If all molecules have basically the same KE at a given temperature, why do lighter isotopes evaporate more readily? One of the proxies used for paleoclimate is the ratio of oxygen-18 to oxygen-16 in ice cores and in sedimentary rocks.  The idea is that water molecules with oxygen-18 generally evaporate less readily than oxygen-16, and condense more readily;  and this proclivity varies with temperature.  One can then use the deviations of oxygen-18 concentration from a standard ratio to say something about Earth's temperature in the distant past.
Why do light water molecules evaporate more readily?  It makes some intuitive sense that lighter molecules would evaporate more easily.  But how does this square with the equipartition theorem and the idea that all molecules have the same amount of kinetic energy at a given temperature?
It would seem to me that because the intermolecular forces between two water molecules are independent of their mass, it should take a certain amount of energy to remove a water molecule from its neighbors in the liquid phase and move it far away.  This energy requirement doesn't depend on the mass of the molecules, only on how the force on it varies with distance ($W = \int \vec{F} \cdot d\vec{r}$.)  So by this argument, molecules with the same amount of kinetic energy would have the same likelihood of evaporation.  And by the equipartition theorem, the translational kinetic energy of a water molecule is only dependent on the temperature, not on the mass of the molecule.
EDIT:  Just to pre-empt the "it's velocity that matters, not energy" argument:  I agree that lighter molecules would, on average, have a higher velocity.  But from classical physics, we know that applying a fixed force to a smaller mass leads to a larger change in velocity, while a larger mass gets a smaller change in velocity.  That would imply that the "escape velocity" required for a heavier molecule would be smaller; since it's heavier, its motion would not be affected as much by the force it feels from the other molecules. This would imply that while heavier molecules are moving more slowly on average, they also don't need to be moving as quickly to escape.  In other words, the velocity threshold for evaporation should be lower for heavier molecules.  These two effects (lower speeds and lower threshold for evaporation) should cancel each other out.
Where does my logic fail?

Evaporation depends on molecular velocity not on the energy
The simple mistake in your assumption is that evaporation is driven by molecules having enough energy to escape the liquid phase. This is wrong. Evaporation depends on whether the molecule is travelling fast enough to escape the liquid.
The molecules in the liquid will have, on average, the same kinetic energy. But this means that, in practice, the molecules will have a range of different kinetic energies across a wide distribution. But, since kinetic energy is a function of both molecular mass and molecular velocity, there will also be a distribution of velocities. And, crucially, the more massive molecules (in this case those with 18O) will have slightly lower velocities at the same average kinetic energy. So the lighter isotopes will be slightly more likely to have a high enough velocity to escape the liquid phase.
You can work through the mathematical details (taking into account the statistical distribution of kinetic energy and converting that to the distribution of velocities and then calculating the proportion of molecules with a velocity above the threshold for evaporation) and check that the proportion of light molecules is higher than the proportion of heavy molecules.

Clarification: why is the force-based view wrong
The argument that applying a fixed force leads to a smaller effect on a heavier molecule so heavier molecules don't need to moves as quickly to escape the liquid is flawed for several reasons.
It sounds relevant but isn't because the vast majority of all collisions don't lead to escape from the liquid. There are so many momentum (and KE) redistributing collisions that all we have to consider is the distribution of energy/momentum across the ensemble of molecules in the liquid. And what we need to know is the average proportion of molecules having enough velocity to escape after their last collision.
Yes, the force argument matters for individual collisions (if they are not purely elastic), but, if you simulated a drop of liquid down to the molecule level, you would find that the distribution of energy/momentum would match the statistical one expected from theory and you would have gained no new knowledge. Follow an individual molecule and, yes, the thinking about force will apply. But follow the distribution of velocities across the ensemble of molecules and what matters is the proportion of molecules having a high enough escape velocity and that is governed by the statistical distribution. Think about the properties of the ensemble of molecules not the individual molecule.
And the force argument ("the escape velocity will also be lower") doesn't work at the ensemble level. All that matters is the velocity after the last collision and the only way to know that is to look at the statistics of the whole system not the individual forces on individual molecules.

The simplest and most direct explanation is that isotopically heavier molecules have higher boiling points. Regular water (R) boils at 100.0$^o$ C. Heavy water (H) boils at 101.4$^o$ C. A mixture of the two (R + H) can be separated by a distillation column. The heavier molecules require more energy to evaporate, to push away one atmosphere.
Any similarity to (astronomical) orbital mechanics is misleading. The orbital metaphor is like throwing a stone, involving one total energy. Liquid evaporation (even at lower temperatures) is like wiggling thru a crowd: an equilibrium of molecules leaving and returning to the liquid, with a net loss of liquid (like diving thru a gap in traffic).
Evaporation of either R or H at lower temperatures relies on diffusion of the escaped molecule to a distance so far away that it is no longer in equilibrium with the liquid. Here the higher velocity of an escaped molecule gets it farther away quicker, and less likely to return to the liquid. Heavier molecules with the same energy move slower, don’t get as far away as fast - thus evaporate slower.
Looking at the macroscopic systems (like pure R and H) rather than individual molecules focuses on measurable properties (bp). It all boils down to differences in boiling point.


Is the t-butyl carbocation more stable than the benzyl carbocation? Various authors have different views regarding stability order of the benzyl and t-butyl carbocations.

$$\ce{PhCH2+ ; (CH3)3C+}$$

In my opinion, resonance effect dominates, so the benzylic carbocation should be more stable. But in the other case, both the inductive and hyperconjugation effects are present, which stabilize the intermediate carbocation.

What should is the correct stability order? Especially when you consider them in SN1 reactions, and what their effect on their rate is.

I am using a very simplistic quantum chemical approach of the following isodesmic reaction:

I have used Gaussian 16 Rev. A.03 and the DF-B97D3/def2-TZVPP level of theory. The summaries of the calculations are included below.

On this level of theory the depicted reaction has an energy change of $\Delta G = \pu{- 37.1 kJ mol-1}.$ Therefore one could assume that the 2-methylpropan-2-ylium cation is more stable than the phenylmethylium cation. These values were estimated at $T = \pu{298.15 K}$ and $p = \pu{1 atm}.$

One certainly can do more calculations, but it is, however, a start.

Calculation summaries

INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               phch2_q1/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -270.5777462860 hartree
zero-point corr.             (ZPE):            +0.114824 hartree/particle
thermal corr.                  (U):            +0.120571 hartree/particle
ther. corr. enthalpy           (H):            +0.121515 hartree/particle
ther. corr. Gibbs en.          (G):            +0.085614 hartree/particle
entropy (total)            (S tot):              +75.561 cal/(mol K)
heat capacity (total)       (Cv t):              +22.578 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               phch3_q0/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -271.4854244270 hartree
zero-point corr.             (ZPE):            +0.125049 hartree/particle
thermal corr.                  (U):            +0.131428 hartree/particle
ther. corr. enthalpy           (H):            +0.132372 hartree/particle
ther. corr. Gibbs en.          (G):            +0.093359 hartree/particle
entropy (total)            (S tot):              +82.110 cal/(mol K)
heat capacity (total)       (Cv t):              +23.831 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               t-c4h10_q0/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -158.4201368210 hartree
zero-point corr.             (ZPE):            +0.128849 hartree/particle
thermal corr.                  (U):            +0.134606 hartree/particle
ther. corr. enthalpy           (H):            +0.135550 hartree/particle
ther. corr. Gibbs en.          (G):            +0.101200 hartree/particle
entropy (total)            (S tot):              +72.297 cal/(mol K)
heat capacity (total)       (Cv t):              +20.521 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               t-c4h9_q1/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -157.5167928240 hartree
zero-point corr.             (ZPE):            +0.114033 hartree/particle
thermal corr.                  (U):            +0.120557 hartree/particle
ther. corr. enthalpy           (H):            +0.121501 hartree/particle
ther. corr. Gibbs en.          (G):            +0.083653 hartree/particle
entropy (total)            (S tot):              +79.659 cal/(mol K)
heat capacity (total)       (Cv t):              +20.965 cal/(mol K)

(Compiled with tools-for-g09.bash, which somewhat surprisingly works for g16.)

Optimised geometries

14
phch2_q1/b97d3tzvpp.freq.xyz
C          0.08663       -0.19347        0.09384
C          0.05691       -0.11475        1.46208
H          1.02702       -0.19311       -0.45239
H         -0.82919       -0.25883       -0.48899
C         -1.20323       -0.11721        2.16356
C         -1.22180       -0.03857        3.53614
C         -0.00341        0.04387        4.24119
C          1.24453        0.04873        3.58467
C          1.28552       -0.02918        2.21254
H         -2.12409       -0.18167        1.59151
H         -2.16067       -0.03932        4.07940
H         -0.02695        0.10566        5.32621
H          2.15902        0.11360        4.16439
H          2.23039       -0.02773        1.67720

15
phch3_q0/b97d3tzvpp.freq.xyz
C          0.00007        0.00000       -0.04694
C          0.00682       -0.00000        1.46143
H          1.01682       -0.00000       -0.44951
H         -0.51968       -0.88199       -0.43936
H         -0.51968        0.88199       -0.43936
C         -1.19805       -0.00000        2.17854
C         -1.20438       -0.00000        3.57251
C          0.00015        0.00000        4.27958
C          1.20614       -0.00000        3.57927
C          1.20606       -0.00000        2.18275
H         -2.14093        0.00000        1.63572
H         -2.14998        0.00000        4.10835
H         -0.00257        0.00000        5.36594
H          2.14940        0.00000        4.11910
H          2.15082        0.00000        1.64397

14
t-c4h10_q0/b97d3tzvpp.freq.xyz
C          0.00369       -0.00664        0.00284
C         -0.01764        0.03020        1.53527
C          1.44139        0.03012       -0.52791
C         -0.74698       -1.23338       -0.52790
H         -1.04360        0.03406        1.92005
H          0.49077       -0.85020        1.94859
H          0.49182        0.92075        1.92008
H          1.46219        0.03417       -1.62347
H          1.97408        0.92055       -0.17569
H          2.00042       -0.85041       -0.18649
H         -0.76068       -1.24961       -1.62346
H         -0.26410       -2.15775       -0.18616
H         -1.78452       -1.24933       -0.17586
H         -0.51485        0.89146       -0.36388

13
t-c4h9_q1/b97d3tzvpp.freq.xyz
C         -0.00184       -0.00056        0.03904
C          0.01681        1.45949        0.05366
H         -0.46367        1.80346       -0.88071
H          1.00698        1.90257        0.15089
H         -0.66770        1.83073        0.83269
C          1.25179       -0.74789        0.02785
H          1.76765       -0.48991       -0.91708
H          1.93393       -0.35706        0.79738
H          1.13840       -1.82916        0.09679
C         -1.27605       -0.71190        0.03180
H         -1.24998       -1.55839       -0.66820
H         -1.35257       -1.20345        1.02233
H         -2.15046       -0.07921       -0.11824

(g09.chk2xyz does not work for g16.)

