Miller indices. What determines the appropriate axis? So, first of all, what's up with this "plane" I see two planes!

The basis of my question is that to determine miller indices the plane cannot lie on your origin, so you have to make a translation. If I move my origin to the right then I get the miller indices that are indicated in the figure. But if I move backwards then I don't get the miller indices in the figure.

So my question then is: what determines the appropriate translation?

There is no "appropriate" translation. You must remember that everything includes translational symmetry, including the Miller planes.

The Wikipedia page on crystal structures and Miller indices go into much more detail, but some key points:

The unit cells possess translational symmetry, so there are a set of Miller planes that also differ by translational symmetry.
The Miller indices (hkl) (which yes should be expressed in smallest possible integers) define the normal of the Miller plane.

  That is, (hkℓ) simply indicates a normal to the planes in the basis of the primitive reciprocal lattice vectors. Because the coordinates are integers, this normal is itself always a reciprocal lattice vector. The requirement of lowest terms means that it is the shortest reciprocal lattice vector in the given direction.

So in your case, you have to remember if you "move the origin", you have to also consider the symmetrically identical unit cells around the one in your picture. As I said above, there are an infinite set of identical Miller planes that simply differ by the translational symmetry of the lattice.


Is Hexafluorocyclohexane the "Most Polar" Small Molecule? Earlier this year, there was a lot of attention when all-cis 1,2,3,4,5,6-hexafluorocyclohexane was synthesized*:

Notably, C&E News quoted the lead author:

  This compound is remarkable for being the most polar nonionic compound now known to exist

Now, the original paper reports a molecular dipole moment of $\pu{6.2 D}$, which is extremely high for an alkane. It also claims the:

  unusual property of a facially polarized ring in organic chemistry

Certainly having one side of the cyclohexane ring completely substituted with fluorines is unusual, so I agree with the facial-polarity property being interesting.

But there are many small molecules (e.g., under 20-30 atoms) with large dipole moments, e.g., p-nitroaniline:

A quick PM7 calculation gives the predicted dipole moment of $\pu{7.93\pm1 D}$. So I dispute the claim from C&E News - nitroaniline is clearly more polar and non-ionic.

So my question is this: What's the largest dipole moment for a known non-ionic organic molecule under 30 atoms?

*: O'Hagan, et. al, "All-cis 1,2,3,4,5,6-hexafluorocyclohexane is a facially polarized cyclohexane" Nature Chemistry 7, 483–488 (2015)

At first, I thought, that those 6.2 D of O’Hagan et al. have been measured somehow but it as later described, they calculated it (1):

  A molecular dipole value of 6.2 D for 1 was calculated at the M11/6-311G(2d,p) theory level using natural bond orbital (NBO) analysis with NBO 6.0.

I was using your given database for a little search, starting with nitroaniline, and came up with a few small molecules with higher dipole moments. Those seem to be PM7 calculated values and are shown in the following table:

\begin{array}{lll}
\hline
\text{Name} & \text{Dipole moment in D} & \text{Source}\\
\hline
\text{4-Nitroacetanilide} & 8.36 \pm 1.08 & (2)\\
\textit N\text,\textit N\text{-Dimethyl-4-nitroaniline} & 9.53 \pm 1.08 & (3)\\
\textit P\text ,\textit P\text{-Dimethyl-}\textit N\text{-(4-nitrophenyl)phosphinic amide} & 10.89 \pm 1.08 & (4)\\
\hline
\end{array}

As there were some comments, “it seems that this level of calculation does not provide accurate estimates of dipole moments”, I did some further calculations on levels that might provide better estimates.

There is a quite new paper of Hickey and Rowley (5) which says:

  CCSD, MP2, or hybrid DFT methods using the aug-cc-pVTZ basis set are all able to predict dipole moments with RMSD errors in the 0.12–0.13 D range and polarizabilities with RMSD errors in the 0.30–0.38 Å³ range.

Within the paper, there is later on a comparison between cc-pVTZ, Sadlej-pVTZ and aug-cc-pVTZ:

  The Sadlej basis set yields results that are improved over the cc-pVTZ basis set and have effectively the same accuracy as the aug-cc-pVTZ basis set.

That’s why I chose Sadlej’s basis set at first … but I wanted to look if it could not be easier hence faster. Therefore I also chose the def2-SVPD and looked, what an improvement a def2-TZVPD single point calculation on this geometry would yield.

I use ORCA 3.0.3 for calculations – the command line is like the following:

! B3LYP Sadlej-pVTZ TightSCF TightOpt

Table 1: "Pure" dipole moments (in Debye) straight from the calculations
\begin{array}{lllll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{4-Nitroacetanilide} & 9.18261 & 9.00380 & 9.07922 & 8.6738\\
\textit N\text ,\textit N\text{-Dimethyl…} & 7.39418 & 7.29188 & 7.30862 & 9.4209\\
\textit P\text ,\textit P\text{-Dimethyl…} & 8.58913 & 8.44430 & 8.51591 & 11.8352\\
\text{all-}\textit{cis }\ce{C6H6F6} & 6.17422 & 5.97013 & 6.04794 & 5.6298\\
\hline
\end{array}

What I did then, was to think of how to produce values that could be properly compared to the experimental values from the paper. As they said, with their methods they can yield RMSDs of around 0.12 D. So I took their values for the Sadlej basis set and made my own calculations on their test molecules for the two def2 basis sets (and also for PM6 with Gaussian 09 Rev. A.02).

Table 2: Some analytical values to compare with the publication values; the units are in Debye
\begin{array}{lllll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{MAE} & 0.08 & 0.08 & 0.08 & 0.40\\
\text{RMSD} & 0.13 & 0.12 & 0.12 & 0.61\\
\hline
\end{array}

So for my two ways, the MAE and the RMSD are not too far away from the triple zeta values but are calculated way faster. Now I have made a linear regression between the experimental and the calculated dipole moments from Hickey and Rowley’s test molecules to be able – as said before – to calculate proper error ranges at a given probability.

Figure 1: Regression functions for the four tested methods (blue), prediction intervals at 95% (gray); from top left to bottom right: def2-SVPD, def2-TZVPD//def2-SVPD, Sadlej-pVTZ, PM6

Now let’s turn the “pure” values from Table 1 into somehow more real values:

\begin{align}
\text{def2-SVPD} \to \text{exp}&: -0.0097439128 + 0.9763275\ x\\
\text{def2-TZVPD//def2-SVPD} \to \text{exp}&: -0.0045050428 + 0.99095691\ x\\
\text{Sadlej-pVTZ} \to \text{exp}&: -0.012922547 + 0.98812009\ x\\
\text{PM6} \to \text{exp}&: -0.16057241 + 0.89545049\ x\\
\end{align}

Prediction Intervals (for the error):

$$\small\begin{align}
\text{def2-SVPD} \to \text{exp}&: 0.24473369 \sqrt{1.0217391 + 0.010348758 (x-1.4590217)^2}\\
\text{def2-TZVPD//def2-SVPD} \to \text{exp}&: 0.23719452 \sqrt{1.0217391 + 0.010656662 (x-1.4321957)^2}\\
\text{Sadlej-pVTZ} \to \text{exp}&: 0.23333979 \sqrt{1.0217391 + 0.010593477 (x-1.4448261)^2}\\
\text{PM6} \to \text{exp}&: 0.98446665 \sqrt{1.0217391 + 0.0097476423 (x-1.7592391)^2}
\end{align}$$

Table 3: Converted dipole moments (in Debye) from Table 1 including error ranges at 95% probability

\begin{array}{llll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{4-Nitroacetanilide}  & 8.96 \pm 0.31 & 8.92 \pm 0.30 & 8.96 \pm 0.30 & 7.61 \pm 1.20 \\
\textit N\text ,\textit N\text{-Dimethyl…}     & 7.21 \pm 0.29 & 7.22 \pm 0.28 & 7.21 \pm 0.27 & 8.28 \pm 1.24 \\
\textit P\text ,\textit P\text{-Dimethyl…}     & 8.38 \pm 0.30 & 8.36 \pm 0.29 & 8.40 \pm 0.29 & 10.44 \pm 1.40 \\
\text{all-}\textit{cis }\ce{C6H6F6} & 6.02 \pm 0.27 & 5.91 \pm 0.26 & 5.96 \pm 0.26 & 4.88 \pm 1.06 \\
\hline
\end{array}

What can be seen from this now? The smaller B3LYP/def2-SVPD method yields nearly the same values as the more expensive Sadlej version. Putting the def2-TZVPD single point on top of them, yields slightly deeper results with actually no accuracy improvements. PM6 is no real good option as the error ranges are way higher, with values of about $1-1.4$ D.

In the end, the dipole moment of all-cis 1,2,3,4,5,6-hexafluorocyclohexane is probably somewhere around $6$ D but there are probably many other molecules with higher dipole moments … “my” three molecules shown above go at least up to somewhere around $9$ D.

As long as no experimental values are available, everything here is only a good guess – but you will know this too good.

Sources:

O’Hagan et al., Nature Chemistry, 2015, 7, 483–488
http://pqr.pitt.edu/mol/NQRLPDFELNCFHW-UHFFFAOYSA-N
http://pqr.pitt.edu/mol/QJAIOCKFIORVFU-UHFFFAOYSA-N
http://pqr.pitt.edu/mol/FICBIFYYTQXSEY-UHFFFAOYSA-N
A. L. Hickey, C. N. Rowley, J. Phys. Chem. A, 2014, 118 (20), 3678–3687


How can I balance the redox reaction of sulfur dioxide with oxygen? How do I balance this reaction using half-redox equations?
$$\ce{2SO2 + O2->2SO3}$$

My work is:

Oxidation: $\ce{SO2 + H2O -> SO3 + 2e- + 2H+}$

Reduction: $\ce{H2O + 2e- + O2 -> SO3 + 2H+}$

Total: $\ce{SO2 + 2H2O + O2 -> 2SO3 + 4H+}$

BUT it's false. What did I do wrong?

WELL I think you put something erroneous in your reduction formula.

You will never get SO3 out of just O2 and water. Instead of SO3

The reduction formla will be 4e- + O2 + 4H+ --> 2 H2O

Multiply oxidation formula by 2 and add it to new reduction formula and you got your answer


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

To find the oxidation state using Lewis structure, I have to compare the electronegativity of the bonded atoms and assign the bonding electrons to the more electronegative atom. Then, the difference between the number of electrons in element state and bonded state will give the oxidation number.

Here, the bonding electrons of nitrogens bonded with a double bond will stay in between both the atoms will not be assigned to both. The $\ce{N-H}$ bonding electrons will be assigned to $\ce{N}$. What about the two $\ce{N-N}$ bonding electrons?

This answer uses electronegativities for the calculation of oxidation states as proposed in the Expanded Definition of the Oxidation State by Hans-Peter Loock in 2011.

Comparing Electronegativities

The following table shows an excerpt from Pauling electronegativities ($\chi_{\mathrm{Pauling}}$):

$$\begin{array}{cc}
\hline
\text{Element} & \chi_{\mathrm{Pauling}}\\
\hline
\ce{H} & 2.20\\
\ce{N} & 3.04\\
\hline
\end{array}$$

Looking at $\ce{N-H}$ Bond (heteroatomic)

The table shows, that the nitrogen atom is more electronegative than the hydrogen atom. It will therefore attract the electrons more than hydrogen does, and thus polarize the bond. Imagining the gedankenexperiment of an hypothetical heterolytical bond cleavage - presumed by applying pulling force to the atoms, until they get separated - it is assumed, that the electrons are allocated to the nitrogen atom during separation.

