysmic event. Pretty much all of the elements (except $\ce{H}$) here on Earth are star-forged of course, but the elements in $\ce{LiF}$ crystals thus may be remnants of the supernova event itself. That would... cool!)

-- And speaking of unique, wearable crystals: I've always wondered if there's a market out there for diamond crystals synthesized using carbon that was captured as carbon dioxide during the cremation of a loved one. If there is, feel free -- that is definitely not my bailiwick!


Doubts about IUPAC nomenclature of 4-ethyl-1-fluoro-2-nitrobenzene Why is 4-ethyl-1-fluoro-2-nitrobenzene (structure is in the attached picture) called so ?

Shouldn't it be called as 1-ethyl-4-fluoro-3-nitrobenzene since the precedence of ethyl is greater than halogens?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Furthermore,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

Therefore, the example is named as 4-ethyl-1-fluoro-2-nitrobenzene rather than 1-ethyl-4-fluoro-3-nitrobenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.


Name of round piece of glassware with thin neck? Found in the lab, it looks like a round bottom flask with an extremely thin neck. Effectively a glass sphere about 60 mm in diameter with what looks like a capillary tube as the neck. Not sure whether it is a standard COTS item or custom glass.

We used to have glassware like that in our lab.  It is basically an "unfinished" piece of glassware. In a nutshell, it was used to handle (collect, transport) gases.

It was easy and inexpensive for the glassblower to make several of those bulbs at a time; he'd make a variety of sizes.  We'd keep them in a drawer and then before we used them we'd do some glassblowing ourselves and add the expensive part - a stopcock with a ground glass joint.  Then, having already measured the volume, we could attach one to a vacuum line (no vacuum) with the stopcock open and feed a gas into the vacuum line, close the stopcock and remove the bulb.  Knowing the pressure, temperature and volume, we knew how many moles of gas were in the bulb.  We could then attach the bulb to a reaction flask and open the stopcock, so that a known amount of gas was used in a reaction.

Here is a simple picture, imagine the tube on the bottom is the vac line.

These bulbs were also used to trap gaseous reaction products by attaching the bulb with its stopcock and joint to a vac line along with a reaction flask.  The bulb was cooled to the desired temperature and when the reaction produced gases, they could be collected in the cooled bulb.


What is the net ionic equation for the reaction between Ba(OH)2 and H2SO4? I know its either:

1.) Ba + SO4 = BaSO4

2.) Ba + 2OH + 2H + SO4 = BaSO4 + 2H2O

But can someone explain why water is or isn't part of the net ionic?

The reaction between $\ce{Ba(OH)2, H2SO4}$ is known as acid-base neutralisation, as $\ce{Ba(OH)2}$ is a relatively strong base and $\ce{H2SO4}$ the strong acid.
The molecular reaction is given as,
$$\ce{Ba(OH)2 + H2SO4 -> BaSO4 (s) + 2H2O}$$

Before moving to net-ionic reaction, let's see the reaction of $\ce{BaCl2, Na2SO4}$ [1]:
$$\ce{BaCl2 + Na2SO4 -> BaSO4 (s) + 2NaCl}$$
So, breaking up into ions, we see:
$$\ce{Ba^{2+} + 2Cl- + 2Na^{+} + SO4^{2-} -> BaSO4 (s) + 2Na^{+} + 2Cl^{-}}$$
$\ce{Na^{+}, Cl^{-}}$ remains completely dissociated in the aqueous solution as they're salts of strong acid and strong base (i.e., they're strong electrolytes). Hence, $\ce{Na^{+}, Cl^{-}}$ are spectator ions as they don't participate in ionic reaction. Let's cancel them out, and we get the net ionic reaction:
$$\ce{Ba^{2+} + SO4^{2-} -> BaSO4 (s)}$$

Now, let's see the reaction given in the question. In this, $\ce{BaSO4}$ precipitates out, so $\ce{Ba^{2+}, SO4^{2-}}$ can't be spectator ions. Also, H2O has a very low dissociation contant ($\ce{K_W = 10^{-14}}$, at 25°C), so it can't dissociate into ions. Hence, $\ce{H+, OH-}$ are too not the spectator ion. Therefore, this reaction has no spectator ion and hence ${H_2O}$ will be included in the ionic reaction, which is given as:
$$\ce{Ba^{2+} + 2OH- + 2H+ + SO4^{2-} -> BaSO4 (s) + 2H2O(l)}$$

Another example:
[1]: Complete ionic and net ionic equations, Khan Academy
Note: As mentioned by andselisk in comments, $\ce{BaSO4 (s)}$ is the recommended way to write the precipitate, as per IUPAC. Old school symbol for precipitate was "↓".


Why would 1kW power supplies fail around Nitric acid (HNO3) gas? I experienced a situation where some electrical power supplies (rated over 1kW) were exposed to nitric acid ($\mathrm{HNO_3}$) fumes at roughly 9ppm for half a day; these power supplies intermittently failed (perhaps 2 to 3 times per hour) in the presence of the fumes.  Replacing the power supplies restored proper functionality to the equipment they were powering.

At first, I thought that perhaps there was arcing inside the power supply; however, nitric acid doesn't seem to have a flash point.  I'm trying to come up with a reasonable explanation for why these particular components would fail.  Maybe the fumes increased atmospheric conductivity, which shorted across components?

If there isn't a compelling reason from a chemistry perspective, that's also a fair response.

Nitric acid corrodes copper. Drop a penny into some nitric acid (under a hood!) and you'll see that your power supplies really don't stand much of a chance in that environment.
This happens with dilute nitric acid:

$$\ce{3 Cu + 8 HNO3 -> 3 Cu^{2+} + 2 NO + 4 H2O + 6 NO3^{−}}$$

When you open up one of the failed power supplies, do you see bluish crusty white stuff anywhere? That's nitric acid corrosion.

Addendum: Here are some quantitative corrosion rates:
Karl Hauffe and Roman Bender: Copper. In Corrosion Handbook. Wiley-VCH, 2008.

Addendum 2: Even though the concentration of nitric acid in the air is low, your power supplies are sucking a lot of air through them in 12 hours.

Let's suppose you have a power supply that dissipates 300 W of heat and you can tolerate a 10 degree Celsius temperature rise. Then according to this reference you need an airflow rate of about 53 cubic feet per minute.

That's 1.5 cubic meters of air per minute.

The density of air in a warmish (300 K) room is about 1.177 kilograms per cubic meter. So you're pulling

$$\frac{1.5~\mathrm{m^3}\ \text{air}}{1~\mathrm{min}} \frac{1.177~\mathrm{kg}\ \text{air}}{1~\mathrm{m^3}\ \text{air}} \frac{9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{kg}\ \text{air}} = 15.9~\frac{\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}}$$

which over 12 hours is

$$\frac{15.9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}} \frac{60~\mathrm{min}}{1~\mathrm{hour}} (12~\mathrm{hours}) = 11.4~\mathrm{g}\ \ce{HNO3}$$

If all of that reacted with copper in the power supply, this is how much copper would be converted into corrosion:

$$11.4~\mathrm{g}\ \ce{HNO3}\frac{1~\mathrm{mol}\ \ce{HNO3}}{63.01~\mathrm{g}\ \ce{HNO3}}\frac{3~\mathrm{mol}\ \ce{Cu}}{8~\mathrm{mol}\ \ce{HNO3}}\frac{63.546~\mathrm{g}\ \ce{Cu}}{1~\mathrm{mol}\ \ce{Cu}} = \fbox{4 g Cu }$$

...that's a lot of copper to lose (a penny is 2.5 g!). Even if only 10% of all the $\ce{HNO3}$ that's being sucked through the power supply reacts, it's still 0.4 grams of copper, certainly enough to significantly corrode electrical contact surfaces.


Exceptions to the Nitrogen Rule in mass spectrometry The Nitrogen Rule in mass spectrometry states that (organic) molecules containing no or an even number of nitrogen atoms will have even masses, and molecules containing an odd number of nitrogen atoms will have odd masses.

Examples (all monoisotopic masses):
No nitrogen, even mass:
H2, 2 Da
HBr, 80 Da
C2H6, 30 Da
One nitrogen, odd mass:
NH3, 17 Da
C5H5N, 79 Da
 etc.

Often there is a little disclaimer saying that the nitrogen rule is only valid for molecules that do not contain "exotic atoms." What are some examples for molecules do not follow the nitrogen rule?

The reason for the rule is that, outside of transition metal complexes, stable free radicals are relatively rare. If all electrons are paired and we are talking about molecules, not ions, there must also be an even number of protons. Thus an odd number of neutrons must be present to arrive at an odd molecular weight, and the only elements with the most abundant isotope having an odd number of neutrons are beryllium and nitrogen, although dysprosium is kinda close.

In an undergraduate lab you wouldn't be given an unknown with beryllium in it because it's so dangerously toxic, so that leaves you with nitrogen unless you have a deuterated sample.

But there are stable free radicals like nitric oxide (neurotransmitter) or chlorine dioxide (fumigant) that violate the rule. Molecular oxygen doesn't count because it's a diradical and so has an even number of electrons.

EDIT: I spoke too soon about abundances of nuclides with an odd number of neutrons. Here is a table of such nuclides with abundance $> 20\%$ taken from http://atom.kaeri.re.kr/nuchart/ :
$$\begin{array}{rl} \text{Nuclide} & \text{Abundance} \\
\hline
\sideset{^{9}}\ {Be}& 100\%\\
\sideset{^{14}}\ {N}& 99.636\%\\
\sideset{^{105}}\ {Pd}& 22.33\%\\
\sideset{^{129}}\ {Xe}& 26.4006\%\\
\sideset{^{131}}\ {Xe}& 21.232\%\\
\sideset{^{163}}\ {Dy}& 24.896\%\\
\sideset{^{167}}\ {Er}& 22.869\%\\
\sideset{^{195}}\ {Pt}& 33.78\%\\
\sideset{^{207}}\ {Pb}& 22.1\%
\end{array}
$$
Of these $\sideset{^{195}}\ {Pt}$ is the most abundant isotope of $Pt$ in addition to $\sideset{^{9}}\ {Be}$ and $\sideset{^{14}}\ N$ previously mentioned.


Which has higher priority according to Cahn-Ingold-Prelog priority rules, vinyl or secondary butyl? According to my textbook, vinyl has higher priority than isopropyl, because an atom which is connected by a double bond can be treated as two identical atoms (one of them is just an imaginary atom).

