Is it possible to recycle an alkaline battery? Is it possible to recycle an alkaline battery?

The application I am thinking about is that in 3rd world countries often access to western markets is limited. You can't just pop into a store and buy batteries. Would it be possible for a small factory to be in these places and the locals could bring their batteries in for recycling?

The reactions for an alkaline battery are the conversion of zinc to zinc oxide and MnO2 to Mn2O3. So, to recycle a battery of this type you would have to reverse these reactions. Is that easy to do?

Is it possible to recycle an alkaline battery?

Yes, in theory. Everyready has 4% recycled materiel in some of its batteries.

  The application I am thinking about is that in 3rd world countries often access to western markets is limited. You can't just pop into a store and buy batteries. Would it be possible for a small factory to be in these places and the locals could bring their batteries in for recycling?

No. The chemistry and processing involved would be far to complicated for some sort of rudimentary factory. There is also the economic factor. Even with a well educated work force such a factory would be very expensive to run and need a massive number of used batteries to recycle.

Part of the expense that recycling would offset is the cost of disposing of the used batteries. For example NiCd batteries in a landfill are bad news because of the toxicity of cadmium. Third world countries generally have much poorer environmental regulations than the developed countries.

To answer the additional question...

  The reactions for an alkaline battery are the conversion of zinc to zinc oxide and MnO2 to Mn2O3. So, to recycle a battery of this type you would have to reverse these reactions. Is that easy to do?

No it isn't easy. You can't just mechanically separate the zinc component of an used battery from the manganese component. You'd have do something like dissolve the mess with acid then separate and process the two elements.


Will NaF + CaCO3 precipitate much CaF2? If I mix sodium fluoride, calcium carbonate powder, and water, under what conditions (if any) would there be an equilibrium?

I.e. assume we start out with fully dissociated $\ce{Na+(aq) + F-(aq)}$, because it's fairly soluble. But when I add the $\ce{CaCO3}$, it's only slightly soluble, so assume an excess of solid $\ce{CaCO3}$ and a small amount dissolved.
Will that be stable with $\ce{NaF}$ concentration much greater than the saturated concentrations of $\ce{CaCO3}$, or will $\ce{CaF2}$ start to precipitate, and will it continue until exhaustion of a reactant, or reach equilibrium?

I'm wondering if it's the relative solubilities of $\ce{CaCO3}$ and $\ce{CaF2}$ that will drive the equilibrium (if that's what it is) one way or the other. I can't work out how to apply the theory of an equilibrium constant because of the excess solid reactant and corresponding possible precipitate messes up how concentrations balance out in a normal equilibrium… And I'm wondering if the absolute concentration of $\ce{NaF}$ effects things as well.

Any pointers appreciated. Or even some jargon to say what this kind of reaction is called.

Yes, the calcium ion could lead to precipitation. The solubility of $\ce{CaCO3}$ in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate.

The solubility constant for $\ce{CaF2}$ is about $4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's well below the solubility of NaF.

However, the presence of additional carbonate could be used to reduce the calcium concentration. Similarly, any addition of acid will convert carbonate to bicarbonate and increase the maximum calcium concentration.

UPDATE:
In the above answer, I assumed a basic familiarity with solubility products. For those not familiar with those, here's more detail.

The key quantitative measure of solubility of ionic compounds is the solubility product, usually indicated as $K_{sp}$, which is the product of the concentrations of the separate ions.

For $\ce{CaCO3}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{CO3^2-}]$. Reported values vary somewhat, but are typically around $2\times 10^{-8}$.

Likewise, for $\ce{CaF2}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{F-}][\ce{F-}]$, and the reported values are around $4\times 10^{-11}$. Thus, the condition for keeping fluoride in solution is

$[\ce{Ca^2+}][\ce{F-}][\ce{F-}]<4\times 10^{-11}$.

If $\ce{CaCO3}$ is dissolved in distilled water to maximum solubility, the calcium ion concentration is $\sqrt{K_{sp}}\approx 0.15$ mM, which is the basis of the calculation above.


Elementar cell of Y2O3 I think this picture from wikipedia is wrong because I count:

\begin{align}
\left.
\begin{array}{ccccr}
1&×&1 &= &\ce{1~Y} \\
6&×&\frac{1}{8} &= &\ce{\frac{3}{4}O}
\end{array}
\right\} \text{ gives }\ce{Y4O3}
\end{align}

How would be the correct elementary cell of $\ce{Y2O3}$?

(I didn't found any correct elementary cell using google)

The picture in Wikipedia doesn't give you the primitive cell. In Yttria, $\ce{Y2O3}$, 8 metal ions are in the special positions $\frac{1}{4}$, $\frac{1}{4}$, $\frac{1}{4}$; the remaining 24 metals occupy the sites $u$, $0$, $\frac{1}{4}$. The 48 oxygen ions are in general positions $x$, $y$, $z$, and are arranged in distorted octahedra around the metal ions, the metal-oxygen bonding distances being unequal. So, the stoichiometry is $\ce{Y32O48}$ which gets simplified to $\ce{Y2O3}$. This structure is generally referred to as the C-type metal oxide structure. All the sesquioxides of the rare earths belong to this system, as do $\ce{Ga2O3}$, $\ce{In2O3}$, $\ce{Tl2O3}$ and the mineral bixbyite, $\ce{(Fe, Mn)2O3}$. The primitive cell looks like this (oxygen ions are red, metal ions are gray)

Here another view which might show more easily that indeed the positions of all the 48 oxygen atoms are different in such a way that no smaller primitive cell can be found.

Reference

M. G. Paton, E. N. Maslen, Acta Cryst. 1965, 19, 307-310


Why do phosphorus and sulfur have variable valences? I have learnt that phosphorus and sulfur have valences 3, 5 and 2, 4, 6 respectively because of vacant d orbitals. But how?
Phosphorus' configuration is $\mathrm{[Ne]3s^23p^3}$. There are 3 unpaired electrons that can get paired with bonds, so 3 makes sense (similar to nitrogen) and a similar argument for 2 with sulfur (similar to oxygen), but how do the vacant d orbitals help in giving valence 5?
Do 2 electrons from the s orbitals go to the d orbitals and then there are 5 unpaired electrons available for bonding and similarly for sulfur?

For a modern answer (with some hand-waving), we can simply say that valence electrons occupy more space as we go from first period to second period to third period. So hydrogen makes one bond, second period elements up to four, and third period elements up to six.
If you want to stick to Lewis structures and hybridization, you can explain $\ce{PCl5}$ with resonance:

Source: https://www.cusd80.com/cms/lib6/AZ01001175/Centricity/Domain/1108/003-center%204-electron%20bond.pdf
If you like molecular orbital diagrams, it is pretty straightforward to expand a $\ce{EX3}$ MO diagram to a $\ce{EX5}$ diagram without using d-orbitals, e.g. http://www.chembio.uoguelph.ca/educmat/chm364_preuss/5_5contd2%20MO%20Theory_ICl4%20PCl5.pdf. Like in the valence bond model, the bond order will decrease as you increase bound atoms.


Energies of d orbitals relative to the barycentre Rather than referring to the energy of an isolated metal atom, the weighted mean of these two sets of perturbed orbitals $(\mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$ and $\mathrm d_{x^2 - y^2}$, $\mathrm d_{z^2} )$ is taken as zero. This sometimes is called the barycentre. The difference in energy between the two $\mathrm d$ levels is given either of symbols $\Delta_\mathrm O$ or $10~\mathrm {Dq}$. It follows that $\mathrm{e_g}$ orbitals have $+0.6 \Delta_\mathrm O$ above average level and $\mathrm{t_{2g}}$ orbitals are $-0.4\Delta_\mathrm O$ below the average.

It might be basic but why is orbitals split in $-0.4$ and $0.6$? If the barycentre is at the average then the orbitals should be split in $-0.5$ and $0.5$ below and above average, no?

The passage you quoted is perfectly correct and contains one word which you may have overlooked:

  the weighted mean of these two sets of perturbed orbitals is taken as zero [...]

In this case the weight corresponds to the degeneracy of the set. The lower-energy $\mathrm{t_{2g}}$ set contains three orbitals and the higher-energy $\mathrm{e_g}$ set contains two. If we place the $\mathrm{t_{2g}}$ set at $x \Delta_\mathrm{O}$ and the $\mathrm{e_g}$ set at $y\Delta_\mathrm{O}$, then

$$\begin{align}
y - x &= 1 & &\text{(difference must be equal to }\Delta_\mathrm{O}) \\
\frac{3x + 2y}{3+2} &= 0 & &(\textit{weighted}\text{ mean is equal to zero)}
\end{align}$$

and $(x,y) = (-0.4, 0.6)$.


Nitrobenzene and other monosubstituted benzenes 1H vs 13C NMR assignments The reported 1H NMR spectrum for nitrobenzene has three signals corresponding to H2/H6, H3/H5 and H4, respectively. The signals (ppm) appear at 8.25 (H2/H6), 7.71 (H4), and 7.56 (H3/H5). This is, the ortho protons are the most deshielded, followed by the para, and, finally, the meta. This makes sense when rationalizing the inductive effect of the nitro group by resonance, which makes the ortho and para carbons deficient in electrons, hence deshielding them. Furthermore, the ortho protons are closer to the nitro group, hence they are the most deshielded of all, so the from more to less deshielded: ortho > para > meta.

However, the reported 13C NMR spectrum shows signals at 148.3 (ipso), 134.7 (para), 129.4 (meta), 123.5 (ortho). So from more to less deshielded, the positions are now ipso > para > meta > ortho. Why is the trend so different for the carbon-13 NMR?

Also the trends for other monosubstituted benzenes are:

Benzonitrile. 1H NMR: ortho > para > meta; 13C NMR: para > ipso > ortho > meta. Here the trend changes a bit as well, but now only between ortho and para.

Benzaldehyde. 1H NMR: ortho > para > meta; 13C NMR: ipso > para > ortho > meta.

Aniline (EDG). 1H NMR: meta > para > ortho; 13C NMR: ipso > meta > para > ortho. Here the trend is inverted, but why would the para still be more deshielded than the ortho as seen in all the other cases?

As I wrote in the comments, there is not really a good way of arriving at these conclusions for a general aromatic molecule. The energy differences at play in these systems are so small that qualitative arguments will often be insufficient to get the order right. The diamagnetic term, which can be qualitatively rationalized by arguments based on electron density, for non-H atoms, is dominated by the paramagnetic term, which is, in some sense, an orbital response term due to the application of the magnetic field of NMR. Qualitatively reasoning about the paramagnetic term is possible in some special systems like PhLi or pyridine, but the accuracy of these qualitative arguments is insufficient for general substituted benzenes. The paramagnetic term can be calculated by electronic structure packages, but I assume that is not really the question.

I wrote some lecture notes a few years ago for a lecture I gave on the subject of diamagnetic and paramagnetic terms in NMR. I tried to pare it down a bit but they are still rather long. I will start with the qualitative quasiclassical conclusions and then have the derivation.

