When to use chromatographic separation vs vacuum-assisted rectification distillation? I'm trying to understand a bit more about industrial scale separation processes. One particular element I'm interested in is the question as to when chromatography (HPLC or Simulated Moving Bed) may be preferable to vacuum-assisted rectification distillation.

My first thought would be that chromatography sees its greatest industrial use when processing & separating easily damaged compounds, or to select an individual compound for separation through ion exchange chromatography. This, by comparison to vacuum-assisted rectification distillation, where differing vapor pressures allow for high-purity separation -- but requires the addition of heat to material that may destroy easily-damaged compounds. Would love some help on this! Thank you!

The decision to use a particular separation technique industrially is based on multiple factors. To speak generally, this comes down to ease of separation and process economics. The simplest and cheapest techniques would be ones like biphasic separation or filtration.

The distillation you refer to bears more relation to bulk chemical manufacture such as oil refining.  Process HPLC and simulated moving bed (SMB) are more involved techniques,  requiring more sophisticated technology and hence, is a more expensive technique. This puts it into the realms of higher value products which would tend to be pharmaceuticals or ultra high purity display chemicals.

So, for process economics, if we think in terms of Input-Process-Output we have cost of input material,  the cost of processing and the cost of output material.  If Input + Processing are greater than the ultimate commercial cost that the product can bear, then you don't have a product or it would need a key differentiator (which could be purity), to justify the extra cost.

If the costs justify it, then using higher vacuum can offset degradation risk involved in a fractional distillation. Physicochemical factors come into play to select this technique (under the ease of separation banner). Industrial distillation would tend to be used for more complex mixtures of desired volatile and semivolatile components.  A good Quora discussion on the topic is here:

https://www.quora.com/How-are-rectification-and-distillation-related

The technique, obviously relies on boiling point and exploitation of potential synergies of compound (usually liquid/oil) mixtures.

Chromatographic separation relies on adsorption/desorption by judicious selection of stationary phase and solvent (mobile phase). For commercial processes, this needs to be repeated over and over in a completely reproducible way with minimal variation. If the stationary phase is cheap, then less material has to go through the material before being changed. If the stationary phase is expensive (decent spherical functionalise silica can easily be over £2000/kg, chiral phases over 10-20k per kg. Irregular silica can be £500/kg), then you want a faster,  cleaner separation with minimal fouling of the support. Over 200 injections onto the same stationary phase is not unheard of.

These injections would also be maximised on production scale vs the standard analytical technique.
For SMB,  typically four HPLC columns are used and a more sophisticated algorithm is required to inject the right amount of material onto the correct column at the right time.

As commented on, isomeric separation,  particularly if the materials are solids,  then  favours process HPLC rather than high vacuum distillation. But again, the process economics would help decide this.

Other factors are internal expertise in the relevant techniques and/or capability of outsourcing partners (contract manufacturing organisations) in more niche techniques. SMB would probably fall into this category i.e. most manufacturing companies would look to a specialist 3rd party unless they were already engaged with this technology.


How to add two hydroxy groups to 1,5-diphenylpentan-3-ol with as high as possible yield? From 1,5-diphenylpentan-3-ol (above), how to add the two hydroxy groups in red (below)?
Is there any proven successful synthesis route to 1,5-diphenylpentane-1,3,5-triol with higher yield (40 % or above)?

A few possibilities for direct hydroxylation:

$\ce{SeO2}$ immediately comes to mind as one potential option; it is well known to selectively oxidize allylic positions to yield alcohols and carbonyl compounds, and there appear to be references suggesting it reacts at the benzylic position as well.
In Advanced Organic Chemistry, March notes that cerium(IV) triflate "converts benzylic arenes to benzylic alcohols, although the major product is the ketone when > 15% of water is present." Ref: Laali, K.K.; Herbert, M.; Cushnyr, B.; Bhatt, A.; Terrano, D. J. Chem. Soc., Perkin Trans. 1 2001, 578.
Intriguingly, March also reports a variety of oxidations accomplished enzymatically. I have no experience and very limited knowledge of these reactions, so I'll just give the relevant primary literature references:

Adam, W.; Lukacs, Z.; Saha-Moller, C.R.; Weckerle, B.; Schreier, P. Eur. J. Org. Chem. 2000, 2923.
Hamada, H.; Tanaka, T.; Furuya, T.; Takahata, H.; Nemoto, H. Tetrahedron Lett. 2001, 42, 909.
Adam, W.; Lukacs, Z.; Harmsen, D.; Saha-Moller, C.R.; Schreier, P. J. Org. Chem. 2000, 65, 878.