How about vinyl and secondary butyl? Should the imaginary atom be treated as that nothing else is connected to it? Does secondary butyl have higher priority?

The assignment of CIP priorities has some intricate details when it comes to multiple bonds.

First, the general procedure (Wikipedia):

  If there is a tie, we must consider the atoms at distance 2 from the stereocenter — as a list is made for each group of the atoms bonded
  to the one directly attached to the stereocenter. Each list is
  arranged in order of decreasing atomic number. Then the lists are
  compared atom by atom; at the earliest difference, the group
  containing the atom of higher atomic number receives higher priority.
  If there is still a tie, each atom in each of the two lists is replaced with a sub-list of the other atoms bonded to it (at distance
  3 from the stereocenter), the sub-lists are arranged in decreasing
  order of atomic number, and the entire structure is again compared
  atom by atom. This process is repeated, each time with atoms one bond
  farther from the stereocenter, until the tie is broken.

Now, concerning double bonds in particular:

  If an atom A is double-bonded to an atom B, A is treated as being
  singly bonded to two atoms: B and a ghost atom that has the same
  atomic number as B but is not attached to anything except A. In turn,
  when B is replaced with a list of attached atoms, A itself is excluded
  in accordance with the general principle of not doubling back along a
  bond that has just been followed, but a ghost atom for A is included
  so that the double bond is properly represented from both ends.

I have constructed a table of these atom lists for the vinyl, isopropyl and sec-butyl rests:
$$\small
\begin{array}{llll}
\hline
\text{Dist} & \text{Vinyl} & \text{Isopropyl} & sec\text{-Butyl} \\
\hline
1 & \text{C} & \text{C} & \text{C} \\
2 & \text{C} & \text{C} & \text{C} \\
  & \text{(C)} & \text{C} & \text{C} \\
  & \text{H} & \text{H} & \text{H} \\
3 & \text{(C)} & \text{H} & \text{C} \\
  & \text{H} & \text{H} & \text{H} \\
  & \text{H} & \text{H} & \text{H} \\
  & & \text{H} & \text{H} \\
  & & \vdots & \vdots \\
\hline
\end{array}
$$

The sec-butyl rest has thus the highest priority of the three, followed by vinyl and then isopropyl.

The ghost atom at distance three is the ghost carbon at distance one, as seen from the ghost carbon that was doubly bonded to it. If the distance 4 was to be considered, no more ghost atoms would appear, as the ghost atom at position 1 does not have any more ghost atoms bonded to it (atom 1 however is bonded to the ghost atom at position 2).


Is temperature just the phenomenon of particles colliding against an object? This is just something that I thought up while doing busy work at my job. Is the sensation of heat simply coming from particles colliding with your, say, hand? Are they transferring energy more to objects because they have higher velocities, and thus heat can hurt? (Sorry, this isn't a more cohesive question...).

Heat transfer occurs when particles collide against an object (which is made of other particles) or when photons impact on the object.

Temperature is a measure of the average amount of kinetic energy of the particles (in a gas this will be simple related to how fast the particles of gas move but in a solid it will be about vibrations of the particles).

The sensation of heat is a product of both thermal radiation (infra-red photons are emitted by the hot object and absorbed by your hand) and contact with something that can transfer some of that energy directly (hot gas particles or a hot solid directly touching your hand). Living things have evolved to sense heat and cold because they only survive if they avoid extremes of both.


Hyperconjugation in ethane conformers I refer you to this set of presentation slides. On slide 8, the author explains the preference for the staggered conformation of ethane by saying that there is a favourable hyperconjugative interaction between $\sigma^{}_{\ce{C-H}}$ of the $\ce{C-H}$ bond in one methyl group and $\sigma^{*}_{\ce{C-H}}$ of another $\ce{C-H}$ bond in the other methyl group.

However, I am puzzled by the diagram. Since both $\ce{C-H}$ bonds are similar in terms of bonding, their bonding and antibonding $\sigma$ molecular orbitals should have the same energies. Instead, the diagram shows that one of the $\sigma_{\ce{C-H}}$ bonds is lower than the other in energy. How can this be right? I have shown a screenshot of the diagram below.

Also, how can a $\sigma_{\ce{C-H}}$ donate electron density to a $\sigma^*_{\ce{C-H}}$ effectively? After all, the energy difference between them is rather significant since the orbital interaction between the atomic orbitals of $\ce{C}$ and $\ce{H}$ is large.

I am not entirely sure what purpose the lower sigma bonding orbital in the MO-diagram serves, because an ethane molecular orbital would, as you have surmised correctly, have six degenerate $\sigma_{\ce{C-H}}$ MOs and a $\sigma_{\ce{C-C}}$ MO, lower in energy.

My best guess would be the position of the $\sigma_{\ce{C-H}}$ on the right hand is quite arbitrary and the basic purpose of the diagram is to show the possible overlap between the $\sigma_{\ce{C-H}}$ of one $\ce{C-H}$ bond and the $\sigma^{*}_{\ce{C-H}}$ of $\ce{C-H}$ bond anti to it - you don't really need to be worried about the $\sigma_{\ce{C-H}}$ of that anti bond because it cannot overlap with the other $\sigma^{}_{\ce{C-H}}$.

That brings us to the second part of your query. A $\sigma_{\ce{C-H}}$ can donate electron density to an empty $\sigma^{*}_{\ce{C-H}}$ anti to it, because the latter is oriented correctly (anti) to accept the electron density from it. It's a question of orbital symmetry and not energy difference.


Bromination on the aromatic ring vs aliphatic chain I thought that the answer would be a simple electrophillic aromatic substitution and as the carbonyl is a meta director, the bromine would be present meta to the group.

But the given answer was the α-brominated compound and is also found out an article which says that this happens in case a catalytic amount of $\ce{AlCl3}$ is used while the bromine is substituted in the ring if a larger amount is used. How does this happen?  Why does the amount of $\ce{AlCl3}$ matter?

The key in this reaction is ketone in propiophenone is acting as Lewis base to $\ce{AlCl3}$ Lewis acid. If $\ce{AlCl3}$ is in catalytic amount, it activate $\alpha$-$\ce{H}$ in side chain sufficiently enough (similar to acid would do to a carbonyl compound) and progress to produce $\alpha$-substituted product. If $\ce{AlCl3}$ is in excess, it seemingly deactivates aliphatic side chain completely (Ref.1), and ring substitution predominates to give $meta$-substituted product exclusively as a result (Ref.2 and 3):

The reaction with excess $\ce{AlCl3}$ should necessarily be done in solvent free conditions, and excess $\ce{AlCl3}$ is needed such that one equivalent is used to complex with carbonyl oxygen and another equivalent is used to activate $\ce{Br2}$ molecule. As a result, benzene nucleus would be substituted and 3-bromoacetophenone would be obtained exclusively (Ref.1). The crude liquid product is purified as an oil after fractional distillation in 70-75% yield (Ref.2). The reaction with trace amount of $\ce{AlCl3}$ has done in dry ether and solid crude yield has been 88-96%. When recrystallized by methanol, the yield has been reduced to 64-66% (Ref.3). Another stark different between these two reactions are addition order. In the excess reaction, acetophenone has been added to $\ce{AlCl3}$ while in the catalytic reaction, $\ce{AlCl3}$ has been added to etherial acetophenone solution.

References:

D. E. Pearson, H. W. Pope, W. W. Hargrove, W. E. Stamper, "The Swamping Catalyst Effect. II. Nuclear Halogenation of Aromatic Aldehydes and Ketones," J. Org. Chem. 1958, 23(10), 1412–1419 (https://doi.org/10.1021/jo01104a003).
D. E. Pearson, H. W. Pope, W. W. Hargrove (Checked by B. C. McKusick, D. W. Wiley), "3-Bromoacetophenone," Org. Synth. 1960, 40, 7 (DOI: 10.15227/orgsyn.040.0007) (Organic Syntheses, Coll. Vol. 5, p.117 (1973)).
R. M. Cowper, L. H. Davidson (Checked by Lee Irvin Smith, E. W. Kaiser), "Phenacyl Bromide ($\alpha$-Bromoacetophenone)," Org. Synth. 1939, 19, 24 (DOI: 10.15227/orgsyn.019.0024)(Organic Syntheses, Coll. Vol. 2, p.480 (1944)).


How do I calculate the hydronium ion concentration if I know the strontium hydroxide concentration? This is my first question so I would appreciate feedback on how to ask better questions. Forgive me if I don't know how to properly format math/chem. The guide is confusing.

I was taking a practice test on Brønstead–Lowry acids and bases and I came across a question that has totally stumped me:

  Calculate the $\ce{[H3O+]}$ in a $\pu{0.010 M}$ solution of $\ce{Sr(OH)2}$.

Here are my steps:

Write a balanced equation:

$$\ce{Sr(OH)2 <=> Sr^2+ + 2OH-}$$

Initial concentrations:
$$\ce{[Sr(OH)2]} = \pu{0.010 M}$$

$$\ce{[Sr^2+]} = 0$$
$$\ce{[OH-]} = 0$$

Change in Concentration:
$$\ce{[Sr(OH)2]} = (0.010 - x)~\pu{M}$$

$$\ce{[Sr^2+]} = +x$$
$$\ce{[OH-]} = +2x$$

I am not sure whether $\ce{OH-}$ loses 2 moles to every one of $\ce{Sr(OH)2}$.

Assume $x\ll\pu{0.010 M}$. Now I will set up a $K_\mathrm{a}$ expression:

$$K_\mathrm{a} = \ce{\frac{[H3O][OH]}{[Sr(OH)2]}}$$

$$\frac{K_\mathrm w}{K_\mathrm b} = \frac{x\cdot 2x}{0.010}$$

I don't know what to now.  I feel like I have probably started the problem wrong.

Strontium hydroxide is a strong base, so you can calculate $[\ce{OH-}]$ as $\pu{0.02 M}$, then use

$$K_\mathrm{w} = \ce{[OH-][H3O+]}$$

$$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$


Is the cathode and anode on the cell notation of an Electrolytic cell on the same side as if how you will write it for a Galvanic cell? I was reading an online material then I came upon a cell notation for an electrolytic cell (I determined that it was an electrolytic cell because of the absence of a salt bridge). Usually, I will proceed on reading but then I realized, since cathode and anode is switched for an electrolytic cell, which side of the cell will be the cathode and the anode?