Looking at $\ce{N-N}$ and $\ce{N=N}$ Bonds (homoatomic)

Looking again at the previous gedankenexperiment, now with two identical Atoms bound to each other. Since they have, of course, the same electronegativity, there is no polarization of the bond. It is assumed, that separation would therefore result in an homolytical bond cleavage, allocating always one electron to each of the two nitrogen atoms for each bond respectively.

Summing up

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

Last thing to do is calculating the atoms hypothetical charge after separation, which is to be equatable with the oxidation state:

$$ \text{Oxidation state} = N_\mathrm{i}(\ce{e-}) - N_\mathrm{f}(\ce{e-}) $$

With $ N_\mathrm{i}(\ce{e-}) $ representating the number of electrons in a free atom, and $ N_\mathrm{f}(\ce{e-}) $ the one after separation (One should not forget the lone pairs).

One will end up with the following oxidation states for the different (nitrogen) atoms, with the last entry representating the mean oxidation state of all nitrogen atoms:

$$\begin{array}{cccc}
\hline
\text{Atom} & N_\mathrm{i}(\ce{e-}) & N_\mathrm{f}(\ce{e-}) &\text{Oxidation state}\\
\hline
\ce{H} & 1 & 0 & +1\\
\hline
\ce{N-1} & 5 & 5 & 0\\
\ce{N-2} & 5 & 5 & 0\\
\ce{N-3} & 5 & 6 & -1\\
\hline
\ce{N} & - & - & -\frac{1}{3}\\
\hline
\end{array}$$

And the structure with assigned oxidation states:


Equating emissions to mass loading I want to estimate aerosol mass loading by dividing emissions of the particular aerosol by the associated area. Assuming the air had no other particles in it to begin with - is this procedurally correct?

Given the context of your question, I assume you must mean "the associated volume", not "the associated area".

Other than that, you are exactly correct.  For example, given the following parameters:

  aerosol emissions rate = 0.010 mg $\mathrm{hr^{-1}}$
  duration of aerosol emissions = 1 $\mathrm{hr}$
  container volume = 1 $\mathrm{m{^3}}$

Then the aerosol mass loading would be given as:

  total aerosol mass = 0.010 mg $\mathrm{hr^{-1}}$ $\mathrm{x}$ 1
  $\mathrm{hr}$ = 0.010 mg

  aerosol mass loading = 0.010 mg / 1 $\mathrm{m{^3}}$ = 0.010 mg
  $\mathrm{m{^{-3}}}$

If you really meant area and not volume, as in surface area of the aerosols, then your statement would be over-simplified and you would need to include other assumptions made concerning particle size and density etc. in order to determine whether the way your statement is correct as written.


Standard electrode potential of a galvanic half cell is zero at equilibrium at standard conditions? At standard conditions(at 1atm pressure and unit activity(1molal or 1molar concentration)of all dissolved compounds),the electrode potential is equal to the standard electrode potential by the Nernst equation. $$ \Delta E = \Delta E^0 $$
At the same time,while the electrochemical reaction is at equilibrium,the electrode potential of the cell, $$ \Delta E =0 $$

So this must mean,that when ANY reaction is at equilibrium under the standard conditions, $$ \Delta E = \Delta E^0 = 0 $$
Which is quite surprising.Is this correct?please explain.

There is a difference in the meaning of $\Delta E^\circ$ and $E^\circ$. $\Delta E^\circ$ actually denotes the change in standard electrode potential while $E^\circ$ denotes standard electrode potential.

$E^\circ$ is always constant for a given redox couple (eg. $\ce{Zn^2+|Zn}$). So no matter what  concentration of $\ce{Zn^2+}$ you take in a certain electrode, the standard electrode potential of $\ce{Zn^2+|Zn}$ is always $-0.76\ \mathrm{V}$.

Since $E^\circ$ is always constant, it never changes. So the value of $\Delta E^\circ$ (change in $E^\circ$) would always be zero and so it is quite a useless quantity to consider.

For $\Delta E$ however, consider the Nernst equation,
$$E = E^\circ - (RT/nF)\ln Q$$
Change in $E$
$$\Delta E = \Delta E^\circ - \Delta(RT/nF)\ln Q$$
therefore
$$\Delta E = -(RT/nF)\Delta\ln Q$$

At equilibrium $\Delta E$ is zero because there is no change in the reaction quotient $Q$ of the reaction (therefore $\Delta \ln Q$ is zero).


Regioselectivity in radical chlorination of a carboxylic acid From an online textbook, the first step in the preparation of an amino acid:

Can we be sure that the chlorine will connect exactly at that spot, in the alpha-position? Won't we get an acyl chloride instead? Or a molecule where Cl is connected in the beta-position, or further down the line?

I am going to say: no, this does not work in practice. In fact it does not even work in theory.

In simple theory you would say that that the α C-H bond is the weakest because the resulting radical is stabilised by conjugation into the C=O. But under radical conditions, we don't observe α-chlorination in the lab. Is there a better way to explain this?

The chlorine radical

The problem is twofold. Firstly, the chlorine radical, $\ce{Cl.}$, is much less stable and therefore extremely unselective in radical chlorinations. In physical organic chemistry, we quote something called the Hammond postulate, which says that in this case, the transition state for hydrogen abstraction resembles the starting material more than the product. Therefore, the stability of the product does not significantly affect the activation energy, and the rate, of the hydrogen abstraction.

On top of that, it is what we would consider an "electrophilic radical"; it has a very low-energy SOMO, and therefore has a tendency to abstract a more electron-rich hydrogen atom. Essentially, it means that the radical behaves more like an electrophile than a nucleophile. In the transition state for hydrogen abstraction, the chlorine wants to have a significant build-up of negative charge on itself (remember, the product of hydrogen abstraction is going to be $\ce{HCl}$). It therefore goes for the hydrogen that is most willing to give it this negative charge, and that is the most electron-rich hydrogen atom.

An example

Consider the radical chlorination of propionic acid, i.e. $\ce{R} = \ce{CH3}$ in your diagram above.

The α-hydrogen, compared to the β-hydrogen, is comparatively electron-poor because of the electron withdrawal by the $\ce{-COOH}$ group. So, it turns out that the rate of abstraction of the α-hydrogen is much slower: $k_\alpha/k_\beta = 0.03$ (Moody & Whitham, Reactive Intermediates, p 12). Here's an article which discusses the selectivity observed in the chlorination of propionic acid under different conditions: Tetrahedron, 1970, 26, 5929.

As written in Carey & Sundberg 5th ed., Part A, p 1022:

  Radical chlorination shows a substantial polar effect. Positions substituted by EWG [electron-withdrawing groups] are relatively unreactive toward chlorination, even though the substituents are capable of stabilizing the radical intermediate. [...] Because the chlorine atom is highly reactive, the reaction is expected to have a very early TS [transition state] and the electrostatic effect predominates over the stabilizing effect of the substituent on the intermediate. The electrostatic effect is the dominant factor in the kinetic selectivity of the reaction and the relative stability of the radical intermediate has relatively little influence.

As an example, they use the chlorination of butyronitrile, which is the same as propionic acid except that the EWG is now $\ce{-CN}$ instead of $\ce{-COOH}$:

As @Loong mentions in the comments, in the specific case of propionic acid, there is a further problem: there are three β-hydrogens for it to abstract, and only two α-hydrogens. Therefore, even in the absence of everything else I have described, we would expect there to be 1.5 times as much chlorination at the β-position. This certainly does not help at all with the selectivity! The importance of statistical factors is most clearly seen in the radical chlorination of simple alkanes, where all the hydrogens are pretty much similar. I shan't go into detail here, but you can look at this answer (by Loong again), and there are examples on Clayden 2nd ed., pp 987-8.

The solution

If you wanted to selectively chlorinate propionic acid at the α-position, you would use a polar mechanism that goes via the enol. The paper above has some examples of conditions that work. All of them involve suppressing the radical reaction by adding radical scavengers.


IUPAC Nomenclature of organic compounds Can the prefixes like iso,neo be used in the IUPAC nomenclature of organic compounds and can all the common names of aromatic compounds be used in their IUPAC name?

Prefixes -  I think the prefixes neo, iso etc. are used only in common names. There's a different system in IUPAC for naming things having a branched chain.
Aromatic Compounds - The only aromatic compound that I know of having same common and IUPAC name is benzene.


why does milk of magnesia have an expiration date? I noticed an old bottle of milk of magnesia in my refrigerator and saw that it had an expiration date printed: January 2015.  Checking the ingredients, I saw only three: magnesium hydroxide, purified water, sodium hypochlorite.  I understand that milk of magnesia is more or less a mineral solution.  How could this require an expiration date?  What would be the consequences of taking milk of magnesia past the expiration date?

(I see that this page discusses expiry dates on chemicals in the abstract; I don't know which of these arguments apply to milk of magnesia in particular, or any other over-the-counter dietary supplement or drug.  As an IT professional, I have absolutely no experience or expertise in the physical or life sciences.)

Many countries require anything marketed as edible to have an expiry date. You can speculate about what might spoil your Magnesia (see comments above), but likely it will be good four or fourty times as long as indicated.

Of course the lid might not be perfectly tight, and it could dry up or get contaminated by bacteria, but this can happen anytime. And certainly after some time it will separate into clear water and a thick suspension of magnesia particles and you'll have to shake or stir it very thoroughly before consummation.

Anyway these expiry dates are 99% baloney. Very few things really get bad before you see the mould grow on them.

Fun fact and (german) exemption to that rule: After six months at the latest, the taste of beer (the non-pasteurised stuff) starts to deteriorate. Yet, according to the date on the bottle, you have 12 more weeks to drink the remainder of that crate with increasing disgust before it "expires".


Molybdenum/Tungsten alloy armor piercing round As we all know, the military's use of depleted uranium in armor piercing rounds and shells is very dangerous as far as health concerns, and an alternative is needed. Tungsten is already a very dense metal, but molybdenum is widely used to harden metals, and also has a very high density. Would a molybdenum and tungsten alloy work for armor piercing capabilities? If so, why not?

Probably not. Molybdenum and Tungsten are both hardening alloys to steel, they perform the same function, though differently. I suspect a Mo - W alloy would have the fragility of both molybdenum and tungsten while not hardening each other especially much nor increasing density.

The answer is: Sure, it would be hard, heavy and likely armor piercing, but very inefficient use of both Mo and W. Just use W alone, or alloy it with elements that are lighter, but melt within reasonable ranges. A quick glance at wikipedia confirms that Fe, Ni and Co are usual alloying elements. Tungsten carbide is also used.

Creating this Mo-W alloy would also be very difficult, usually you solvate the high mp material in the liquid low mp material but these both do not melt within reasonable limits so you would have to use some seriously high temperature equipment, we are talking beyond bright white heat - or rely on very, very slow processes (diffusion - and these are large atoms...).


Based on their respective number of degrees of freedom, what's the most efficient fuel; hydrogen or gasoline? The reaction of octane's combustion is:

$$\ce{2 C8H18 + 25 O2 -> 18 H2O + 16 CO2}$$

I am asked for computing the number of degrees of freedom of the octane given that its specific heat is $\pu{256 JK^{-1}mol^{-1}}$.