From my knowledge, the benzylic carbocation should be more stable. This is due to the fact that it is aromatic as well. Aromaticity further enhances stability.


What is the structure of CF3 radical? In the class, I was told that $\ce{H3C^.}$ has a trigonal planar structure with the unpaired electron in $\mathrm{2p_z}$ orbital. But $\ce{H3C -}$ has a trigonal pyramidal structure.

But why does this happen?
Why does $\ce{F3C^.}$ also doesn't have a trigonal planar?

Edit : I have seen one possible answer in a different question.  But is there any answer that is explained with the help of VSEPR or MOT or VBT?

CH3 radical is Sp2 hybridised carrying the lone electron in a pure p orbital. This is so because Sp2 hybridised orbitals are lower in energy than Sp3 hybrid orbitals (The more the S character the lesser is the energy of the orbital). When there is a lone electron present, it is more efficient to keep the bond pair electrons in lesser energetic Sp2 hybrid orbitals and lone electron in higher energy pure P orbital than keeping all bonding pairs and the lone electron in Sp3 hybrid orbitals. So most of the radicals are Sp2 hybridised.

But in situations like •CF3 where the atoms/groups attached to C are of high electronegativity things may change. According to Bent's rule more electronegative atoms/groups will favour less electro negative orbitals i.e. orbital s with lesser s character(Sp3 has lesser S character compared to Sp2). This is quite obvious because the more the S character of the C center the more is the tendency of the C center to attract the bond pair towards itself and in cases like electronegative atom, it will also try to attract the bond pair towards itself. This makes some unhappy situation. This is reflected by the Bent's rule. That is why •CF3 radical favours Sp3 hybridisation.

It's not that only such radicals like •CF3 can acquire Sp3 hybridisation. In many other cases like bridge head of adamentum also acquire Sp3 hybridisation becauseof opposition by its rigid structure to acquire planarity on generation of radical though it's not that much easy process to form radical here.

In conclusion, a radical will favour that hybridisation which will be energetically favourable for it.


Addition of oxygen molecule to given aromatic compound I did figure out that the hydrogen atom is reactive as it has three electron donating groups attached to it, I know it reacts. But the answer given is that 6 different compounds can be formed (minor and major; including stereoisomers). But I need some help figuring them out.

The three hydrogen bond at -OCH3 would be reactive and
-CH3 at attached directly to ring would react so 6 products possible.


Why can sulfur have an expanded octet but not oxygen? In my search to understand the bonding in structures like $\ce{SF6}$ I found many sources that said it was because sulfur has d orbitals to accommodate an expanded octet, which made sense to me. But I also found sources like the paper by Reed and Weinhold (1986) that say d orbitals contribute very little to the bonding in $\ce{SF6}$. An alternative explanation is that $\ce{SF6}$ has three-center/two-electron bonds. What I don't understand is why compounds like $\ce{SF6}$, $\ce{ClF5}$, $\ce{PCl5}$ exist for third row element centers, but not $\ce{OF6}$ if it has nothing to do with d orbitals? References would be great.

sulfur has 2 electrons in its 3s subshell , and 4 in 3p subshell. On absorbing energy the paired electrons get excited to to the next higher energy level which is the 3d subshell. So after exciting it's electrons six unpaired electrons are present which can take part in bonding via overlapping of orbitals. Other compounds can also be explained similarly. Also the d- orbitals contribute to only two bonds out of six bonds( which would have been meant in the paper by Reed and Weinhold (1986) ). Oxygen has no empty orbitals with it . It only possess orbitals upto 2p which contains 4 electrons. So it cannot expand its octet by exiting it's electrons.


Why doesn't iron(III) iodide exist? $$\ce{Fe2O3 + 6HI -> 2FeI2 + I2 + 3H2O}$$

Why don't we get $\ce{FeI3}$? After all, iron's oxidation state is $+3$ in the reagent.

Should one just memorize that up to bromine, it's $\ce{FeX3}$, and below it's $\ce{FeX2}$?

The standard reduction potentials for the following half reactions can be found here.

$$
\begin{align}
\ce{Fe^3+(aq) + e- &-> Fe^2+(aq)} &\quad E^\circ &= \pu{+0.77 V} \\
\ce{I2(s) + 2 e- &-> 2 I-(aq)} &\quad E^\circ &= \pu{+0.54V} \\
\ce{Br2(l) + 2 e- &-> 2 Br-(aq)} &\quad E^\circ &= \pu{+1.07V} \\
\ce{Cl2(g) + 2 e- &-> 2 Cl-(aq)} &\quad E^\circ &= \pu{+1.36V}
\end{align}
$$

You can see from this that only iodide is a strong enough reducing agent to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$ at standard conditions. Even with non-standard concentrations it will be very difficult to get bromide to do the reduction because the difference in electrode potential is large.

The trend in electrode potentials for the halogens can be explained in terms of the increasing electronegativity going from iodine to chlorine which increases the first electron affinity. It just so happens that the crossover point with the iron reduction is between iodine and bromine.

Actually, some have voiced an opinion that FeI3 can apparently be created with difficulty, and is reputedly very unstable, decomposing into FeI2 and I2. Similarly, with CuI2, which is likewise unstable. To quote a source reference.
$$
FeCl3 + 3KI = FeI3 + 3KCl ;  FeI3 = FeI2 + I2
$$

Observation-
Large anions diminish the lattice energy of the higher halide to such an extent that the higher halide may be thermodynamically unstable.
Sample Solution-
Iodide ion is a good reducing agent and hence reduces metal in higher oxidation state into lower Ones.

Even, more interesting is the fact that FeI3 is extremely photosensitive! See this ebook, which describes FeI3 as "very difficult-to-prepare" and as "intensely black" in appearance, which is somewhat definite for a compound arguably that some claim does NOT exist.
So, in light, I would expect:
I- + hv -> .I + e-(aq)
Fe(III) (aq) + e- (aq) -> Fe(II)
.I + .I = I2
which means, in light, an even shorter half-life with potential transient atomic iodine presence and creation of elemental iodine.
So per reference above, one may be able to prepare (albeit briefly) some FeI3 (aq) by adding FeCl3 to a source of iodide (like aqueous KI). Here also is an interesting (but still speculative) supporting source, to quote from the abstract:

This study first reports that ferric chloride (FeCl3) can lead to the formation of iodinated coagulation byproducts (I-CBPs) from iodide-containing resorcinol solution or natural waters. The unwanted I-CBP formation involved the oxidation of iodide by ferric ions to generate various reactive iodine species, which further oxidize organic compounds. Although the oxidation rate of iodide by FeCl3 was several orders of magnitude slower than that by chlorine or chloramine, most of the converted iodide under the ferric/iodide system was transformed into iodine and iodinated organic compounds rather than iodate. Formation of four aliphatic I-CBPs was observed, and four aromatic I-CBPs were identified by gas chromatography mass-spectrometry and theoretical calculation. Coagulation of iodide-containing waters with FeCl3 also produced I-CBPs ranging from 12.5 ± 0.8 to 32.5 ± 0.2 μg/L as I. These findings call for careful consideration of the formation of I-CBPs from coagulation of iodide-containing waters with ferric salts.

The liberation of iodine together with the formation of iodinated coagulation byproducts (supportive of possible atomic iodine formation with any light exposure from the highly photo-sensitive FeI3) possibly suggests a photo-assisted breakdown of FeI3 (aq), in my speculation.


Reason for enthalpy being non-zero at constant temperature Suppose a reaction is occuring at constant temperature. Then $\mathrm dH=nC_{p}\mathrm dT = nC_{p} × 0=0$.

Please explain what the mistake is that I have done. Clearly, there are many reactions which take place at constant temperature and have a definite enthalpy change. But according to the equation $\mathrm dH=nC_{p}\mathrm dT$, if temperature $T$ is constant then $\mathrm dT =0$ and $\mathrm dH = 0$. But this is clearly not true. There are reactions with constant temperature and $\mathrm dH≠0$. Please explain what is wrong?

The only way that an exothermic reaction can occur with "constant temperature" is if the heat generated is constantly removed in some way. Even if the heat removal is essentially instantaneous, it is incorrect to say that there is no temperature change. Rather there is an infinite number of infinitely small temperature changes as the system heats up by an amount dT and then is returned to the set T. If you were able to measure these infinitely small changes and also to know the instantaneous Cp (which changes as the reaction composition changes), you could sum them up and get a non-zero value for the enthalpy change. Obviously, that's not a practical approach.
Instead, to determine the enthalpy change experimentally, it is easiest to perform the reaction within an insulated apparatus such that the temperature is not constant and the change in temperature can be used to determine the enthalpy change.

[OP] But according to the equation $dH=nC_{p}dT$, if temperature $T$ is constant then $dT =0$ and $dH =0$.

Even if this equation would describe a system completely, a constant temperature does not imply that $dH$ is zero. Instead, you can have a temperature bath of infinite size (n approaches infinity) so that the average temperature does not increase. Of course, there would be temperature fluctuations while the reaction goes on, even if it is a glacially slow reaction.

[OP] Clearly, there are many reactions which take place at constant temperature and have a definite enthalpy change.

From the discussion above, it might be clear that it is not so easy to run an exothermic reaction at constant temperature. Instead of having a huge water bath, you can also buy an instrument for isothermal calorimetry (ITC) that monitors the temperature and cools or heats the reaction mixture to keep it at constant temperature. In this instrument, the amount of heat tranfered is logged, and that is how you measure the enthalpy change. This is distinct from measuring with a traditional calorimeter, where you monitor the temperature change and estimate the heat capacity of the calorimeter. Estimating the heat capacity is difficult because it is affected by the reactants and products you are measuring.
What this means is that there are multiple sources and sinks of thermal energy. If you are successful in balancing them perfectly, it will be an isothermal process. Often in thermodynamics, you are making arguments that rely on an ideal system. The real systems might approach that ideal, so the observation will also approach that of the ideal system.


Reversible & irreversible adiabatic processes Work done in reversible is more than that of irreversible

Suppose we are going from state $(P_1, V_1, T_1)$ to some other state say $(P_2, V_2, T_2)$ for reversible and $(P_2, V_2, T_2')$ for irreversible. Which should be greater, $T_2$ or $T_2'$?

My attempt

Let us say heat supplied is same i.e q for both reversible and irreversible. We already know that $U = q - W$

Reversible as work done is more so $U$ becomes less and in irreversible $W$ is less so $U'$ is more than $U$ i.e $U' > U$

As $U'$ is greater $T_2'$ should be greater. Is this correct explanation?

For a given number of moles, once $P_1$ and $V_1$ are specified, that determines state 1 and $T_1$.  Once $P_2$ and $V_2$ are specified, that determines state 2 and $T_2$.  It doesn't matter what process is used to get from state 1 to state 2.  Only 2 intensive variables are required to specify the state of the system (assuming no phase change).  So $T_2=T_2"$.

Maybe you want to remove either the $P_2$ or the $V_2$ from the problem specification.

I believe your answer is correct.