The example given was:

  $$\ce{Ag | AgCl(s), Cl- (\pu{0.200 M}), Cd^2+ (\pu{0.00500 M}) | Cd}$$

A good mnemonic to remember which is the anode or the cathode is to remember that Right side represents Reduction in a cell diagram. This is fixed by those who invented cell notation.. Also, by definition, reduction occurs at the cathode. Whether it is a spontaneous cell or not, is another question.

Now as EdV said, calculate the cell potential as written i.e., consider cadmium half cell as the cathode, and the other silver electrode as the anode. If the Ecell=Ecathode-Eanode turns out to be positive, everything is fine, all that means that the cell, as written, is a spontaneous cell.

If Ecell is negative, then the cell is non-spontaneous.


Estimation of CO2 by titrimetric method using phenolphthalein indicator We were estimating dissolved $\ce{CO2}$ in water by American Public Health Association (APHA) method. It was a titrimetric method using phenolphthalein indicator.

Titrant used was $\ce{NaOH}$ and analyte was sample water. Reactions involved:

$$\ce{CO2 (g) ⇌ CO2 (aq)}$$
$$\ce{CO2 aq + H2O ⇌ H2CO3 ⇌ H+ + HCO3^2-} \label{rxn:1}\tag{1}$$

(Only $10\%$ of dissolved $\ce{CO2}$ get converted into $\ce{H2CO3}$.)

On addition of $\ce{NaOH}$:

\begin{align}
\ce{2NaOH + CO2 &-> Na2CO3 + H2O} \label{rxn:2}\tag{2} \\
\ce{Na2CO3 + H2O + CO2 &-> 2NaHCO3} \label{rxn:3}\tag{3}\\
\ce{NaHCO3 &-> Na+ + HCO3-} \label{rxn:4}\tag{4} \\
\ce{\underset{from \eqref{rxn:1} and \eqref{rxn:4}}{HCO3-} + H2O &-> H2CO3 + OH-} \\
\ce{OH- + H+ &-> H2O} \\
\ce{HIn &⇌ H+ + In-}
\end{align}

(The equilibrium shift to the right and when $[\ce{HIn}]:[\ce{In^-}]=1:10$ pink colour appears.)

But when we calculate the concentration of $\ce{CO2}$ we consider only reaction $\eqref{rxn:2}$, which means we are calculating only a portion of the molecular $\ce{CO2}$ while the $\ce{CO2}$ that was present in the solution and reacted in step 3 and the $\ce{CO2}$ (the $10\%$) that was directly converted  into $\ce{H2CO3}$ (Rxn. $\eqref{rxn:1}$) are not taken into account.

So it means we are calculating only a part of the free $\ce{CO2}$ ($45\%$) and not all.

From this can we conclude that APHA's method is not very accurate at estimating the free $\ce{CO2}$ in water?

P. S. I read the procedure and also the calculations in a practical book.

But when we calculate the concentration of $\ce{CO2}$ we consider only reaction 2 which means we are calculating only a portion of the molecular $\ce{CO2}$ while the $\ce{CO2}$ that was present in the solution and reacted in Step 3 and the CO2 (the $\pu{10\%}$) that was directly converted into $\ce{H2CO3}$ (eq 1) are not taken into account.

Almost $\ce{H2CO3}$ in the sample will be converted to $\ce{HCO3-}$ as you proceed with the phenolpthalein endpoint. Search for "Speciation of Carbonate as function of pH" (Sorry I do want to immediately link the graph but it is only a http and stackexchange won't accept it).

At ~8.3 or so, all carbonate species is at the form of bicarbonate. All carbonic acid in the solution will be transformed as bicarbonate. The amount of carbonate ion is almost nonexistent. The $\ce{H+}$ released will be neutralized by NaOH, so aside from reacting directly with $\ce{CO2}$, it will also react with the $\ce{H+}$ released by $\ce{H2CO3}$, so (1) is accounted on the amount of NaOH that will be used.

Also, the sum of all equations (2) and (3) is just

$$\ce{2 NaOH + 2CO2 -> 2NaHCO3}$$

So in essence, for 1 mole of NaOH you react with one mole of $\ce{CO2}$ to produce one mole sodium bicarbonate.

In totality, for every one mole $\ce{H2CO3}$, you react with one mole NaOH. For every direct reaction of NaOH with $\ce{CO2}$, the ratio is $1:1$.

So I guess that's all reactions accounted for.


Confusion over the rate law I am really confused about the rate law. From my understanding, the rate law gives an equation that tells us the rate of the disappearance of the reactants. So, for example, the rate law of the reaction $\ce{2 NO + O2 -> 2 NO2}$ is as follows
$$\text{rate of disappearance of reactants} = k \, [\ce{NO}]^2 [\ce{O2}]$$
My biggest confusion stems from the fact that the rate law gives one uniform rate for the disappearance of all the reactants, which makes no sense to me. For every $\ce{O2}$ used, two $\ce{NO}$ are used, so how can they be disappearing at the same rate?
Hopefully, it makes sense where I am coming from, and I would appreciate any help trying to clarify this for me.

Rodrigo de Azevedo's answer can be stated with words by saying that the rate of the reaction is the rate of disappearance of $1$ mole $\ce{O2},$ and also the rate of disappearance of $2$ moles $\ce{NO}$. That is why the coefficient $2$ appears in the law written for $\ce{NO}:$
$$\frac{d}{dt}[\ce{NO}] = - 2k[\ce{NO}]^2[\ce{O2}]$$


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

A simple QM explanation I can think of would invoke Hückel theory. It would go this way:

Suppose you have a two-state system (analogous to having two orbitals) with eigenvalues of the Hamiltonian $\lambda_1$ and $\lambda_2$. These correspond to the orbital energies. So just for argument's sake, let's say that

$$\begin{equation}
\mathbf{H} = \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}
\end{equation}$$

then obviously you have $\lambda_1 = 1, \lambda_2 = 2$. Let's say this system has two electrons. They would both go into the orbital with the lower energy, i.e. $\lambda_1$. In the context of organic chemistry, this would typically mean that the first orbital would be bonding in nature (hence filled), and the second orbital would be antibonding (hence empty), although that's not always the case.

Now, you perturb the system by making the non-diagonal elements of the Hamiltonian negative (analogous to allowing the orbitals to overlap, i.e. allowing delocalisation of electron density between the two). Just for argument's sake, let's change $\mathbf{H}_{12}$ and $\mathbf{H}_{21}$ to both be $-0.1$. If you recalculate the eigenvalues you get

$$\lambda_1 = 0.99, \lambda_2 = 2.01$$

The new eigenvectors will be linear combinations of the old eigenvectors – one in-phase and one out-of-phase combination. The "bonding" or in-phase combination of the two orbitals will have decreased in energy, whereas the out-of-phase combination of the two orbitals will have an increased energy. If you put both electrons into the bonding orbital, you get a net stabilisation.

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

On the other hand, interactions between two filled orbitals are destabilising, because you would have four electrons and you would have to fill both of the new orbitals. If you add up the two new eigenvalues, the total energy is greater than it was before when $\mathbf{H}$ was unperturbed.

This is not evident in my example above. The reason is that I made the assumption that $\langle i | j \rangle = \mathbf{S}_{ij} = \delta_{ij}$. In extended Huckel theory you would solve the generalised eigenvalue equation, $\mathbf{Hc} = E\mathbf{Sc}$. However with my assumption, $\mathbf{S}$ is simply the identity matrix and this reduces to $\mathbf{Hc} = E\mathbf{c}$. A theorem from linear algebra states that the sum of the eigenvalues of $\mathbf{H}$ is simply equal to the trace of $\mathbf{H}$. Therefore, as long as we assume $\mathbf{S}_{ij} = \delta_{ij}$, merely changing the off-diagonal values will not change the sum of the orbital energies.

In general, the basis vectors are not orthogonal. If you solved the full equation, $\mathbf{Hc} = E\mathbf{Sc}$ then you would find that the sum of the new eigenvalues is greater than the sum of the old, unperturbed eigenvalues.

Obviously this is ignoring other important aspects, e.g. electron-electron repulsions (that is a problem intrinsic to Huckel theory), but it's a start.


What determines the coefficients in the wavefunction of a hybrid orbital? I was studying the hybridization of carbon atoms and I came to figure out that the $sp$ hybridization wavefunction is given as
$$|sp_{+}\rangle =\frac{1}{\sqrt{2}}\left(\psi_{2s}+\psi_{2p_x}\right)$$
$$|sp_{-}\rangle =\frac{1}{\sqrt{2}}\left(\psi_{2s}-\psi_{2p_x}\right)$$

From a simple mathematical point of view, Why don't I have states that look like the ones below?
$$\frac{1}{\sqrt{2}}\left(-\psi_{2s}+\psi_{2p_x}\right)$$
$$\frac{1}{\sqrt{2}}\left(-\psi_{2s}-\psi_{2p_x}\right)$$

Even for the $sp^2$ hybridization, the states are
$$\frac{1}{\sqrt{3}}\psi_{2s}-\sqrt{\frac{2}{3}}\psi_{2p_y}$$
$$\frac{1}{\sqrt{3}}\psi_{2s}+\sqrt{\frac{2}{3}}\left(\frac{\sqrt{3}}{2}\psi_{2p_x}+\frac{1}{2}\psi_{2p_y}\right)$$
$$-\frac{1}{\sqrt{3}}\psi_{2s}+\sqrt{\frac{2}{3}}\left(-\frac{\sqrt{3}}{2}\psi_{2p_x}+\frac{1}{2}\psi_{2p_y}\right)$$
I don't know how the construction of the hybrid orbitals are done mathematically...all that I know is that in case of the $sp^2$ hybrid orbitals, the formulation should be
$$\psi_{sp^2}=\alpha\psi_{2s}+\beta\psi_{2p_x}+\gamma\psi_{2p_y}$$
where $\alpha^2+\beta^2+\gamma^2=1$. But, mathematically there should be infinite solutions to this equation with this one constraint, where do this extremely precise values of $\alpha$, $\beta$ and $\gamma$ arise from?

P.S. :  I am not a student of chemistry, but just an engineer.

For the $sp$ case, the other wavefunctions you listed would be equally valid. However, these are just $sp_+$ and $sp_-$ with the signs flipped. Since we square wavefunctions to get the probability density, these would give the exact same physical observables.