Then I have to compare the number of degrees of freedom obtained for octane with those for $\ce{H2}$. Based on this I have to argue what's the most efficient fuel.

What I've done

I assumed octane stores thermal energy only in quadratic degrees of freedom and these do not depend on $T$. Using the equipartition theorem one gets:

$$C_V = \frac{\partial U}{\partial T} = \frac{Nfk_\mathrm B}{2} = \frac{nfR}{2} $$

Where: $k_\mathrm B = \pu{1.381 \times 10^{-23}JK^{-1}}$ and $R = \pu{8.315 JK^{-1}mol^{-1}}$

Then for octane:

$$C_V = fR $$

$$f = 31$$

Aren't these too many degrees of freedom?

For $\ce{H2}$ there are $3$ translational + $2$ rotational degrees of freedom; $f = 5$

What about efficiency?

I have worked with Otto's engine model (real heat engines; internal combustion):

$$\eta = 1 - \left(\frac{V_2}{V_1} \right)^{\gamma -1}$$

Usually the compression ratio is of the order:

$$\frac{V_1}{V_2} = 8$$

Where $\gamma$ is the adiabatic exponent:

$$\gamma = \frac{f + 2}{f}$$

Thus, the efficiency of a gasoline engine:

$$\eta = 0.12$$

While the efficiency of an hydrogen engine:

$$\eta = 0.56$$

My point is that the efficiency I get for gasoline is not what I expected (at least $0.20$). So I must be missing something on calculating the degrees of freedom for octane.

Thus hydrogen is a more efficient fuel.

It sounds like most of your confusion is coming from the fact that you are assuming that the vibrational degrees of freedom are completely frozen out at moderate temp. This is a common misconception. To see that this assumption is not true, we can look at the simple example of carbon dioxide. It is a linear molecule, so it has three degrees of translational freedom and two of rotation. If vibrational modes do not contribute at all, we expect that $\overline{C}_v=\frac52 R$ and $\overline{C}_p=\frac72 R$, so $\gamma=\dfrac{C_p}{C_v}=1.4$. If we include the 4 vibrational degrees of freedom, each of which contributes 1R to $\overline{C}_v$, we would have $\dfrac{C_p}{C_v}=1.15$. The experimental value at $20$ $^\circ C$ is about 1.3, indicating that the vibrational modes are already contributing to the heat capacity at this temperature. A plot of the heat capacity as a function of temp shows a gradual increase across a very wide range of temperature.

More rigorously, the contribution of a vibrational degree of freedom to the molar heat capacity is

$$R\cdot\left(\frac{\Theta}{T}\right)^2\cdot\frac{e^{\Theta/T}}{(e^{\Theta/T}-1)^2},$$

where the "vibrational temperature" $\Theta=\dfrac{hcv}{k_B}$. For the lowest energy vibrational normal modes of $\ce{CO2}$, $v=667 cm^{-1}$, so $\Theta = 961 K$. Although this is quite a bit higher than typical temperatures of interest, the above expression tells us that these vibrations contribute $\sim 0.45 R$ to $\overline{C}_v$ at 300 K. Since there are two modes at this frequency in $\ce{CO2}$, we expect that $\overline{C}_v$ and $\overline{C}_p$ are both increased by 0.9R at 300 K relative to the values excluding vibrations. This correction gives a value of $\dfrac{C_p}{C_v}$ very close to observed experimental value.

Going back to your case of octane, there are a number of normal modes with frequencies around 720 cm$^{-1}$, certainly low enough to contribute at 300 K. We have 18 + 8 = 26 total atoms, so we expect a maximum of 3*26=78 degrees of freedom. Since octane is not linear (in the strict sense of all atoms in a single line), there are 6 non-vibrational DOF. Thus, we expect that the number of degrees of freedom calculated using your method based on the experimental specific heat should be somewhere between these numbers.

Your calculation, however, needs to be corrected to account for the fact that vibrational degrees of freedom contribute 1R rather than R/2, so the equation becomes

$$\overline{C}_v = 3 R + (f-6)R.$$

From that, I get $f=33.8$, well within our expected range. You can repeat the rigorous calculation using the exact frequency of each normal mode of octane and see how close the result is to the experimental value.

UPDATE: To address the efficiency calculation, note that including the vibrational modes changes your equation for $\gamma$. Now we have that $\overline{C}_v=R(3+f-6)$ for (nonlinear) octane and $\overline{C}_v=R(\frac52)$ for hydrogen (ignoring its one vibrational mode). If we use the apparent degrees of freedom of octane of 33.8, $\gamma_{octane}=1.0325$ and $\gamma_{H2}=1.4$. [Note also that there is really no reason to calculate those apparent degrees of freedom except for curiosity, since we can just add R to the given $\overline{C}_v$ to get $\overline{C}_p$ and divide to get $\gamma_{octane}$.] From that I get an efficiency of 6.5% for the octane and 56% for the hydrogen.


Actinides and Lanthanides group I would like to ask which group do actinide and lanthanide belong?
I had a question in my assingment that which group has most number of elements?
if they belong to 3rd group then the 3rd group should be the answer.
Am I wrong or Right? Please Help Me.

The periodic table actually looks like this:

...you can see that the first 14 elements in either series don't belong to any group from the main body of the table.

You could say that the lanthanides and actinides form fourteen new 2-element groups, with Lu and Lr belonging to group 3. But IUPAC does not number these groups:


What happens to chemical shifts of protons close to aromatic systems in ¹H-NMR? I have two questions regarding chemical shifts on aromatic systems.

First, what does being near an aromatic system do to protons? If we look at the spectra of 1-napthalenemethanol and then 9-anthracenemethanol, the peak of the two aliphatic protons of the benzylic methylene group get shifted way up. Can someone explain why adding in that third aromatic ring does this?
The spectra:

9-anthracenemethanol
1-naphtalenemethanol

I was also under the impression that protons near electronegative elements would be deshielded and be shifted higher. However in the spectrum of Benzil, the protons para to the carbonyl group are higher shifted than the protons meta to the carbonyl. Why is this? Is it to do with conjugation of the carbonyl with the aromatic ring? I'm having trouble finding a proton nmr in the literature, so there's nothing to compare it to.

The aromaticity found in certain ring systems (e.g. benzene, see Hückel's rule) causes a magnetic field because of the circular motion of the electrons in the ring (see picture).

(Source: Clayden et al. Organic Chemistry 2nd edition, p. 277)

This additional field enforces the magnetic field of the NMR machine inside the ring, while countering the field outside the ring, causing the signals of the outer hydrogens to shift downfield (i.e. to the larger shifts).

Now for your other questions:

  First, what does being near an aromatic system do to protons? If we
  look at the spectra of 1-napthalenemethanol and then
  9-anthracenemethanol, the peak of the two aliphatic protons of the
  benzylic methylene group get shifted way up. Can someone explain why
  adding in that third aromatic ring does this?

Lookig at the difference in shift between the benzylic methylene protons of 1-Naphthalenemethanol and 9-Anthracenemethanol we must conclude that the induced magnetic field of the second is stronger at that position than that of the former, as there is no other appreciable difference. Intuitively this seems to make sense as the field of the 9-Anthracenemethanol is more uniform at and closer to (with respect to the "right" side of the molecule as in the pictures) the discussed proton. Of course this could/should be confirmed by quantum mechanical calculations, but I do not have the tools or the knowledge to do that (if someone has, please feel free to expand this answer!)

(1-Naphthalenemethanol Proton NMR)

(9-Anthracenemethanol Proton NMR)

  I was also under the impression that protons near electronegative
  elements would be deshielded and be shifted higher. However in the
  spectrum of Benzil, the protons para to the carbonyl group are higher
  shifted than the protons meta to the carbonyl. Why is this? Is it to
  do with conjugation of the carbonyl with the aromatic ring? I'm having
  trouble finding a proton nmr in the literature, so there's nothing to
  compare it to.

Yes, conjugating effects and inductive effects can counter each other, but there is no easy general way to say which will win out in a particular situation, although usually it is the conjugating effect.


Is there a relationship between Free Radicals and Beta Particles? This question is contingent on my understanding of the two terms in the title. As I was taught, a Free Radical is a reactive (high energy?) valence electron released for stability's sake because it is unpaired. I was taught that a Beta Particle is a high energy and fast moving radioactive positron or electron released to stabilize an atom. If both of these premises are true, is there a relation between the two terms, and if so, what is it?

There is high-energy in the sense of chemical reactivity, and then there is high energy in the sense of nuclear reactivity. The energy associated with the latter is many orders of magnitude greater than the energy associated with the former, so that there is generally no direct interaction between the two (where the free radical is chemically reactive, and the beta particle arises from nuclear reactivity).

A free radical is a chemical species, whose electron configuration is such that it is highly reactive towards other chemical species. However, this energy is far too low to affect any nuclear reactivity.

A beta particle is a high-energy electron or positron emitted by the radioactive decay of an atomic nucleus. As these particles are emitted from the nucleus of the atom, the electron configuration (i.e. a free radical) is irrelevant; only the stability/reactivity of the nucleus determines whether an atom will emit a beta particle, not the chemical species.

If any of this seems incorrect or unclear, please do not hesitate to ask for clarifications in the comments.


Which of the following hydrocarbons react with sodium in liquid ammonia? Problem

The hydrocarbon which can react with sodium in liquid ammonia is
(1) $\ce{CH3CH2CH2C#CCH2CH2CH3}$
(2) $\ce{CH3CH2C#CH}$
(3) $\ce{CH3CH=CHCH3}$
(4) $\ce{CH3CH2C#CCH2CH3}$

Solution

 Solve it as a case of terminal alkynes, otherwise all alkynes react with sodium in liquid ammonia.

Question
I know sodium in liquid ammonia is used to produce trans-alkenes from alkynes. But from the options, it seems all of them react with sodium in liquid ammonia.
The solution doesn't make any sense to me. If all of the alkynes given react with sodium in liquid ammonia, then why the question? And what difference does it make if the alkyne is terminal?

Although this is more of a case of communication rather than actual chemistry, I'll explain exactly what the author meant.

They basically want you to ignore the Birch reduction reaction. They want you to find the compound that reacts with sodium in liquid ammonia, but not via Birch reduction.

I however, do stand by that the question is very poorly framed. It is definitely ambiguous, and relies on strange, unrealistic assumptions.


Geometries of SCl2(OCH3)2 and SF2(OCH3)2 I am not able to understand the difference between structure of $\ce{SCl_{2}(OCH_{3})_{2}}$ and $\ce{SF_{2}(OCH_{3})_{2}}$.

I am already well versed with Molecular Orbital and VSEPR theories. I found the geometries,
but don't understand why they are:

Why is the $\ce{F}$ at axial positions and $\ce{Cl}$ on the other hand in equatorial positions? (I am not able to understand Bent's Rule to be straight-forward, I have read Wikipedia already.)

Check the bond angles. The larger the angles around a bond, the more S character.
Electronegative elements prefer higher P character(Therefore less the bond angles).
Axial position has lesser bond angles.Hence higher electronegative element takes axial position.( F>O>Cl)

This concept however may not be applicable to every compound.


Can iproniazid be prepared by reacting isoniazid with 2-chloropropane? I'm asking as I fear that the 2-chloropropane might react with the pyridine ring to form k-isopropylpridine-4-carbohydrazidine where k=2,3,5. Or that no reaction will take place.