Because we are doing work in this case, we can imagine that this is simply a piston being expanded against the atmosphere.

As you have said, a reversible process corresponds to the maximum amount of work, which in this case also corresponds to the largest change in energy.

Because we are only interested in comparing the temperature $T_2$ and $T_2'$ because the pressures and volumes change by the same amount, we must ask ourselves if the temperature is going to increase or decrease?

Well, without any mathematical proof, I will say that while we are doing work, and thus giving out energy to the atmosphere, the temperature must decrease because our system must lose energy. Thus, while outputting work, the more work we do, the greater the change in temperature.

The above paragraph depends on the process being adiabatic though because if the process were not adiabatic, we would have to consider the relative amounts of heat flow and things could get weird.

But, we have said that $$W_{rev}>W_{irrev}$$thus, $$|\Delta T_{rev}|>|\Delta T_{irrev}|$$but, since temperature is decreasing as we give out energy, $$T_2<T_2'$$

So, you are correct unless we are both making a stupid mistake.

This answer makes sense though, because the for the reverse process, we would be compressing a piston and thus must get back as much energy as possible in the form of a temperature increase. With changes in pressure and volume being the same, we know immediately that the process which begins at the lower temperature and gets back to the original temperature $T_1$ has the largest change in internal energy which must be true for the reversible process.


Why is entropy favorable? I cannot seem to grasp the logic behind it. We say that more entropy (or more disordered system) is favorable over less entropy. But why? Why is randomness preferred over proper arrangement of atoms/molecules? The only thing I can think of is more degrees of freedom. Molecules in a more random system would have more degrees of freedom, and would thus be favorable. But, again, why? Why would more degrees of freedom be favorable? I've already asked my teachers and looked up internet, but haven't found any satisfactory answer. Also, is there any way you could also explain diffusion by the same concept (diffusion is entropically favorable, right?)?

PS: if this question is better suited for physics stack exchange, I'd be more than happy to have it migrated.

Thermodynamics. The second law of thermodynamics states that entropy always increases in an isolated system. This is taken as a fundamental postulate---we simply accept this statement as a fact regarding how the world works, and our justification is that no experiment has ever shown the second law incorrect. In the framework of macroscopic thermodynamics, there is no deeper principle from which we can hope to derive the second law.

Statistical mechanics. Statistical mechanics brings a microscopic perspective into thermodynamics. We talk of macrostates---macroscopic states of our system, characterized by (macroscopic) observables like pressure, temperature, volume,  etc.---which may be realized by a number of microstates---a complete microscopic description of our system, consisting of the positions and momenta of each particle. Many different microstates can correspond to the same macrostate, since a macrostate only deals with macroscopic properties.

In statistical mechanics we can go one level deeper, taking as a fundamental postulate the principle of equal a priori probabilities, the idea that the system has an equal probability of being in any given microstate. This implies that the most likely macrostate for the system is that with the greatest number of corresponding microstates. At this point, knowing also from statistical mechanics that the entropy is a monotonically increasing function of the number of microstates (specifically the logarithm of the number of microstates in an isolated system, but the functional form is unnecessary here), we can motivate the second law: the entropy of a macrostate of an isolated system is a measure of the number of its corresponding microstates; the greater this number, the more likely this macrostate will be observed. The equilibrium state of a system is simply its most probable macrostate.

Example. (Microscopic diffusion.) Let us consider a microscopic model of diffusion: a $10$-by-$10$ grid of particles, $50$ red and $50$ blue. The particles are indistinguishable except by color. In addition, we will consider two sets of macroscopic states, 'ordered' and 'disordered'. Take as initial state the configuration with the two groups of particles separated by color into two equal halves by a horizontal boundary, the blue particles on the top and the red on the bottom. The diffusion process is akin to removing this artificial boundary.

There is one initial ordered microstate and $100!/(50!)^2$ final microstates. (Note that indistinguishable microstates should be considered the same microstate.) The entropy of our system has increased, because we have relaxed a constraint and allowed more microstates into our system. Most of these final states look disordered. Diffusion is therefore an entropically favorable process that brings an ordered system into a disordered one.

Additional remark. I find statements like "entropy drives the system towards a particular state" to be somewhat misleading, because they imply that entropy behaves like a force. But entropy, as we've seen, is really just statistics.

If we look at why a chemical bond occurs maybe that might shed light on to this question. If we take diatomic hydrogen which has one electron from each hydrogen, the question arises why do two nuclei come together to form a bond since Coloumbic forces are increased. And the answer is because the electrons have a larger volume to exist in as they will be able to exchange between nuclei(I'm referring to the exchange integral). Thus we can summarize that a chemical bond occurs due to purely entropic reasons and as a result the energy of the system is decreased.


Regioselectivity in Carbometallation I am a student of organic chemistry and I frequently watch online open course lectures by various professors around the globe. Of these, in https://www.youtube.com/watch?v=OYAMls5x4EI, at about 27:10, I had a question, as mentioned in the image below.

Mechanism

What factors determine where the palladium will attach?

My initial thoughts were to find out the electron-rich regions of the $\pi$ bond, and the palladium atom in the $\mathrm{+II}$ oxidation state will act as an electrophile. But I wasn't able to figure it out exactly.

I understand that the ratio of products shown in the figure can be controlled by using various ligands, especially based on their steric effects, in this case.

Your mechanism is missing an important step: between the oxidative addition of palladium into the $\ce{C-Br}$ bond and the alkene insertion there should be an alkene annealing forming a palladium-alkene complex. This is displayed in the mechanism scheme of a generic Heck reaction below, which I sheepishly borrowed from Wikipedia.

Scheme 1: Catalytic cycle of a Heck reaction. Image taken from Wikipedia, where a full list of authors is available. Note that the β hydrode elimination in step C should be syn and not anti or gauche as displayed.

The annealing step and the π complex are important because they influence the Heck reaction’s selectivity. Due to the avoidance of steric stress, both residues will adopt an anti-type conformation as depicted in structure 3. On the (syn-selective) insertion of the alkene into the $\ce{C-Pd}$ bond, this geometry is kept resulting in the palladium catalyst ending up on the higher-substituted alkene carbon. For the inverse insertion, both organic residues — in your case two phenyl groups — would have to occupy the limited space on one side of the catalyst.

The subsequent β hydride elimination can theoreticly occur in both directions in your molecule. However, eliminating towards the hydroxy group allows keto-enol tautomerism with further stabilisation. Since β hydride elimination is a reversible process, the reaction will select the ketone over the allylic alcohol that would be formed upon opposite β hydride elimination.

The alkene insertion into the $\ce{Pd-C}$ bond, is not reversible. Two $\ce{C-C}$ single bonds are formed at the expense of a palladium-π interaction resulting in a notably downhill process.

Just have a look at the reductive elimination step of this Heck reaction:

In case of the second reaction the resulting allylalcohol is far less stable than the enol tautomerizes to the ketone. The other thing are sterics aspects that make the migratory insertion step far less favourable for the second product as well.


Neutralizing sulfur-like smells in car A full bottle of jewelry cleaning (links product and material safety data sheet below) broke in my car and went unnoticed for a few days.  Now the car has a sulfur-like smell whenever I go into the car.  It dissipates within a few minutes, but then returns very quickly once I leave the car.

The primary chemical components are alcohol ethoxylates, and can decompose into carbon monoxide, carbon dioxide nitrogen Oxides, sulfur oxides and other toxic gases.

I have cleaned the area and had the carpet (rear driver) shampooed from a car detailer, but still the smell remains.

What chemical compound can I use to neutralize the smell permanently (enter equilibrium)?

Product Link (Amazon)
Product Link (Manufacturer)
Material data safety sheet

I guess that it actually was a silver cleaner, not one for gold or precious stones.

The latter mostly contain nonionic surfactants, which are made by the reaction of water-insoluble long chain alkanols (C12 - C15, from natural fatty acids) with oxirane. These washing agents dissolve well in water and help to remove grease and fingerprints from the jewelery. Their action isn't much different from that of a dishwasher for manual cleaning.

These cleaners don't smell the way you described and the car retailer probably used similar cleaners for the car.

Silver cleaners are a different league. The do contain the same surfactants (washing agents), but in order to actually remove the tarnish (silver sulfide) from the silverware, they do also contain a bit of acid (typically sulfuric acid) and thiourea, which probably is responsible for the smell.

Repeated shampoeing probably is the best method to get rid of the thiourea without ruining the carpet.

Homemade Carpet Fresh
I would recommend a home made dry cleaning using diatomaceous soil compound. This will eliminate the problem of toxins and erosion of the metal interior under the carpet. Use some DAP mixed with your favorite dry detergent and vacuum it up. If plaster of Paris is not available basically any spackle will work. Mix 1/1


How to calculate the Gibbs energy for the vaporisation of solid ammonia? Given that $P_{\ce{NH3}}=\mathrm{10~atm}$ at $298\ \mathrm K$, calculate the $\Delta G^\circ$ of the reaction $$\ce{NH3(s) <=> NH3(g)}$$

I'll begin solving the problem...

I'm using the formula $$\Delta G=\Delta G^\circ+RT\ln Q.$$ The vaporisation of $\ce{NH3}$ is an equilibrium so  $\Delta G=0$ and $Q=K_c$ . Also $K_c=\left[\ce{NH3}\right]$. So $\Delta G^\circ=-RT\ln K_c$.

How do I continue? Is it $K_c=K_p=10$ or $K_c=K_p/\left(RT\right)$?

Actually for this reaction $\Delta G=\Delta G^\circ+RT\ln Q$ holds only when $Q$ is used in terms of partial pressures. This is because the activity of an ideal gas is measured by its pressure, not its concentration. This equation can be derived from the definition of the Gibbs energy.


Hydrogen Bond Without Hydrogen Surely, by the same logic, any atom with a moderate electronegativity (carbon, phosphorus, silicon) can form a polarise covalent bond with an atom with a high electronegativity (fluorine, oxygen). Giving that carbon, phosphorus or silicon a partially positive charge (just like hydrogen), allowing it then to form a 'hydrogen bond' with another atom with a high electronegativity (fluorine, oxygen). Is it called a hydrogen bond because it most commonly occurs with hydrogen? Or is it that a hydrogen bond with carbon, phosphorus or silicon is instead called a carbon bond, phosphorus bond and silicon bond? Since you don't need a hydrogen to make an atom high electronegativity partially negatively charge, surely those atoms around atom would be paritally postive to balance the charge?

Just looking at the electronegativities, there seems to be no reason why atoms such as boron, carbon or silicium — which all have similar electronegativities to hydrogen — should not undergo hydrogen-bond–like intermolecular interactions.

There is one key difference that separates hydrogen from all the other nonmetals that readily form 2-electron-2-centre (i.e. standard) bonds at ambient temperature and pressure: the lack of any core electrons.

Atoms like boron and carbon (or worse: silicon) have a low-lying 1s atomic orbital that is generally not considered to take part in bonding. However, its spherical structure surrounds the nucleus and effectively shields its positive charge from the outside world. Hydrogen, on the other hand, has a basically unshielded nucleus consisting only of a positively charged proton.