As for $sp^2$, the reason for the particular choice of $\alpha,\beta,\text{ and }\gamma$ comes precisely from the name of this hybridization: we want an orbital that is 2 parts p orbital for every 1 part s orbital. In other words, we want a mixture that is 1/3 s and 2/3 p which happens with exactly the normalization coefficients you listed. This is a second constraint. Now this still will lead to an infinite number of different solutions, as we can consider rotating these orbitals by any amount in the xy plane, which would give another valid solution. One of the nice things about the choice you listed is that one of the MOs is aligned with the y-axis and the other two are mirrored across the y-axis, so it is easy to consider the effect of axis symmetry operations on this choice of orbitals. If we include this third constraint of having one of the MOs aligned with an axis, we get a unique solution (besides possible sign flips and that we could align the orbital along the x axis instead) for $sp^2$.


Metal extraction by pyrolysis There is a process named solar furnace pyrolysis which is used to reduce the oxides of various elements by concentrating solar energy. Oxygen is liberated but an oxide slag is left.

$$\ce{SiO2 ->[pyrolysis] SiO + 1/2O2}$$

How can the element can be extracted from the oxide produced?

I am making a project for NASA on space settlements in solar furnace pyrolysis. Can solar furnace pyrolysis be used for obtaining both metals as well as oxygen from the metal containing oxides left after pyrolysis? Can it be done by electrolysis? If possible, please provide with a link with your answer.

The answer appears to be forthcoming in the wikipedia article on silicon monoxide and references therein — to quote:

  $\ce{(SiO)_{n}}$ irreversibly disproportionates into $\ce{SiO2}$ and $\ce{Si}$ in a few hours between 400 and 800 °C and very rapidly between 1000 and 1440 °C, although the reaction does not go to completion.

The article implies that $\ce{(SiO)_{n}}$ generates a passivating layer of $\ce{SiO2}$ in contact with dioxygen, so presumably the slag would be encapsulated in a thin layer of that.

Additionally, silicon is not generally considered to be a metal, but rather a semimetal or metalloid.


Why does nitrogen have a maximum covalency of 4? As nitrogen has $1$ lone pair and $3$ electrons, either it should have maximum covalency of $5$ or $3$. But why does it have a maximum  covalency of $4$ instead?
Why did it leave $1$ electron? Why did it have to break a lone pair?

Recall that covalency is the number of shared electron pairs formed by an atom of that element. Nitrogen's maximum covalency is indeed $4$. And no, it does not break up its lone pair.

I'll give you a simple example. Have a look at the Lewis structure of the ammonium ion:

(source)

Notice that nitrogen's octet is complete as soon it bonds with three $\ce{H}$ atoms (aka forms ammonia). The fourth covalent bond is actually a coordinate covalent bond, formed when that nitrogen atom's lone pair gets donated to a proton.

This is also the maximum covalency for the nitrogen atom, since it has no more unpaired electrons that could be paired up with other atoms, to form more covalent bonds.

Homework exercise: Can you now deduce the maximum covalency of nitrogen's elder brother, oxygen?

Caution: extra stuff ahead this will help those with knowledge beyond high school level. If you're at or below high school level, go back home and play with your cat.

After-thought: Valency is a useless term that doesn't help you do any chemistry. Different sources define it differently (two definitions on Wikipedia). It just helps fill high school textbooks with more pages, but it becomes irrelevant with the introduction of better terms like coordination number, which actually tells you something about the structure of the molecule. While "valency" maybe  useful in basic beginners course at getting a grip, there are limitations to this point of view.

Here's an example of such a contradiction. You may expect elements of period $3$ and above to display higher covalencies. One such example is phosphorus. Though it belongs to the same group as nitrogen, it can form compounds like $\ce{PCl5}$, (apparently) increasing its maximum covalency to $5$ instead. The reasons for these are usually attributed to hypervalency/octet-expansion, but they are wrong and obsolete concepts, having been superseded by newer concepts. In fact, $\ce{P}$ still has a covalency of four in $\ce{PCl5}$, since there are only four shared pair of electrons (the non-bonding electrons don't count). This reinforces the idea that coordination numbers are a better and more useful term than valency. ($\ce{P}$ now has a coordination number of $5$ in $\ce{PCl5}$)


Could blue flames arise within the hull of a burning ship? In Joseph Conrad's Youth: A Narrative, set in 1876, a 400-ton barque ship, the Judea, is sent to Java Head to transport 600 tons of coal. Although various delays occur during its journey, the coal it is carrying spontaneously combusts while off the coast of Western Australia. The crew attempt to smother the fire, unsuccessfully. Then attempt to flood the fire, but this is also unsuccessful. Finally, the gases in the hull explode and blow up the deck, and the crew hail a steamer to tow them to an intervening location. However, during their towing, the speed of the steamer's ship cause the fire currently present on the Judea to smoulder into flames, which the narrator describes thusly:

The speed of the towing had fanned the smoldering destruction. A blue gleam appeared forward, shining below the wreck of the deck. It wavered in patches, it seemed to stir and creep like the light of a glowworm

My questions:

Given the amount of fuel and material of the ship available, is it possible that a blue fire of 2300-3000 degrees Celsius (from a quick Google search), could be produced within the hull of the Judea?

Or is it the case that due to the nature of coal gas, a lower temperature than aforementioned need only be met for flames to appear blue, similar to how the flames of a gas stove appear blue?

Additional information, if helpful:
At its last port to Java Head, the Judea had been recalked and copper sheathed before embarking on the last stretch toward Java Head. The Judea explodes and is towed at night.

In a Bunsen burner, the gas (methane or propane) may burn under two practical regimes : reductive or oxidizing conditions. In the reductive conditions, the flame is yellow and unstable, because the amount of oxygen is low with respect to the stoichiometric ratio. In the oxidative conditions, the flame is noisy, strong and stable, and it is due to an excess of oxygen in the mixed gases before the combustion. The oxidative flame is nearly colorless, but it has a characteristic blue cone at its base. This blue cone has a temperature of about $900°$C, and its spectrum has been analyzed by Hertzfeld in $1935$ : it shows a line spectrum typical of the diatomic "molecule" $\ce{C_2}$, with plenty of vibrational and even rotational sublevels. Maybe that was the origin of the blue flame described by Joseph Conrad. Why not ?


Why is ice sticky between −20 to −8 degrees Celsius but not below −30 °C? I was watching an Air crash investigation video where the cause of the plane crash was determined to be due to blockage of fuel pipes by ice crystals (from the water in the jet fuel) which would have formed while the plane was cruising over Siberia where temperatures are as low as −74 °C. Now the engines were working fine until the fuel temp. was −30 °C. But as soon as it started getting warmer near London, Heathrow and the fuel temp hit −20 °C the ice crystals coalesced to form a mesh that completely blocked the fuel pipes and caused the engines to lose power and end up crashing.

My Question is: What caused the ice to not stick at a colder temp (−30 °C) when I believe the molecules would be vibrating slower than those at −20 °C? It seemed from the investigation's experiment that below −20 °C big ice crystals don't form at all, if they do they're tiny. So why does the increase in temperature help ice sticking?

I think I remember that episode. Basically, crystals of water in kerosene were not a problem during the whole flight (at minus thirty something) but crystals were too sticky when the temperature raised to -20°C and prevented the fuel to go to the engines just a few millimeters before entering the "warming cell" which was supposed to prevent any such blockage.

Crystals at low temperature (let's say -35°C) will be quite hard and quite big, even if some molecules of the fuel will be trapped in the crystals of ice (because there is much more fuel than water). If the temperature goes up to a certain level (let's say -20°C), crystals will break where there was molecules from the fuel. As a consequence, the big crystals will break into very small crystals. Ice crystals do not like their environment (i.e. the fuel) because they cannot find any stabilizing effect with it, but they love other ice crystals. Moreover, the overall surface greatly rises when the size of the crystals diminishes. As a consequence, they will try to stay together and form a "slush".

In that accident, it was not the only factor. The other two problems were that acceleration increased the flux of kerosene (so more small crystals of ice were available to create a slush), and there was a misconception in the heat exchanger which was aimed at avoiding water crystals.


Why is the bond length of CO+ less than that of CO? According to molecular orbital theory, the bond order of $\ce{CO}$ is 3. When $\ce{CO+}$ is formed, the bond order decreases to 2.5, and thus the bond length should increase.

However, the bond length of $\ce{CO+}$ is found to be less than $\ce{CO}$. How is that so?

Unfortunately, the arguments presented by buckminst and Uncle Al aren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ in buckminst's diagramm, $\sigma_{3}$ in Uncle Al's diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the right symmetry (in this respect the MO scheme of Uncle Al is better than the one of buckminst because it indicates this mixing).

So, the original problem still persists.
One possible explanation for the bond shortening after ionization is that the ionization leads to a shift of the $\ce{CO}$-electron-polarization (on ionization an electron is lost from the mostly $\ce{C}$-centered HOMO-$\sigma$ orbital and this leads to a formation of a positive partial charge on the $\ce{C}$ atom).
This strengthens the covalence of the $\ce{CO}$-bond and thus reduces the bond length (since the HOMO-$\sigma$ orbital is only slightly bonding there is not much bonding lost by taking away one of its electrons - the lost bonding is outweighed by the gain in covalency).
You can think of this strenghtening of the bond covalence in the following way: Two atomic orbitals can interact better (form stronger bonds) if their energies are close. Without the positive partial charge on $\ce{C}$ the AOs of $\ce{O}$ lie energetically a lot below the AOs of $\ce{C}$ (this is most pronounced for the $\ce{s}$ AOs which form the  HOMO-$\sigma*$ orbital). But with the positive partial charge on $\ce{C}$ the AOs of $\ce{C}$ are shifted down in energy and thus their energies are closer to the related AOs of $\ce{O}$ which leads to a stronger interaction when bonds are formed.


Does HCl kill algae? I live in the tropics, and have a water tank that sits in the sun, forming algae. Every year, I need to scrub the thing out. The plumber I’ve hired for the job wants to flush my water pipes with HCl to kill any residual algae. The pipes are mostly PVC, with braided metal hose and metal fittings here and there. Should I let him? If not, why not?

Are they metal pipes?  If you want to kill algae as such, flush with a diluted quaternary ammononium disinfectant, let sit a day, then flush everything very well indeed, and again.  If you have rubber or polymer fittings, don't do that, either, for the stuff diffuses in.


Elimination of bromide and acetate under reducing conditions Here is the reaction that I didn't find the mechanism of:

The conditions are zinc dust, acetic acid (solvent) at $\pu{11 °C}$
for 1 h.
The reagents are sodium acetate; 2,3,4,5-tetraacetoxy-1,6-bromohexane.
The product is 3-4-diacetoxy-1,5-hexadiene.
There is also some evolution of gas (I don't know what it is,
$\ce{HBr}$ or $\ce{CO2}$).