Qualitative Explanation of Diamagnetic and Paramagnetic Terms

The diamagnetic and paramagnetic terms in NMR shifts arise from considering how the electron motion is affected by an applied field. The terms diamagnetic and paramagnetic only indicate the sign of the magnetic response, opposed to the field and with the field, respectively (it does not have anything to do with unpaired electrons, which is the place most chemists have heard these words before). The diamagnetic term can be thought of being like Lenz's Law in classical electromagnetism; applying a field induces a current which opposes the field. This is sort of (with excessive handwaving) why electron density correlates with how shielded the nuclei are: with less electron density around, the amount of current is less, the induced field opposing the applied field is less, and the effective field is larger.

For the paramagnetic term, think of an electron moving (remember this a quasiclassical picture, but it's not wrong) in a $p$ orbital. It has angular momentum $l=1$, including an $m_l$ value of $+1$ which would generate a huge field and an $m_l$ value of $-1$ which generates the same field and in the opposite direction (also $0$ but that's not important here). Hence, in the absence of a field there is no field due to this orbital angular momentum being ``quenched'', that is $m_l=\pm 1$ are represented equally in the ground state wavefunction and exactly oppose each other. If a field is applied though, the $m_l$ aligned with the field (let's say $+1$) is stabilized while the $m_l=-1$ is destabilized and so there is more of one than the other in the ground state wavefunction at that angular momentum is unquenched. This results in the addition of a field in the same direction as the applied field, making it larger. This is the origin of the much larger range of shifts for $\ce{C}$ as opposed to $\ce{H}$. Calculating the magnitude of the paramagnetic term is not super straightforward and so qualitative arguments to try to rationalize the order of simple aromatic molecules is hard to do (examples like PhLi or pyridine are possible because there is an obvious paramagnetic perturbation, the lone pair). The amount of paramagnetic shifting tells us something about the low-lying vacant orbitals of the system and so can be a useful tool for predicting reactivity in some cases, but predicting those shifts without calculating them explicitly is not readily possible.2

In hydrogen, which is essentially only an $s$ orbital (which has an angular momentum of 0), there is no paramagnetic term (because there is no orbital angular momentum to unquench) and all of the shifts are due to the diamagnetic term. For heavier elements, like C, where $p$ orbitals get involved, there is a paramagnetic term and it is dominant.

Derivation of the Diamagnetic and Paramagentic Terms

Classical electromagnetism tells us that the magnetic field $\textbf{H}$ generated by a charge $q$ moving with velocity $\mathbf{v}$ at a point $\mathbf{r'}$ away is

$$\textbf{H}=\frac{q}{c}\frac{\mathbf{v}\times\mathbf{r'}}{r'^3}$$

Since we care about the field at the nucleus, we set the coordinate system origin at the nucleus and consider the field at the origin of a charge a distance $\mathbf{r}$ away. Then $\mathbf{r'}=-\mathbf{r}$ and we have

$$\textbf{H}=\frac{q}{c}\frac{\mathbf{r}\times\mathbf{v}}{r^3}=\frac{q}{mc}\frac{\mathbf{r}\times\mathbf{mv}}{r^3}=\frac{q}{mc}\frac{\mathbf{L}}{r^3}$$

We'll need this later. Let's consider though: the above implies that $\textbf{H}$ is on the order of $\approx \beta/r^3$, where $\beta$ is the Bohr magneton and $(1/r^3)$ for a 2p orbital is about 0.25 Å and so $H$ is about 600,000 gauss, which is massive! This should bring up the question: if non-zero $\mathbf{L}$ leads to huge magnetic fields why don't these fields completely destroy our NMR signal? Are there magnetic fields present in diamagnetic molecules when there are no magnetic fields present? I've sort of already hinted at the answer. No, there aren't such huge fields. Loosely, this is because the field generated by the electrons circulating in the $m_L = +1$ level exactly cancels the field of the electrons in the $m_L=-1$ because both are equally represented in the ground state (as they must be without a magnetic field present). This quenching of angular momentum occurs in all cases where the orbitals are real valued, such as if they are solutions of a Hamiltonian in the absence of a magnetic field (so nothing with unpaired spins, which carry a magnetic field). If we turn on an $\textbf{H}_0$ field though, one of the directions of circulation will be favored over the other (the $m_L=-1$) and the wavefunction will adjust so that the ground state has a slight net circulation in the favored direction. In this way, an applied magnetic field unquenches a small amount of the orbital angular momentum. This gives rise to the paramagnetic contribution.

Let's develop this mathematically. First, we should show that our semi-classical picture is a valid way to consider this quantum mechanical system. Fundamentally, a magnetic field $\textbf{H}$ is given by the curl of some vector potential $\textbf{A}$. We ultimately have two fields to consider: $\textbf{H}_0$, the applied field, and $\textbf{H}_n$, the field generated by the nucleus of the atom, each with their associated vector potentials.

\begin{align}
\textbf{H}_0&=\nabla\times \textbf{A}_0\\
\textbf{H}_n&=\nabla\times \textbf{A}_n
\end{align}

We're going to neglect the magnetic fields of the nucleus for this discussion. Obviously, they are important, since the perturbation of the nuclear spin energy levels are mediated by the coupling of the nuclear magnetic fields to the electronic magnetic fields, but we first need to determine what those electronic fields are.

As you may recall from classical electromagnetism, there is more than one vector potential that produces a given field. Specifically, if $\textbf{H}=\nabla\times\textbf{A}$, another vector potential $\textbf{A}' = \textbf{A}+\nabla\phi$, where $\phi$ is a scalar function, gives the same field, since the curl of a gradient of a function is always 0. The transfomation $\textbf{A}\rightarrow\textbf{A'}$ is called a gauge transformation. We need all physical results that we calculate to be gauge-invariant.

So now, how does an applied magnetic field show up in our Hamiltonians. It appears by changing $(\hbar/i)\nabla\rightarrow(\hbar/i)\nabla-(q/c)\textbf{A}$. So the Hamiltonian becomes

$$\mathcal{H} = \frac{1}{2m}\left(\mathbf{p}-\frac{q}{c}\textbf{A}\right)^2+V$$

If we were to calculate expectation values of $\hbar/i\nabla$ with different gauges, we would find that they aren't equal and so $m\mathbf{v}$ is no longer $\hbar/i\nabla$ but is $\hbar/i\nabla-\frac{q}{c}\textbf{A}$. Similarly, $\mathbf{L}$ is now

$$r\times\left(\frac{\hbar}{i}\nabla-\frac{q}{c}\textbf{A}\right)$$

We define something that will be useful to us throughout this discussion, the current density $\mathbf{j}(\mathbf{r})$.

\begin{align}\label{current}\tag 1
\mathbf{j}(\mathbf{r})=\frac{q}{2m}\frac{\hbar}{i}[\psi^*\nabla\psi-\psi\nabla\psi^*]-\frac{q^2}{mc}\textbf{A}\psi^*\psi
\end{align}

Note that it is a function of position and real-valued. It is also $q$ times the quantum mechanical probability current and gauge-invariant. If we assume $\psi$ is a solution of the time-dependent Schrodinger equation, then

$$\text{div} \mathbf{j}+\frac{\partial \rho}{\partial t} = 0$$

where $\rho = q\psi^*\psi$ (that is, the charge distribution). For pure states, $\psi^*\psi$ is time-independent and so $\text{div} \mathbf{j}=0$, and hence behaves as a classical current density. One can do some more math and show that it is possible to express the chemical shift, a quantum mechanical phenomenon, entirely classically as the coupling of the the nuclear magnetic moment to the effective magnetic field at the nucleus due to the circulating density $\mathbf{j}_0$.

\begin{align}\label{chemshiftE}\tag 2
E_\text{pert}=-\mathbf{\mu}\cdot\left[\frac{1}{c}\int{\frac{\mathbf{r}\times\mathbf{j}_0(\mathbf{r})}{r^3}d\tau}\right]
\end{align}

So if we can calculate $\mathbf{j}_0$ we have done the hard part of finding the chemical shift. Let's turn to that now.

Now, the Hamiltonian for our electron is

$$\mathcal{H} = \frac{1}{2m}\left(\mathbf{p}-\frac{q}{c}\textbf{A}_0\right)^2+V$$

We multiply out the binomial and rearrange to get

$$\mathcal{H} = \frac{p^2}{2m}+V-\frac{q}{2mc}(\mathbf{p}\cdot\textbf{A}_0+\textbf{A}_0\cdot\mathbf{p})+\frac{q^2}{2mc^2}A_0^2$$

Let's assume we have the eigenfunctions of $\mathcal{H}_0=\frac{p^2}{2m}+V$ in the absence of the field (i.e. $\mathcal{H}_0\psi_n=E_n\psi_n$). Our goal is to calculate $\mathbf{j}_0$, which has a term linear in $\textbf{A}_0$, which looks like

$$\textbf{A}_0=\frac{1}{2}\textbf{H}_0\times\mathbf{r}$$

That's one choice of gauge, but in any gauge $\textbf{A}_0$ is proportional to $\textbf{H}_0$. So if we want $\mathbf{j}_0$ right to first order, we need $\psi$ at least accurate to first order in $\textbf{H}_0$ (the $\psi^*\psi$ in the last term of $\mathbf{j}_0$ can be the unperturbed wavefunction since that term already has a $\textbf{A}_0$). So let's do some first-order perturbation theory! In case you forgot your first-order PT, we first need to define the perturbing Hamiltonian. We'll use the linear part of our Hamiltonian above.

$$\mathcal{H}_1=-\frac{q}{2mc}(\mathbf{p}\cdot\textbf{A}_0+\textbf{A}_0\cdot\mathbf{p})$$

Our first-order corrected ground-state wavefunction (this is just standard first order PT) is

$$\psi'_0=\psi_0+\sum_n{\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}\psi_n}$$

where $E_0$ is the ground state energy, $E_n$ is the energy of the $n$th excited state $\psi_n$. Let's define

$$\epsilon_{n}=\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}$$

So that

$$\psi'_0=\psi_0+\sum_n{\epsilon_{n}\psi_n}$$

Substituting this $\psi'_0$ into the equation for $\mathbf{j}$ (equation \ref{current} above), we obtain

$$\mathbf{j}_0(\mathbf{r})=\frac{\hbar q}{2mi}(\psi_0^*\nabla\psi_0-\psi_0\nabla\psi_0^*)+\frac{\hbar q}{2mi}\sum_n\left[{(\psi_0^*\nabla\psi_n-\psi_n\nabla\psi_0^*)\epsilon_n+(\psi_n^*\nabla\psi_0-\psi_0\nabla\psi_n^*)\epsilon_n^*}\right]-\frac{q^2}{mc}\textbf{A}_0\psi^*\psi$$