Figure 1: Isoniazid

Figure 2: Iproniazid

Figure 3: 2-chloropropane

The $\ce{NH_2}$ group in given compounds is very similar to regular amine fragment and is more active than acylated $\ce{NH}$. However, alkylated amines are more active than non-substituted ones. So, I'd expect a very significant amount of double-alkylated product.

To avoid purification issues I recommend to consider synthesis of acetone hydrazone followed by reduction with $\ce{Na[BH_3(CN)]}$ or similiar low-active hydrogen donor (hi-active ones will reduce amide fragment)


How do I calculate the degree of dissociation in equilibrium? In my textbook, for calculating the percentage dissociation of $\ce{HF}$ for the given equation:
$$\ce{HF + H2O <=> H3O+ + F-}$$

The solution is:

Initial Concentrations
$$[\ce{HF}] = 0.08~\mathrm{M}, \: \ce{[H3O+]} = 0, \:\ce{[F- ]}= 0$$
Equilibrium concentrations
$$[\ce{HF}] = 0.08~\mathrm{M} - x, \: \ce{[H3O+]} = x, \:\ce{[F- ]}= x$$

I am not able to undestand why $x$ is subtracted from 0.08 and not $cx$ [ where $x$ is the degree of dissociation]

I tried solving the same problem taking $cx$ but not able to get the solution, can anybody explain the difference to me? I have tried asking a similar question earlier too but it is really hard for me to get my head around this concept.

Let us first define the terms needed here.

Degree of dissociation (DOD)
Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by $\alpha$.
$$\alpha = \frac{\text{amount of substance of the reactant dissociated}}{\text{amount of substance of the reactant present initially}}$$
Number of moles dissociated
It is defined as the product of the initial concentration of the reactant and the degree of dissociation

Now suppose you have a reaction like this
$$\ce{A->B + C}$$

The initial state of A is always the concentration of A (should be given in the question) while initial moles of B and C are zero (if anything else is not specified).
The final state of A is always defined as $\text{(amount of substance initially present) - (amount of substance dissociated)}$ while for B and C it is just $\text{amount of substance of A dissociated)}$.

Writing our equation again,
\begin{array}{lcccc}
 & \ce{A &-> &B &+ &C}\\
\text{Initial amount of substance}& a && 0 && 0\\
\text{Final amount of substance}&   a - a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD})\\
\end{array}


Adjustment of calcium, magnesium and alkalinity in drinking water? It's been a long time since I did any molar chemistry in school and despite doing a lot of reading on the subject this week, I am making a mess of some basic calculations. I hope that someone who understands water chemistry & molar equations a lot better than I do can point me in the right direction.

I am trying to add mineral salts to soft water to adjust the calcium, magnesium and alkalinity in a predictable manner. But I can't get past the basic equations so far.

The source water I am working with is:

12.6 mg/L Hardness as CaCO3 equivalent
11.3 mg/L Alkalinity as CaCO3 equivalent
  4.8 mg/L Calcium
  0.2 mg/L Magnesium
  1.5 mg/L Sodium

And the desired target water is:

70    mg/L Hardness as CaCO3 equivalent
40    mg/L Alkalinity as CaCO3 equivalent
20    mg/L Calcium
  4.9 mg/L Magnesium
  8   mg/L Sodium

I would prefer to have most of the hardness derive from calcium, rather than magnesium if possible. I plan to try

adding sufficient baking soda, NaHCO3 to the source water to obtain 8 mg/L sodium, then
add sufficient calcium carbonate (chalk), CaCO3 to the source water to make 50/50 GH/KH water (general hardness/carbonate hardness) i.e. 50 mg/L hardness and 50 mg/L alkalinity, both expressed as CaCO3 equivalents
add sufficient lactic acid (88%  CH3CH(OH)COOH) to neutralize some carbonate and (I hope) get to 50/40 GH/KH
add sufficient epsom salt, MgSO4*7H2O to obtain a final result of 70/40 GH/KH.
This assumes that adding lactic acid won't remove (precipitate) calcium hardness.
The solubility of calcium carbonate in water shouldn't be an issue, I can put it under pressure with carbon dioxide, C02 if necessary to dissolve it.

If the above is not feasible, then alternately

add sufficient baking soda, NaHCO3 to the source water to obtain 8 mg/L Sodium
add sufficient calcium carbonate, CaCO3 to reach the alkalinity target of 40 KH (40/40 GH/KH) i.e.. 40 mg/L hardness and 40 mg/L alkalinity, both expressed as CaCO3 equivalents
Add non-carbonate hardness by adding sufficient epsom salt, MgSO4*7H2O to obtain a final result of 70/40 GH/KH

Any help, guidance as to molar calculations and best practices in water chemistry or well honed levity welcome

Edit to add molar amounts
Source water
12.6 mg/L Hardness as CaCO3 equivalent
11.3 mg/L Alkalinity as CaCO3 equivalent
Ca 2.4 x 10-4 M
Mg 1.7 x 10-5 M
Na 1.4 x 10-4 M

Target water
70 mg/L Hardness as CaCO3 equivalent
40 mg/L Alkalinity as CaCO3 equivalent
Ca 1.0 x 10-3 M
Mg 4.1 x 10-4 M
Na 7.3 x 10-4 M

The parallel thoughts:

The source water I am working with is:

12.6 mg/L or 0.126 mmol/L GH as CaCO3 equivalent
11.3 mg/L or 0.113 mmol/L KH as CaCO3 equivalent
  4.8 mg/L or 0.12 mmol/L Ca
  0.2 mg/L or 0.0082 mmol/L Mg
  1.5 mg/L or 0.065 mmol/L Na

And the desired target water is:

70    mg/L or 0.7 mmol/L GH
40    mg/L or 0.4 mmol/L KH
20    mg/L or 0.5 mmol/L Ca
  4.9 mg/L or 0.2 mmol/L Mg
  8    mg/L or  0.35 mmol/L Na

What is needed to add:

0.574 mmol/L GH, as Ca+Mg
0.287 mmol/L KH  as Ca+Mg as bicarbonates, I.e. 2x0.287=0.574 mmol/L of bicarbonate
0.38 mmol/L Ca
0.192 mmol/L Mg
0.285 mmol/L Na

Reaction of 0.285 mmol/L sodium bicarbonate
with 0.285 mmol/L lactic acid
releases 0.285 mmol/L carbon dioxide,
dissolving 0.285 mmol/L calcium carbonate
forming 0.285 mmol/L calcium bicarbonate.

( Not exact due equilibrium reactions but good enough )

What is needed to add:

0.287 mmol/L GH, i.e Ca+Mg

0.093 mmol/L Ca
0.192 mmol/L Mg

Adding 0.192 mmol/L MgSO4.

What is needed to add:

0.093 mmol/L GH, i.e Ca only

0.093 mmol/L Ca

This can to be added as 0.093 mmol/L calcium carbonate(optionally supported by pressured carbon dioxide) + 0.186 mmol/L lactic acid.


Why is platinum denser than gold? The atomic masses of gold and platinum are 196.96657 u and 195.084 u respectively, meaning that (on average) an individual gold atom is heavier than an individual platinum atom.
At the same time, the van der Waals radius of gold is 166 pm, whereas that of platinum is 175 pm. So an individual gold atom is smaller than an individual platinum atom.
Given these data, how is it possible that platinum metal can actually be denser than gold (21.5 g/cm3 vs 19.3 g/cm3)?

Since OP is still in the high school, I'll try to explain it simply as possible using mathematical manipulation (hoping OP is more familiar with mathematics than chemistry).

Both gold and platinum consist of same crystal packing called body-centered cubic, which is illustrated in following diagram:

Crystal studies of both gold and platinum has revealed that the lengths of their cubic unit cells ($a$) are $\pu{407.82 pm}$ and $\pu{392.31 pm}$, respectively. If you inspect closely, you would realize that each of eight corner atoms shares with eight unit cells while each of six face-centered atoms shares with only two unit cells. Thus, total atoms per unit cell  $= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$. Keep in mind that $a$ is depend on the atomic radius of each metal and how closely they packed against each other against some concerning forces (e.g., van der Waal's). For example, the experimental finding has been reported that the closest $\ce{Au-Au}$ separation is $\pu{288.4 pm}$ while that of $\ce{Pt-Pt}$ is $\pu{277.5 pm}$. These values imply that the metallic radius of gold is $\pu{144.2 pm}$ and that of platinum is $\pu{139 pm}$. This would cover the quote "the radii of the atoms and the bond lengths within the crystal lattice."

Based on the value of $a$ and number of atoms per unit cell for each metal, you can calculate the molar volume of each metal ($V_\mathrm{Metal}$):

$$V_\ce{Au} = a_\ce{Au}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\
= (\pu{407.82 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{10.21 cm3mol-1}$$

and similarly,

$$V_\ce{Pt} = a_\ce{Pt}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\
= (\pu{392.31 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{9.09 cm3mol-1}$$

Once we know the molar volume of each metal, we can calculate the density ($d$) by using $d=\frac{\text {molar mass}}{\text {molar volume}}$:

$$d_\ce{Au} = \frac{\pu{196.97 gmol-1}}{\pu{10.21 cm3mol-1}} = \pu{19.29 gcm-3}$$

$$d_\ce{Pt} = \frac{\pu{195.08 gmol-1}}{\pu{9.09 cm3mol-1}} = \pu{21.46 gcm-3}$$

Therefore, it is true that $d_\ce{Pt} \gt d_\ce{Au}$.

Relevant resources:

John W. Arblaster, In Selected Values of the Crystallographic Properties of Elements; ASM International: Material Park, OH, 2018 (ISBN-13: 978-1627081542).
Arnulf Maeland, Ted B. Flanagan, “Lattice Spacings of Gold–Palladium Alloys,” Canadian Journal of Physics 1964, 42(11), 2364-2366 (https://doi.org/10.1139/p64-213).
https://www.webelements.com/gold/crystal_structure.html
https://www.webelements.com/platinum/crystal_structure.html


How to find the amount of moles given mass percentage data A hydrated aluminium sulphate, $\ce{Al2(SO4)3.xH2O}$, contains 8.10% of aluminium by mass. Find the value of x.

My attempt: (Assuming 100g sample)

Calculate the amount of aluminium using the formula $\mathrm{amount = \frac{mass}{molar~mass}}$, yielding $\mathrm{0.300~mol}~\ce{Al}$ having a mass of $\mathrm{8.10~g}$.

Determine the amount of sulphate ($\ce{SO4^2-}$) by recognising a ratio between the aluminium and sulphate. 2 moles of aluminium to 3 moles of sulphate, 2:3.

Using the ratio, I determined the amount to be $\mathrm{0.450~mol}~\ce{SO4^2-}$, which has a mass of $\mathrm{43.2~g}$.

I can now calculate the mass of the anhydrous compound, aluminium sulphate ($\ce{Al2(SO4)3}$), to be $\mathrm{51.3~g}$, leaving $\mathrm{48.7~g}$ of anhydrous substance, or $\ce{H2O}$.

Using the figure of $\mathrm{48.7~g}$, I can calculate the amount of $\ce{H2O}$ using the formula $\mathrm{amount = \frac{mass}{molar~mass}}$ which produced a figure of $\mathrm{2.71~mol}~\ce{H2O}$.