What reaction can lead to this elimination when it seems that no proton is leaving the molecule (instead of bromide) and acetate is leaving?

What reaction can lead to this elimination when it seems that no
  proton is leaving the molecule (instead of bromide) and acetate is
  leaving?

This is a reductive elimination reaction.  The classic textbook example is the reductive elimination of trans-1,2-dibromocyclohexane to cyclohexene in the presence of zinc dust and acetic acid.

It is basically a redox reaction where the zinc metal is oxidized and the dibromocyclohexane is reduced.  As with many reactions involving electron transfer from metal surfaces, the mechanism of this reaction is not fully understood.  Perhaps the zinc inserts itself into the carbon-bromine bond as a first step, but in any case, electron flow proceeds in the direction shown above.

The reaction has found many uses and has been widely modified to include cases such as yours where a bromine and another group are eliminated to form a double bond.  Here is a link to a paper where one of the bromines is replaced by an alkoxy group and the same zinc-mediated reductive elimination occurs.  Here is a link to another paper where zinc is used to perform a reductive elimination in a series of compounds where both bromines have been replaced, one by acetate and the other by hydroxyl.


Dissolving of non polar gases in water (liquid) How does dissolving happen for non polar gases in water? Is it like being in the space between the water molecules or does there occur some kind of interaction occur between gas molecules and water molecules?

Also if the gases dissolve in water, how can their solubility be increased? I have also heard what's called Henry's law accounted for this?

Answers on molecular scale would be more appreciated.

Certainly there are attractive interactions between water molecules and nonpolar solutes, there are always dispersion attractions.  Additionally, water will "layer" against a solute  particle, producing solvation shells, even in the complete absence of attractions, just because of the excluded volume forces packing  everything together.   However, for nonpolar gases the new gas-water interactions are significantly weaker than the water-water interactions they replace, so the enthalpic change on dissolve is strongly positive (unfavorable).  Hence the only reason gases dissolve is because the entropy increases, and that is simply because the volume available to the gas molecules is larger if it includes the volume of the water as well as the volume of the gas above it.

Another way to put that is that molecules of nonpolar gas are found in the water simply because they can be.  Id est, those molecules will zoom around in the gas phase, occasionally strike the surface of the water, plunge in, and wander around randomly for a while before exiting.  Net result: a low concentration of nonpolar gas dissolved in the water.  The concentration will increase with pressure simply because a higher pressure means a higher density of the gas, which means more gas molecules  will strike each cm2 of the water surface per second, and at equilibrium that must be matched by a higher rate of exit from the surface, which requires a higher equilibrium concentration in the bulk water.  For small pressures, the effect is approximately linear in pressure, and the coefficient is called the Henry's Law coefficient.

The dissolved gas particles do not occupy spaces between the  water molecules, if by that you mean preexisting spaces.  They shove the water molecules out of the way, create a hole, and live in that.  The volume of the solution may not increase, however, or may not increase as much as one might think, because as I mentioned above, the water will form solvation shells around the dissolved particle, and will moreover disrupt the hydrogen-bonding network in the water.  Both effects could result in an increase in density in the regions near the hole created by the solute particle.  So predicting whether and by how much the volume of the water expands (or contracts) is in general difficult.


Formation of hydrogen bromide: Does the concept of "order" apply to this reaction? Started reading Levine's Physical Chemistry [1] yesterday, and I hit upon (another) wall.

On page 517, Chapter 15, she includes a reaction resulting in the formation of $\ce{HBr}$ from $\ce{H2}$ and $\ce{Br2}$ (it's a homogeneous, gas-phase reaction) along with the corresponding rate equation:

$$\ce{H2 + Br2 -> 2HBr} \qquad r= \frac{k[\ce{H2}][\ce{Br2}]^{1/2}}{1 + j[\ce{HBr}]/[\ce{Br2}]},$$

where $k$ is the rate constant and $j$ is...um...just a(nother) constant O:)

Levine goes on to mention:

  [For this reaction] the concept of order does not apply.

Now I'm still new to the whole Chemical Kinetics thing... But what's this?!

How come the "concept of order does not apply" to this reaction? I've never come across this situation in any book so far. Well, the reaction seems trivial (though the rate equation says otherwise...); I fail to understand why an order can't be assigned to this (overall) reaction.

(Note: Until I know otherwise, I'm assuming "concept of order doesn't apply", is the same as saying "this reaction has no order". Do correct me if I'm wrong in doing so...I tend to miss out on subtleties...)

Even having an order of zero still counts as having an "order". But what could Levine possibly be talking about when she says that the idea of an "order" doesn't apply to this reaction? What are the (physical) implications of a reaction lacking an "order"?

My questions, explicitly put:

Why can't an order be assigned to the above-mentioned reaction? Or can it?
What are the physical implications of a reaction having "no order" (as compared to reactions with an order)?

References

Levine, I. N. Physical chemistry, 6th ed.; McGraw-Hill: Boston, 2009. ISBN 978-0-07-253862-5.

We don't assign orders. An order either simply is there, or it isn't. In particular, this expression doesn't look like $\ce{\underbrace{(something)}_{\substack{\text{may include }\ce{[H2]},\\ \text{but not }\ce{[Br2]}}}\cdot[Br2]^a}$ for any $a$, hence there is no order.

Alternatively, if we look at the reaction at its initial moments when $\ce{[HBr]}\approx0$, the denominator simplifies to just $1$, and we may say that the reaction has order $1\over2$ with respect to bromine.

Having or not having an order doesn't have any particularly far-reaching physical consequences. Moreover, it is not even an intrinsic property of a reaction; it may depend on the approximation used, as you saw above.

The scary rate equation comes from the so-called steady state approximation and the detailed reaction mechanism with intermediates and stuff, which is about two times longer than my answer, so I'd better refrain from including it. Well, here is the first line, just to give you the taste of the whole thing:
$$\ce{Br2}\underset{k_{-1}}{\stackrel{k_1}{\longleftrightarrows}}\ce{2Br^.}$$

To tell the truth, most real reactions are like that. The well-known $r=k\ce{[A]^a[B]^b}$ expression with stoichiometric coefficients for exponents is but a great oversimplification. It applies to reactions with no intermediates, which are quite rare.


Why is microcosmic salt called so? I really see no relation between the terms microcosmic and the salt's structural formula, $\ce{Na(NH4)HPO4.4H2O}$.

Devoting sufficient time on Google didn't help at all.

Information about the name of this compound is what I'm searching for, thanks in advance for sharing.

Devoting sufficient time on Google didn't help at all.

Hmmm... I found this PDF in just a few seconds, where we read the following.

The origin of the name microcosmic salt.
Beads of sodium metaphosphate
saturated with some oxides cool to a clear glass. Small crystals grow
in the glass on repeated heating and cooling. A microscope with a 80
to 100 x magnification is needed to inspect these crystals with
various morphologies. Such beads resemble a microcosmos. Some authors
tried to use the effect for analytical purposes in the 19th century,
but these efforts failed. Borax beads sometimes show the same effect.

Microcosmos here is understood (as usual) as a "small world", a "world in miniature".


How do we get NaClO3 in the reaction between Cl2 and NaOH? From problem no. 4080 in a Russian example exam set:

$$\ce{6NaOH + 3Cl2 -> NaClO3 + 5NaCl + 3H2O}$$

How do we get that $\ce{NaClO3}$? I can't seem to guess. I found this explanation on YouTube, but there the products are different:

$$\ce{Cl2 + 2NaOH -> NaCl + H2O + NaClO}$$

In both reactions, there is twice as much sodium chloride as there is chlorine. How do the two additional oxygens connect to the chlorine?

P.S. A related question - thanks, Nilay Gosh.

I find that when I'm faced when problems having to do with why the products are formed I make lewis structures.

if you make a lewis diagram of the $$\ce{NaClO3}$$

First off, you will see that a single Cl will be bonded to three oxygen molecules whom all carry full valance shells which were striped from the hydroxide ions in the reactants. (Remember that in solution the Na and OH will be around as free spectators.)

From the structure you will see the reason why there was a formation of the three waters. Since there were three oxygen stripped from the hydroxides and then recombination with more OH to make a reaction like this:

$$\ce{OH- + H+ -> H2O}$$

Also notice that on the products side for every molecule of NaClO3 made there will be a molecule for NaCl made. From there you can speculate that when the Cl-Cl bond breaks three oxygen atoms and a sodium atom will surround one of the Cl while another is tied to a Na molecule simultaneously.


Why do acids ionize in solution if they are covalently bonded? Maybe I am wrong, but I thought acids were covalently bonded since hydrogen doesn't form ionic compounds. For example, wouldn't $\ce{HCl}$ look like:

Since the electrons are shared, why does it break into $\ce{H+}$ and $\ce{Cl-}$ in solution?

Because it is energetically favourable ($\Delta{G}<0$) for hydrogen chloride to react with water to form hydronium( $\ce{ H3O+}$) and chloride ions.

Remember that $\ce{ H+}$ does not exist as $\ce{ H+}$ in water, but rather as $\ce{ H3O+}$.


how to add a methyl group to a straight carbon chain I mean like how can I convert ethane into propane, ethene into propene and like this further on

Why would you want an expensive process to convert cheap molecules into other cheap molecules?  Brominate ethane, make the organo-lithium, displace methyl iodide. Epoxidize ethylene, react with methyl lithium, eliminate to the olefin. Silly.

Doing something cheap, easy, and fast on a huge scale with unwanted methane from oil wells to form a high energy density liquid for pipeline conveyance will make you rich. Methanol is marginal - and still can't be done.  Perhaps methane plus air into titanium-silicalite (plus hydrogen peroxide, or cobalt- or vanadium-doped plus air) then that mixed product through ZSM-5 into aromatic gasoline.  That could be exciting in various ways.

http://dx.doi.org/10.1016/j.apcata.2008.04.020
http://chemelab.ucsd.edu/methanol/memos/final.html


Which elements have no liquid form at atmospheric pressure? Carbon has no liquid form at atmospheric pressure. Is this unique amongst naturally occurring elements?

The more interesting and vague question is, is this property somehow important or related to the many other amazing properties carbon has on Earth or just a coincidence?

Have a look at a generic phase diagram, for a compound with one solid phase, one liquid phase and a gas phase:

For the solid to go directly into the vapor phase (sublimation) at ambient pressure upon heating, you thus need its triple point to have a higher pressure than atmospheric pressure, i.e. ~1 bar.