The positively charged hydrogen nucleus can now interact well with other orbitals on other atoms (typically part of other molecules). Any in-between core electrons will repulse the electrons that the electropositive element is trying to interact with. Hence, the stabilisation is greatest in hydrogen’s case.

To be honest, I am unable to explain why hydrogen is in a special category when it comes to potential bonding, it just is. (The quantum mechanics would tell you that it does hydrogen-bond, and a thorough analysis of the QM might shed some light on it - if you want to go there. Personally, I've always been comfortable with just saying that nothing else comes close to the non-covalent (intermolecular) bonds it can participate in. Two other notes:1. Chemistry is "all about" electron density. Sure, it's quantum mechanical wave function probability distributions, but its all about electron density changes. Electronegativity is a convenient term to "invoke" to "explain" the effects of electron density on changes in bonding. This is very true. But, electronegativity is the "wrong" term. Technically, electronegativity is the property of the element in its neutral ground state. That is Co° and Co(+3) have the same electronegativity...even though we KNOW that the electron density is vastly different around them (both can participate in bonding with organics (organometallics)). So, the term is misused probably as often as it is not (if not more). Especially by beginning students (and their teachers!). I see no real harm in it - except when kids (or adults) go to the electronegativity tables and claim that say C in methane has a different electronegativity than C in ethane or formic acid. It's a slippery slope. 2. Not too long ago (10 years or so, couldn't (I hope!) have been 20), a paper was published showing that halogen bonding was a "thing". It remains to this day to be controversial. The Wikipedia article on h-bonding says it typically has bond strengths of 1-5 kcal/mole. The Wikipedia article on halogen-bonding says the bonds range between 5 and 180 kJ/mole. What's wrong with this picture?!! Well, 1 kcal ~ 4.2 kJ. So, 180 kJ is ~43 kcal. There's no f**king way halogen bonding is 8 times stronger than h-bonding. It's nonsense. X-bonding is so weak, that there's real debate about the usefulness of the concept (if it isn't significant then there would be no use in invoking it). My point is, separating out hydrogen-bonding for special consideration has proved itself to be a USEFUL thing. This is just not so with C, P, Si, O, or F (although the latter two often participate in h-bonding). It's just not very useful to consider say the Si in Me2C=Si interacting strongly with the O in Me2C=O  While if you take a look at the difference in Electronegativity between them (using the tables) you'd not come to that conclusion, I think. Hence: not useful.


Which molecule is more acidic: CHF3 or CHBr3? I know that fluorine is more electronegative than bromine. However, because of the size of bromine, it is more stable with a negative charge. In the case of $\ce{HF}$ vs. $\ce{HBr}$, to me, $\ce{HBr}$ is without a doubt the stronger acid. When it comes to comparing $\ce{CHF3}$ with $\ce{CHBr3}$, we are supposed to compare the stability of their conjugate bases.

In $\ce{CF3^-}$, the fluorine can hold the electronegative charge quite well. It also exerts a strong inductive effect.

In $\ce{CBr3^-}$, the bromine is less electronegative than the fluorine but is more polarizable, so I think it can hold the negative charge more efficiently.

With this understanding, I am concluding that $\ce{CBr3^-}$ is more stable which would mean $\ce{CHBr3}$ is the stronger acid. However, I've been told otherwise.

I'm aware that the $\ce{C-Br}$ bond length is longer than the $\ce{C-F}$ bond length. I'm wondering if that short bond length causes $\ce{CF3^-}$ to be more stable than $\ce{CBr3^-}$. If this is true, is it okay for me to think about $\ce{CF3^-}$ dissociating vs. $\ce{CBr3^-}$ dissociating? In that scenario, I believe $\ce{CBr3^-}$ dissociates more easily than $\ce{CF3^-}$, meaning $\ce{CF3-}$ is more stable.

Note: This logic wouldn't apply to the instance of $\ce{HO-}$ vs. $\ce{HS-}$ because $\ce{HS-}$ is more stable than $\ce{HO-}$ (even though $\ce{H-O}$ bond length is shorter than $\ce{H-S}$ bond length).

The following table contains some relevant data, the $\mathrm{p}K_\mathrm{a}$'s of the various haloforms along with the Pauling electronegativity of the corresponding halogen.

\begin{array}{|c|c|c|} \hline
\text{Haloform} & \mathrm{p}K_\mathrm{a} & \text{Electronegativity}\\ \hline
\ce{CHF3} & 25\mathrm{-}28 & 3.98\\ \hline
\ce{CHCl3} & 15.7 & 3.16\\ \hline
\ce{CHBr3} & 13.7 & 2.96\\ \hline
\ce{CHI3} & \sim 25 & 2.66\\ \hline
\end{array}

Looking at the $\mathrm{p}K_\mathrm{a}$'s, the first thing we notice is that the two haloforms at the beginning and end of the series are weaker acids than the two haloforms in the middle of the series. This behavior is often seen when there are at least two factors affecting acidity and they are in opposition to each other.

Let's consider inductive, steric and resonance effects and examine their expected effect upon haloform acidity.

Inductive effects should correlate with electronegativity.  Based only on inductive effects we would expect fluoroform to ionize and produce the most inductively stabilized cabanion ($\ce{CF3^{-}}$), while iodoform would ionize to produce the least inductively stabilized carbanion ($\ce{CI3^{-}}$).
Steric effects are often discussed in terms of bulky groups coming too close to one another, but what we are really saying is that the electron clouds around the groups are coming too close to one another causing electron-electron repulsion. Iodoform is a very bulky molecule.  The iodine substituents are very large and there are many lone pair-lone pair repulsions.  Upon ionizing iodoform can relax from a crowded $\mathrm{sp^3}$-like geometry to a more open and spread out $\mathrm{sp^2}$-like geometry.  As the halogen gets smaller there is less of a steric \ electron repulsion driving force for haloform ionization.
Resonance effects are not a significant stabilizing factor in the haloform series. Significant low energy resonance structures cannot be drawn. Thanks to TanYongBoon for noting this.

Conclusion: Inductive effects and steric effects (steric meaning electron-electron repulsion) control acidity in the haloform series. The two effects are in opposition to one another.  The inductive effect strongly stabilizes the fluoroform anion and plays a lesser role as we move down the series.  Steric \ electron repulsion effects strongly stabilize the iodoform anion and plays a lesser role as we move up the series.  Said differently, only one effect stabilizes the fluoroform (inductive) and iododform (steric) anions, but both inductive and steric effects play a role in stabilizing the chloroform and bromoform anions making chloroform and bromoform the most acidic molecules in the series.

To compare the acidic strengths of the haloforms, we need to check the relative stabilities of their conjugate bases.
CF3- , CCl3-, CBr3-.

Here, only CCl3- has effective backbonding and hence the negative charge partially gets stabilised by back donation to the vacant 3dπ orbitals of Cl. Thus, CHCl3 is a stronger acid among them.

The acidic strengths of the other two csn be compared their inductive effects.
F is very electronegative and hence stabilises the negative charge on the C atom. So, CHF3 is a better acid than CHBr3.


Is the Henderson–Hasselbalch equation volume independent? Looking at this equation $\mathrm{pH} = \mathrm{p}K_\text{a} +\log\frac{[\ce{CB}]}{[\ce{A}]}$ makes me think that it is concentration independent.

What I mean is, when we look at $\frac{[\ce{CB}]}{[\ce{A}]}$, the volumes cancels out, so we are only left with the molar ratio of the conjugate acid and conjugate base. So does this mean that the pH is only dependent on the molar ratios? Obviously not, because $0.003\ \mathrm{mol/L}$ of acid $\ce{A}$ and $0.003\ \mathrm{mol/L}$ of its conjugate base is going to have a different $\mathrm{pH}$ from $0.3\ \mathrm{mol/L}$ of acid $\ce{A}$ and $0.3\ \mathrm{mol/L}$ of its conjugate base right?

What am I missing here? This is seriously confusing me. Please help, I want to use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of a weak acid as it’s being titrated with a strong base, but this is making me reluctant about the results.