By using the principle of relative amounts, I find that x=9.

The answer is incorrect. Where did I go wrong?


Why doesn't my stainless steel Thermos get really really hot? Why doesn't my stainless steel Thermos get really really hot? Seems like a fairly straightforward question.

A few notes:

I was looking and found (obviously different values because steel is an alloy) that $$0.35~\mathrm{\frac{J}{g\,K}}\leq C_{p,\,\mathrm{steel}}\leq 0.55~\mathrm{\frac{J}{g\,K}}.$$
Obviously, that is just a rough range, but that is a very small heat capacity as is usual of metals.

Why then, does my 200 degree Fahrenheit coffee not heat up the stainless steel mug super quickly!

And even more baffling to me: Why in the world would it be a really good insulator?

It's entirely possible that I am devoid of any understanding of how insulation works, but I thought things which are bad conductors of electricity made good insulators. While stainless steel isn't praised for its electrical conductivity, I thought a material had to be about as bad as a gas at conducting electricity to be crowned with the title of "good insulator." I would just imagine that stainless steel, being metallic in many ways, would be a better conductor than that.

So what's going on? Why isn't my Thermos reaching 200 degree Fahrenheit?

I have a solid steel pint glass and I can tell you that it pretty much instantly heats up to the temperature of the liquid you put inside. You probably have some sort of double-wall vacuum flask, which insulates by using a vacuum between the steel.

As you can see, there are two walls to the flask: an inner wall and an outer wall. The inner wall is probably very close to the temperature of whatever is inside the flask. However, since the two walls are separated by a vacuum (which is a fairly good insulator), the outer wall, i.e. the one that you touch, remains pleasantly near ambient temperature*.
*Unpleasantly near ambient temperature, if you've been outside in a hot summer for a while.


Why is there an upper limit to the photon energy that can be observed in the discrete emission spectrum of the hydrogen atom? I think it must have to do with the fact that an incident photon with a sufficiently high energy will ionize the hydrogen atom (i.e. the electron will be ejected). Therefore it won't show in the emission spectrum. Is this correct?

The closest an e- can get to nucleus is the 1st, it can't get closer and so can't get to lower E. If an e- is beyond the influence of the nucleus and falls to 1s it can't fall any further  (drop in E any more ). This represents the largest E emission.  Once an e- is beyond the influence of the nucleus it doesn't matter how much further it goes it is still beyond its influence and hence the energy of the transition doesn't increase


Averaging of dipolar coupling in solution due to tumbling The angular part of dipolar splitting depends on $\theta$ as:

$$3\cos^2\theta - 1$$

In the liquid state or in solution molecular tumbling occurs which causes the NMR signal to become an average of all $\theta$ which is expressed mathematically as:

$$\int_0^\pi(3\cos^2\theta-1)\sin\theta\,\mathrm{d}\theta$$

where $\sin\theta$ is described, in my the book that I'm reading, as a weighting factor to take into account the probability of finding the system with an angle $\theta$

It is this weighting factor that confuses me. Surely all angles are equally likely? It looks to me like the $\theta$ part of the triple integral one would do to integrate over a sphere using polar coordinates (notice the limits on the integral). However, even if this is correct, I am still finding it difficult to picture this volume element $\sin\theta\,\mathrm{d}\theta$.

Please help me to rationalise why the weighting factor is there and also to visualise this $\sin\theta\,\mathrm{d}\theta$ volume element.

The $\sin(\theta)\,\mathrm{d}\theta$ comes from the surface element of a sphere of unit radius where $\theta$ is the polar angle measured from the north pole, as it were, to the equator. The surface element is $\mathrm{d}s = \sin(\theta)\,\mathrm{d}\theta\,\mathrm{d}\phi$ where $\phi$ is the azimuthal angle measured is if longitude, i.e around the equator. The integral is
$$\frac{1}{4\pi}\iint f(\theta)\sin(\theta)\,\mathrm{d}\theta\,\mathrm{d}\phi$$
where $f(\theta)$ is the function you want to average.

The sketch shows the idea


How is chemical and dynamic equilibrium related? I have been told that dynamic equilibrium is attained when the reactants and products of a system has achieved the same concentration or when their rates of formation are the same. And I also know that even though nothing seems to go on in a dynamic equilibrium there's still so much activity.
Now assuming I've got that right. I want to know how is chemical equilibrium any different from what I mentioned abt dynamic equilibrium.
Also can chemical and dynamic equilibrium coexist??

Background: It is generally the case that there is an energy barrier between reactants and products, sometimes called an activation energy, and this is smaller for reactants than products so that a reaction proceeds towards products.

Suppose now that there is a reaction such that $\ce{ A \to B}$ and initially only A is present so that the rate of forward reaction is large, i.e. lots of A and almost no B make the forwards reaction faster than the reverse reaction. As soon as some B is present it can react back to make A, but $\ce{ B \to A}$ will be slow compared to $\ce{ A \to B}$ because there is less B (smaller concentration) and the activation barrier is larger so the 'rate constant' is smaller.

Eventually the amount of A is reduced and B increased until both rates are the same and equilibrium is achieved. Note that the concentrations need not be the same, in fact they usually are not. Also note that molecules of A continuously convert to B; $\ce{ A \to B}$ and similarly B convert to A; $\ce{ B \to A}$ which is to say that the normal state of affairs is the 'dynamic equilibrium' and simply using the word 'equilibrium' implicitly assumes that this is the case.


Are bones generally harder to corrode than human skin? I am making a fantasy universe and really I am also curious to know if bones generally are harder to corrode than human skin in our world and at what rate?

I was thinking to compare how a skeleton would react to corrosion compared to a regular human. Maybe some chemists had any insight to this. I see many fantasy universes incorporate "acid" as an element but neglect bases and corrosions as a whole.

Bonus question, is it possible to find a list on how different metals react to strong acids and bases. I have heard of the Reactivity series, but not much beyond that, or perhaps I just don't understand how to use it.

Acids and bases are considered corrosive in industrial chemical hygiene. While it is true on a technical level that harder tissue (like bone tissue) will be a bit harder to corrode compared to soft tissue (like skin), a corrosive substance that can cause damage to soft tissue will cause enough damage to harder tissue to be a medical emergency; this is certainly true when you have a highly concentrated substance. If in your world, things that are worn on the person are made from either soft or harder tissue, then the harder tissue would provide slightly better protection. But to be absolutely clear, both types of tissue will be eaten away, it's just a matter of how much.

Metal reactivity with acids and bases is nuanced because it isn't just about hydrogen ion reactivity. The alloy known as hastelloy (trademarked name) and tantalum alloys are metals used in industry for corrosive environments. Silver and gold are resistant to some common acids, but not all (e.g. nitric acid will dissolve silver).


Particle size vs dissolving rate of sucrose I’m required to make a lab with this information only (particle size vs dissolving rate of a solid (molecular covalent compound-sucrose)? so far what I’m thinking is using three different kinds of sugar like table sugar, sugar cubes, and one of them grinding it but the issues is how do you find how they dissolving rate?

I have found in Germany sugar crystals that are about 5 mm thick. If you are able to get some, you may drop one of these crystals in a beaker with 50 mL water and an stirring bar. If not take a standard sugar cube. Using a watch, measure how long it takes for the crystal to be dissolved. Now start again with another rotating speed. Now break a new crystal in two or three parts with a hammer. And start again with smaller crystals (to be weighed in advance). Repeat the measurements. Report these values on a graph vs rotating rate, or vs. weight of sugar, or vs. temperature. Now start again with powder, and even with the sugar powder used in bakery which is made of extremely fine grains. This is just ideas.


How to explain this acid-catalysed acetal rearrangement mechanistically? I have spotted a protecting acetal group and thought this would undergo hydrolysis … My working is such:

But I cannot get to the desired isomer. Could someone please help? (Diagrams would be very much apprecited thanks!)

You have managed to draw the correct acyclic hydrolized compound with a decent mechanism.

This is a problem that can be solved by numbering the carbon atoms. Number the carbon atoms sequentially along the chain, starting with the acetal carbon (which has to be the carbonyl group). Likewise number the unpacked acyclic molecule. Use what you get to figure out which of the three hydroxy groups are used to make the second acetal.

In general, numbering your atoms using common structural features as anchors can help unravel complicated mechanism problems.


Why is cis-1,2-dichloroethene more stable than trans-1,2-dichloroethene? Why is cis-1,2-dichloroethene more stable than trans-1,2-dichloroethene?

Usually, trans compunds are more stable than cis ones, due to less strain and its non-polarity. But, in this case it's quite the opposite. I tried to search the reason behind it, and found that it's supposedly due to the cis effect?

Here's a pre-MOT rationalisation of the cis form being more stable than the trans form for 1,2-dihaloethylenes:

The lone pair of chlorine atoms is involved in resonance with the double bond, as it does so positive charge appears on one chlorine and negative on another. A cis geometry allows for a stabilizing interaction (attraction) between these positive and negative charges, which makes this resonance form a greater contributor to the resonance hybrid than it is for the trans isomer.

MOT-based explanation:

Eyring's work suggests that there is maximum delocalisation possible in trans form.

Quoting, from the same paper:

  In a recent paper Binghams proposes that the main effect responsible for the “cis effect” of ethylenes is an effect which tends to destabilize the trans form rather than provide extra stabilization for the cis form. In short, this is an unfavorable conjugation effect which stems from the fact that the conjugation in these cases is between two filled orbitals (halogen lone pair and $\pi$ bond). The interaction gives rise to two new orbitals, one bonding and one antibonding, the latter being more antibonding than the former is bonding. Consequently, the total interaction is energetically unfavorable, in contrast to interaction in hydrocarbon polyenes. This would be equally true for both the cis and trans forms but, following Eyring’s argument, the delocalization is greater in the trans form.
  Consequently this form will be the less stable one.

Reference: Structure of 1,2-difluorocyclopropane and the "cis effect"


Material with refraction index similar to Si I am looking for a material with a refractive index over the visible spectrum , which is close to the refractive index of Si, but which does not oxydise in air.

Do you know of any such material ?

Thank you.

Edit:
Some more information. A optical reflector should be build from Si. I will not go into the detail, since it is irrelevant to the question, but I will provide all necessary information. The list is ordered according to importance:

The complex refractive index should be close to Si over the entire visible spectrum
The material should not oxydize in air
The working conditions should be between -30°C to 60°C.
Not necessary, but it would be great if epitaxy on single-crystal silicon is possible.

Over the range of the visible spectrum, ~1.8 - 3.1 eV, or 700-400 nm, the refractive index (nD) of Si varies from about 3.5 - 5, though it is also absorbent and somewhat reflective over that range. Gallium(III) phosphide, $\ce{GaP}$ is one material that comes close to that. It should be fairly stable in air at STP.

Of course, it all depends on your intended use. At what wavelength should nD match that of Si? At what temperature and in what atmosphere will the material be? Does the crystal structure of the material need to match that of Si, so that a regular layer can be deposited? More information is needed for the answer to be of any use.


What happens to gunpowder when it gets wet? What happens to the gunpowder and other explosives when they get wet? I know that they stop "working" (won't explode anymore) but for what reason? Is it anyhow similar to the fact, that wet wood is burning worse than the dry one?

In general, putting explosives in water will not prevent them from exploding, because most explosives already contain all of the necessary reactants to produce an explosive chemical reaction.  They do not need to draw oxygen from the air, and the reaction happens much too fast for the water to act as a heat sink.