Looking at this partial table of element triple points, arsenic is another element that sublimes at ambient pressure. Iron would seem to qualify, but it has more phases that reported in the table, and the full phase diagram indicates that it does not sublime at ambient pressure.


Work is not a state function From Work-Energy Theorem we have,

Work = Change in kinetic energy
         = Final K.E - Initial K.E

This shows that work only depends on the final and initial value of K.E independent of the path.

I'm lost. Can someone clear this up for me?

Your definition of work is not quite correct.  A better definition is**

Work w, is the energy change that drives net shift in the distribution of mass in the system.

Thus energy here means total energy which is kinetic plus potential. Imagine going to the top of a building. You gain potential energy by virtue of being at the top (m.g.h, mass m, acceleration due to gravity g, and height h). If you are not sure that you have gained energy, think about what happens if you were to walk of the top of the building!

You can walk up the stairs and gain this potential energy but you could also run up, and if you could manage to run, (huff-puff-pant-groan) the amount of work you would have to do is far more than walking, but the potential energy gained is exactly the same. Therefore in general the work depends on the 'path' taken, not just the starting and ending conditions. The word 'path' used in thermodynamics generally means the sequence of all the changes made to go from start to end. Hope this helps to clarify.

** ref A. Cooksey Physical Chemistry


Analytical solution for kinetics of bimolecular reaction Consider two chemicals, $\ce{A}$ and $\ce{B}$ that react with each other to make $\ce{C}$ with a reaction rate $k$. The reaction can be expressed as $$\ce{A + B->C}$$ The equation expressing the rate of the reactions can be expressed as $$\frac{d[\ce{A}]}{dt}=\frac{d[\ce{B}]}{dt}=-\frac{d[\ce{C}]}{dt}=-k[\ce{A}][\ce{B}]$$

I can separate this equation to make a system of differential equations.$$\frac{d[\ce{A}]}{dt}=-k[\ce{A}][\ce{B}]$$ $$\frac{d[\ce{B}]}{dt}=-k[\ce{A}][\ce{B}]$$

With these two equations, I note that they are similar and will only work with one of these equations for the time being. Therefore, we can write one of these equations as $$\frac{d\ln([\ce{A}])}{dt}=-k[\ce{B}]$$ and by taking another derivative $$\frac{d^2 \ln([\ce{A}])}{dt}=-k\frac{[d\ce{B}]}{dt}=-k\frac{[d\ce{A}]}{dt}$$

I solved this equation using Wolfram Alpha (QED) $$[\ce{A}](t)=\frac{c_1 \exp[c_1(t+c_2)]}{k \exp[c_1(t+c_2)]-1}$$ Therefore the rate of reaction can is $$[\ce{A}]'(t)=\frac{c_1^2 \exp[c_1(t+c_2)]}{k \exp[c_1(t+c_2)]-1}-\frac{k c_1^2 \exp^2[c_1(t+c_2)]}{(k \exp[c_1(t+c_2)]-1)^2}$$

I observed that the rate of change can be written as $$[\ce{A}]'(t)= c_1 [\ce{A}](t)-k [\ce{A}](t)^2$$ so that $c_1$ may be solved, given the initial conditions of $[\ce{A}](0)$ and $[\ce{A}]'(0)$ such that $$c_1=\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}$$

Substituting the definition of $c_1$ into the equation of $[A](t)$ and $[A]'(t)$ an equation for $c_2$ can be found.

$$c_2=\frac{1}{c_1} \ln(1-\frac{c_1 k}{[\ce{A}](0)}) $$$$
c_2= \frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2}  \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k) $$

Using the equations for $c_1$ and $c_2$ an explicit equation for $[\ce{A}](t)$ can be found.

$$ [\ce{A}](t)=\frac{\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)} \exp[\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}(t+\frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2}  \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k))]}{k \exp[\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}(t+\frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2}  \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k))]-1} $$

Side note: since $[\ce{A}]'(0)=[\ce{B}]'(0)= -k[\ce{A}](0)[\ce{B}](0)$ then $c_1$ can be rewritten as $$c_1=\frac{-k[\ce{A}](0)[\ce{B}](0)+k[\ce{A}](0)^2}{[\ce{A}](0)}=k([\ce{A}](0)-[\ce{B}](0))$$

this simplifies $c_2$ to

$$c_2=\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)})$$

which simplifes the equation for $[\ce{A}](t)$ to

$$[\ce{A}](t)=\frac{k([\ce{A}](0)-[\ce{B}](0)) \exp[k([\ce{A}](0)-[\ce{B}](0))(t+\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)}))]}{k \exp[k([\ce{A}](0)-[\ce{B}](0))(t+\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)}))]-1}$$

with a similar equation for $[\ce{B}](t)$

My quesion is: Is this a valid mathematical model for a bimolecular reaction? If not, what is commonly used?

Assuming that the bimolecular chemical reaction $\ce{A + B ->[\kappa] C}$ has mass action kinetics, we have the following pair of coupled ODEs

$$\begin{array}{rl} \dot a &= - \kappa \, a \, b\\ \dot b &= - \kappa \, a \, b\end{array}$$

where $\kappa > 0$ is the rate constant, $a := [\ce{A}]$ and $b := [\ce{B}]$. Since $\dot a = \dot b$, we have $\frac{\mathrm d}{\mathrm d t} \left( a - b \right) = 0$ and, thus, integrating, we obtain

$$a (t) - b (t) = a_0 - b_0$$

where $a_0 > 0$ and $b_0 > 0$ are the initial concentrations. Since $b (t) = a (t) - (a_0  - b_0)$, the 1st ODE can be decoupled from the 2nd, as follows

$$\dot a = - \kappa \, a \, \left( a - (a_0  - b_0) \right)$$

which can be rewritten in the form

$$\frac{\mathrm d a}{a \, \left( a - (a_0  - b_0) \right)} = - \kappa \, \mathrm d t$$

Assuming that $a_0 \neq b_0$, we have the following partial fraction expansion

$$\left( \frac{1}{a - (a_0  - b_0)} - \frac{1}{a} \right) \mathrm d a = - \kappa \, (a_0  - b_0) \, \mathrm d t$$

Integrating, we obtain

$$\ln \left( \frac{a (t) - (a_0  - b_0)}{a_0 - (a_0  - b_0)} \right) - \ln \left( \frac{a (t)}{a_0} \right) = - \kappa \, (a_0  - b_0) \, t$$

which can be rewritten as follows

$$\ln \left( \frac{a (t) - (a_0  - b_0)}{a (t)} \right) = \ln \left( \frac{b_0}{a_0} \right) - \kappa \, (a_0  - b_0) \, t$$

Exponentiating both sides, we obtain

$$\frac{a (t) - (a_0  - b_0)}{a (t)} = \frac{b (t)}{a (t)} = \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)$$

and, eventually, we obtain

$$\boxed{\begin{array}{rl} &\\ a (t) &= \dfrac{a_0  - b_0}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}\\\\ b (t) &= \dfrac{(a_0  - b_0) \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}\\ & \end{array}}$$

Taking the limit,

$$\lim_{t \to \infty} a (t) = \begin{cases} a_0 - b_0 & \text{if } a_0 > b_0\\\\ 0 & \text{if } a_0 < b_0\end{cases}$$

$$\\$$

$$\lim_{t \to \infty} b (t) = \begin{cases} 0 & \text{if } a_0 > b_0\\\\  b_0 - a_0 & \text{if } a_0 < b_0\end{cases}$$

What if $a_0 = b_0$?

Previously, we assumed that $a_0 \neq b_0$. If $a_0 = b_0$, then

$$\frac{\mathrm d a}{a \, \left( a - (a_0  - b_0) \right)} = - \kappa \, \mathrm d t$$

becomes

$$-\frac{\mathrm d a}{a^2} = \kappa \, \mathrm d t$$

Integrating, we obtain

$$\frac{1}{a (t)} - \frac{1}{a_0} = \kappa \, t$$

and, eventually, we obtain

$$\boxed{ a (t) = \frac{a_0}{1 + a_0 \, \kappa \, t} = b (t)} $$

In this case, both reactants are eventually exhausted

$$\lim_{t \to \infty} a (t) = \lim_{t \to \infty} b (t) = 0$$


Does nitric acid release hydrogen only when reacting with Mg and Mn? $\ce{HNO3}$ always reduces its anion to give nitrogen containing compounds ( $\ce{NO2}$, $\ce{NO}$, and even $\ce{N2O}$) when reacting with metals, but with $\ce{Mg}$ and $\ce{Mn}$ it releases $\ce{H2}$. Is there a specific reason for this variation?

Actually, it usually doesn't. It produces all types of products ranging from $\ce{NO2}$ to $\ce{NH4NO3}$ with exact products depending on concentration and conditions.

The reason for it is high oxidation power of nitric acid, exceeding that of hydrogen cations, so nitric acid reduces first. As result, to produce some nitrogen, the oxidation power must be brought below that of hydrogen cations. I'm not aware if it is actually possible, but possibly high dilution may help.

As for Mg, it reacts with hot water, so it's unclear if the reaction of $\ce{Mg}$ with diluted nitric acid counts. Not aware about $\ce{Mn}$ though.


Normalization of a free particle in 1 dimension I am having difficulty normalizing the wavefunction for a free particle in 1 dimension.
I have sent an attachment of my working out.
My attempt at the normalisation
Free electron:
$$-\frac{\hbar^2}{2m}\frac{\mathrm d^2 \psi(x)}{\mathrm dx^2} = E\,\psi(x)$$
\begin{align}
\psi(x) &= A\sin(kx) + B\cos(kx) \\
\psi(x) &= \psi^*(x)
\end{align}
$$\therefore \int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(ks)\right]^2 = 1$$
$$\int^\infty_{-\infty}\left[A^2\sin^2(kx) + AB\sin(kx)\cos(kx) + AB\cos(kx)\sin(kx) +B^2\cos^2(kx)\right]\mathrm dx  = 1$$
Since, $\sin^2 x + \cos^2 x = 1$,
$$\implies \int^\infty_{-\infty}\left(A^2 + B^2 + 2AB\sin(kx)\cos(kx\right)\mathrm dx) = 1$$
Since, $\sin(2x) = \sin x\cdot\cos x$,
$$\int^\infty_{-\infty}\left(A^2 + B^2 + 2AB\sin(2kx)\right)\mathrm dx = 1$$
$$\int^\infty_{-\infty}\left(A^2 + B^2\right) + AB\sin(2kx)dx = 1$$
\begin{align}
A^2 + B^2 + AB\int^\infty_{-\infty}&\sin(2kx)dx \\
&=A^2+ B^2 +AB\left[-\frac{1}{2}cos(2kx)\right]^\infty_{-\infty} \neq 1
\end{align}
Could someone please let me know of my mistake?