For a much shorter and clearer explanation of this problem, look here.
The Henderson-Hasselbalch (HH) equation is not to blame here. It is an approximate equation, with a certain region of validity. By its nature, it does not take into account the self-dissociation of water, which becomes increasingly important in dilute solutions. When concentrations reach somewhere around $\pu{10^-5 mol L-1}$ or lower, the true $\mathrm{pH}$ will deviate significantly from the value predicted by the HH equation. However, this is not the problem most of the time.
The reason the HH equation might produce poor predictions when calculating buffer $\mathrm{pH}$ is really because of an oft-made assumption which has nothing to do with the HH equation itself: the weak acid (or weak base) is assumed to be so weak that its ionization contributes almost no conjugate base (or conjugate acid) in comparison to the dissolution of the buffer salt. In other words, we assume the formal concentrations $C_X$ of species in the buffer are equal to their actual concentrations $[X]$; we replace
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{A^-}]}{[\ce{HA}]}
$$
for the worse approximation
$$
\mathrm{pH}
  = \mathrm{p}K_\mathrm{a} + \log \frac{C_{\ce{A^-}}}{C_{\ce{HA}}}.$$
This assumption is incorrect, but it is used to make the $\mathrm{pH}$ calculation much easier at the cost of accuracy. The resulting approximation breaks down in sufficiently dilute solution, and is already quite noticeable far before reaching water self-dissociation issues. It is also a noticeably poor approximation for weak acids/bases with relatively high ionization constants (say, $K > 10^{-2}$).
So how can we visualize the effect of diluting a buffer without approximations, and where do the approximations start to break down? Let us consider the simple case of solutions of a weak monoprotic acid $\ce{HA}$ with acid dissociation constant $K_\mathrm{a}$ where the concentration of acid and conjugate base are formally equal. This can be done by titrating half of the weak acid with a strong monoprotic base (e.g. $\ce{KOH}$). This problem can be solved exactly (assuming all activity coefficients are equal to $1$, which is generally a good approximation for solutions below about $\pu{1 mol L-1}$), and I have previously outlined the solution here. The equation relevant to this problem is the following:
\begin{multline}
0 =
  [\ce{H^+}]^3
    + \left(K_\mathrm{a}
      + \frac{C^\circ_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}}
    \right)[\ce{H^+}]^2 \\
    + \left(
      \frac{C^\circ_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}} K_\mathrm{a}
      - \frac{C^\circ_\ce{A} V_\ce{A}}{V_\ce{A} +  V_\ce{B}}
      K_\mathrm{a} - k_\mathrm{w}
    \right)[\ce{H^+}]
    - K_\mathrm{a}  k_\mathrm{w}
\end{multline}
$V_\ce{A}$ is the volume of the weak acid solution being titrated and $C^\circ_\ce{A}$ is its formal initial concentration (before mixing and reacting with the base), while $V_\ce{B}$ is a variable volume of strong base solution added with formal initial concentration $C^\circ_\ce{B}$ (before mixing and reacting with the acid). The presence of the self-dissociation constant of water $k_\mathrm{w}$ shows that it is being taken into consideration.
We can now substitute values as desired to obtain a polynomial in $[\ce{H^+}]$. I'll set up the volumes and concentrations arbitrarily so we can reach the acid and conjugate base formal concentrations of $\pu{0.3 mol L-1}$ as mentioned in the question. We can start off with $\pu{100 mL}$ of weak acid solution at a concentration of $\pu{0.9 mol L-1}$ with a dissociation constant $K_\mathrm{a} = 10^{-3}$ (this constant can be changed at will). We can reach the target concentrations of acid and conjugate base by adding $\pu{50 mL}$ of strong base at the same concentration $\pu{0.9 mol L-1}$ (neutralizing half the original acid). You can check the resulting concentrations after mixing the solutions, if you wish. Solving this polynomial, the resulting buffer solution containing formal concentrations $C_{\ce{HA}} = \pu{0.3 mol L-1}$ and $C_{\ce{A^-}} = \pu{0.3 mol L-1}$ has $[\ce{H^+}] = \pu{9.93399E-4 mol L-1}$. This results in $\mathrm{pH} = 3.00288$, which is very close to the value predicted by the approximations made at the start ($\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log 1 = \mathrm{p}K_\mathrm{a} = 3$). For now, they work.
After looking at this first case, we shall investigate the effects of dilution by factors of 10. This can be performed simply by diluting both the initial weak acid and strong base concentrations by 10 (i.e., $\pu{100 mL}$ of $\pu{0.09 mol L-1}$ $\ce{HA}$ plus $\pu{50 mL}$ of $\pu{0.09 mol L-1}$ $\ce{KOH}$ result in a buffer with acid/conjugate base concentrations of $\pu{0.03 mol L-1}$, and so on). Before going into the calculations, it's quite simple to see an issue with the approximation in the second paragraph. Independent of the level of dilution, so long as the concentration of acid and conjugate base remain equal, then according to the approximation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 3$. However, consider the case when starting with a solution of $\ce{HA}$ with initial concentration equal to $\pu{9E-4 mol L-1}$. This is a weak monoprotic acid, so the initial solution must have $[\ce{H^+}] < \pu{9E-4 mol L-1}$ and thus $\mathrm{pH} > 3$. Clearly it's impossible to add any amount of strong base to this acid and get a buffer solution with $\mathrm{pH} = 3$! The approximations at the start have broken down at this point.
Now, for the calculations. To get the proton concentration, I replaced every value of $0.9$ in the previous Wolfram Alpha link with $0.09$, then $0.009$, and so on. With the proton concentration determined, I used equations shown in my other linked answer to calculate the actual concentrations $[\ce{HA}]$ and $[\ce{A^-}]$ for comparison with the formal concentrations $C_\ce{HA}$ and $C_{\ce{A^-}}$. Note that they are almost equal at high concentration, but the error increases quite suddenly at lower concentrations. While the ratio $\frac{C_\ce{A^-}}{C_\ce{HA}}$ is always equal to $1$, the ratio $\frac{[\ce{A^-}]}{[\ce{HA}]}$ rockets away.
$$ \scriptsize
\begin{array}{c|ccc|cc}
\hline
C_\ce{HA} = C_\ce{A^-} / (\pu{mol L^-1})  &
[\ce{HA}] / (\pu{mol L^-1}) &
[\ce{A^-}] / (\pu{mol L^-1}) &
\frac{[\ce{A^-}]}{[\ce{HA}]} &
[\ce{H^+}] / (\pu{mol L^-1}) &
\mathrm{pH} \\\hline
0.3 & 0.29901 & 0.30099 & 1.00664 & 9.93399\times 10^{-4} & 3.00288 \\
0.03 & 0.029061 & 0.030939 & 1.06464 & 9.39282 \times 10^{-4} & 3.02720 \\
0.003 & 0.00235421 & 0.0036458 & 1.54858 & 6.45751 \times 10^{-4} & 3.18993\\
3\times 10^{-4} & 1\times 10^{-4} & 5\times 10^{-4} & 5 & 2 \times 10^{-4} & 3.69897 \\
3\times 10^{-5} & 1.6539\times 10^{-6} & 5.83461\times 10^{-5} & 35.2778
 & 2.83464 \times 10^{-5} & 4.54750 \\
3\times 10^{-6} & 1.78596\times 10^{-8} & 5.98214\times 10^{-6} & 334.953
 & 2.98549×10^{-6} & 5.52498 \\
3\times 10^{-7} & 1.97992\times 10^{-10} & 5.99802\times 10^{-7} & 3029.42
 & 3.30096×10^{-7} & 6.48136 \\
3\times 10^{-8} & 6.96609\times 10^{-12} & 5.99939\times 10^{-8}  & 8612.15
 & 1.16115×10^{-7} & 6.93511 \\
3\times 10^{-9} & 6.09004\times 10^{-13} & 5.99939\times 10^{-9} & 9851.15
 & 1.01511×10^{-7} & 6.99349 \\
\hline
\end{array}
$$

Try plugging in 0.3 and 0.003 for each case. Either way $\ce{pH = pK_{a}}$ because the ratio of conjugate base concentration to acid concentration is 1. Therefore, the Henderson-Hasselbalch equation is independent of volume.


Bonding in diatomic C2, a carbon-carbon quadruple bond? Carbon is well known to form single, double, and triple $\ce{C-C}$ bonds in compounds. There is a recent report (2012) that carbon forms a quadruple bond in diatomic carbon, $\ce{C2}$. The excerpt below is taken from that report. The fourth bond seems pretty odd to me.

  $\ce{C2}$ and its isoelectronic molecules $\ce{CN+}$, BN and $\ce{CB-}$ (each having eight valence electrons) are bound by a quadruple bond. The bonding comprises not only one σ- and two π-bonds, but also one weak ‘inverted’ bond, which can be characterized by the interaction of electrons in two outwardly pointing sp hybrid orbitals.

  According to Shaik, the existence of the fourth bond in $\ce{C2}$ suggests that it is not really diradical...
  If $\ce{C2}$ were a diradical it would immediately form higher clusters. I think the fact that you can isolate $\ce{C2}$ tells you it has a barrier, small as it may be, to prevent that.

Molecular orbital theory for dicarbon, on the other hand, predicts a C-C double bond in $\ce{C2}$ with 2 pairs of electrons in $\pi$ bonding orbitals and a bond order of two. "The bond dissociation energies (BDE) of $\ce{B2, C2}$, and $\ce{N2}$ show increasing BDE consistent with single, double, and triple bonds." (Ref) So this model of the $\ce{C2}$ molecule seems quite reasonable.

My questions, since this is most definitely not my area of expertise:

Is dicarbon found naturally in any quantity and how stable is it? Is it easy to make in the lab? (The Wikipedia article reports it in stellar atmospheres, electric arcs, etc.)
Is there good evidence for the presence of a quadruple bond in $\ce{C2}$ that wouldn't be equally well explained by double bonding?

Okay, this is not so much of an answer as it is a summary of my own progress on this topic after giving it some thought. I don't think it's a settled debate in the community yet, so I don't feel so much ashamed about it :)
A few of the things worthy of note are:

The bond energy found by the authors for this fourth bond is $\pu{13.2 kcal/mol}$, i.e. about $\pu{55 kJ/mol}$. This is very weak for a covalent bond. You can compare it to other values here, or to the energies of the first three bonds in triple-bonded carbon, which are respectively $348, 266$, and $\pu{225 kJ/mol}$. This fourth bond is actually even weaker than the strongest of hydrogen bonds ($\ce{F\bond{...}H–F}$, at $\pu{160 kJ/mol}$). Another point of view on this article could thus be: “valence bond necessarily predicts a quadruple bond, and it was now precisely calculated and found to be quite weak.”

The findings of this article are consistent with earlier calculations using other quantum chemistry methods (e.g. the DFT calculations in ref. 48 of the Nature Chemistry paper) which have found a bond order between 3 and 4 for molecular dicarbon.

However, the existence of this quadruple bonds is somewhat at odds with the cohesive energy of gas-phase dicarbon, which according to Wikipedia is $\pu{6.32 eV}$, i.e. $\pu{609 kJ/mol}$. This latter value is much more in line with typical double bonds, reported at an average of $\pu{614 kJ/mol}$. This is still a bit of a misery to me…

Carbon 100% could form a quadruple bond in C2. Each and every non-noble gas atom is trying to get an outer shell of 8. Diatomic elements are elements that can bond with themselves (the three most common being Oxygen, Nitrogen, and Hydrogen). An acronym to remember the Diatomic elements is H O Br F I N Cl. As you can see, "C" isn't one of them. But, these elements can only bond with themselves because  if they do, they get a full outer shell. 2 Nitrogens share 3 electrons with each other, and have 2 left over on each one. 3+3=6+2=8. Carbon has 4 valence electrons and needs to gain 4. It is light enough to be able to form a quadruple bond, and if it does, it will share all four of its electrons with the other carbon! So each carbon will have a total of 8 valence electrons.


Which one is nitric Acid, HNO3 by itself, or when it is included in H2O? This is a quote from my Chemistry textbook:

  Nitric acid ionises according to the following equation:
  $$\ce{HNO3(l) + H2O(l) -> H3O+(aq) + NO3- (aq)}$$

I know that the formula for nitric acid is $\ce{HNO3}$. But I cannot understand what is the product of this equation. What is $\ce{H3O+(aq) + NO3- (aq)}$? Is it a diluted acid with less concentration?

This is another quote (from my textbook) that confuses me:

  Nitric acid and sulfuric acid are not quite as strong as hydrochloric acid. Chemical and spectroscopic studies have shown that concentrated solutions of both these acids do contain some unionised molecules. For our purpose, however, their ionisation is effectively complete. This is particularly true in dilute solutions.

I cannot understand the part that is in the bold. What are these unionised molecules? I know that these acids are considered strong acids which completely ionise in water solution. If that is the case, then why does the text indicate that there are some unionised molecules?

In your first reaction, $\ce{H2O}$ acts as a base to abstract an $\ce{H+}$ from the nitric acid. The resulting $\ce{H3O+}$, called hydronium, is the conjugate acid, while the $\ce{NO3-}$, called nitrate, is the conjugate base (this is the molecule of nitric acid, but stripped of its $\ce{H+}$). You could view this as a dilution, in the sense that the resulting solution is less acidic than highly concentrated nitric acid would be, but the process of dilution of acid by mixing with water inevitably involves an acid-base reaction.