Traditional "black powder" is an exception; it doesn't explode when wet.  I don't know exactly why this is, but two plausible reasons are:

Unlike most other explosives, black powder is shock-insensitive; it has to be initiated with a flame or spark.  Water can easily soak up all the heat energy from a small flame, like the one you get from a percussion cap.  This is the same as how it's harder to set fire to a living tree than a dry log, but if the fire gets hot enough it will still catch.
Black powder is a "low" explosive, meaning that the chemical reaction happens relatively slowly (for an explosion), which gives the water more time to absorb heat energy and quench the reaction.

Worse, black powder can be permanently ruined by water.  This is because it's a mixture of saltpeter (potassium nitrate, $\ce{KNO3}$), sulfur, and charcoal, and saltpeter is water-soluble.  If you get it wet enough, the saltpeter can be completely washed out of the mixture.  This doesn't always happen; the 19th century industrial manufacturing process for black powder involved deliberately dampening it, both to make it less dangerous during manufacture, and to make saltpeter crystals form inside each particle of charcoal.  But they were careful to use only just enough water.


Molecular orbital diagram for BF3 I'm trying to build a molecular orbital diagram for BF3 and I'm running into problems with irreducible representations on the F side.

2s for B has an irreducible representation of A1. 2p for B has an irreducible representation of E' and A''2.

2s for F considered non bonding

2p (along the bond axis) for F has an irreducible representation of E' and A'
2p (z) for F Has an irreducible representation of E'' and A''2
2p (the remaining one) has an irreducible representation of E' and A'2

From this I built A1 bonding and antibonding orbitals, A'2 bonding and antibonding orbitals, 1 A'2 and 2 E''2 non bonding orbitals.

I'm now stuck because I have 4 E' on the F side and only 2 E' on the B side.

I found the following diagram for BF3 online but it doesn't generate the E' anti bonding and also doesn't generate enough molecular orbitals.

Does anyone know how to mix 4 E' with 2 E'

I think you are most of the way to the answer, but I will start the process from scratch for the sake of a full explanation for future readers.
First, we determine the symmetry group of $\ce{BF3}$, which by quick inspection we can determine to be $D_{3h}$ (since it has a principle $C_3$ axis, $3\perp C_{2}$ axes, and a horizontal mirror plane of symmetry).

Knowing this, we can now make use of the $D_{3h}$ character table (thank you @orthocresol for compiling these in mathjax format) to determine the irreducible representation of the atomic orbitals (or their symmetry adapted linear combinations SALCS) within the group.
$$\begin{array}{|c|cccccc|c|c|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & \text{linear/rotations} & \text{quadratic}\\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 & R_z & \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 & (x,y) & (x^2-y^2,xy) \\
\mathrm{A_1''} & 1 & 1 & 1 & -1 & -1 & -1 & & \\
\mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 & z & \\
\mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$
Using this, we can quickly determine the irreps of boron's orbitals using the 2nd column from the right.

$\ce{B}$: $2s=a_1'$, $2p_x,2p_y=e'$, and $2p_z=a_2'$.

Next, we will determine the irreps for the $3\ce{F}$ atoms, which will require that I consider combinations of $s,p_x,p_y, \text{and } p_z$ orbitals.
Below, I show how the symmetry operations of the group affect each set of orbitals and show the orbitals that result (image source).
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_s & 3 & 0 & 1 & 3 & 0 & 1 \\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{20ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_x} & 3 & 0 & 1 & 3 & 0 & 1 \\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{12ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_y} & 3 & 0 & -1 & 3 & 0 & -1 \\ \hline
\mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{20ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_z} & 3 & 0 & -1 & -3 & 0 & 1 \\ \hline
\mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 \\
\mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0  \\ \hline
\end{array}$$
$\hspace{21ex}$
These reductions to irreducible representations can be done via the reduction formula $$n_i=\frac{1}{h}\sum_c g_c \cdot \chi_{i_c} \cdot \chi_{r_c}$$ where $n_i$ is the number of times a particular irrep $i$ occurs in the reducible representation, $g_c$ is the number of symmetry operations $c$, $\chi_{i_c}$ is the character of the irrep $i$ for the symmetry operation $c$, and $\chi_{r_c}$ is the character of the reducible representation $r$ for the symmetry operation $c$.
This allows us to write the irreps of the three $\ce{F}$'s orbitals.

$3\ce{F}$: $a_1'+a_2'+a_2''+2e'+e''$

With all the formalities out of the way, we can construct the qualitative MO diagram for $\ce{BF3}$. Like much of this post, my source for the MO diagram is the pdf of lecture notes for a Dartmouth chemistry course located at http://www.dartmouth.edu/~chem64/64%20pdf%20files/PS3A.pdf

Working our way up through this diagram, we see that we start by assuming that the SALCs of the $s$ orbitals of the three $\ce{F}$'s are assumed to be low enough in energy where they do not interact with the orbitals of $\ce{B}$.
The next point of interest is the $e' (\sigma)$ MO, which we form from the $p_x$ SALC of the $\ce{F}$ atoms and the $p_x$ and $p_y$ of the $\ce{B}$. We could additionally consider the interaction of the $e'$ $p_y$ SALC to form a $\pi$ bonding interaction, but we will assume that, with the $\sigma$ interaction already present, this $\pi$ bonding interaction will be weak and thus can be neglected.
This previous point leads us to have a set of nonbonding orbitals $a_2'+e'(y)+e''$. We then also have antibonding orbitals for each of the bonding orbitals below.
As a check on our work, we can compare this to the somewhat more quantitative diagram given in Principles of Inorganic Chemistry by Brian PFennig using the Student Editon of Spartan:

The only noticeable difference here is that the nonbonding MOs are now placed with respect to their computed energy rather than being lumped together. The classification of orbitals as bonding, nonbonding, and antibonding seems to fit with the diagram we have above, with the caveat that this more quantitative diagram does suggest some $e'(y)$ bonding contribution, as its energy is somewhat lower than the p orbitals it came from.
Now that we have the correct diagram, we can then compute the bond order for each $\ce{B-F}$ using: $$\frac{\text{Bonding electrons } -\text{Antibonding electrons}}{2\cdot\text{number of bonds}}$$ which for this diagram gives a value of $\frac{4}{3}$ per bond. This gives a sense of why $\ce{BF_3}$ is a strong Lewis acid, since it can accept a pair of electrons into its antibonding LUMO and still leave a compound with a BO of $1$ per bond.
As a final point, it is worth noting that this method of classifying orbitals is far from perfect and in general we should really look at individual bonds to classify if the orbital is anti/bonding with respect to that bond. However this energy based approach is what is commonly taught in school and happens to work well in this simple case; just be aware that it can fail for even seemingly simpler molecules like $\ce{CO}$.


Ballistic behavior of molecules on potential energy surfaces I came across the following sentence in a publication$^1$: "At this point we have to remember that the computed MEP path [sic] represents the reaction in the absence of any kinetic energy, while in reality the true reaction path, to some degree, will be ballistic." Further, they write that, "This path is accessed because after the transition state the molecule is 'ballistic'."

A minimum energy path (MEP) is calculated "step-wise", taking the previous geometry as a starting point, and the program follows the gradient. In classical terms, this might be to place a ball on a tilted plane, releasing it, and stopping it after a given distance, and repeat.

When they say that, in a real reaction, the molecules will have kinetic energy and be ballistic, I think they mean that the potential energy surface (PES) may be explored in directions slightly different than the gradient. This can be compared to releasing a ball on the top of a changing surface and allowing it to roll freely, where the ball will not follow the gradient when its velocity is high.

I cannot find a decent explanation of this anywhere, and it would be nice to get some input.

$^1$ De Vico, Liu, Krogh, and Lindh, 2007. J. Phys. Chem. A 111:8013-8019. doi:10.1021/jp074063g.  Cited passages are on p. 8016, Section 3.3.

I think the description you lay out in your question has it basically right.

MEPs, IRCs, etc. all assume that the geometric rearrangements that occur over the course of a reaction strictly follow the gradient uphill from reactants to transition state, and then downhill from TS to products (or intermediates).  Energetically, this provides the "path of least resistance."

In the paper you cited, it looks like the MEP would require some atoms in the system to "take a right-angle turn." Quoting a broader excerpt around the first sentence you include (emphasis added):

  As can be seen in Figure 3, after TS‘ S$_0$ the MEP takes a turn on the PES:  the O−O‘ distance increasing and the torsion around the O−C−C‘−O‘ dihedron nearly stopping, while the increase of the C−C‘ distance comes into action.
  ...
  At this point we have to remember that the computed MEP path represents the reaction in the absence of any kinetic energy, while in reality the true reaction path, to some degree, will be ballistic.
  ...
  What would be the consequences if the reaction continued along the torsional mode? Section 3.2 already suggested that this is of interest—the torsional mode is the principal internal coordinate of the T$_1$ MEP.

Nuclei have mass (obviously), and thus have inertia. While Newton's relevance at the atomic scale can be fuzzy at times, classical treatment of atomic motion has sufficient validity that acceleration of nuclei can be considered to require a suitable accompanying force (viz., PES gradient).  If the system has enough "inertia" and the PES gradient provides too small an acceleration to overcome it, the progress of the reaction can absolutely move away from the MEP.

Alternatively, at least one additional mechanism can introduce  deviations from the MEP: molecules at finite temperature are always vibrating. Thus, even in a case where the key internal molecular coordinates (bond lengths, angles, dihedrals) for a given reaction strictly follow the IRC, the "uninvolved" vibrational motion is still occurring, and so the  system is actually always oscillating around the IRC in a high-dimensional sense.  If the energy of the system is high enough, these oscillations have the potential to be of sufficient magnitude to "bump" the reaction coordinate away from the gradient-following pathway.

Rotational motion may also contribute to this, but I would think in a more subtle way than vibrational motion.

For further reading: Steven Bachrach of the Computational Organic Chemistry blog has an entire tagged category, "Dynamics," dedicated to this phenomenon.  In no particular order, some posts of possible survey value include:

An approach towards identifying dynamic effect without trajectories
Insights into dynamic effects
Bifurcating organic reactions

The entirety of Chapter 8 of his book is also dedicated to this topic.

Addendum: FWIW, as a chemical engineer, the (imperfect) analogy that I like to use is to liken the MEP to the 'reversible process' of thermodynamics:

Start at the beginning state
Take an infinitesimal step
Let the system infinitesimally relax
Repeat 2 & 3 until you reach the end state

As I (very approximately) understand it, irreversibility comes from changing the state of a system "too much, too fast", leading to net generation of entropy as the system relaxes back to the reversible pathway, and it's "kinetic energy" in a qualitative, generalized sense that allows the system to depart from the reversible path.


How to solve constants out of the internal energy equation? Imagine we deal with a new kind of matter, whose state is described by:
$$PV = AT^3$$
Its internal energy is given by
$$U = BT^n \ln\left(\frac{V}{V_0}\right) + f(T)$$
where $A$, $B$ and $V_0$ is a constant and $f(T)$ is a polynomial function.
Find $B$ and $n$.