The first term

$$\frac{\hbar q}{2mi}(\psi_0^*\nabla\psi_0-\psi_0\nabla\psi_0^*)$$

is the current flowing in the absence of a magnetic field. If our angular momentum is quenched as we described above (as in closed-shell molecules), then this is 0. If we take the $\psi_n$ as real, then

$$\mathbf{j}_0(\mathbf{r})=\frac{\hbar q}{2mi}\sum_n\left[{(\epsilon_n-\epsilon_n^*)(\psi_0\nabla\psi_n-\psi_n\nabla\psi_0)}\right]-\frac{q^2}{mc}\textbf{A}_0\psi^*\psi$$

We're gonna see that the first term is the paramagnetic current, related to $\epsilon_n$, which recall is the matrix element coupling the ground state to excited states by the angular momentum operator, and the second term is the diamagnetic current, always present and independent of $\textbf{L}$. Let's go back to that choice of the vector potential $\textbf{A}_0$ we made before

$$\textbf{A}_0=\frac{1}{2}\textbf{H}_0\times\mathbf{r}$$

Subbing that into the perturbing Hamiltonian gives

\begin{align}
\mathcal{H}_1&=-\frac{q}{2mc}(\mathbf{p}\cdot\frac{1}{2}\textbf{H}_0\times\mathbf{r}+\frac{1}{2}\textbf{H}_0\times\mathbf{r}\cdot\mathbf{p})\\
&=-\frac{q}{2mc}(\textbf{H}_0\cdot(\mathbf{r}\times\mathbf{p}))\\
&=-\frac{q}{2mc}(\textbf{H}_0\cdot\mathbf{L})
\end{align}

Where that $\mathbf{L}$ is in the absence of a field. Then we have that

$$\epsilon_{n}=\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}=\frac{qH_0}{2mc}\langle n|\mathbf{k}\cdot\mathbf{L}|0\rangle$$

where I assumed that $\textbf{H}_0$ is some magnitude $H_0$ along some unit vector direction $\mathbf{k}$. Now, going back to our chemical shift energy (equation \ref{chemshiftE}), we want the field that this current $\mathbf{j}_0$ generates. This is

$$\textbf{H}=\frac{1}{c}\int{\frac{\mathbf{r}\times\mathbf{j}_0(\mathbf{r})}{r^3}d\tau}$$

Substituting $\mathbf{j}_0$ and $\mathcal{H}_1$ in gives

$$\textbf{H}=H_0\frac{q^2\hbar}{2m^2c^2}\sum\frac{\langle 0|\sum_j\frac{\mathbf{L}_j}{r_j^3}|n\rangle\langle n|\mathbf{k}\cdot\mathbf{L}|0\rangle+\langle 0|\mathbf{k}\cdot\mathbf{L}|n\rangle\langle n|\sum_j\frac{\mathbf{L}_j}{r_j^3}|0\rangle}{E_n-E_0}-\frac{q^2}{2mc^2}H_0\langle 0|\sum_j\mathbf{r}_j\times(\mathbf{k}\times\mathbf{r}_j)|0\rangle$$

where $\mathbf{k}$ is the direction of the applied field and the $j$ subscripts denote electron number in the wavefunction.

For an $s$ orbital, all matrix elements with $L$ are 0 and so the first term is 0. This leaves only the second term which is opposed to the applied field. For anything with higher angular momentum than $s$ orbitals, the first term will be nonzero and aligned with the field. This term arises due to unquenching of the orbital angular momentum in the presence of a magnetic field and hence the direction of the paramagnetic field was determined by the projection of angular momentum which aligns with the field, rather than arising out of Lenz's Law. Said yet a different way, the orbital was already a magnet, it's just that we couldn't see it before. The applied field does not induce the quantum mechanical current flow or paramagnetic field (in the sense of classical EM), merely reveals it by mixing more of the aligned magnet into the ground state wavefunction than the anti-aligned one.

Putting some reasonable numbers into these equations gives that the paramagnetic term is larger at distances less than 1 Å from the nucleus, while the diamagnetic term dominates outside this range. As we mentioned above, the distance important for chemical shifts (average $r^3$ for a $2p$ orbital) is about 0.25 Å and so the paramagnetic term will be larger for NMR related problems.  On the other hand, these paramagnetic terms play practically no role in macroscopic magnetic moments of molecules.

References:

Slichter, Charles P. Principles of Magnetic Resonance (Harper and Row Publishers, 1963).
Gordon, C. P., Yamamoto, K. , Liao, W., Allouche, F., Andersen, R. A., Copéret, C., Raynaud, C., Eisenstein, O. Metathesis Activity Encoded in the Metallacyclobutane Carbon-13 NMR Chemical Shift Tensors. ACS Cent. Sci., 2017, 3 (7), 759–768. DOI: 10.1021/acscentsci.7b00174


What undergoes reduction in a zinc-copper voltaic cell? In a $\pu{Zn-Cu}$ voltaic cell, the electrolyte is sodium chloride solution. The question was:

  In this cell, which of the following undergoes reduction?

My answer was copper, but why is the exact answer "a hydrogen ion"?

Copper cannot undergo reduction because, in the list of components of the cell you listed, it is only present in its most reduced state, Cu(0). Thus, it cannot be copper that is reduced. See here for additional hints about your system.

You may have been confused by the fact that, in the historical (and widely taught) Daniell cell, copper is indeed reduced: it is present as Cu(+II) in a solution of copper sulfate.


How to measure corrosion rate of an iron nail in salt water? I want to measure rate of corrosion of an iron nail. One way to calculate the corrosion rate is to use the weight loss of the iron nail. However rust does not come off easily. Do have any other ways I could measure the rate of corrosion? Can I measure the weight lost somehow?

There is an ASTM standard(G01) that gives detailed descriptions of cleaning procedure for corrosion measurements.

For iron and iron oxide, the recommended way is to leave the sample in a cleaning solution(e.g a dilute acid solution), followed by a light brushing repeatedly until the weight stabilizes.

You could also try sonication in a solvent to see if that helps.

In any case, to make sure you are not losing significant amounts of the base metal, you should repeat the cleaning procedure on some uncorroded control sample. This will help you dial in the proper procedure for your samples.


H-NMR Equivalent Protons and Coupling of Glycerol I've been searching for explanations to understand why the labeled protons on carbons 1 and 3 are not equivalent in the below figure.

Another figure that shows the same results:

SDBS confirms the same, although the assignments of the signals show that all 3 hydroxyl Hs are equivalent, which I don't understand.

I would think that all 4 hydrogens on C1 and C2 would be in identical environments but it seems this is incorrect. In fact, the figures show that the two hydrogens on C1 (and C2) are not equivalent to each other. Why is it that the hydrogens on the terminal carbons are not equivalent, but rather each hydrogen on one terminal carbon has an equivalent hydrogen partner on the other?

Also, what is the explanation for why SDBS shows all 3 hydroxyl protons as equivalent? Why would the -OH on C2 be equivalent to the ones on the terminal carbons?

Edit: As for my first question, I think I found the correct explanation. I believe it's because the hydrogens pairs are either diastereotopic or enantiotopic.

This might be confusing since both ends of the molecule are the same. The difference is that the alcohol points to only one side of the molecule.

So, while the hydrogen atoms on carbons 1 and 3 on paper have the same environment, they are not chemically equivalent when looking at the whole molecule.

Concerning your second question: The chemical shift of protons on heteroatoms such as Oxygen, Sulfur, or Nitrogen is highly dependent on the recording conditions (concentration, temperature, solvent) of the NMR. Frequently, the signals are broadened and show no coupling (broad singlett). The reason for this is the rapid intermolecular proton exchange. If there is a rapid exchange between two chemically non-equivalent groups within a molecule (like in this one), only a single, averaged signal is observed.


Why do boiling points of alkanes and alcohols become closer with increased chain length I came across this graph, which clearly shows a large difference between the boiling points of alkanes and their equivalent alcohols.

However, it is clear that as chain length increases, the difference in boiling point between alkanes and alcohols decreases.

What is the main reason for this?

I have read that it is due to the dispersion forces increasing in both alcohols and alkanes, but wouldn't this not really have an effect on the hydrogen bonding of the alcohol? (meaning that boiling point should be consistently higher)

It is simple. Longer the chains higher the similarity between them. In other words the OH group are getting "diluted".
It is more difficult for the donor to find the accepting counterpart, and the number density of the hydrogen bonds, both for volume and mass, gets lower.
The interaction between molecules becomes dominated by VdW forces.

To visualize this think of methanol and of a very very long primary alcohol. In the second case it is somehow "more difficult" to spot the difference versus the parent alkane.


Spontaneity of Reactions- Gibbs Energy or Entropy change I just wanted to check if I am thinking right:

The reason why we consider the Gibbs energy to be the measurer for spontaneity of a reaction is because it does not depend whether the reaction was exo- or endothermic.

Is it arising because:
 if the change in entropy were less than zero, the Gibbs change in energy is less than zero? So if that were true, then entropy may not be a measurer for spontaneity, correct?

Neither enthalpy nor entropy alone can decide the spontaneity of a process.

For example, evaporation of water from oceans is an endothermic process $(\Delta H \gt 0)$, which clearly violates the fact that a system aims for minimum energy, yet this process is spontaneous (or naturally occurring).

To determine the spontaneity of a process the Gibbs-Hemholtz equation was introduced. This combines the effects of both enthalpy and entropy for determining the spontaneity of a process.

$\Delta G=\Delta H-T\Delta S$

A system generally aims for two things; minimum energy and maximum probability to be found in the universe (or more entropy). Therefore, for a spontaneous process, $\Delta H \lt 0$ and $\Delta S \gt 0$, Which gives an overall negative change is Gibbs energy.

For a spontaneous process, $\Delta G$ is negative.


Do chemicals have an expiry date? A bonus question would be, if so, why? Is it because it decomposes and hence the expiry date is calculated out?

They do indeed. It could be for different reasons:

A chemical might undergo chemical changes due to different storage conditions: it could react with other chemicals present in the air or in the environment. This could change its nature (reaction to create other compounds), properties and/or purity.
A chemical might undergo physical changes as well: the temperature and pressure vary, as well as other conditions like humidity, light, electromagnetic fields. This could change it as stated above.
Certain chemicals are simply unstable and decay or degradate over time to something else, without the need of externals input / changes. This could affect the chemical as stated above.
Certain chemical suffers transpiration losses.
Human factors, like contaminations, spillings, loose caps, etc.

However, it might be noticed that these reasons could be divided to produce two different dates: shelf life vs expiration date.


Overlap between computer science and chemistry? This isn't particularly a question about chemistry itself, but more-so the prospect of chemistry.

What kind of opportunities arise when you overlap chemistry and computer science? I'm a computer science major, but prior to going to university I had (and still have) a deep passion for chemistry. My highest marks were always in chemistry, but I decided to go down the compsci route because likewise it is a passion of mine. An opportunity has presented itself to double major in compsci and chemistry and I'm curious to know whether it's worth it.