If those approaches aren't appealing, there are various less direct routes. Radical bromination (with, e.g., NBS under photochemical conditions), followed either by direct substitution with hydroxide, or alternatively elimination followed by subsequent acid-catalyzed hydration, would likely work well. If you need stereoselectivity, there are are numerous epoxidation reactions (that is, after bromination and subsequent elimination) that would likely be appropriate (Sharpless, Jacobsen, Shi, and certainly numerous others). Conveniently, your molecule would already have the allylic alcohol necessary for Sharpless asymmetric epoxidation, should you want to go that route. In any case, it should be possible to open the epoxides at the less hindered carbon with a suitable hydride donor (e.g., LAH) to yield the alcohols after work-up.


Why doesn't nitrogen monoxide dimerize? Why doesn't Nitrogen monoxide dimerize even though there is an odd electron present whereas nitrogen dioxide does (because of the odd electron on nitrogen)?

Highstaker's answer is almost correct.  The antibonding orbitals that are occupied in NO are in fact pi symmetry, but when the dimer forms that is no longer relevant.  It is a sigma bond.

The enthalpy of the newly formed sigma bond in the dimer is weak because the net gain in bond is off set by the loss of a very odd set of single-electron resonance forms available for NO monomer.

Given $\Delta G= -17 \:\mathrm{kJ/mol}$ , and that dimerization is entropically disfavored, when the total free energy is considered there is no gain since entropic effects are on the order of $10-30 \:\mathrm{kJ/mol}$.  Thus any small gain in enthalphy is offset by the loss of entropy.

Always look at these things through Gibbs free energy and not just enthalpy.


How DFT-D3 incorporates coordination number (CN) into C6 I'm reading Grimme's DFT-D3 paper
but really puzzled by how the $\mathrm{CN}$ is considered in the calculation of $C_\mathrm{6}$ coefficients. (Stefan Grimme, Jens Antony, Stephan Ehrlich, and Helge Krieg. J. Chem. Phys. 2010, 132 (15), 154104.)

According to Fig. 5 in the paper, it seems the $\ce{AB}$ pair is put in some reference system and then based on the precomputed $C_\mathrm{6}$ values in the reference system we could get the $C_\mathrm{6}$ in the environment.

Especially, I'm confused by the fact in the figure that $\mathrm{CN}^\ce{A}$ and $\mathrm{CN}^\ce{B}$ are in different dimension, what does it mean?

I will quote the paper quite a bit, but I'll try and summarize a bit after the quotes and equations. You might want to start at the bottom and work backward, a lot of this is just for later reference.

First, some background. Equation 3:

$$
E^{(2)} = \sum_{AB} \sum_{n=6,8,10,\dots} s_{n} \frac{C_{n}^{AB}}{r_{AB}^{n}} f_{d,n}(r_{AB})
$$

  Here, the first sum is over all atom pairs in the system, $C_{n}^{AB}$ denotes the averaged (isotropic) $n$th order dispersion coefficient (orders $n=6,8,10,\dots$) for atom pair $AB$ (...)

So $A$ and $B$ are nuclear indices, and together specify an arbitrary pair.

  Although the $C_{6}^{AB}$ values can be compute easily for any pair of free atoms by using Eq. (5) [the Casimir-Polder integral] in principle, this would lead to a rather inconsistent treatment of dispersion in and between molecules.