What I know
The given expressions remind me of adiabatic compression/expansion. If we assume quasistatic adiabatic compression/expansion we know that heat won't get out/in the system.
$$\Delta U = -W$$
And work is:
$$W = -P\Delta V$$
Some thoughts on how to solve the problem
We notice here that we are dealing with a non-ideal gas. Assuming that the above equations are correct and using first thermodynamics law one gets:
$$\mathrm{d}U = \left[nBT^{n-1}\ln\left(\frac{V}{V_0}\right) + f'(T)\right]\mathrm{d}T$$
$$\left[nBT^{n-1} \ln\left(\frac{V}{V_0}\right) + f'(T)\right]\mathrm{d}T = \frac{AT^3}{V}\mathrm{d}V$$
I do not see a solution to this differential equation. There has to be a easier way to get both $B$ and $n$, but how?

You need to use the equation $$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$


Is Ammonia (NH3) base? In my text book, it is written that $\ce{NH3}$ is a base. They mentioned the reason for this is that it produces $\ce{OH-}$ ions in water. They stated this reaction:

$\ce{NH3 + H2O -> NH4OH}$

So That's why, they said that it is a base.
But in a neutralization reaction, mentioned by our teacher,
Acid + Base = Salt + water
So, if $\ce{NH3}$ is a base then,
it should be,
Acid + $\ce{NH3}$ = Salt + Water
But it does not happen. When reacting with acid, it doesn't produce water.
For example,

$\ce{NH3 + HCl -> NH4Cl}$
--  Reference

Again,

$\ce{NH3 + H2SO4 -> (NH4)2SO4}$
--  Reference

So it is not producing water in the neutralization reaction. Hence, I'm confused whether it is base or not.

Your problems are caused by using the Arrhenius theory, which is based on electrolytic dissociation.

According to the Arrhenius definition, acids are compounds that dissociate and release hydrogen ions $(\ce{H+})$ into the solution:
$$\ce{HCl -> H+ + Cl-}$$
Bases are defined as compounds that dissociate and release hydroxide ions $(\ce{OH-})$ into the solution:
$$\ce{NaOH -> Na+ + OH-}$$
The products of a neutralization reaction of an acid with a base are a salt and water:
$$\ce{HCl + NaOH -> NaCl + H2O}$$
This definition works for metal hydroxides such as $\ce{NaOH}$, which can exist as pure compounds. However, it doesn’t work for bases such as $\ce{NH3}$.

In order to solve this problem for $\ce{NH3}$, the fictitious compound “ammonium hydroxide” $\ce{NH4OH}$ was proposed (see also this related answer). Assuming this formula, the Arrhenius definitions of bases and neutralization reactions work again:
$$\ce{NH4OH -> NH4+ + OH-}$$
$$\ce{HCl + NH4OH -> NH4Cl + H2O}$$
However, ammonium hydroxide cannot be isolated; the fictitious solid compound does not exist. Nevertheless, the misleading traditional name “ammonium hydroxide” is still widely used for solutions of ammonia in water.

In order to really solve your problems with $\ce{NH3}$, you need to switch from the definition of bases according to the Arrhenius theory to the definition of bases according to the Brønsted–Lowry theory.


Dissociation degree hell, Ostwald's law of dilution Greetings dear chemists!

I got a nice exam problem on wich i am stuck for hours, well maybe the problem is with me. I know the problem can be solved. So here we go.

We got a weak monoprotic acid, we don't know its concentration, dissociation degree nor the dissociation constant. If we dilute it 3 times its volume, then the dissociated ions grow with an unknown constant, lets call it $y$. If we dilute the diluted solution again by 5 times its volume, then the dissociated ions grow 4 times. Whats the dissociation degree of the undiluted solution?

Yeah woah that looks tough. So lets give it my try.

Lets call the concentrations in order of dilution by $c_0, c_1, c_2$ and the dissociation degree by $\alpha_0,\alpha_1,\alpha_2$, and the dissociation constant of course $K_s$.

I managed to get three equations for this problem, but i feel i missed something or messed up my algebra someway i dont understand.

$$c_0=?,\alpha_0=?,K_s=?$$
$$\alpha_0=\frac{[A^-]}{c_0}$$
$$HA=H^++A^-$$
$$c_1=\frac{c_0}{3},\qquad c_2=\frac{c_1}{5} \to c_2=\frac{c_0}{15}$$
$$\alpha_1=y\alpha_0,\qquad \alpha_2=4\alpha_1 \to \alpha_2=4y\alpha_0$$
We call forth Ostwald;
$$K_s=\frac{c_0\alpha_0^2}{1-\alpha_0}\qquad \qquad \qquad \qquad (1)$$
$$K_s=\frac{c_1\alpha_1^2}{1-\alpha_1}=\frac{c_0y^2\alpha_0^2}{3(1-y\alpha_0)}\qquad \qquad(2)$$
$$K_s=\frac{c_2\alpha_2^2}{1-\alpha_2}=\frac{c_016y^2\alpha_0^2}{15(1-4y\alpha_0)}\qquad \qquad(3)$$
Equating the first and the second equation, i could get an expression for $y$. By setting $\beta=\frac{3}{1-\alpha_0}$ and ignoring the negative root
$$y^2+y\beta\alpha_0-\beta=0$$
$$y=\frac{-\beta\alpha_0+\sqrt{\beta^2\alpha^2+4\beta}}{2}$$
My problem starts when i equate the second and the third equation;
$$\frac{c_0y^2\alpha_0^2}{3(1-y\alpha_0)}=\frac{c_016y^2\alpha_0^2}{15(1-4y\alpha_0)}$$
$$5-20y\alpha_0=16-16y\alpha_0$$
$$11+4y\alpha_0=0$$
From wich i clearly see that im on the wrong track, since $\alpha_0$ and $y$ cant be negative, please help me on solving this nightmare. Maybe i missed something obvious, i dont really know that im on the right method to solve this either.

You are on the right track according to the directions given to you ( = the problem statement), but the directions themselves are wrong.

Say you had an equilibrium with $\ce{{[H^+][A^-]\over[HA]}=K}$. You dilute it five-fold and tell your particles to wait a bit with dissociation until you are done with the math. That makes all concentrations 5 times smaller, and consequently, the LHS is now $1\over5$ of what it used to be. Now you tell them to increase their dissociation by a factor of 4. In doing so, each of $\ce{[H+]}$ and $\ce{[A-]}$ increases four-fold, thus growing the LHS to $16\over5$ of its initial value, or more than 3 times greater than $K$, and they do that at the expense of the remaining $\ce{[HA]}$, decreasing the denominator and increasing the LHS further yet. Just how is it going to be the same $\ce{K}$ again?

I suppose it is some subtlety of wording that we are misinterpreting here. It might have been lost in translation, but can't we at least make a reasonable guess? Could it be that the 4-fold increase is compared to the initial state, that is, to $\alpha_0$ and not to $\alpha_1$? No, that would not suffice. Still, there might be other possibilities...


Is anode actually ever negative? Disclaimer: This question comes from a physics student.

I was studying electrostatics and I was wondering how to charge a conductor at home without any fancy furs etc. So an idea came to me: take a battery, touch for example the negative terminal with the conductor and then remove it. Since the negative battery terminal is at a negative potential and the conductor was previously grounded ($V = 0$) then as their potentials must become equal, it means that electrons will flow from the battery terminal to the conductor for a short period of time. Afterwards, I disconnect the conductor and the electrons remain 'trapped' in it hence the conductor is now negatively charged.

At least that was the idea. Now suppose that the battery is a galvanic cell ($Zn, Cu$). The question is whether there are free electrons floating inside the $Zn$ anode when it is not connected to the cathode?

The whole system (battery, wires) is electroneutral while the current is flowing e.g. Zinc atoms at anode dissolve and electrons stay at the anode while the $Zn^{2+}$ ions go to the solution, but in the same instant other electrons leave the anode.

When the electrodes are not connected, I'm not sure would the redox reactions happen. In my understanding, $Zn$ oxidizes because $Cu$ on the other side has greater electron affinity, but $Zn$ would have to be able to 'sense' $Cu$ on the other side somehow, otherwise the oxidation reaction happens anyways for a short time until it reaches some equilibrium point. This is what I think actually happens and that's why the anode would actually be negative until connected to the cathode.

So am I on the right track with this and ultimately, will my attempt to charge the conductor a bit succeed?

It is a very good questions from a student who seems to be thinking deeply. I have been thinking to write on this topic for the last 7-8 years. In short the poles of the batteries are electrostatically charged and one can actually "sense" this charge with the help of sensitive electroscopes which were known in the time of Volta.

The label anode or cathode is not defined with respect to the electrostatic sign as explained in the post by Maurice.

Think of a large parallel capacitor. Connect one terminal of the plate to one positive terminal and the other one the negative terminal. The capacitor plates would show a electrostatic charge.

By all means, using one single cell you would not be able to detect the electrostatic charge, like a charged comb. Only specialized electroscopes can sense that.


Determining the more acidic proton in cyclopenta-1,3-diene I am in the David Klein textbook in Chapter 3 going over rule 2 of finding the more acidic proton: if the electron pair in the conjugated base is delocalized, then the conjugate base is more stable, ergo, the proton at that location is more acidic. Here's the practice problem I am stuck on:

I generated two different conjugated bases. The book indicates the red proton is the more acidic proton. Is this the proper resonance structure for the red proton's conjugate base?

If so, then this explains why the red proton is more acidic, because it has a delocalized pair of electrons, while the blue proton does not have a delocalized pair of electrons.

Figure c does not look like a real compound. If it represents the conjugate base of cyclopentadiene, it is lacking a negative sign, and all five protons around the ring are equivalent, so the question of whether the red or blue proton is more acidic makes no sense.

If Figure c represents cyclopentadiene, the acid, it does not clearly show that the red position really has 2 hydrogens. However, the double bond structure shown is consistent with 2 hydrogens in the red position, and the question becomes very simple: in cyclopentadiene, is a methylene proton or a vinylic proton more acidic? Removing a vinylic proton puts a negative charge in the localized sigma system - no aromaticity develops. However, removing one of the methylene protons causes that carbon to go from sp3 to sp2p, and the negative charge goes into the pi system where it can be delocalized over the 5-membered ring. And aromaticity is a characteristic feature of cyclopentadienide ion.


Which alkene reacts faster with HCl? Which alkene reacts faster with HCl?

The logic tells me the most stable alkene 2 which is the most substituted would react the slowest.
On the other hand, it would form the most stable tertiary carbocation which would mean it's best for this specific reaction which would make it react faster. Also, one could think about rearrangements because these could take time too. How should I answer this question?

This is an exercise in applying the Hammond postulate (multiple times). To accurately compare reaction rates, we need to have the activation energy. We don't have this, and the actual energy of the transition state is unknowable without experimental data.
Hammond's postulate states that the structure of the transition state is more similar  to the intermediate to which it is closer in energy. For these reactions, the formation of the carbocation is endergonic, so the transition states look more like the carbocation than the alkene. We can then make some assumptions about the relative energies of the transition states from the energy change of the mechanism step if some things remain the same.
For example, alkene 3 can be protonated to form two different carbocations: A and B.

Using PM6 calculations in MOPAC, I calculated the heats of formation of alkene 3 and cations A and B.
$$\begin{array}{l|r}
& \mathrm{\Delta_r H^\circ \ (kJ/mol)} \\ \hline
3 & -42.1 \\
\mathrm{A} & 710.7 \\
\mathrm{B} & 702.6\\
\end{array}$$
Since we are starting from the same place (alkene 3), we can postulate that cation B forms more quickly than cation A because it is slightly lower in energy.
You need to do similar analysis for all your alkenes and the cations they can form. In some cases you have two alkenes that might form the same cation. In that case, the higher energy alkene would likely translate into the more rapid reaction.
You do not need to go and calculate all the energies. You should have learned some rules by now to assess the relative stability of alkenes and carbocations based on substitution pattern.