My knowledge of chemistry is fairly barebones (a limitation of highschool education), but as I said, it was my strongest subject and for some reason it just clicks with me. I'd often spent sleepless nights perusing chemistry textbooks and online periodicals because it fascinates me to no end.

I'm not looking for greater job prospects (I already work in both compsci as a programmer and chemistry in an electronic cigarette company), I'm more-so looking for a more fulfilling pursuit of knowledge. While irrelevant, my university recently opened a new building dedicated primarily to chemistry, so facilities are recent and faculty is adept and education is modern. I'd very much like to take advantage of the opportunity, but I'm just unsure if a university-level chemistry program will scratch that itch for me.

Have a look at list of computational chemistry software. It gives you overview, how the computers are used in chemistry. Many of them fall more into the chemical physics field, some into the chemoinformatics, to name just few. So definitely there are lot of places to contribute to use of computers in chemistry. If you are able to mix the software engineering and the science itself in a good way, possibilities are immense.

As for the last question - almost whatever field of chemistry you choose, you can contribute to the programs used within it. There is no such thing like completed scientific code. And university level is a necessary prerequisite to do reasonable science. Just concentrate on the models and mathematics behind the scenes and everything will follow.


Can we use liquid hydrogen to power a conventional engine? Hydrogen has the best fuel efficiency, so if we were to use it in our engines wouldn't their efficiency increase as well.

The engines would require changes and all but would it be plausible to have that?

The efficiency that matters is the efficiency of the vehicle not the efficiency of the engine

The problem of designing more efficient transport is bigger than the problem of designing better engines. Conventional combustion engines can be adapted to burn a wide variety of fuels (alcohols, natural gas and hydrogen are not too big a challenge). And there are alternative ways to move vehicles than combustion engines (fuel cells, for example, which run on hydrogen). So it looks like we can solve the engine problem.

But that isn't enough. To have a viable vehicle we have to solve other problems as well. How to store and deliver the fuel to the engine, for example. And how to distribute the fuel across the country so vehicles can be refilled. The second problem is probably easier to solve if the first problem–a viable vehicle–has a solution. Unfortunately, it isn't that easy.

To have a hydrogen powered car we would need a way to store enough hydrogen in the vehicle to allow it to drive several hundred miles.

This is easy to do with liquid hydrocarbon fuels in conventional vehicles (petrol/gasoline and diesel have very high energy densities and can be stored in a simple unpressurised tank). The weight of the fuel alone is larger then the weight of the tank that contains it and the volume is small as the energy density is very high.

It isn't so easy with hydrogen. While there are several existing ways to store hydrogen and some proposed solutions that are not yet operational, all of them involve big issues with volume and weight. Compressed hydrogen will involve very heavy thick-walled tanks and even then the energy density of the hydrogen gas alone will be lower than a tank full of petrol. The weight and volume of the tanks required to store as much energy as a tank of petrol will very large. Liquid hydrogen has a different set of compromises but isn't much better as it requires cryogenic system to keep the liquid hydrogen cool. There is considerable interest in clever chemical ways to store hydrogen (clathrates, metal organic frameworks and various molecular systems than can reversibly store hydrogen are all being researched).

To get an idea of the scale of the challenge it is worth looking at the mass and volumetric characteristics required to compete with conventional fuel. Diesel (slightly more efficient than petrol) carries about 12kWh/kg and about 10kWh/L. The current targets for research on hydrogen storage for the Fuel cell Technology Office are 1.5kWh/kg and 1kWh/L. This is about 10 times worse than diesel even ignoring the fact that fuel-cell power vehicles might have other advantages partially compensating the overall vehicle when compared to combustion engines. And the best technology solutions are much better than liquid hydrogen storage.

So you can use liquid hydrogen (or other ways of storing hydrogen) to power an conventional engine or vehicle (and such vehicles exist). But the cost and design compromises to make this work for a whole vehicle given that you need 10 time the volume and weight to store the equivalent energy, make such solutions not competitive for mass market vehicles.


Is the behavior of a dilute strong acid the same as the behavior of a weak acid? My assumption is yes, as the $\mathrm{pH}$ levels of each will be similar and thus they have the same acidity. Is this correct or am I missing something?

The pH is the same, yes, but the weak acid has undissociated molecules "in reserve", so to speak, and thus can neutralize a lot more of any base.

Indeed, this concept of weak acids and their conjugate bases having molecules "in reserve", reacting only when called upon, is the principle behind acid-base buffers.


Does bicyclo[3.2.0]heptane have a plane of symmetry? Most stable form of cyclobutane is the "envelope form" and most stable
  conformation of cyclopentane is either the "envelope form" or the "twist"
  form.

In the above compound do the individual rings exist in their most stable conformation? Or is the actual situation different?
What is the most stable conformation of the above compound? And does it have a plane of symmetry?

It turns out that bicyclo[3.2.0]heptane does have a plane of symmetry (courtesy PubChem):

According to The Molecular Structures of Bicyclo[3.2.0]heptane and $\Delta^6$-Bicyclo[3.2.0]heptene. An Electron-diffraction Study of Gaseous $\ce{C7H12}$ and Molecular Mechanics Calculations on $\ce{C7H12}$ and $\ce{C7H10}$. Robert Glen, Grete Gundersen, Peter Murray-Rust, and David W. H. Rankin, here are the bond angles in degrees:

The link explains the experimental procedure in deduction of the actual structure.There are two possible stable structures/forms for the molecule namely endo and exo.The endo form is supposed to be more favourable than the exo form.

From here, we can conclude Cs symmetry.


Could Na+ ions in water used with a humidifier cause dust everywhere? We have very dry air in our small family house.

So we bought a humidifier ( https://www.klarstein.com/All-products/Monaco-Digital-Ultrasound-Humidifier-White.html )

But here comes the issue: the tap water is very hard and this causes that dust is everywhere.. due to the humidifier + hard tap water.. like everywhere, even the notebooks fan are full of the white dust, overheating, brutal.

So we need to buy a "water-softener" ex.: "BlueSoft S30 SVR34" ( http://euro-clear.eu/uploads/content/catalog/EC_residential_product_catalog.pdf )

But: I can read that this "water-softener" device "replaces the Ca++ and Mg++ to Na+ ions"

The big question to chemistry: if the water will have many "Na+ ions" and not "Ca++ and Mg++" ions, would it solve the "dust" issue? So we will won't have any more white dust everywhere, if the humidifier is turned on?

We already asked the vendor, but they are not sure that this will solve the issue.

In the end, the solution was that instead of using an "ultrasound" based humidifier, we bought "evaporative humidifier".

delete this line if it cannot be said, but we bought a "trotec aw 10 s"


Understanding FHF- molecular orbital diagram? I need help understanding the molecule FHF-'s molecular orbital diagram:
So I know that "ag" refers to the 2pz and 2s orbitals of fluorine atoms...but I am not sure why "ag" is labeled for the topmost orbital level? I don't really know what that is...but it's interacting with both hydrogen's 1s orbital and fluorine's 2p orbital, so does it imply that it's either antibonding or bonding orbital? And the one right below the topmost orbital--would that be the antibonding sigma 2pz orbital?

Thanks, I'd appreciate the help!

So we are treating the two fluorines as ligand groups on the right hand side and we introduce a Hydrogen on the left. We observe that each fluorine has a 2S, 2px, 2py and 2pz. If we take the 2pz as being along the internuclear axis then we assign it with sigma symmetry, while the x and y p orbitals have pi symmetry.

It is the 2pz that can interact with the Hydrogens 1S because it has the correct symmetry. It turns out that the symmetry of that orbital is ag. This creates as you say the bonding and anti bonding pair with the H's 1S. It is important to note that the bonding and anti bonding pair are of the same symmetry.

The hydrogen only has the 1S orbital (symmetry ag) and no others, so every other orbital of a symmetry that is not ag is non-bonding since it has no respective partner in the FHF environment.

With regard to the assignation of symmetry to an orbital you must have an understanding of SALCs (symmetry adapted linear combinations of molecular orbitals), if you perform a projection operator method then you will generate the symmetrys ... however this is a bit advanced and not something expected at undergraduate level (to my knowledge).

I hope that helps :)


What's the most effective way to reliably trigger and observe hard water scale buildup at a micro level? We live in an area that's got some really hard water. It's so hard that it kills most of the fish we try to care for in our aquariums and shrimp?... forget it. While this is sad and challenging for aquarium lovers, this constant quest my spouse is on got one of my kids really interested the concept of "hard water." One particular point that I've been struggling with was to figure out how to "see" hard water. They asked me if we could build an apparatus that would show them the build up of the scales more quickly so they could look at it under a microscope and do various experiments.

I'd really like to encourage their curiosity and so I wanted to ask, what options exist to:

trigger buildup (some kind of a "seed" on a glass pipette or...?)
capture or isolate buildup so that it can be observed with a
microscope
what kind of microscopic resolution would / should be
used to observe this (preferably semi-realtime, perhaps via
stop-motion capture, with a photo/video attachment as it
accumulates)?

You have to find out what kind of hardness is present in your water. Is it permanent hardness or temporary hardness? The latter goes away with boiling.

The easiest and visual demo for children is to show the behavior of hard water with a soap bar. Soap when added to hard water does not lather very well. Watch the video here:

https://www.youtube.com/watch?v=jDgSohKVLio

https://www.youtube.com/watch?v=bBYsfEHKYCk

As a control, you would need to buy distilled water.

Does your electric kettle develop crusts with time?


What is the amount of limiting reagent in this chemical reaction? In the reaction involving $\pu{124 g}$ of aluminium and $\pu{601g}$ iron (III) oxide, what is the amount of substance in moles of the limiting reagent?

Here is what I have so far:

Balanced equation:
$$\ce{2Al + Fe2O3 -> Al2O3 + 2Fe}$$

$\ce{Al}$ has MW = $27$, mass = $\pu{124 g}$ and determined its amount using $n=\frac{\text{mass}}{MW} = 4.59$

$\ce{Fe2O3}$ has MW = $160$, mass = $\pu{601 g}$ and determined its amount using n=mass/MW = 3.756

I was able to identify limiting reagent by:

g $\ce{Al}$ --> n $\ce{Al}$ -->n $\ce{Fe2O3}$--> g $\ce{Fe2O3}$

$124 g --s1-> 4.59 --s2--> 2.295 -s3--> 367 g$

$S1 = n = 124/27 = 4.59$

$S2 = \frac{\text{want}}{\text{have}} \times \ce{Al} = 1/2 \times 4.59 = 2.295$

$S3 = m = n\times MW = 2.259 x 160 = 367 g$

Therefore as all $\ce{Al}$ did not use all the $\ce{Fe2O3}$ that means the $\ce{Al}$ is the limiting reagent.

The amount of substance in moles of limiting reagent $\ce{Al}$ (answer to the question) is $4.59$ OR $9.18$ (which is moles multiplied by the stoichiometric coefficient, which is $2$).