For reference, here is equation 5:

$$
C_{6}^{AB} = \frac{3}{\pi} \int_{0}^{\infty} \alpha^{A}(i\omega) \alpha^{B}(i\omega) \, d\omega
$$

Instead they calculate the integrals over pairs of hydrides, subtracting the effect of molecular hydrogen (equation 10), in order to capture the effect of being bonded:

$$
C_{6}^{AB} = \frac{3}{\pi} \int_{0}^{\infty} d\omega \frac{1}{m} \left[ \alpha^{A_{m}H_{n}}(i\omega) - \frac{n}{2} \alpha^{H_{2}}(i\omega) \right] \times \frac{1}{k} \left[ \alpha^{A_{k}H_{l}}(i\omega) - \frac{l}{2} \alpha^{H_{2}}(i\omega) \right]
$$

  At first sight this new approach seems to be a disadvantage because it leads to reference molecule dependent (ambiguous) coefficients. However, as will be discussed in detail below (...), it opens a route to system [coordination number (CN)] dependent "atomic" $C_6$(CN) coefficients. (...) We propose here a radically new "geometric" approach that is based on the concept of a fractional coordination number. (...) We base our ansatz on a fractional CN for an atom $A$, that is a sum of a "counting" function over all atoms in the system,

Equation 15:

$$
\text{CN}^{A} = \sum_{B\neq A}^{N_{at}} \frac{1}{1 + e^{ -k_1 ( k_2 (R_{A,\text{cov}} + R_{B,\text{cov}}) / r_{AB} - 1) }}
$$

  where $R_{A,\text{cov}}$ is a scaled covalent (single-bond) radius of atom $A$ (and analogously for $R_{B,\text{cov}}$).

So the CN$^A$ is some bond distance-dependent scaling factor for the specific atom $A$ under the influence of all other atoms $\{B\}$. Here is how they incorporate their coordination number scaling into the dispersion coefficients. Following emphasis mine:

  Turning back to the dispersion coefficients we propose the following general approach. For each element in the Periodic Table, at least one reference molecule is used to compute $\alpha(i\omega)$ values in Eq. (10). (...) If an element is commonly investigated, and/or often found in different chemical environments, we suggest to use several representatives for which values are computed. In the case of carbon for example, ethyne, ethene, and ethane can be used for CNs between 2 and 4. Values computed for $\ce{C-H}$ and the carbon atom complete the set. For the atom of interest in any of the reference systems also the CN is calculated and stored. Using Eq. (10) the $C_{6,\text{ref}}^{AB}$(CN$^A$,CN$^B$) values are computed for this pair with the two atoms in their specific environments. These serve as supporting points in an interpolation procedure to derive the coefficient for any combination of fractional CN values. Because we base our approach on pair-specific coefficients $C_{6}^{AB}$, a two-dimensional (2D) interpolation scheme must be used. After extensive testing we propose a simple Gaussian-distance ($L$) weighted average,

Equation 16:

$$
C_{6}^{AB}(\text{CN}^{A},\text{CN}^{B}) = \frac{ \sum_{i}^{N_A} \sum_{j}^{N_B} C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B}) \times L_{ij} }{ \sum_{i}^{N_A} \sum_{j}^{N_B} L_{ij} }, \\
L_{ij} = e^{ -k_3 [ (\text{CN}^A - \text{CN}_i^A)^2 + (\text{CN}^B - \text{CN}_j^B)^2 ] },
$$

  where $N_A$ and $N_B$ are the number of supporting points (= number of reference molecules) for atoms $A$ and $B$, respectively. The CN$^A$ and CN$^B$ are coordination numbers for the atom pair $AB$ in the system of interest, and the CN$_i^A$ and CN$_j^B$ are those for the two reference systems $i$ and $j$, for which $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ is the precomputed value. (...) For most of the elements only two reference values are necessary to cover the typical bonding situations. For "important" elements such as carbon, however, more values (5 in this case) seem appropriate. Currently we have computed in total 227 reference systems for the elements up to $Z=94$ which leads to about $2.6\times10^4$ difference values for $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$. They are computed once and read at the beginning of any computation.