Structure of diamond "diamond structure. Every carbon atom is bonded
to four other carbon atoms forming a three-dimensional lattice of chair conformations."

SENTENCE FROM A BOOK

What does this mean?
Where I am struck is that I knew carbons occupy zinc blende like lattice so, how can this line be true.

Where I am struck is that I knew carbons occupy zinc sulfide like
  lattice so, how can this line be true.

Zinc sulfide can exist in two different crystalline forms, Zincblende (sphalerite) and wurtzite.  Zincblende is face-centered cubic (also known as diamond cubic), each ion is tetracoordinate and has local tetrahedral geometry - just like the carbons in diamond.  The wurtzite structure is hexagonal close-packed with interconnected 6-membered rings.  You may have been thinking of the wurtzite structure, while it is the zincblende structure that is related to diamond.  You can see pictures of the two structures here.


Alternative Program for Orbital Population Analysis Gaussian has Orbital-by-Orbital Population Analysis.

From the manual:

# UHF/6-311+G(d) Pop=Orbitals=3
Here is the resulting output from a calculation on FeO+ quartet:

Atomic contributions to Alpha molecular orbitals:
Alpha occ 16 OE=-0.923 is Fe1-d=1.00
Alpha occ 17 OE=-0.699 is O2-p=0.88
Alpha occ 18 OE=-0.690 is O2-p=0.68 Fe1-s=0.21
Alpha vir 19 OE=-0.253 is Fe1-s=0.70 Fe1-p=0.27
Alpha vir 20 OE=-0.188 is Fe1-p=0.71 O2-p=0.29
Alpha vir 21 OE=-0.133 is Fe1-p=1.04

Atomic contributions to Beta  molecular orbitals:
Beta  occ 13 OE=-0.801 is O2-p=0.79
Beta  occ 14 OE=-0.783 is Fe1-d=1.00
Beta  occ 15 OE=-0.758 is O2-p=0.89
Beta  vir 16 OE=-0.241 is Fe1-s=0.81 Fe1-p=0.17
Beta  vir 17 OE=-0.139 is Fe1-p=0.91 Fe1-d=0.14

This can be used to provide a more quantitative characterization of the nature of canonical molecular orbitals.

Due to certain restrictions on the use of Gaussian (http://www.bannedbygaussian.org/), I am unable to use Gaussian. Are there other programs that can perform such a calculation? NBO only seems to provide such a breakdown in terms of NBOs, and I can't find the feature available in any other quantum chemistry program. I realize that I can do this myself by extracting the C matrix, figuring which coefficients go with which orbital, and summing up myself. However, this would probably take me a day to program.

Note: My calculations use B3LYP, ECPs, and the COSMO solvent model. So the program would need to be able to at least produce single points with such a method.


Why is the IUPAC definition of adsorption so different from other definitions? NB: This question has been edited. Originally, some of the confusion was caused by me not knowing that condensed phase could refer to solid phases as well. This has been removed, but the confusion caused by the other elements still remains and even the first answerer to this post was unable to clear that up.
The Gold Book's definition of adsorption

An increase in the concentration of a dissolved substance at the interface of a condensed and a liquid phase due to the operation of surface forces. Adsorption can also occur at the interface of a condensed and a gaseous phase.

What I've learned is that adsorption is when a fluid, or the solute(s) within a liquid, adhere to a surface. The IUPAC definition however, only recognizes the case when the solute(s) within a substance adhere to a surface. A more limited, and to my knowledge, less useful definition. Furthermore, the formulation is very ambiguous. It does not state if it is the condensed phase or the fluid phase that contains the solute(s), or if either can contain the solute(s). The middle option would necessarily be in contradiction to the IUPAC's definition of a solution:
The Gold Book's definition of solution (shortened):

A liquid or solid phase containing more than one substance, ...

This definition does not count gaseous phases as solvents. The fluid phase can be a gas, but if the dissolved substance is always within the fluid, then in cases where the fluid phase is a gas, it would be a solvent gas. This is assuming that dissolved substance is the same as solute.
Now, for part two of this mess:
Furthermore, the IUPAC definition of absorption encroaches onto the semantic territory of adsorption, at least the territory that I've seen and been taught.
The Gold Book's definition of absorption

The process of one material (absorbate) being retained by another (absorbent); this may be the physical solution of a gas, liquid, or solid in a liquid, attachment of molecules of a gas, vapour, liquid, or dissolved substance to a solid surface by physical forces, etc.

The last part there is the, or a part of the, definition of adsorption. It described surface attraction, which I thought was clearly within the domain of adsorption. The first part of the definition says that the absorbate and absorbent create a physical solution. I cannot find any definition of that, but I assume they are referring to a homogeneous mixture where the solute isn't disassociated, but still kept within a solution complex. However, an absorbate isn't (necessarily) homogeneously distributed within the absorbent. This means they're violating their own (and countless other) defintion(s) of solution.
As mentioned, a solution is a single phase containing more than one substance, and a phase, as defined by the IUPAC, is homogeneous.
The Gold Book's definition of a phase:

An entity of a material system which is uniform in chemical composition and physical state.

Are these IUPAC definitions outdated? If so, what is the most authoritative definition of adsorption and absorption. I like to use words consistently, but given how inconsistent and/or ambiguous scientific literature can be, I want to pick the best definition in order to influence the literature, if however minutely, towards more consistency and away from ambiguity.

The IUPAC defintion isn't more limited than the ones you've read. Instead, it is more general.
Your confusion results from not being familiar with the term "condensed phase". You think a condensed phase is a 'more condensed liquid', and consequently think IUPAC is referring to a surface formed between a liquid and a 'more condensed liquid':  "where there is an interface of two liquids, one containing a solute, and one being more condensed".
No.  A condensed phase means a liquid or a solid.  Thus, when IUPAC says the following, what they mean is that adsorption can occur at liquid-liquid, liquid-solid, gas-liquid, and gas-solid interfaces.

An increase in the concentration of a dissolved substance at the
interface of a condensed and a liquid phase due to the operation of
surface forces. Adsorption can also occur at the interface of a
condensed and a gaseous phase.

Is the IUPAC definition completely general?  No.  It doesn't account for adsorption between supercritical fluids and solids. But it at least covers all the routine phase combinations.
As to the issue of whether IUPAC's defintion of solutions (which limits the solvent to be a condensed phase) contradicts its defintion of adsorption: While it could have been written more explicitly, a close read indicates there's not necessarily a contradiction.  Let's take each sentence in turn:
The first sentence concerns a liquid phase in contact with a solid or liquid surface.  Here the adsorbent is referred to as a "dissolved substance", which would make it the solute within a liquid solution:

An increase in the concentration of a dissolved substance at the
interface of a condensed and a liquid phase due to the operation of
surface forces.

Note that the solute can be (prior to its dissolution) a solid, liqud, or gas.
The second sentence concerns a gaseous phase in contact with a solid or liqid surface.  Here no explicit mention is made of a dissolved substance:

Adsorption can also occur at the interface of a condensed and a
gaseous phase.

Thus we could parse IUPAC's language to mean that adsorption involves either the solute within a liquid solution that is in contact with a liquid or solid surface, or a gas that is in contact with a solid or liquid surface (where that gas could be pure, or part of a mixture).


Decarboxylation under acid, water, and extreme heat I am asked whether dimethyl malonate or methyl acetoacetate decarboxylate when hit with acid, water, and extreme temperatures.

I believe the answer is both because both are esters, and esters, upon being hit by such conditions, return to their respective carboxylic acids.

From there, both carboxylic acids can undergo decarboxylation because both resulting carboxylic acids are beta-keto acids. Technically, dimethyl malonate becomes malonic acid, which can still undergo decarboxylation (but malonic acid isn't of the beta-keto class; rather the dicarboxylic acid class).

Acid and heat treatment of methyl acetoacetate initially gives 3-oxobutanoic acid but this can decarboxylate under the heat as well.

Is my analysis correct?

Yes, your analysis is correct. This is a common sequence in the malonate synthesis of carboxylic acids and acetoacetate synthesis of methyl ketones. After taking advantage of the acidic, doubly alpha-proton and using the resulting enolate as a nucleophile, the ester(s) are hydrolyzed with acid. A carboxylic acid with a carbonyl in the beta-position can be thermally decarboxylated to give an enol, which typically tautomerizes to the keto form.


Why is the Lucas test not recommended to differentiate higher alcohols? I read about the Lucas test for alcohols in which a cloudy precipitate is produced with a tertiary or secondary alcohol. However, on a side note it was mentioned that this test is useful only for compounds having less than six carbon atoms. Why is this so?

The problem is that the alcohols themselves might be insoluble. Ethanol is soluble, yes, but as the alkyl chain length increases, the hydrophobic nature due to the chain starts dominating over whatever hydrophilicity the $\ce{-OH}$ group had been providing for lower homologues.

From source [1]:

  .. any alcohol that is insoluble in the reagent might appear to be giving a positive 3° test.

The paper goes on to properly quantify several things:

  Based on the results of this study, using 1 drop of alcohol added to 10 drops of reagent, it can be said that, at the very least, all saturated acyclic monofunctional alcohols having six or fewer carbons are soluble in the Lucas reagent.

It is clear that the relative amounts of the sample and the reagent become important in such borderline cases (~6 carbon long chains). If too much alcohol is added, it may cause an illusion of positive lucas test.

From source [1]: (emphasis mine)

  That statement may not be true when 3 or 4 drops are used
  per 10 drops of reagent. In fact, sometimes when 10 drops of
  reagent are added to 3 or 4 drops of some of the alcohols, an
  interesting event occurs after a delay of a couple of minutes
  or so. Apparently, some undissolved alcohol that has been
  clinging to the bottom of the test tube suddenly breaks loose
  and, by streaming to the surface of the very dense Lucas
  reagent, gives a momentary illusion of a positive test. The
  real cloudiness, alkyl chloride formation, appears later—all
  the more reason to use a high reagent/alcohol ratio.

References:

[1] "A study of the Lucas test"
R. A. Kjonaas and B. A. Riedford
J. Chem. Ed. 1991 68 (8), 704
DOI: 10.1021/ed068p704


Carbonate + acid reactions - How to recognize these type of reactions? So I've got $$\ce{CaCO_3 + HCl \rightarrow CO_2 + H_2 O + CaCl_2}$$

I'm wondering how I can recognize these types of reactions? I know that when I mix an acid and a base I get salt and water but in this reaction there is also some $\ce{CO2}$ which is confusing me. When can I expect this type of salt base reaction and should I expect more like these with other by products other than $\ce{CO2}$?

The reaction you showed is actually a salt "exchange" reaction (also called salt metathesis), in which a salt is transformed into another one during the course of the reaction. In your example when calcium carbonate is treated with hydrochloric acid, calcium chloride is formed. At the same time, carbonic acid is generated and it spontaneously decomposes into carbon dioxide and water. The equilibrium is shifted towards the products due to the release of carbon dioxide gas from the reaction mixture, thus driving the reaction to completion.