$4.59$ is not "something" (you haven't specified the unit) of an $\ce{Al}$ mole, but your amount of $124\mathrm{~g}$ of $\ce{Al}$ is exactly $4.59$ moles of $\ce{Al}$, it's just a unit change. In reaction you wrote, every two moles of $\ce{Al}$ reacts with one mole of $\ce{Fe2O3}$. This may help you to find the solution.


If double and triple bonds are stronger than single bonds then why are unsaturated hydrocarbons less stable than saturated hydrocarbons? I have learnt that unsaturated hydrocarbons are more reactive and less stable.  However if double and triple bonds are stronger and shorter bonds than single bonds, aren't they supposed to be more stable and inert?

When I was a first year, this concept was hard for me to grasp as well. The idea is that, even though the bond itself (double or triple) is stronger than a single bond as a whole, we must consider the components of a double or triple bond. A double bond consists of one sigma and one pi bond.

Let's take ethene for example. The $\ce{C=C}$ double bond has one sigma and one pi bond. The sigma exists because of two, sp2 hybridized orbitals overlapping head-to-head. This type of covalent bonding is very strong. But since it is only sp2, there is one p orbital leftover that was not hybridized. So, each carbon takes their left over p orbital, and forms a pi bond from the p orbitals overlapping side-to-side (the p orbitals are parallel). The unhybridized p orbitals are perpendicular to the sp2 orbitals.

Pi bonds, because of the way they are constructed, are much weaker than a sigma bond due to side-to-side overlap, versus head-to-head. Having said, that pi bond can be easily cleaved by an incoming group. This properties of the pi bond is what makes an unsaturated hydrocarbon so much more reactive and unstable compared to a saturated hydrocarbon.


Why does formic acid give positive Fehling's test? $$\ce{HCOOH ->[\text{Fehling's solution}] CO2 + H2O}$$

The points discussed in a previous question do not account for the fact that formic acid also gives positive Fehling's test since it is not enolisable. Then what is the reason/mechanism for the reaction?

Here is a short simple available answer provided online by Prachi Sawant in 2016 to quote:

  Formic acid has both aldehydic (-CHO)  and carboxylic (-COOH) functional groups.
  Hence, it gives both Tollen's and Fehling's test positive.

  Reaction of formic acid with Tollen's reagent:

  $\ce{HCOOH + 2[Ag(NH3)2]+ +2OH- → 2Ag + CO2  + 2H2O  + 4NH3 }$

  Reaction of formic acid with Fehling's reagent:

  $\ce{HCOOH + 4OH- + 2Cu^{2+}(in complex)  → CO2 + 3H2O + Cu2O }$

Now, here is my attempt on the possible underlying mechanic, which is likely complex. I start with the above net reaction, which in reality, is probably not completely accurate, but nevertheless, may provide some insight into actual half-cell reactions, beginning with the reduction of the organic cupric complex:

$\ce{Cu^{2+} +  e^-  ->  Cu^+ }$

where the insoluble Cu2O (the test indicator) is formed. The corresponding other half-cell is less clear accept for the liberation of an electron:

$\ce{COOH^- -> .HOCO + e^- }$

which may speculatively may be followed by:

$\ce{.HOCO + .HOCO -> H2O + CO + CO2 }$ (speculation)

And/or, this paper favoring non-conventional ionized species in the context of formic acid as possessing more stability as demonstrated, for example, by:

$\ce{HCOOH -> [.HCOOH]^+ + e^- }$

With its source cited below, followed by:

$\ce{[.HCOOH]^+ -> H2O .CO^+ }$  (a distonic ion, per 2nd cited source)

$\ce{OH^- + H2O .CO^+ -> H2O + COOH^- }$ (speculation)

To quote a cited source above :

  Rate constants for the radical-induced hydrogen abstraction from formic acid, HCOOH, are presented here. Only those reactions leading to the formation of $\ce{.HOCO}$ were investigated.

Per a source: 'The gas phase chemistry of the formic acid radical cation $\ce{.[HCOOH]^+}$ Mechanism for exchange of the hydrogen atoms: a quantum chemical investigation', to quote:

  Earlier labelling experiments had indicated that prior to the loss of H the hydrogen atoms in ionized formic acid, $\ce{.HCOOH^+}$, can become positionally equivalent via an unknown pathway. Using ab initio molecular orbital calculations we have located a minimum energy reaction pathway for hydrogen exchange reactions which takes place close to the dissociation level and proceeds via flexible ion/dipole complexes. Along this path three equilibrium structures of different atom connectivity are found: $\ce{.HCOOH^+}$, 1, ionized formic acid; two forms of the surprisingly stable ion/dipole complex $\ce{.HOHCO^+}$, 2, 3; and the distonic ion $\ce{H2O CO.^+}$, 4. Ion 1 is separated from ion 2 by a barrier of 142 kJ mol−1 and this barrier corresponds to a 2A′ → 2A″ surface crossing. The barrier for the proton transfer 3 → 4 is only 25 kJ mol−1 relative to 3; the transformation $\ce{.HCOOH^+ -> H2O .CO^+}$, formally a 1,2-hydrogen shift, takes place via the ion/dipole complex $\ce{HO· HCO^+}$, $\ce{H2O .C^+}$ is the reacting configuration for decarbonylation and this reaction is calculated to take place at the dissociation limit in full agreement with experiment.

Where ‘decarbonylation’ can also be effected by several transition metal complexes, (source), as in the current context with an organic copper complex.


What is the difference between the esterification and transesterification? What is the difference between an esterification and a transesterification. Esterification objective is the formation of an ester by reacting an alcohol and a carboxylic acid. I am more concerned transesterification or reference to a journal article.

You are getting too specific as though these are biological processes. They do not have to be... ANY ester made is an esterification reaction.

So any ester can be transesterified into a different alkyl alkanoate and any carboxylic acid or derivative (even nitriles) can be esterified. See the diagram:


Why is the discovery of chiral molecules in space a big accomplishment? Today, I came across the discovery of a chiral pair of molecules in interstellar space. I know that a chiral molecule is a molecule that is not superimposable on its mirror image. But why is this discovery such a big deal?  I think that the existence of chiral pairs should be obvious where multiple molecules exist. So why is this discovery significant?

The story is a big deal because it might help explain why molecules in living creatures have a single chirality

One of the great mysteries of living systems is that, when they use chiral molecules, they only use molecules of one chirality. There is no obvious mechanism to kick-start this preference and, while there are many esoteric theories, there is very little evidence that explains why one chirality was chosen over another.

Proteins are, for example, made from amino acids which are almost all chiral. And they are all left-handed (technically they are all levorotatory, see wikipedia). Most sugars used in biological systems are right-handed. There are simply no convincing theories that explain the origin of this preference.

One modern theory suggests that there is a subtle energetic preference for one chirality caused by the interaction of molecules with the electrons emitted in certain types of radioactive decay (the weak nuclear force involved in radioactive decay is one of the few fundamental forces to have a mirror-image preference). But other experts think the preference is arbitrary.

Resolving this would be big news for biology.

The reason why the observation of chiral molecules in space (the news story only reports the observation of a molecule that is chiral not that we can observe the chiral preference) is that it creates hope than we can observe directly whether the chiral preferences of life are inherent in the universe or are arbitrary.

We don't yet know whether propylene oxide (the molecule in the news story) exists with more of one chirality than the other, but now it has been observed, we can expect to find out as instruments improve. So the big story is that we now have some hope that we can test some of the theories about why life has a handedness preference.


Lemon juice and carbonic acid reaction I was doing a chemistry project to see if the pH of a carbonated solution effects the time taken for it to settle after being disrupted, and I am trying to map out the reaction when the components are combined.

I was able to figure out a few, but I got stuck on carbonic acid and lemon juice.  For some reason, I was getting ethane as a by product of the reaction, and I'm sure that's not right.  Could someone explain what I'm doing wrong?

the formula is $\ce{H2CO3 + C6H8O7}$

These two weak acids have similar disassociation constants in water.  A low Ka indicates a relatively small number of acid molecules will form an anion stabilized by water.  Two weak acids with similar disassociation constants will coexist in a solution without interacting.  If two weak acids have very different Ka then there can be a reaction.  In this case nothing would happen.   For more generalized information search for mixture of weak acids.

  Carbonic Acid

$$\ce{H2CO3 + H2O <=> HCO3- + H3O+}$$
$$K_a = 4.3 \times 10^{-7} = \frac{[\ce{H3O+}][\ce{HCO3^-}]}{[\ce{H2CO3}]}$$

  Citric Acid

$$\ce{H3C6H5O7 + H2O <=> H2C6H5O7- + H3O+}$$
$$K_a = 1.8 \times 10^{-6}  = \frac{[\ce{H3O+}][\ce{H2C6H5O7^-}]}{[\ce{H3C6H5O7}]}$$


Lewis Acidity and Bacteria How does Lewis acidity correlate with bactericidal activity? I read on Wikipedia that the two are positively correlated but no explanation was given.

Does it have to do with the movement of electrons - i.e. a current? I'm guessing that having a current passed through you isn't good for you.

I suppose also that the interactions (i.e. coordination bonds) of metal (ions) with bacterial components can interfere with cellular processes.

A review on $\ce{Ag+}$ quotes the following situations, which may be provoked by the Lewis acidic behavior of a metal cation.

  Silver treated cells exhibited electron-light regions in
  their cytoplasm thus forcing DNA to condense. Authors state that
  condensed DNA molecules lose their ability to replicate.

I presume that $\ce{Ag+}$, acting as a Lewis acid, will reduce the negative charge on DNA, thus inducing DNA aggregation.

  Electron-dense region were also observed around the cell wall inside
  the cytoplasm, and the cytoplasmic membrane experienced  shrinkage
  and  detachment  from  the  cell wall.

  There was also indication  that  silver  ions  interact  with  thiol
  groups leading to the deactivation of enzymatic proteins.

Here one sees the action of $\ce{Ag+}$ on thiols which may be interpreted as an acid-base interaction of the Lewis type.

  Also the effect of $\ce{Ag+}$ on the respiratory electron chain of E. coli was stressed, since $\ce{Ag+}$ is toxic to microorganisms due to inhibition of the
  respiratory chain at multiple sites.

It appears that in many cases, one can link the antibacterial function of metal ions to their acidic properties. And certainly, those discussed above are not the only ones.


How does concentration affect The Position Of Equilibrium? I was taught that
when the concentration of the reactants is greater than the concentration of products then the position of equilibrium is shifted towards the left.

However when I used La'Chateliers Principle then we know that if you increase the concentration of reactants then the system wants to revert that change a hence produce a higher concentration of products (and subsequently a lower concentration of reactants) but wouldn't this then mean that the equilibrium would shift to the right?

You statement of Le Chatelier's Principle is a perversion that isn't true. The equilibrium does not depend on the sum of the concentrations of the reactants relative to the sum of the concentration of the products.