To answer the question, over and above the errors noted in the other answer the integral is just not well defined - it is infinite. There are two ways to see this

The "common sense" method. Consider
$$ \int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(kx)\right]^2dx$$
The quantity under the integral is positive semi-definite - it is always greater than or equal to zero. It describes oscillations all of which are positive. Each oscillation bounds a finite positive area between it and the x axis. As the bounds of the integral are all space there are an infinite number of oscillations. An infinite number of positive finite areas evaluates to an infinite area. Thus your integral is not well defined.

Maths - assuming A and B are real (but removing this restriction makes no difference to the final conclusion)

$$\left[ A\sin(kx) + B \cos(kx)\right]^2 = \left[A^2\sin^2(kx) + 2AB\sin(kx)\cos(kx) +B^2\cos^2(kx)\right]$$
The way to evaluate the integral is to use the identities
$$ 2sin(\theta)cos(\theta) = sin(2\theta) $$
$$ sin^2(\theta) = {{(1-cos(2\theta))}\over2}$$
$$ cos^2(\theta) = {{(1+cos(2\theta))}\over2}$$
Thus we can write the integral as
$$
\int^\infty_{-\infty} \left[A^2{{(1-cos(2kx))}\over2}+B^2{{(1+cos(2kx))}\over2}+ABsin(2kx)\right]dx
$$
The first term contains
$$
\int^\infty_{-\infty} \left[ A^2\right]dx
$$
This is infinite. Similarly the second term for $B^2$. Thus the integral is not well defined [yes this is slightly hand-waving but avoids a discussion of $cos(\infty)$ and similar]. So what you are doing wrong is trying to evaluate an integral that doesn't have an answer.
What you are running into is the fact that not all wavefunctions are normalisable. See Normalizing the solution to free particle Schrödinger equation in the physics stack exchange forum for more details


How do phase transitions occur in the solid phase? For example, how does graphite transform into diamond as pressure increases at a certain (adequate) temperature? I'm not sure if this is a great example to look at the bigger picture, so another example would be pure iron. At atmospheric pressure, as liquid iron cools down it solidifies to $\delta\text{-}\ce{Fe}$, with a bcc crystal structure. Further cooling down brings it to $\gamma\text{-}\ce{Fe}$, which has a fcc. Then, it keeps cooling down and it solidifies to $\alpha\text{-}\ce{Fe}$, again with a bcc structure. I don't really see how do the atoms physically rearrange to form the different crystal structures, neither do I see why would they jump back and forth between these. To add to this, I believe the difference between $\alpha$ and $\delta$ are their paramagnetic properties.

I guess the general question is, how do the atoms in the solid phase rearrange themselves by changes in pressure or temperature? (in contrast to, for example, a liquid-gas phase transition). And of course, why? What drives this change?

Phase transitions in solids occur in the same manner as in liquids or gases, through nucleation and growth. Yes, those processes require atomic motion. Yet,atoms diffuse in solids all the time, through various mechanisms. While potentially slower, the thermodynamics of solid phase transitions are easily seen in calorimetry measurements for example.

Nucleation and growth of precipitates is a key part of many engineering alloys. On the other hand, some phase changes (like diamond to graphite) are very hard to make happen at near-human-tolerable temperatures.

The solid phase also sees the martensitic phase transitions, which are second-order diffusionless atomic rearrangements. The name comes from the martensite phase observed in steels. Shape memory alloys are another example of martensitic phases.

A classic materials science text would be 'Phase Transformations in Metals and Alloys", by D.A. Porter and K.E. Easterling. In my opinion, the second edition is vastly preferable to the more recent third edition where the changes mainly seemed to be reintroducing all the errors that were removed in the second edition (sigh).


Comparison of basicity of o-phenanthroline and ammonia I was trying to compare the basicity of o-phenanthroline and ammonia.
According to me, the factors that are seen in o-phenanthroline are:

o-Phenanthroline has two nitrogen atoms at the 1,10 positions.
It has a free lone pair on one nitrogen that is not delocalized due to resonance.
The nitrogens are $\mathrm{sp^2}$ hybridized which is more electronegative compared to the $\mathrm{sp^3}$ lone pair on ammonia.

However, would there be hydrogen bonding in its conjugate base? If so, then how do we compare its basicity with a base like ammonia?

A quote from here explains and give a good answer to your wondering:

Amines are the most basic of the common organic functional groups, but are still fairly weak bases. Protonation occurs on the non-bonded electron pair exclusively. The basicity of amines is directly dependent on the “electron density” at the nitrogen atom. Both inductive and resonance effects can alter the basicity of a nitrogen atom.
Hybridization on the $\ce{N}$ also affects basicity. An increase in $\mathrm{s}$ character on an atom increases the electronegativity of that atom which favors acidity and therefore disfavors basicity. Hence $\mathrm{sp^3}$-hybridized nitrogen is more basic than either $\mathrm{sp^2}$ or $\mathrm{sp}$ hybridized nitrogen.

The availability of this non-bonding lone pair is a factor of basisity. The 1,10-Phenanthroline is a pyridine derivative. Thus, both lone pairs of two $\ce{N}$ atoms in the ring system are contributed to the system's aromaticity. This make them not 100% available to incoming protons while the lone pair in ammonia is 100% available. Therefore,  in aqueous solutions, the basicity of $\ce{NH3} \ (\mathrm{p}K_\mathrm{a} = 9.3)\gt$ basicity of 1,10-Phenanthroline $(\mathrm{p}K_\mathrm{a} \approx 4.9)$. This is similar to $\mathrm{p}K_\mathrm{a}$ of pyridine, which is $5.2$ in water (comparison with piperidine is depicted in following scheme):

Keep in mind that there are some other factors are effecting the basicity of amine as well. For example, the ring sizes of cyclic amines (Ref.1):

The order is 5-membered $\ge$ 4-membered $\gt$ 6-membered $\gg$ 3-membered.
References:

Scott Searles, Milton Tamres, Frank Block, Lloyd A. Quarterman, "Hydrogen Bonding and Basicity of Cyclic Imines," J. Am. Chem. Soc. 1956, 78(19), 4917–4920 (https://doi.org/10.1021/ja01600a029).


Why is the inside of a cell both more acidic and more negative than its environment? In a very simplistic way of thinking, an acidic solution has more positively charged $\ce{H+}$ ions present, whereas a basic solution has more negatively charged $\ce{OH-}$ ions present.

Can that be simply reconciled with the fact that (1) the intracellular pH of about 6.8 is more acidic than the intercellular pH of about 7.3, and that (2) a typical cellular membrane potential is -40 to -80 mV?

The key is that $\ce{H+}$ and $\ce{OH−}$ are not the only ions at play here.

The selective permeability of different cellular membranes for different anions (i.e. $\ce{Cl−}$) and cations (i.e. $\ce{K+, Na+, Ca^{2+}, Mg^{2+},}$ etc.) results in charge separation. Cellular membrane potentials are thereby created, with or without an intercellular / intracellular pH gradient.


Relative O-O bond lengths in O2X2 molecules (X = H, F, Cl) I am facing problem in the following question:

  What is the order of bond length of $\ce{O-O}$ in the following
  compounds:

  $\ce{H2O2, O2F2, O2Cl2}$

In my view since fluorine is more electronegative than oxygen it will attract the bonded electrons towards itself resulting in low bond length in case of $\ce{O2F2}$ than in case of $\ce{O2Cl2}$. However if the same concept is applied to $\ce{H2O2}$ it should have the greatest $\ce{O-O}$ bond length however my book states it otherwise.

First things first: the shorter a bond, the higher the multiple bond character. I.e. a double bond is shorter than a single bond, and a bond with a bond order of $1.5$ is in the middle of the two. Thus, the shorter the bonds in $\ce{X2O2}$, the higher the corresponding double bond character.

The key to understanding why the bond in $\ce{O2F2}$ has a very high double bond character is the gauche effect. The effect has its name because of the observation that $\ce{O2F2}$ prefers a gauche configuration rather than the expected (due to sterics) anti configuration. The reason is electronic in nature: a lone pair of oxygen A can interact with the $\sigma^*(\ce{O_B-F})$ orbital. This lowers the energy of the populated orbital at the expense of the unpopulated one. Since we have an interaction along $\ce{O_A-O_B}$, this increases the double bond character of the $\ce{O-O}$ bond. At the same time, since we are using the $\sigma^*$ for stabilisation, the $\ce{O-F}$ bond decreases in bond order — it becomes less than a single bond. In Lewis resonance depictions, this can be explained by the following mesomeric structures:

$$\ce{F-O-O-F <-> F-O^+=O\bond{...}F-}$$

This effect is strongest in $\ce{O2F2}$ since fluorine is more electronegative than oxygen and therefore the $\sigma^*$ orbital has a much higher oxygen contribution.

In hydrogen peroxide, a similar mechanism is at play that also contributes to reducing the bond length and increasing the bond order. In $\ce{Cl2O2}$, the effect is weakest.


Why is F2 (fluorine) colourful despite being diamagnetic? According to Molecular Orbital Theory, $\ce{F2}$ must be diamagnetic, then why is it colourful?

Also, $\ce{O2}$ is paramagnetic but it is colourless in gas phase. Why?

The colour of a molecule does not depend on the whether the molecule is diamagnetic or paramagnetic. The reason is that the colour is caused by absorption of a photon in the visible part of the spectrum. The photon has enough energy to excite one of the two (spin paired in a diamagnetic molecule)) electrons  in the highest occupied molecular orbital (HOMO) and into a higher energy unoccupied orbital, usually the lowest unoccupied molecular orbital (LUMO). From spin conservation rules the electron should not change spin. A spin change would correspond to a 'forbidden process' which has a very low chance of occurring so leads to exceptionally weak absorption.

The molecule now has one electron in the lower orbital and one in an upper orbital, these form the excited state, and if the spins are paired it is called a singlet state and the spin multiplicity is 1.

(multiplicity is 2*(total spin) + 1:  if the spins are paired the total spin quantum numbers are +1/2+(-1/2)=0. If the spins are unpaired the total spin is 1/2+1/2 and the multiplicity is 3, a triplet state).