As to your second question, the unionized molecules are the neutral water and nitric acid on the reactant side, while the ions are the products resulting from the acid-base reaction. Nitric acid is relatively strong, meaning it tends to ionize almost completely in aqueous solution. Additionally, the less acidic the solution already is prior to addition of the acid, the more extensively the added acid will dissociate into ions (to understand this, please read up on acid-base equilibria and Le Chatelier's principle, as the topics are too lengthy and complex to cover adequately here). This is why the author emphasizes that the dissociation is more complete in dilute solutions. Statistically, however, some proportion of the nitric acid molecules will inevitably not dissociate into ions. The stronger an acid, the greater the extent of dissociation, but there is always some statistical chance that a proportion of the molecules will remain unreacted. For strong acids, however, the overwhelming preponderance of the molecules will dissociate, while only a tiny fraction remain unreacted. Hence, for many (if not most) purposes, it's reasonable when dealing with strong acids to proceed as if the dissociation is truly complete, since the unreacted portion is typically negligible in quantity.

The product is simply the ionized form of $\text{HNO}_3$; if you prefer it is equally correct to write $$\text{HNO}_3(aq)\rightarrow \text{H}^+(aq)+\text{NO}_3^-(aq)$$

Note that technically nitric acid is $\text{HNO}_3(aq)$and not $\text{HNO}_3(l)$ (which would presumably be the salt hydrogen nitrate in liquid form) As hydrogen nitrate will simply decompose under standard conditions if not dissolved so $\text{HNO}_3$ is sufficient to refer to nitric acid; however, it is understood to refer to $\text{HNO}_3$ dissolved in water, not just $\text{HNO}_3$ alone.


How 'heavy' should an element be, to be a "Heavy Metal"? I've come across the term "Heavy metals" innumerable times in articles, mostly pertaining to environmental issues.

Is there a weight range (of sorts) against which an element (metal) is classified as a "Heavy metal"?

Often I see "Heavy metals" refer to Cadmium (Atomic number 48), Mercury (Atomic number 80) and Lead (Atomic number 82), but does this term also apply to metals of intermediate atomic masses [such as Tungsten (Atomic number 50), Gold (Atomic number 79) and Tin (Atomic number 81)]?

I can't seem to find any chemical literature that deals with this, so I suspect that the term "Heavy metal" has no real serious scientific definition.

There is no true, accepted definition of heavy metal. I was taught to apply the option a metal that has density equal to or over $5.0\ \mathrm{g/cm^3}$.

Other variants include a different density range, specific gravity over density, environmental impact, atomic number, toxicity, or atomic mass, even chemical properties. See here$^{[1]}$ for further information and references.

$[1]$ John H. Duffus. '"Heavy metals"—a meaningless term?'. IUPAC technical report. Pure and Applied Chemistry, $(2002)$, 74(5), pp 793$-$807.

Like almost all technical terms, there is more than one definition depending on context. In general the US EPA has defined certain heavy metals as "toxic" or "of concern", and a lot of their literature uses the term for them. see this and the first reference therein:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4144270/
Please ignore any answers which talk about cosmology, they're just plain wrong.


Is most of the weight you lose actually released as carbon dioxide when you exhale? I've heard that most of the weight you "lose" is actually carbon dioxide released when you exhale. In percent its much more than what is released in pure heat energy and feces. Does anyone have statistics on what's the split between where the energy leaves your body?

How many percentage of food weight is actually released through carbon dioxide when its burned in your body? For example when you eat a hamburger and fries, is all the carbon released later when you exhale? How much is heat energy? Provided it does not go into fat reserves.

Let's start with:

  How much is heat energy?

2000 kilocalories per day, considering $E=mc^2$, is
$9.3 \times 10^{-8}$ grams

  is all the carbon released later when you exhale?

No.  Brainstorming, the others ways would include: feces, urine, perspiration, flatulence, shedding skin, hair and nails, mucous from nose and mouth, for women: menstruation, vaginal discharge, lactation, and giving birth, and for men ejaculation.

According to The Value of Human Feces:

Per day the amount of carbon excreted in feces is 21 grams and in urine is 7 grams.

According to The Composition of Human Perspiration Proc R Soc Med. 1934 May; 27(7): 839–848, a typical amount of perspiration a day is 700mL, and per 100mL the main carbon containing compounds are 20 mg urea, 20 mg glucose and 5mg amino acids.  So clearly less than 1 gram of carbon loss per day from perspiration. If we were considering loss of salts instead of just carbon, that would be about 3 grams per day from perspiration.

For menstruation, blood loss is about 40mL per cycle, and blood is 92% water, so maybe 1 gram of carbon on a heavy day of a period.

Skin loss per day is 1.5 grams according to Cells for life, cells for an instant.

A highly controlled study having a person in a chamber for 24 hours and analyzing the air found 653g of CO2 or 178g carbon "A Respiration Calorimeter with Appliances for the Direct Determination of Oxygen" Carnigie Institution of Washington Publication 42.

So, for a person that is not lactating or giving birth, I would estimate about 30 grams per day of carbon loss is other than from exhaling carbon dioxide.  Therefore:

About 86% of carbon loss is through exhaling carbon dioxide.

However, this will certainly vary depending upon the amount of indigestible carbon such as fiber, in ones diet.

Also, if mass loss through nitrogen and phosphorous (but not hydrogen and oxygen) were also considered the loss through breathing would come down to about 80%.  Inclusion of salts would bring this percentage down further.

The sum of water from feces, urine and respiration is about 1.8 kg per day and the total mass emission of the body is 2.6kg, according to the Carnigie source above.

Sorry but E=mc2 is not applicable. Chemical reactions do not result in a change in mass of the components but a change only in the molecular combinations. Yes, the heat generated in oxidation comes from the potential energy of the chemical bonds so one would have to put a lot of energy into again separating the oxygen and carbon in CO2. A change in mass only occurs in nuclear reactions which also results in a change in the atoms of the elements involved to a new element.


Is there a way to oxidize salicyl alcohol to salicylic acid without using chromium trioxide? I'm doing a project for school and I need to convert salicyl alcohol to salicylic acid, but I'm not allowed to use $\ce{CrO3}$ because of its toxicity. I know it is also possible to oxidize a primary alcohol to a carboxylic acid with potassium permanganate, but I read it might be a too strong oxidizer. Is there any good alternative?

@Waylander: A general Procedure given in J. Org. Chem. 1999, 64(7), 2564–2566 as follows:

  Typical Procedure. A mixture of alcohol 1 ($\pu{40 mmol}$), TEMPO ($\pu{436 mg}$, $\pu{2.8 mmol}$), $\ce{MeCN}$ ($\pu{200 mL}$), and sodium phosphate buffer ($\pu{150 mL}$, $\pu{0.67 M}$, $\pu{pH = 6.7}$) is heated to $\pu{35 °C}$. Then mixtures of $\pu{9.14 g}$ of 80% sodium chlorite ($\ce{NaClO2}$, $\pu{80.0 mmol}$) in $\pu{40 mL}$ of water and 2.0 mol% dilute bleach ($\pu{1.06 mL}$ of 5.25% $\ce{NaOCl}$ diluted into $\pu{20 mL}$ with water) are added simultaneously over $\pu{2 h}$ (Caution! Do not mix bleach and $\ce{NaClO2}$ before being added to the reaction mixture). The mixture is stirred at $\pu{35 °C}$ until the reaction is complete (<2% starting material, $\pu{2−5 h}$ reaction time), then cooled to room temperature. Water ($\pu{300 mL}$) is added, and the $\mathrm{pH}$ is adjusted to $8.0$ with $\pu{2.0 N}$ $\ce{NaOH}$ (about $\pu{48 mL}$). The reaction is quenched by pouring into cold ($\pu{0 °C}$) $\ce{Na2SO3}$ solution ($\pu{12.2 g}$ in $\pu{200 mL}$ of water) maintained at $\pu{<20 °C}$. The $\mathrm{pH}$ of the aqueous layer should be $8.5$−$9.0$. After stirring for $\pu{0.5 h}$ at room temperature, methyl tert-butyl ether (MTBE, $\pu{200 mL}$) is added. The organic layer is separated and discarded. More MTBE ($\pu{300 mL}$) is added, and the aqueous layer is acidified with $\pu{2.0 N}$ $\ce{HCl}$ (~$\pu{100 mL}$) to $\pu{pH = 3−4}$. The organic layer is separated, washed with water ($2 \times \pu{100 mL}$) and brine ($\pu{150 mL}$), and then concentrated to give the crude carboxylic acid 2 in 85−100% yield. The products may be purified by silica gel column chromatography or crystallization.

Sodium Chlorite with TEMPO and bleach as reported in J. Org. Chem. 1999, 64 (7), 2564–2566.


What does the reaction of hydrogen and iodine to hydrogen iodide look like? The reaction, $\ce{H2 + I2 <=> HI}$ is often used as an example for equilibrium reactions. It is however a difficult reaction to show in a school laboratory: we are not equipped to work with flammable gas at $\pu{450 ^\circ C}$ and high pressure.

It would be great if I could show my students what this process looks like in real life. I have a description, with drawings, by Max Bodenstein, from 1896 -- slightly dated.

Does anyone have pictures or videos of the gases involved and of the equipment used to make $\ce{HI}$ from the elements and to measure reactants and products?

To show an example for equilibrium reactions to your students, you can use a reaction which has visual change as shifting the equilibrium position such as color. My best bet is, use of cobalt(II) chloride equilibrium with water:
$$\ce{[CoCl4]^2-(aq) + 6H2O(l) <=> [Co(H2O)6]^2+(aq) + 4Cl-(aq)}$$

The beauty her is The ion, $\ce{[Co(H2O)6]^2-(aq)}$ is pink while $\ce{[CoCl4]^2-(aq)}$ is blue.  According to the Le Chatelier’s principle, at relatively low chloride concentrations (e.g., diluting the solution or adding $\ce{AgNO3(aq)}$ to remove $\ce{Cl-}$ ions), the equilibrium shifted to the right and, therefore, the solution becomes pink. However, if there is a large concentration of excess chloride (e.g., adding $\ce{Cl-}$ ions by means of adding $\pu{12 M}~\ce{HCl}$ drop wise or adding $\ce{CaCl2}$ beads), the equilibrium tends to shift to the left, making the solution blue.

In addition, similar to its sensitivity to the concentration of solutes, the equilibrium is also sensitive to temperature as well (See picture below). At colder temperatures, the equilibrium tends to shift to the right(pink color at LHS of the picture) while at warmer temperatures, it lies to the left (blue color at RHS of the picture):

Reading:

http://www.rsc.org/learn-chemistry/resource/res00000001/the-equilibrium-between-two-coloured-cobalt-species?cmpid=CMP00005957
https://eic.rsc.org/cpd/understanding-equilibrium-a-delicate-balance/2000012.article
http://research.chem.psu.edu/mallouk/ilab/Equilibrium_and_Le_Chatelier.htm

Well you could create extra pressure by adding both gases in a container close the container with a moving top . As long as you pressure the top and the container's volume is decreased according to the law of thermodynamics (Boyle) you can increase the pressure at a significant level .


Why is the first energy of ionization of oxygen lesser than that of nitrogen? The following question arises from a question I found in my book.