Your analysis is correct.


How do we recognize the internal plane of symmetry in meso compounds? How do we recognize the internal plane of symmetry in meso compounds? I read online that the substituted groups attached to a stereocenter could be rotated as such, to obtain an internal plane of symmetry:

Why would rotating the bonds in a molecule not affect its optical activity?

Also, exactly how does the presence of an internal plane of symmetry render a molecule optically inactive, as seen in meso isomers?

I see two issues here:

Rotational conformations do not change whether or not the molecule is meso. For meso compounds, the presence of the symmetry element means that for each conformation that is chiral, we are guaranteed to have the enatiomeric conformation with equal probability. Therefore, the net optical rotation is zero.
Fischer projections are no doubt a usual way to represent chiral molecules, but in general, we use Fischer projections most heavily in the case of carbohydrates. The reason is because carbohydrates have shared homology where we care about the specific placement of hydroxyl groups. In general, they are not a great visualization tool. Wedge and dashes are much better. In this case, for a simple two carbon system, where the two carbons have identical connectivity, you could just as well have assigned Cahn-Ingold-Prelog absolute stereochemistry to each carbon to see that the molecule is meso.


How strict is the "to excite electrons the energy must equal the energy state difference" fact? We are always told that to excite an electron from one state to a higher energy states, for example from the valance band to the conduction band, the energy must equal the energy difference between the two energy states.

The conduction band consists of many allowed energy states, separated from the valance band by an energy gap. For the element Si, this energy gap is (at room temperature) about 1,1 eV. The range of energies for visible light is 1.8 eV to 3.1 eV. Since Si indeed conducts electricity at room temperature, is it then assumed that there is an allowed energy states with exactly the same energy as the energy difference between the two states? The conduction band, or the valance band for that matter, is not continuous, but discrete. Or is it sufficient that the photon energy "roughly equals", or is "close enough"? If so, how close is close enough?

I hope my question is clearly stated.

In fact there are many factors to consider, so the answer to your question depends on what type of material you are analyzing.

First of all you should know that even if the absorption level are discrete the absorption and the emission of a material are never a single value but a peak. If you consider the absorption or the emission of a gas you can assume them discrete value (even is not completely true), and so you can call them lines (see e.g. Lyman series), but if you condense the matter the interaction between the atoms/molecules broad the absorption/emission range and so you have peaks. If you are talking about electrons you are dealing with UV-VIS spectra in this case if you look at solid or a liquid UV-VIS spectra you can see that you don't have a well define peak but a continuous absorption, in this case you have bands. So it really depends of the state of the material you are analyzing.
There are other factors that determine the broadening of the absorption to multiple wavelengths. Form the previous explanation you can derive that more an atom is isolated, less interaction occurs, and the absorption is more limited to determinate wavelength, because you have similar "oscillators". But if you increase temperature and pressure you increase the probability that some atoms/molecules interact, and have different energy levels. The effect is the same: you are broadening the absorption band, respectively you will have a _temperature and a pressure broadening.
So it seems that if you have a single atom this absorbs at a determinate wavelength. Even in this case is not true, the explanation is not so intuitive as the previous ones because you have to understand Heisenberg uncertainty principle but roughly it states that you can't determine precisely the energy states of an atoms. This last broadening is called Natural broadening.

For your specific case you should look at the the UV-VIS spectrum, you will not see a sudden step next to the absorption band but a gently curve. This means that there are less and less atoms absorbing in this region because statistically is less probable that atoms can be perturbed so much to absorb that wavelengths but there are still chances that some atoms absorbs even at wavelength far from the mean.

Have a look to wikipedia spectral lines page for more infos.


in hybridization, can bond angles be different Is there a concept of mixed hybridization where bond angles in a hybridized molecule are different angles?

You have to keep in mind that hybridisation doesn't really exist, its just a convenient way of explaining how x orbitals can form y bonds with a certain geometry even though the constituent atomic orbitals point in wildly incorrect geometries.

Take sp3 hybrid orbitals as an example, we combine s, px, py and pz to get 4 sp3 hybrids which we consider to be in a tetrahedral arrangement (i.e. bond angles of around 109.5).

Most molecules deviate (in some cases significantly) from these values, and we can explain that through a variety of steric and electronic effects. The carbon centre may still be considered sp3 due to the orbitals, even if the geometry doesn't exactly match what you'd expect.

Within various hybridisation states, theres also a situation where we would say that a centre has 'more s-character' or 'more p-character', for instance if you consider the bond angles of NH3 vs PH3 and keep on going down, the bond angles change progesively despite them all having the same hybridisation state.


Could PNP (Silicon) transistor be toxic when exposed to air\light? The story goes as follow:

While I was in the electronic room I got bored and crushed a PNP transistor (Model CDIL 2N2907) using a rotating pipe jaws just like in the picture bellow.

Now, it took me some time to realize that I hear something like a sizzle or a tiny fire sound.
I looked at the crushed transistor and I saw a hole in the bent metal.

I am almost clueless in Chemistry, but I guessed that there was some kind of reaction
with the substance when exposed to air (Normal air in Temperature room about 20C)

I looked at the datasheet of the exact part: CDIL 2N2907
and I couldn't find any Chemistry info apart for the presence of silicon.

I do know that a N-doped semiconductor \ P-doped semiconductor are being used and in this case the semiconductor is silicon.

My question: Does (doped) silicon become toxic when exposed to air? Maybe there are other well known substances (insulator) being used with silicon when creating PNP that could react to air?

Also, the room had light (Fluorescent white light). Could it had any effects?

Thank in advance for any information on the matter.

Edit : The sound of sizziling came after I already crushed the transistor. Of course there was sound while crushing - of metal and more.

Rotrating pipe jaws:

The semiconductor itself is harmless, but, since you heard some sizzling, it is possible that a getter, such as an alkali metal, was in the metal-cased component. There is a patent for getters to be used with semiconductors, http://patentimages.storage.googleapis.com/pdfs/US3259490.pdf, though I've never seen one in use.

More likely, you heard the glass header breaking.


Which is more stable chemically: DNA or peptide Assuming enzyme-free environment, and the DNA and peptide are lyophillized, and then stored in vials under the same conditions (temperature and humidity).

On average, which would be more chemically stable,

in terms of backbone cleavage?
in terms of loss of side chain?

Well, more than 1 year has passed, and now the field of palaeo-proteomics is born, I think the verdict is clear:

The winner is PEPTIDE.
Still found intact in tens of millions old extinct animals, where DNA had degraded for long.

Source: https://www.nature.com/articles/nature14249


Can an SN1 reaction occur in compounds with the leaving group attached to a primary carbon, which can later rearrange? Can SN1 reactions occur in compounds where the carbocation, even though is is initially $1^\circ$, can rearrange to form a more stable carbocation?

Examples cases:

Your thinking is correct that they can rearrange and then form stable carbocation (though your first example is wrong as it would become anti-aromatic after rearrangement) ,
but the answer to your question is NO , they would be substituted by SN2 mechanism being
1 degree . The reason behind this is that the compound will not try to form a carbocation first where it is not that stable , you can think of it in this way that at the initial stage the compound doesn't know that it can rearrange later to form stable carbocation, this is something that we can see but not the compound and thus it will go by normal SN2 mechanism.


Photochemistry of beta, gamma-unsaturated ketones Is there a Norrish reaction for beta, gamma-unsaturated ketones?
I know that Norrish reaction is used for saturated ketones, but in my homework I found this question. I have looked a lot in books and websites but I haven't found anything

Beta, gamma-unsaturated ketones can undergo Norrish photoreactions from the excited singlet state.  For example, in the case (1) below $\ce{R'}$ would be an allyl radical if we had started with a beta, gamma-unsaturated ketone.

From this solvent caged biradical several things can happen.  The acyl radical can decarbonylate (lose carbon monoxide) and then recombine with either end of the allyl radical to produce hydrocarbons.  Also the radical pair could simply recombine.  This would regenerate the starting material, or if the allyl radical has different substituents on each end, then it could recombine at the other end of the allyl radical to produce an isomeric beta, gamma unsaturated ketone.

However, the really interesting photochemistry observed with beta, gamma-unsaturated ketones occurs from the triplet state and is known as the oxa-di-pi-methane rearrangement.  The following figure (2) provides an example showing the "characteristic" 3-membered ring in the final product.

It is just the "oxo" analogue of the all-carbon photo di-pi-methane rearrangement.


Equilibrium of an electrochemical cell I am quite new to the field. So try to figure out what is meant by the equilibrium state of an electrochemical cell?

Is it, where both half cells passing the same number of electrons to either side? if yes, how do the electrons transfer during non-equilibrium state (is it like, there is a net electron transfer due to electron change from both electrodes).

Any explanation would be much appreciated.

You should consider the Nernst equation
$$
\Delta G = \Delta G^{\ominus} + RT\ln Q=- RT\ln Keq + RT\ln Q
 $$
At equilibrium deltaG is zero and Q the reaction coefficient is Keq also note
$$
\Delta G = -nFE
 $$
so E is zero at equilibrium. Outside equilibrium Q will move towards K decreasing E.


MgO as food supplement - solubility in water A number of magnesium food / sports supplements contain it as MgO - which is practically unsoluble in water. The recommended quantity is somewhere around .3g / day, so not an insignificant amount.
How is that supposed to work? Does it rely on the pH shifts in the human gut? I've tried to check MgO solubility in water depending on pH, but the publication I've found is behind a paywall.
Edit: IvanNeretin has pointed out that MgO will react with stomach acid, not dissolve in it. I suppose that's what the manufacturers go for.

MgO will react with stomach acid, not dissolve in it. I suppose that's what the manufacturers go for.

This is not factually right at all. Stating the metal content as an oxide is an century old tradition of quoting the metal content as oxides. Magnesium percentage quoted as percent MgO does not mean that the tablet actually contains magnesium oxide. The links OP provided in the comments shows that the tablet actually contains magnesium stearate. This is done to avoid upsetting the stomach because a very water soluble magnesium salt can induce diarrhea in a few hours.
Magnesium steareate is a salt of a fatty acid, which is rather water insoluble as expected but it has some fat solubility. So when you take it with food which has some fats along with this fatty magnesium salt, there must a be slow absorption of the fatty salt by the body hence the tummy is happy.
Basically, you are eating a magnesium soap!


Could coating silver inside of a cuvette be used for SERS? I understand that molecules adsorbed to a coinage metal surface experience plasmonic enhanced Raman shifts and I have a spectrometer that I would like to perform SERS measurements with but do not have a Raman microscope/probe. I do have a light-tight cuvette cell and I am wondering if I could coat the inside of a cuvette on two sides with silver to obtain SERS spectra of liquids? This is a standard 1x1 cm cuvette and the SMA cable going to the spectrometer has a long pass filter installed for the correct wavelength.