Conceptually, here is how a calculation would work:

Your SCF has converged and it's time to calculate the -D3 correction. This table is read in from a file into some data structure so that all the $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ can be looked up easily.
Loop over the atom pairs $AB$ of your system as part of equation 3. For atom $A$, calculate its coordination number CN$^A$ using equation 15. Do the same thing for atom $B$. In the equation, $B\neq A$ means all atoms other than $A$, not atom $B$ in the pair!
For this pair, the $C_{n}^{AB}$ in equation 3 is really $C_{6}^{AB}(\text{CN}^{A},\text{CN}^{B})$, which requires evaluating equation 16. What equation 16 does is interpolate between those precomputed $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ for a series of coordination numbers. Because there are two CNs, the interpolation is on a 2D grid, rather than a line or some higher dimension. The interpolation function is chosen to be the $L_{ij}$. From the above quote, each $i,j$ sum is only about 2-5 points, so this is very fast; even the $\ce{C-C}$ grid will only total $5\times5=25$ points.

So, to answer your question,

  According to Fig. 5 in the paper, it seems the AB pair is put in some reference system and then based on the precomputed C6 values in the reference system we could get the C6 in the environment.

This is exactly correct. It is done by interpolation on a grid of reference values, where the coordinates/points on the grid are the coordination numbers, and there is a grid for every possible $AB$ pair.

$\tiny{\text{Phew.}}$


How to know it when I see a covalent network? This is a well-known (better said: well-discussed) question in the internet. When you look for answers for popular questions, you usually see them with a variable degree of reliability and complexity. Unfortunately, for this one, I only observed very very crude and general rules of thumb. So let's get a real answer:

  A network solid or covalent network solid is a chemical compound (or element) in which the atoms are bonded by covalent bonds in a continuous network extending throughout the material. In a network solid there are no individual molecules, and the entire crystal may be considered a macromolecule. Formulas for network solids, like those for ionic compounds, are simple ratios of the component atoms represented by a formula unit. Covalent network, wikipedia

Diamond and SiO$_2$ are really great examples of covalent networks-lattices. So enough with stories:

If you face a new chemical formula, how would you assume it's a covalent network? (In case it is) Is it somehow done by drawing the Lewis structure? Is there a rule for this? Or is it only possible to know such thing with experimental data?

Fun fact: I'm preparing for an internal chemistry olympiad. The examiners usually take all the exceptions into account. So general rules such as "if it contains Si, it's a covalent network" aren't really reliable for me. Better phrased, I'm looking for a rule with the least exceptions.

A good source of information on this topic is the website Tetrahedra of Structure, Bonding & Material Type, from which the above image is taken.

An extreme example of the metal corner should be cesium, the ionic corner CsF, and the molecular corner $\ce{F2}$.  At the top is diamond carbon.

The network covalent apex is truncated on the basis that a network covalent material requires an element with the electronegativity of carbon or less (less than or equal to 2.55).

Then the challenge is to distinguish a boundary between network covalent and the other three types

network covalent vs. metal

network covalent vs. ionic

network covalent vs. molecular

Examples of network covalent compounds other than diamond and silicon dioxide include silicon carbide, silicon nitride, boron nitride, aluminum phosphide, gallium arsenide, aluminum oxide.  However, anytime there is an electronegativity difference there is some ionic component.

Tin can exist in metallic and network covalent allotropes.

Selenium can exist in a network covalent allotrope or as molecular $\ce{Se8}$

Going from network covalent to molecular covalent, think of how there is diamond, graphite, nanotubes, fullerenes and even small molecules like $\ce{C4}$


Ortho vs para alkylation of chlorobenzene I have a question to the solution regarding the product formed in a reaction with chlorobenzene with the use of (i) $\ce{CH3CH2Cl}$ and (ii)$ \ce{AlCl3}$. Since the -Cl substituent is ortho- para-directing shouldn't the answer then be a mixture of 1-chloro-4-ethylbenzene and 1-chloro-2-ethylbenzene? - I suppose that the para-orientation is the major product. But how is that determined?

In general para isomera are thermodynamically more stable than ortho isomers (except in cases when ther is hydrogen bonding in ortho isomer). There are several reasons:-
1. Steric hindrance makes para isomer more stable.
2.Benzene substituents rotate and change thier plane due to steric hindrance. It stops delecolaization of lone pair of electrons (if present in substituents) causing elevation in energy.
Both proper orientation (steric hindrance) and appropriate energy (temperature) is required for formation of product.
Para isomer satisfies these conditions readily and hence are major product.