  Experimentally it has been determined that the value of the first
  energy of ionization of oxygen is lesser than the first energy of
  ionization of nitrogen. Select the statement which explains this:

  The higher electronegativity of oxygen
  The lesser electron affinity of nitrogen
  The lesser radius of oxygen
  The higher stability of nitrogen
  The higher electron affinity of nitrogen

According to my book the answer is the fourth option, but I couldn't find a way to justify this. What sort of explanation can do this?.

My guess is that nitrogen forms a triple bond with itself which is higher than the oxygen and this makes the molecule more stable but I don't know if this can be translated into a higher energy of ionization nitrogen rather than the oxygen which does have a sigma and pi bonds. Am I right?

Can someone please illustrate this or bring some sort of diagram with some values to justify this?. It would help a lot if an answer could help me why should I discard the second option?.

Doesn't nitrogen has smaller electron affinity than the oxygen? Wouldn't it mean that having a smaller electron affinity cause it to be easier to strip out an electron from its outer shell rather than the opposite as the question indicates?

Two electrons of the oxygen molecule are located on a $\pi $* level. They are relatively easy to ionize. It is not the case in $N_2$ : all electrons are located in binding orbitals.


Does Friedel-Crafts reaction fail with benzoic acid? I know that Friedel-Crafts reaction  (either alkylation or acylation) fails with phenol and aniline, due to formation of a coordinate bond between Lewis acid catalyst and the lone pair of the respective functional group.

But I have read that even benzoic acid fails to undergo this reaction. I couldn't think of a reason for this. Can anyone explain why is it so?

It could be that $\ce{-COOH}$ group is electron withdrawing hence there is an increased electron density in $\ce{-COOH}$ group due to -R resonating effect of electron from benzene ring.

Since friedel craft catalyst is a Lewis acid (electron deficient) the catalyst (anhydrous $\ce{AlCl3}$) reacts with electron rich COOH group of benzoic acid and gets bonded.

I think Friedel Craft reaction with Benzoic acid fails because the Benzene ring is deactivated. Here electrophile is Carbocation  which is a weak electrophile. So the necessity for Friedel-Craft reaction is activated Benzene ring.


Ozonolysis conditions for benzene What are the precise conditions under which benzene undergoes ozonolysis? All anyone can tell me is that it undergoes ozonolysis under "vigorous" conditions. But, what vigorous conditions?

What are the precise conditions under which Benzene undergoes ozonolysis?

In Über das Ozobenzol, an early German publication on the reaction of benzene with ozone, Ber. Dtsch. Chem. Ges., 1904, 337, 3431-3433 (DOI)
, the authors write

Leitet man in solches Benzol 1–2 Stunden lang bei einer Temperatur von 5–10° einen 5-procentigen Ozonstrom, so scheidet sich eine gelatinöse Masse aus, welche stark opalisirt. […]
Lässt man das unangegriffene Benzol verdunsten, so bildet es eine weisse, amorphe Masse von furchtbar explosiven Eigenschaften. Selbst beim Uebergiessen mit warmem Wasser detonirte diese Substanz heftig.
Uebergiesst man aber das gelatinöse Product mit eiskaltem Wasser,
so entsteht eine krystallinische Modification, die wohl möglich noch
heftiger als die amorphe bei der geringsten Berührung, ähnlich wie
Jodstickstoff, explodirt.

To sum this up:

The authors prepared the primary triozonide of benzene by passing a stream of ozone though it at a temperature of 5–10 °C.

Once the remaining benzene evaporated at room temperature, a white, amorphous mass with frightfully explosive properties was obtained.

When warm water was poured over it, this mass detonated.

When cold water was poured over the amorphous, jelly-like mass, a crystalline modification was obtained, which detonated on the slightest touch, similar to to nitrogen triiodide.

In the 1950s, a group of Dutch researchers published a series of articles about their kinetic experiments on the ozonization reactions.
The second article in the series, Recl. Trav. Chim. Pays-Bas, 1951, 70, 1005–1019, (DOI) notes that the ozonisation of benzene was performed in chloroform at a temperature of 242 K (–31 °C).

In the late 1960s, ozonization of benzene again gained some interest as a possible source for glyoxal in industrial synthesis.
In Glyoxal from Ozonolysis of Benzene, published in Ind. Eng. Chem. Prod. Res. Dev., 1969, 8, 89–92
DOI), the authors describe the ozonolysis of benzene in a mixture of acetic acid and nitromethane at temperatures between –20 °C and 15 °C.

Summary

All anyone can tell me is that it undergoes ozonolysis under "vigorous" conditions.

Apparently, your sources never bothered to check the literature! While the ozonide itself may detonate on touch, claims on the "vigorous" conditions necessary for its synthesis seem to be fully unsubstantiated by the literature!

Ozonolysis of benzene can be done at very high temperature or by its reaction with  Zinc and water.


How do I do stoichiometry for 2 limiting reactants The Drano reaction produces sodium aluminate along with hydrogen gas. If a typical container of Drano weighs 500 g which is 35% by weight by weight sodium (l) hydroxide and 2.5% by weight aluminum, what volume hydrogen gas would be produced if (excess) water was added to an entire can of crystal Drano at 22 °C and 1.2 atm?

My equation:

$$\ce{4 Al + 4 Na(OH) + 12 H2O -> 4 Na[Al(OH)4] + 6 H2}$$

If my balancing correct, I am trying to get the excess for water. Would I need to do the moles left over? Or would 500 x 62.5% (2.5% and 35% added together) suffice?

If I do need to do moles left over, how do I manage with two limiting reactants?

Your balance equation is correct except for not simplifying it. Simplified equation should be:
 $$\ce{2Al + 2NaOH + 6H2O ⟶ 2Na[Al(OH)4] + 3H2}$$

I also think you're getting a little bit confused about your equation. You probably think reacting $\ce{Al}$ and $\ce{NaOH}$ with same mole ratio (2:2) make them two limiting reagents. If that is the case, you are absolutely wrong. As M. Farooq commented elsewhere, the one reactant, which is present in the least amount in a stoichiometric sense, is the limiting reactant. Let's see your question in hand:

The amount of $\ce{NaOH}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{35 g}\; \ce{NaOH}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{40 g}\; \ce{NaOH}}  = \pu{4.38 mol}$

The amount of $\ce{Al}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{2.5 g}\; \ce{Al}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{27 g}\; \ce{Al}}  = \pu{0.463 mol}$

Since $\ce{NaOH}$ and $\ce{Al}$ react with 1:1 ratio according to the equation, $\ce{Al}$ is clearly the limiting reagent.

Hence, $\pu{0.463 mol}$ of $\ce{Al}$ reacts with $\pu{0.463 mol}$ of $\ce{NaOH}$ if you added enough water to $\pu{500 g}$ of Drano container. The leftover amount of $\ce{NaOH}$ would be: $\pu{(4.38 - 0.463) mol} = \pu{3.92 mol} = \pu{3.92 mol} \times \frac{\pu{40 g}\; \ce{NaOH}}{\pu{1 mol}} = \pu{157 g}$

Now, how much water you need? You need more than $\pu{0.463 mol}\times \frac{6}{2} = \pu{1.39 mol}$ of water, which is more than $\pu{25.0 g}$ (to be excess).

Note: Similar calculations would appied to find out amount of $\ce{H2}$ released by the reaction.

I think you're getting a bit confused with your equation. It's probably better to write two distinct reactions, one with $\ce{NaOH}$ and one with $\ce{Al}$.

35% of $\pu{500g}$ is $\pu{165g}$ $\ce{NaOH}$, and 2.5% of $\pu{500g}$ is $\pu{12.5g}$ $\ce{Al}$.


Why does aluminum chloride have a higher lattice energy than aluminum fluoride? From the table below (source: McMurry's Chemistry [1, p. 212]), it is evident that $\ce{AlCl_3}$ has a higher lattice energy than $\ce{AlF3},$ even though $\ce{F}$ is smaller than $\ce{Cl}$. Why is this so?

However, the melting point of $\ce{AlF3}$ $(\pu{1291 °C})$ is larger than that of $\ce{AlCl3}$ $(\pu{192.4 °C}).$ I think the huge melting point difference is because $\ce{AlCl3}$ is more covalent, while $\ce{AlF3}$ is more ionic. That being said, does lattice energy account for the covalent and ionic character of compounds?

References

McMurry, J.; Fay, R. C.; Robinson, J. K. Chemistry, 7th ed.; Pearson: Boston, 2015. ISBN 978-0-321-94317-0.

Lattice energy is a function of the radius ratio, which when tends to 1, improves the packing efficiency of the molecule. Here the radius ratio of the chloride ions and aluminium ions is much closer than the fluoride ions.
You could think of this from the born habers cycle also. Where the electron affinity of chlorine is higher, which leads to overall higher lattice energy. However, the radius ratio logic is much more convincing as the habers cycle logic would require the other parameters like the heat of formation also


About dative covalent bonding and expanded octet rule I'm beginning my studies of chemistry at high school (I'm not a native English speaker, though) and I have some basic questions:

1)  Consider the compound xenon tetroxide. In this molecule, does the xenon have its valence layer expanded? In another words, the oxygen atoms only have double covalent bonds with the Xenon?

I've tried to look for this answer but I'm still in doubt whether xenon does four dative covalent bonds (with each oxygen) or four double covalent bonds. In the first case, does not occur the expanded octet rule for xenon, right?

2) Another question; is it possible for a dative covalent bond to occur with just one electron? I mean, the atom which has more electrons in its valence layer "donates" just one electron instead of two? If the answer is "yes", can you provide me an example, please?

I'm sorry for the English mistakes. I started to learn this "chemistry vocabulary" just now.

To answer your first question: the concept of octet expansion for main group metals is no longer generally considered correct and only upheld by the more traditional organic chemists that don’t deal with inorganic bonding situations on a day-to-day basis — or anybody else who didn’t need to recheck what they were taught decades ago. Nowadays, these seemingly octet-expanding structures such as $\ce{XeO4}$ or $\ce{XeF2}$ are explained by dative bonds and/or four-electron-three-centre bonds; both have the beauty of not requiring d orbitals (whose energy would be comparable to the next shell’s s orbital; remember the aufbau principle!).

To answer your second question: no. A dative bond always requires two electrons that are given to one atom upon supply of the bond dissociation energy. It doesn’t matter how a certain bond is created, it only matters how it dissociates; and if it dissociates and was a bond, it will have to have had an electron pair.

(1) For xenon tetroxide all four bonds are double covalent bonds. The electrons from xenon all come from its valence shell.

(2) The definition of a dative covalent bond is that both electrons come from the same atom. So something like $\ce{H+ + NH3 -> NH4+}$. Both electrons come from the N molecule.


Why does some coffee smell like urine? I can smell a urine like odour from some coffees. I don't seem to be alone in this as there are multiple reports on the internet of people having the same experience.

I have read from a few posts that results of the Maillard reaction create ketoamines that could be causing it.

What's the science behind this?

References:

Reddit
Yahoo Answers

This answer is purely speculative in nature fueled by high school science and a research paper

From medicalnewstoday

  Urine does not usually have a strong odor to. However, occasionally, it will have a pungent smell of ammonia.