Most of the literature I have read points to using a suspension of silver nanoparticle colloids mixed with the analyte liquid within the cuvette to obtain SERS spectra, but I have not been able to find any papers that use what I am proposing.

from theory this should work what you have scatched here. When it comes to SERS a main point is, that you got a surface which is rough enough that you can create mini hot-spots for the electric field. This will give you the enhancement.
Maybe you must put a little bit of work in you set up. The scattering is needed to arrive at the detector. maybe you try a Raman probe or do a other angle for the wall with silver on it.
Best regards!


How can I define alkenes using graph theory? I want to represent alkenes in a mathematical fashion but I want to make sure of what I'm writing. Here is what I tried by defining "alkenic" graphs:

Alkenic graphs are connected non-looped multigraphs where:

nodes are of degree less than 5 (octet rule);
there is at least a double edge and no triple or quadruple ones (edges like in bonds);
cyclic submultigraphs with conjugated single-double edges given in any conjugation order are to be considered isomorphic (anti aromatic or aromatic isomorphism).
For example, the graphs representing "1,4-dimethylcyclobuta-1,3-diene" and "1,2-dimethylcyclobuta-1,3-diene" are isomorphic.

In this case I'm not taking stereoisomers into account.

Perhaps you could just represent them as simple weighted graphs? Edges of weight 1 represent single bonds, while edges of weight 2 represent double-bonds.

*Edit: For condition #3 simply pick a canonical ordering for the vertices/atoms. See

https://en.m.wikipedia.org/wiki/Graph_canonization

http://depth-first.com/articles/2006/08/12/inchi-canonicalization-algorithm/

for examples


How to identify the parent chain if two chains have the same number of skeletal atoms and the same number of double bonds? Please could you tell me the rule to name the given hydrocarbon ... I am just confused in which chain should I take as the parent chain . The chain from A to C or from C to B. There is no rule about it anywhere on the net or in my books .

The most important simplified criteria for the choice of a principal chain in such compounds are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

Since the compound that is given in the question does not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-44.1.1 is not relevant in this case.

The second criterion given in P-44.3.2 is the longest chain. However, the  chains A–C and B–C both have eight skeletal atoms.

Also, both chains have three double bonds. Therefore, the criteria given in P-44.4.1 (a) (greater number of multiple bonds) and (b) (greater number of double bonds) cannot be used to reach a decision.

The criterion given in P-44.4.1 (h) (lower locant for the suffix) is not relevant since the compound that is given in the question does not have any suffix.

Finally, the criterion given in P-44.4.1 (j) stipulates that the principal chain shall have the lower locants for the double bonds. Therefore, the principal chain for the compound that is given in the question is the octa-1,4,6-triene from A to C rather than the octa-2,4,6-triene from B to C since the locant set ‘1,4,6’ is lower than ‘2,4,6’.

Thus, the complete name for the compound that is given in the question is (3E,4E,6E)-3-ethylideneocta-1,4,6-triene.


Determining if reaction is endothermic or exothermic using equilbrium constant For the following reaction
$$\ce{W(s) + 4 Br(g) <=> WBr4(g)}$$
at $\pu{900 K},$ $K_p = 100$ and at $\pu{2800 K}$, $K_\mathrm{p} = 5.$
Determine if the reaction is endothermic or exothermic.

At higher temperatures, the equilibrium has shifted towards the reactants since $K_p$ is significantly smaller. However, how do we determine if a reaction is endothermic or exothermic from this information?
If the reaction was endothermic, an increase in temperature would push the equilibrium to products to counteract the increase in heat. However, the equilibrium has been pushed to the reactants thus signifying the reaction is exothermic. But this assertion is on the assumption that the forward reaction is exothermic.

Think about it this way.  In an exothermic reaction that is at equilibrium the forward reaction has a lower activation energy.  So, if you increase T the forward reaction rate increases but the reverse rate increases even more.  That's because the rates of reactions with higher activation energy are more sensitive to temperature.  With the reverse rate higher more reactants will be made until the concentration reactants reaches a point where the rate of the forward reaction equals the reverse rate.

In an endothermic reaction increasing T increases the forward rate more than the reverse, producing more product until the increased product concentration causes the reverse rate to equal the forward rate.

The equilibrium constant can be thought of as a ratio of products to reactants needed to equalize the rates of two reactions with different activation energies.


Radial node in p or d-orbital While studying atomic orbitals and nodes, the only illustration of a radial node that I could find is in an s-orbital (like here). How do the radial nodes look like in p or d-orbitals?
(I don't know how to plot the distribution myself.)

I find the following picture from Chemistry Libretexts. I don't own the picture, so the credit goes to respective authors.

I believe the d orbitals illustrated on the diagram are only the $d_{z^2}$ orbitals. For the f orbital, I believe it is $f_{z^3}$. Please let me know of any problems/suggestions.


Cis and trans electrodes I have come across use of the terms cis and trans electrodes in the context of nanopores and driving charged biomolecules through a pore using an electric field [1]:

FIG. 1. Drawing of the cis negative electrode and trans positive electrode chambers. The arrows show the direction of electric field lines. The pore diameter $d$ and length $l$ and the capture radius $r$ are also shown.

I wondered what the origin of the term is.
I know about cis and trans isomers, but I have not come across the term before in terms of electrodes. Is there a simple explanation?
I can just assume that cis/trans are just another terms for cathode and anode, but I think I am missing something. If the cis electrode is the cathode and the trans electrode is the anode, then why not just call them that? Is there any additional meaning?
Reference

Grosberg, A. Y.; Rabin, Y. DNA Capture into a Nanopore: Interplay of Diffusion and Electrohydrodynamics. The Journal of Chemical Physics 2010, 133 (16), 165102. DOI: 10.1063/1.3495481.

Sometimes authors use buzz words in their paper in order to make their work novel or fancier than it should be. However, such newly coined terms never get popular and vanish with the author into oblivion. No, cis and trans can never mean cathode or anode. It refers to the chambers in which the electrode is place.
I looked up the Oxford's Latin Dictionary:
cis has three meanings

(in time) before, within
(in space) on the near side of, on this side of
(expressing motion) to the near side of

trans

to the other side of

Now look at the cell configuration in your link. The very first lines of the paper you linked says

DNA translocation through protein or synthetic solid state nanopores
is a two stage process that involves capture of a DNA coil by the pore
including its delivery to the pore and insertion of one of its ends
into it, followed by its motion through the nanochannel. The process
may be driven by voltage $\Delta$V typically a few hundred millivolts
applied between two macroscopic electrodes positioned on both sides of
the partition separating the cis where DNA is initially placed and the
trans to where it translocates through the pore chambers, as shown
schematically in Fig. 1


Method for regioselective hydroalkylation of terminal alkyne I'm trying to figure out a method to add a linear alkyl chain R and a hydrogen atom H across the triple bond of a terminal alkyne. The alkyl chain should end up on the internal carbon (C-2):

My attempt :
First, we can hydrolyse the alkyne to produce a methyl ketone. Addition of a Grignard reagent produces the tertiary alcohol. But, now I'm stuck as I can't treat it with $\ce{H2SO4}$, because then ring-expansion will ruin the whole matter.

How to move on?

Here is a potential solution using old-school chemistry, though with several steps.

Hydroborate the alkyne and work up with $\ce{H2O2}$ to give the aldehyde.

Alkylate the aldehyde enolate with $\ce{RI}$, either directly (LDA or LiHMDS) or via enamine/silyl enol ether

Reduce aldehyde to alcohol ($\ce{NaBH4/EtOH}$)

Form triflate/mesylate and eliminate (DBU/Toluene, reflux)

If you wish to explore @orthocresol's suggestion of enol triflate coupling then coupling the triflate with the cuprate of your alkyl chain looks to be viable according to this Tet. Lett. article from McMurry and Scott1
References

John E. McMurry, William J. Scott, “A new method of olefin synthesis. Coupling of lithium dialkylcuprates with enol triflates,” Tetrahedron Letters 1980, 21(45), 4313-4316, DOI: 10.1016/S0040-4039(00)77845-2


Equilibrium change in sealed container Question:

Decreasing the product's concentration seemed to follow the logic of Le Chatelier's principle, so a decrease in volume will force the reaction to the left to increase the pressure (increasing the number of molecules). However, decreasing nitrogen gas concentration also has the same effect. I chose volume compression over the other because I assumed. I am probably stretching what it means by "sealed"  though. Apparently I got the question incorrect.

The given graph indicates that $\ce{[NH3]}$ remains constant till the 15 minute mark which implies that the equilibrium was established till then. Now, after the 15 minute mark, $\ce{[NH3]}$ decreases and reaches a constant value which means, the equilibrium was disturbed and re-established.
$$\ce{N2 + 3H2 <=> 2NH3 + heat}$$
Since we observe a decrease in $\ce{[NH3]}$, the reaction must have shifted in the backward direction according to the given equilibrium.
Out of the five options, only match for this is (C) removing some amount of $\ce{N2}$. Let us analyse the other four options,
(A) Decrease in temperature: Note that the above reaction is exothermic and hence according to Le Chatlier’s principle, the reaction shifts in the forward direction for a decrease in temperature.
(B) Decrease in volume: Since we are dealing with a gaseous(ideal) equilibrium, a decrease in volume implies an increase in pressure and hence, according to Le Chatlier’s principle, the reaction shifts in the direction where there is less no. of moles ie., forward direction here.
(D) Removal of $\ce{NH3}$: As Gaurang pointed out in the comments, removal of $\ce{NH3}$ will cause a sharp fall along a vertical line, but the graph indicates a gradual fall in $\ce{[NH3]}$. Hence this is also not valid.
(E) Increase in volume: Assuming a sudden increase in volume, say n times, $\ce{[NH3]}$ should also fall by n-times. Clearly, the graph doesn’t match with our condition. Hence this is also incorrect.

So (C) must be the correct answer, as removing some $\ce{N2}$, shifts the reaction in the backward direction, thereby decreasing $\ce{NH3}$ gradually till equilibrium is re-attained.


X-Ray Diffraction: Phase Identification Strategy I am having some trouble with qualitative phase identification using x-ray diffraction (XRD). Here follows some basic information about the experiment:

Instrument: PANalytical X'Pert$^3$ Powder

Software: PANalytical Highscore Plus

Database: PAN ICSD

Sample: respirable fraction aerosol samples from ferro- and silica manganese plants.

The goal: which manganese phases are present, and which dominate?

Grasping the software is not what I am concerned about. I lack an analytical strategy. I know how to determine the background, but I am not sure when the background is appropriately determined. Too high? Too low? Does this strongly affect the phase identification?

Then I do a peak search, and 50 peaks are detected. Should I focus just on the major peaks? Are the minor ones important too? When I do a search-match analysis, I get 400 suggested phases, where 20 phases share the highest score. Then another 20 phases share the highest score minus 1. I lack criteria needed to accept a phase as present, and discard other phases as not present.