In oxygen, which has a triplet ground state, on absorbing a photon an electron is still excited into an upper unoccupied orbital, the spin multiplicity of the excited state produced depends on the relative spins of all the unpaired electrons just as in the case of fluorine.

(As it happens fluorine seems to be a complicated  example as it has a low dissociation energy so absorption of a photon puts the molecule into a repulsive excited state leading to two F radicals as the molecule dissociates. It also absorbs weakly at longer wavelengths than needed to dissociate and this may be due to  a forbidden, spin changing, transition as mentioned above).


Volume of gas in containers If gases occupy all the volume in a container then how can we define 'volume percent' as all different gases will have same volume?

We can and do use volume fractions to express the level of a gas of vapour in air, it is volume ppm.  For example air to a first approximation is 20 % oxygen and 80 % nitrogen.

This means 20 % of the molecules in air are oxygen while 80 % are nitrogen. If we treat the gases as ideal gases then air is by volume 20 % oxygen.


Testing a sodium aluminate solution I prepared some sodium aluminate by reacting a strong $NaOH$ solution with aluminium foil and filtering the result. I now have a clear solution and I'm not sure what it is. I stopped adding aluminium when the reaction rate showed down dramatically in a hot water bath. Basically what I'm asking is what are the properties of sodium aluminate since Wikipedia is severely lacking in information on sodium aluminate

There are multiple reactions going on simultaneously. Aluminum metal reacts with the oxygen in surrounding air in order to create a plating of aluminum (III) oxide.

$$\ce{4Al(s) + 3O2(l) -> 2Al2O3(s)}$$

Upon addition of sodium hydroxide, the oxide layer reacts with the base, leaving the aluminum free to attack by water for instance.

$$\ce{2 Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$
$$\ce{Al(OH)3 + NaOH -> Na+ + [Al(OH)4]-}$$

In summary, we can write the sum of the reactions as:

$$\ce{2Al(s) + 2NaOH(aq) + 6H2O(l) -> 2Na+(aq) + 2[Al(OH)4]- + 3H2(g)}$$

The reason why the reaction proceeded slowly at first was due to the aluminum (III) oxide layer covering the aluminum metal.

From this link, http://www2.uni-siegen.de/~pci/versuche/english/v44-10.html:

  This reaction is used in drain cleaners. They are mostly made out of strong alkalis, to which alumunim or zinc has been added. The alkalis break down organic residues chemically. In addition, the formation of hydrogen leads to a bubbling effect which adds an additional mechanical cleaning mechanism.

Considering the properties of aluminates, one can go to PubChem hosted by the NIH which has a lot of information for almost every single compound necessary as well as literature linked to the specific compound in question:

http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=14766&loc=ec_rcs


How was water formed if oxygen and hydrogen exist as molecules? I have learnt that both hydrogen and oxygen  exist as molecules. If so how was $\ce {H2O}$ formed?

In water there is two hydrogen molecule and one oxygen atoms but how is this possible? Oxygen only exists as a molecule in our earth then how does the molecule oxygen react with molecule hydrogen to form water.

Moreover, if water is formed by combining of hydrogen and oxygen molecules, then the chemical formula of water should be $\ce{H2O2}$, shouldn't it?

Water does not have two oxygen atoms (has one) and one hydrogen (has two) hence the molecular formula of water is H2O. Water forms when there is enough energy to break the molecular bonds within the Hydrogen and Oxygen diatoms, they then combine and form water. As Mathew said, look at the Standard Gibbs free energy of formation. It will explain that H2O is much more favourable to be formed than H2O2.


will mercury vaporize in boiling water in a flask? Will mercury vaporize with the water vapours if boiled in a flask full of water?

Also if we store mercury under water in the beaker will it still give vapours?

If some mercury is bound to vaporize , will there be any mercuric oxide formation or not ??

Mercury evaporates to some extent even at room temperature: Vapor pressure: 0.002 mm Hg at 250C (~300 K). "Although the vapor pressure of elemental mercury is low [at room temperature], an atmosphere that is fully saturated with mercury vapor contains approximately 18 mg/m . The levels attainable in
indoor airs at room temperature can therefore greatly exceed
safe levels and result in poisoning." Mercury metal stored under water in an open container will still release Hg vapor into the air, as it is insoluble in water. Since mercury vapor, and worse yet, mercury compounds that might be formed are so toxic, I would not suggest experimenting with it except with proper guidance and with proper equipment, such as a negative-pressure hood feeding into a scrubber and a containment area for possible spills.

At one time, mercury was used to treat STD's and was used by hat-makers to preserve furs, with serious neurological damage to the users. It accumulates through the food chain to cause some fish to be toxic. If feasible, avoid using mercury in the lab and substitute a somewhat safer chemical.


short/extensive training on computational chemistry I want to take a training on computational works in chemistry. It should be of a short duration. Where in Europe I can find it?

Did you check the offerings of the national chemical societies and their special interest groups?

In Germany, that would be the CIC - Chemie Information Computer group of the GDCh - Gesellschaft Deutscher Chemiker.

Cosmologic provides TURBOMOLE workshops.

In 2014, the Max Planck Institute for Chemical Energy Conversion organized a Summer School during which a course on QM/MM with Frank Neese's ORCA was given. I don't know if they plan something similar this year.

The NSCCS has some upcoming events.


Flame temperature for cooking food Which has hotter temperature, blue flame or yellow flame?

Blue flame has a hotter temperature. In physics there is a relationship called Wien's displacement law, which states
$$\lambda_{max}T=2.898$$
where $\lambda_{max}$ is the peak wavelength of light emitted (which is the wavelength you see) and T is temperature in kelvins.

This relationship can be rewritten $$T=\frac{2.898}{\lambda_{max}}$$, or just $T\propto \frac 1 \lambda$.

From a table we find that the wavelength of blue light is about 450-500 nm and yellow light is 570-590 nm. Since temperature is inversely proportional to wavelength, the light with the smaller wavelength has the higher temperature.

(Now you might quibble that Wien's law is for idealized blackbodies but it can be used to model most things which emit light because of their temperature regardless.)


Why is it necessary to avoid even traces of moisture from a Grignard reagent? This fact was written in my text book. I wonder why that should be so.

Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.

$\ce{RMgX + H2O -> RH + Mg(OH)X}$

Where, $\ce{R}$ is a hydrocarbon.


What is the influence of an alkyl-hydroxy group on a benzene ring? We all know that an $\ce{NO2}$ group attached to a benzene is a deactivator, and it directs and substituents that would be adding to the meta position to the $\ce{NO2}$.  For an alkyl group attached to a benzene it is an activator, and it directs to the positions ortho/para to the alkyl group.

I haven't found any information on how a $\ce{-CH2-OH}$ group would affect the reactivity of the benzene ring shown below

According to Hansch, Leo, and Hoekman (Exploring QSAR: Hydrophobic, Electronic and Steric Constants 1995) the Hammet sigma value for $-CH_2-OH$ is 0 (zero) in both either the meta or para position.  Positive values indicate deactivators or electron withdrawing (+.778 for nitro in para position) and negative values are activators or electron donating (-0.268 for methoxy group).  The Hammett relationship gives a way to compare electronic effects of substituents on a benzene ring.  Further info on Wikipedia.


Is CH₃CH(CH₂CH₃)₂ the same as CH₃CH(CH₃CH₂)₂ The carbon in front of the parenthesis is bonded to 2 CH2 which then bonded to CH3, it cannot bond directly to CH3. I assume the correct way of writing the formula of this molecule will be CH₃CH(CH₂CH₃)₂.
However, the university lecturer says its ok to write CH₃CH(CH₃CH₂)₂ because it is a common knowledge that CH3 cannot bond to both CH2 and C at the same time which indicate that the carbon in front of the parenthesis is bonded to CH2.
Trying to understand the lecturer's explanation, does it mean that in a parenthesis, it doesn't matter whether one person writes the order of bonding forward or backward and it is up to the reader to figure out whether the carbon in front of the parenthesis is bonded to the first or the last carbon in the parenthesis?
I am a non-native English speaker and I hope my question is explained clearly. Tried my best to phase this. Please let me know if any part of the question does not make sense.

In chemistry, as in any "language", there is what must be said to be correct: it is the 1st way your molecule is written.
Your second way of writing it is not wrong but it does not correspond to the globally institutionalized standard. The goal for any language is to be understood no matter where you are.
The conventions indicate that it should be written $\ce{CH3-CH-(CH2-CH3)2}$ which does not induce any ambiguity


Is it possible to create a solution with superoxide? I am not a chemist, and somehow I cannot find any material on this topic, so I decided to ask here.

I am looking to create a solution with a specified amount/concentration of superoxide ($\ce{O2^.-}$). Can you suggest some compounds that will enable me to "isolate" superoxide in a solution?

Potassium superoxide (Aldrich) and an equimolar
  amount of 18-crown-6 ether used to increase solubility were
  dissolved in dry DMSO.

Role of Hydrogen Bonding in the Active Site of Human Manganese Superoxide
Dismutase Biochemistry 2004, 43, 7038-7045


Solubility and heat release of ytterbium (III) sulfate in water Question:

  The plot of solubility of some salts against temperature is given on the right. A salt like $\ce{NaCl}$ can be recovered fully, when dissolved in a solvent and the solvent is removed by evaporation. Choose the correct statement(s):

  No heat is absorbed or released when $\ce{NaCl}$ is dissolved in water.
  More heat is released when the ions of $\ce{Yb2(SO4)3}$ are solvated than that absorbed when the ions are separated.
  The heat of solution is negative for $\ce{H3BO3}$.
  When $\ce{AlCl3}$ is dissolved in water and the water is evaporated, solid $\ce{AlCl3}$ is recovered.

My approach: (1) is clearly correct. (3) and (4) are clearly wrong. My confusion lies with (2).

When the solubility of a salt decreases with temperature, this indicates that the enthalpy of dissolution is exothermic, i.e. heat is released during this process. This is the case when the lattice energy, which is at least required to separate the ions by breaking up the crystal lattice, is smaller than the energy which is released when the ions are solvated. The solubility curve of ytterbium(III) sulfate therefore proves the statement (B).


Electric conductivity vs potential What is the relationship between electric potential and conductivity?

For example, gold has a more negative electric potential than silver (-1.100 vs -0.800), but silver is a better conductor.

I would not expect to find any relationship.  The potential you are referring to involves breaking metallic bonds and ionizing the resulting atoms while conductivity involves electrons moving through the solid metal.  Copper's is -.34 and it is between gold and silver in conductivity.  Aluminum's is 1.66 and it is #4 behind gold in conductivity.