Googling coffee + ammonia led me to this little article Filter coffee nanoparticles to turn ammonia sensors.

The article  says that the researchers at IISC Bangalore were using coffee powder for detecting ammonia because coffee powder contains carbon and nitrogen in abundance and they've used that to make carbon nanotubes whatnot.

Then I hit gold when I found the research paper itself
- Understanding the ammonia sensing behavior of filter coffee powder derived N-doped carbon nanoparticles using the Freundlich-like isotherm.

The detecting technology is based upon adsorption which is a surface phenomenon basically the adsorbate(ammonia) gets concentrated only at the surface of the adsorbant(coffee powder).

Now, there are two types of adsorbtion.

Chemisorption
Physisorption

Now comes the speculation I'm assuming that the process over here is physisorption because physisorption is inversely proportional to temperature.

How it works?.

Filter coffee is essentially a powder this provides a large surface area for adsorbtion.

Intially when the coffee in your kitchen at room temperature some amount of ammonia gets adsorbed on it. Now, when you make the coffee it is exposed to high temperatures and as you can recall I just discussed that in physisorption the amount of gas adsorped on the surface is inversely proportional to temperature so the hot coffee can hold less amounts of ammonia on its surface as compared to the room temperature coffee.

So that's how ammonia is released on making coffee. Presumably.


Why doesn't water burn? Hydrogen is flammable, and for any fire to burn it needs oxygen. Why does a compound made of hydrogen and oxygen put out fires instead of catalyzing them? I understand that hydrogen and water are chemically different compounds, but what causes water to be non-flammable?

You can think of water as the ash from burning hydrogen: it's already given off as much energy as possible from reacting hydrogen with oxygen.

You can, however, still burn it.  You just need an even stronger oxidizer than oxygen.  There aren't many of them, but fluorine will work,

$$
\ce{2F2 + 2H2O -> 4HF + O2}
$$

as will chlorine trifluoride:

$$
\ce{ClF3 + 2H2O -> 3HF + HCl + O2}
$$

Water is already burnt hydrogen:
$$\ce{H2 + 1/2O2 -> H2O}$$

The reaction of oxidizing water ever more (to peroxide, for example) is non favorable and the peroxide will decompose as:

$$\ce{H2O2 -> H2O + 1/2O2}$$

with  ΔH of −98.2 kJ/mol and a ΔS of 70.5 J/(mol·K)


What is the point group of hexachloridotitanate(III)? I found the answer to this question to be $D_\mathrm{3h}$ but I am not convinced. Can someone explain me with a diagram if possible and explain why it doesn't belong to the octahedral point group $O_\mathrm{h}$?

Getting straight to the point, I believe whichever reference you have that claims $D_\mathrm{3h}$ is incorrect.

Titanium(III) complexes were the first example presented in my coordination chemistry class to examine the energy difference between the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ orbitals, explain why it exists and where the colour of coordination compounds derives from including carefully examining the UV spectrum of titanium(III). Our lecturer, Professor Klüfers of the LMU Munich, explained that the hexaaquacomplex is practically octahedral in shape. However, there is not just a clean absorption band but rather a band with a shoulder. This is because the excited state of $\ce{[Ti(H2O)6]^3+}$ is Jahn–Teller distorted, having an uneven $\mathrm{e_g}$ population, meaning that two slightly different excitation energies exist.

One might go a step further and argue that even in the ground state a certain distortion should exist. If one were to pull (not push) the ligands in $z$-direction slightly closer to the central metal, this would destabilise any orbitals with $z$ contribution meaning that $\mathrm d_{xy}$ suddenly becomes the single most stable d orbital. This slight distortion which one might call anti-Jahn–Teller (since it is opposite to the classic Jahn–Teller distortion) could explain a reduction of symmetry from $O_\mathrm{h}$ to $D_{\mathrm{4h}}$.

$D_\mathrm{3h}$ symmetry is not consistent with octahedraloid coordination spheres at all. It is very common for pentacoordinated centres such as $\ce{PF5}$. The only way to have six ligands — as in hexachloridotitanate(III) — around a central metal in this point group would be a trigonal prism. That would be a very uncommon coordination sphere and I am sure I would have heard of it in the context of titanium(III) in said course above should it exist for this species. Indeed, the source orthocresol found and linked in the comments makes no mention of any distorted geometry at all:

  The low-temperature structure results from rotations of the $\ce{[TiCl6]^3-}$ octahedra around a 4-fold axis. (Emphasis and square brackets mine)

Source: Amels, R.; Kremer, S.; Reinen, D. Jahn-Teller effect of titanium(3+) in octahedral coordination: a spectroscopic study of hexachlorotitanate (TiCl63-) complexes. Inorg. Chem. 1985, 24, 2751–2754. DOI: 10.1021/ic00212a009.

Hexachloridotitanate(|||) has D4h point group because structure distorted by Jahn -Teller distortion, two axial bonds get elongated.
Point group can be determined as following..


What is the canonical reference for studying stereochemistry? I am well versed with all the named reactions and all of organic chemistry.  I am quite weak in stereochemistry.

What canonical text can I use to study this topic and connect with other parts of the chemistry?

The classical book on stereochemistry is Eliel: Stereochemistry of Organic Compounds, first edition appeared 1962.

You will find overwhelming amount of information there. I would recommend it, regardless your age and level. It is so well written, that it is worth at least of fast reading over most important parts.
You should find it in any reasonable library, no need to get the latest edition, even the older ones should cover what you need.

http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
   Start here.  Now, the fun stuff off-syllabus.

It is possible to create chiral centers - including planar chiral centers - whose handedness cannot be named, even in principle, e.g., $\ce{CR4}$ and $\ce{R2>C=C<R2}$. The pictured acyclic carbons are chiral centers.  Show their non-superposable enantiomers that illustrate the anomaly.  (Answer to be posted end of the week - though you already know it, but may not know you know it.)

Optical rotation is not a measure of chirality, or even an indication of chirality. Solution optical rotations ignore atomic mass distribution[1].  Silver thiogallate, $\ce{AgGaS2}$ in achiral space group ${I}$-42${d}$, rotates 522°/mm along [100] at 497.4 nm[2].  ${P3_121}$ quartz and ${P3_221}$ berlinite, with opposite sense three-fold screw axes ($3_1$ right-handed, $3_2$ left-handed), are both levorotatory[3].  CIP notation flips with composition at constant geometry (L-cysteine versus most other L-amino acids).

[1] http://www.mazepath.com/uncleal/norbors.gif
http://www.mazepath.com/uncleal/norone.png
[2] Appl. Cryst. 33, 126 (2000)
[3] J. Appl. Crystallogr. 19, 108 (1986)

J. Math. Phys. 40, 4587 (1999)
http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM
Quantitative geometric chirality can be calculated in any number of dimensions for any collection of a countable number of anonymous points with overall finite principle moments of inertia.  Mathematically perfectly chiral arrays are rotationally symmetric but lacking all ${S_n}$ improper rotation axes.


What is hydrogen-rich water? My question is: Is hydrogen-rich water true? I am aware of the alkaline water. I have read several articles in the Internet regarding this kind of water but I cannot seem to be convinced. Wikipedia isn't even talking about it.

I am puzzled with how can a water be enriched with hydrogen. It's definitely $\ce{H2O}$ and everybody knows it. It's not water anymore if it's not $\ce{H2O}$.

This link is one of the examples of websites that I read about hydrogen-rich water.

Hydrogen water is a solution of molecular hydrogen $\ce{H2}$ in water, which shouldn't be confused with acidic or basic solutions.  The first scientific reference to hydrogen water appears to be in the journal, Nature Medicine which can be accessed here if you have a personal or institutional subscription to the journal. The claim is that molecular hydrogen can serve as a mild reducing agent and therefore has antioxidant properties; namely, it can reduce hydroxyl radicals:
$$\ce{H2 +2 OH\cdot -> 2H2O}$$
The article has generated a considerable amount of research, with well over 300 journal articles citing this work originally published in 2007.  There seems to be some merit to the claim although in my brief scan of the abstracts, there does not appear to be a large body of evidence suggesting that commercial products are beneficial.  One such article, which is publicly accessible is here.  As mentioned in another answer, the commercial product approach to generating hydrogen water appears to be the oxidation of magnesium in water, which generates $\ce{Mg^2+}$ ions.  As with any other nutritional supplement, I would be very wary of entering this type of treatment without consulting a professional.  Magnesium is an important nutrient, however at high doses it does show signs of toxicity.

Great info above, I'm glad you took time to look into H2! The initial ? here was basically the difference between Alkaline water & H2 Water.
Alkalinity can be easily manipulated by the addition of alkaline minerals although this does not increase the molecular Hydrogen content. Much research on Hydrogen gas as a powerful antioxidant and its endogenous antioxidant enzyme enhancing mechanism (see chart #4 in the 2nd study posted below) exists today. Hydrogen is unique in that it has the rare ability to make it past the BBB / Blood Brain Barrier making especially useful for Neuro-degenerative & neurodevelopmental conditions.

Magnesium sticks can create H2 water in small amounts. Most common today is the process of ionization or Electrolysis more commonly recognized in scientific terms as ERW / Electrolyzed-Reduced Water made by quality ionizers. Some call it Kangen Water, a trade name given to the ionized water by one particular Japanese manufacturer whose machines are recognized as home use medical devices.  Here are just 2 of such studies of which the 2nd study continues on to explain the potential of molecular H2 in various prominent disease conditions.

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3285010/ and http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660246/ and because of the great importance this ERW has had in my child's life as it greatly reduced her seizure activity, Autism & related issues which is why I study it.


Reaction between chlorobenzene and methyl chloride What will be the product of the reaction between chlorobenzene and methyl chloride in the presence of anhydrous $\ce{AlCl3}$?

I think we should get 1-chloro-4-methylbenzene as the major product, as we'll have Friedel–Crafts alkylation. Is this correct? Or will we get toluene?

You will actually get a mixture of different products, with the product with the highest yield being p-chlorotoluene, or 1-chloro-4-methylbenzene. The other products resulting from the reaction include o-chlorotoluene and m-chlorotoluene, with a significantly higher amount of the former being produced*. In fact, there would also be products with more than one methyl substituent being produced in various proportions depending on the precise conditions being used.

Yes, the reaction will proceed by Friedel-Crafts alkylation: Firstly, the Lewis acid catalyst abstracts the chloride ion from methyl chloride, giving a reactive, electrophilic methenium ion. Then, electrons from the $\pi$ system of the benzene ring of chlorobenzene then attack the electrophilic ion. Finally, a proton is lost to restore aromaticity of the ring, giving the product.

*Refer to the following post to understand why the chlorine atom as a substituent would be ortho/para-directing: Why are halogens ortho para directing even though deactivating?

1-chloro 4-methyl benzene because chlorine is an ortho para directing group so methyl group eithermust attack at ortho position or para position if chlorine. But chlorine shows negative inductive affect.so in order to increase stability methyl group must be attached at para position of chlorine