I am not able to find good literature on this subject. I did find an article called "Hints on Phase Identification Using Powder X-ray Diffraction", which certainly raised a lot of potential problems and procedures, but which did not necessarily solve these problems or explain why the procedures would help. Does anyone know of any books or other literature that could help me?

My problem boils down to lack of "strategy" or "analytical procedure", and maybe someone here could offer some advice. I hope my question is "answerable", and not too broad; I know there is a "specific-and-precise-question" policy in here.

I don't have much information on your particular system: are they nanocrystals?  Are you expecting just one or two phases?  Dozens of phases? Do you have a lot of amorphous material present?

I have one publication you can look at which addresses a rather complicated sample with amorphous, nanocrystalline, and crystalline objects.  It is (hopefully) overkill for your sample, but it will give you ideas on how to partition the problem (and some of the references may help too):

Gainsforth, Z., Brenker, F. E., Simionovici, A. S., Schmitz, S., Burghammer, M., Butterworth, A. L., et al. (2013). Stardust Interstellar Preliminary Examination VIII: Identification of crystalline material in two interstellar candidates. Meteoritics and Planetary Science, 49(9), 1645–1665. doi:10.1111/maps.12148

You may also find it helpful to play with simple fitting softwares such as CrystalMaker/CrystalDiffract where you can try out different mixtures and fits.  Sometimes the peak-finding softwares are a little simplistic with a large number of false positives.  If you try manually combining the possible phases, you can probably rule many of them out and simplify your list.

A few tips:

1) Look for sets of peaks with FWHM that are similar.  This works with nanocrystalline materials.  If you have a nanocrystal with a larger crystal, then the larger crystal will have sharp peaks and the nanocrystal will have broad peaks.  Then you can select which peaks apply to Phase A and Phase B and simplify your search space drastically.

2) If Peak A is present in your experimental data, and you get a "hit" for phase X, but Phase X also predicts Peak B but you don't see it in your experimental data, then you can rule Phase X out.  You may need to understand the characteristics of the phase in a little detail because you can have situations where one atom substitutes for another and you get a change in peak intensities or positions.

3) Do you know the chemical composition of your sample?  If one of your "found" phases is a Yttrium based superconductor, and you're pretty sure you don't have Yttrium, then you can drop that phase.

4) Try looking at it with an electron microscope (SEM or TEM).  This will provide basic constraints as to whether you have many phases, or just a couple, and how big the crystals are.  This is a lot of work, but maybe less work than spending the next month chasing false positives.  TEM can also give you diffraction on any phase, no matter how small, and will be pretty unambiguous.

5) You can also provide constraints using techniques such as XANES, NMR, FTIR, etc., depending on what kind of sample you have.  Which of these are useful is also dependent on the resources you have present at your university/company/facility.

Hope that helps!


How does work from other sources affect Gibbs free energy of the system? In most physical chemistry textbooks, Gibbs free energy is given like this:
$$\Delta_\mathrm rG = \Delta_\mathrm rG^\circ + RT\ln Q\tag{1}$$
and Gibbs free energy is part of this three-way relationship:
$$\Delta_\mathrm rG^\circ = -RT\ln K = -nFE^\circ\tag{2}$$
which gives rise to the Nernst equation.
However, in many cases, $\Delta_\mathrm rG > 0,$ so it is not thermodynamically favourable. In these cases, there would be an external source of work.
First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra $-VIt$ term or something else which would cause $\Delta_\mathrm rG < 0?$
Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause $\Delta_\mathrm rG < 0?$

I will treat the first and second question as if they were part one and part two of a single question. I think the OP wants to know how non-PV work plays a role when looking at the Gibbs energy of a reaction, and its interpretation.
To start off, the Gibbs energy for a given reaction is a state function, so it does not depend on how much work or how much heat are exchanged between system and surrounding. The total change in entropy, however, is affected by what is going on in the system and in the surrounding, so work vs heat is relevant for the second law of thermodynamics and the direction of the reaction.

First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra $-VIt$ term or something else which would cause $\Delta_\mathrm rG < 0?$

The easiest way to deal with the power source is to place the system/surrounding boundary such that the power source is in the surrounding. This shows it does not affect the Gibbs energy of the reaction at all. The Gibbs energy is positive, but the reaction can go forward because it is offset by the work done on the system. As a result, the combined entropy increases, and we are not breaking the second law.
In general, the criterion is not that the Gibbs energy of reaction has to be negative, but that
$$\Delta_r G < w$$
When there is no work done on the system or by the system, this simplifies to the requirement that the Gibbs energy be negative.
As an aside, when the Gibbs energy of reaction is negative, we can not have the reaction do an arbitrary amount of work on the surrounding, but that amount is limited to the Gibbs energy. That's why the Gibbs energy is sometimes called the maximal work (this is a good term for cases when it is negative; when it is positive, it is the minimal work that needs to be done on the system for it to be possible for the reaction to go forward).

Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause $\Delta_\mathrm rG < 0?$

The hydrolysis of ATP usually does not do work on the surroundings. Instead, the uncatalyzed reaction has a positive Gibbs energy of reaction. To use up the reactants and make the products, the cell runs a different reaction that includes the hydrolysis of ATP. It is not sufficient that ATP be hydrolized somewhere while the desired reaction goes on somewhere else. Instead, the two reactions are coupled so that ATP hydrolysis can not occur without the other part of the reaction happening. Here is an example how that might look for the reaction $\ce{A + B -> C}$:
$$\ce{A + ATP -> A-P + ADP}\tag{1}$$
$$\ce{A-P + B + H2O -> C + P_i}\tag{2}$$
In this scheme, A-P refers to a phosphate group bound to A, and $\ce{P_i}$ refers to phosphate. Both steps can have a negative Gibbs energy of reaction, even thought the Gibbs free energy of reaction for $\ce{A + B -> C}$ is positive. As net result, A and B turn in to C, without work done on the system.
There are other cases where there is work done on the system or by the system. These include proton pumps, and myosin hydrolyzing ATP to contract muscle fibers. You could also discuss photosynthesis in terms of work done to drive the synthesis of ATP.


Enthalpy of solution If NaCl dissolves in water, it is said that it happens because the hydration energy is greater than the lattice energy. If so, the enthalpy of solution attained is about +3.88kJ/mol, implying that it is endothermic.

From what I have understood, the hydration enthalpy is exothermic and thus its effect should be noticeable. That is, when we add up the lattice and hydration, the negative sign should have come in our final answer.
Why does this not happen? Where have I gone wrong?

For any salt dissolving in water, its not compulsory that enthalpy change needs to be negative. Yes, if it is negative then its better dissolving for sure, But it doesn't mean the enthalpy change has to be compulsorily negative for salt to dissolve in water.

Enthalpy change when we dissolve Nacl in water is positive. This is because breaking of ionic bonds in the crystal lattice of Nacl requires more energy than that of provided by hydration of Nacl. Yes, hydration energy for Nacl is less than the energy required to break the Nacl apart. Hence the extra energy which can add up with the hydration energy to break the bonds of Nacl apart comes from the heat of the water. So the extra amount of energy which hydration could not provide comes from the heat of the water. Hence water looses heat, becomes slightly cold and also hydration energy (exothermic process) is less than the required energy to break the bonds, hence the net enthalpy change is positive.

So till now we learnt that for Nacl hydration enthalpy is not greater than enthalpy required to break the bonds and hence the extra energy required to satisfy the bond breaking comes form the heat of the water, and as hydration enthalpy is less than required energy to break the bond of Nacl, the net enthalpy change is positive, hence endothermic process.

Now the question arises how comes then Nacl dissolve in water?
As already said its not compulsory that enthalpy change needs to be negative for a salt to dissolve in water. Remember my friend, dissolving depends upon ΔG. ΔG is the most important factor which decides weather the reaction will happen or not. If ΔG is negative and ΔH is positive, salt will for sure dissolve because dissolving of salt (or any reaction) depends on ΔG. So if ΔG is negative for the dissolving of salt then salt would dissolve. Infact for Nacl ΔG is negative. Lets see how.

ΔG= ΔH −TΔS.

In dissolving Nacl in water ΔS is very much positive because is dissolving Nacl, Na and Cl gets apart, increasing the disorderedness in the system, hence ΔS is highly positive. So practically TΔS term is higher than ΔH term and hence we end up having ΔG negative.

Hence as ΔG is negative, Nacl dissolve. Remember dissolving of Nacl depends on ΔG. Negative ΔG makes Nacl dissolve in water.


Organic Chemistry - Effect of methyl substituent in Aromatic Substitution via Benzyne The following reaction has been taken from my study material. This shows that the benzene ring is undergoing a nucleophilic aromatic substitution via benzyne mechanism (presence of a strong base like $\ce{NaNH2}$). My question is:

Why is substitution preferred at the ortho position?

The following link explains the mechanism quite well and gives an example of ortho-chlorotoluene undergoing a similar reaction. Since both proceed via the same benzyne intermediate, why is there a difference in the products formed by meta-chlorotoluene? Or is my study material wrong in saying that o-aminotoluene is a major product? Any help regarding this is appreciated!

Attaching the related reaction's image:

The Me group has no directing effect in the deprotonation so there are 2 protons (o & p) that can be removed with equal probability by the NaNH2 to give the benzyne. These give 2 possible benzynes (2 and 3 relative to Me). The nucleophile will add with equal facility to either end of either benzyne giving a mixture of 3 isomers in 1:2:1 ratio. The major isomer is the m- aminotoluene because it can arise from both of the possible benzynes.


Solid state and Bragg's Equation A frst order reflection from (111) plane is observed for LiX with 2theta =24.6 degree (X-ray of wavelength 1.54 angstrom). Assuming LiX to be a cubic crystal system, calculate the length of the side of the unit cell in angstrom.
This is a question of JAM 2014, here, the length is considered as a=2a for ionic system in a solution provided by Arihant's book. Why is it so? Is it correct?

From Cubic crystal system on Wikipedia:

  In the rock-salt or sodium chloride (halite) structure, each of the two atom types forms a separate face-centered cubic lattice, with the two lattices interpenetrating so as to form a 3D checkerboard pattern.

If LiX crystallizes like NaCl, then the unit cell length is twice the distance of the shortest Li - X distance. The magnitude of the answer given in the book looks like it could be an interatomic distance, but the calculation does not make any sense to me (it does say $a = 2a$, as @Ivan_Neretin already pointed out in the comments. Also, when a calculation drops the units midway and sticks them back in the result, it demonstrates a certain nonchalance that does not instill confidence in the source.

The question asks about the unit cell parameter a (in a cubic system, this is all you have to specify to describe the dimensions of the cell because the other two lengths are the same and the angles are all right angles). If that's all you want, the internal structure of the unit cell has no bearing on the calculation (there might be systematic absences in a particular space group, but the question says that the 111 reflection was observed, so there is no issue there either). Why not just plug in the angle, get d, and calculate the unit cell length from that?