Calculating numbers of moles at equilibrium given the volume of mixture and initial concentration of reactants The equilibrium between three substances, A, B and C Is shown below.
$$\ce{A(g) + B(g) <=> C(g)}$$
Initially there were $\pu{0.1 mol}$ of A and $\pu{0.2 mol}$ of B in the reaction mixture. A and B reacted together to produce an equilibrium mixture containing $\pu{0.04 mol}$ of C. The total volume of the mixture was $\pu{2.00 dm3}$. Calculate the number of moles of A and B at equilibrium.

I honestly don't know how the answers are $\pu{0.08 mol}$ and $\pu{0.18 mol}$. I got the answers to be $\pu{0.06 mol}$ and $\pu{0.16 mol}$ by subtracting $\pu{0.04 mol}$ from each of $\pu{0.1 mol}$ and $\pu{0.2 mol}$.
Can someone please help?

Following a discussion that took place in the comments:

If given answers are correct, as Poutnik pointed out, your equation should be :
$$\require{cancel}\ce{A(g) + B(g)<=>2 C(g)}$$ - Mathew Mahindratne
Could you explain how you arrived at that conclusion? - Ollie

This answer solves this question for a general case:
$$\ce{a A(g) + b B(g) <=> c C(g)}$$
Let's assume that initial concentrations of A, B, C are $[A_0]$, $[B_0]$ and $[C_0]$ and the final concentration be $[A]$, $[B]$ and $[C]$.
Now, the given variables are $[A_0]$, $[B_0]$, $[C_0]$ and $[C]$, assuming similar data to the question. Let $\mu$ be change for every mole of C
\begin{align}
\therefore\ &\text{Change}=[C]-[C_0] \tag{1}\\
\implies &\text{Change/mol of C} = \frac{[C]-[C_0]}{c}\tag{2}
\end{align}
Now, building a RICE table we get:
\begin{array}{lc} \hline
 &\ce{a A &+ &b B &<=> &c C} \\ \hline
I & [A_0] && [B_0] && [C_0] \\
C & -a\cdot\frac{[C]-[C_0]}{c} && -b\cdot\frac{[C]-[C_0]}{c} && +c\cdot\frac{[C]-[C_0]}{c} \\
E & [A_0] - a\cdot\frac{[C]-[C_0]}{c} && [B_0] - b\cdot\frac{[C]-[C_0]}{c} && [C_0] + [C]-[C_0] \\ \hline
\end{array}
Therefore, the final concentrations of A and B would be:
\begin{align}
[A] &= [A_0] - a\cdot\frac{[C]-[C_0]}{c} \\
[B] &= [B_0] - b\cdot\frac{[C]-[C_0]}{c}
\end{align}

Now, onto the specific solution, how to figure out $a$,$b$ and $c$ given the above convoluted generalisation?
Point 1: The change in $[A]$ and $[B]$ are both equal. Looking at change in the above RICE table, we deduce:
\begin{align}
a\cdot\cancel{\frac{[C]-[C_0]}{c}}&=b\cdot\cancel{\frac{[C]-[C_0]}{c}} \\
\implies a&=b
\end{align}
Point 2: We notice that for the change of $\pu{0.04 mol}$ of C, we observe a change of $\pu{0.02 mol}$ in A (or B, both work). Similar to point one, we equate the equations of change of A and C.
\begin{align}
2a\cdot\cancel{\frac{[C]-[C_0]}{c}}&=c\cdot\cancel{\frac{[C]-[C_0]}{c}} \\
\implies 2a&=c
\end{align}
Now, we substitute the values of $a$,$b$ and $c$ in terms of $a$, we get:
$$\ce{\cancel{a} A + \cancel{a} B <=> 2\cancel{a} C}$$
which implies:
$$\ce{A + B <=> 2C}$$
Therefore, your textbook is clearly in the wrong, since it states $\ce{A + B<=>C}$.


What's the mechanism of the reaction of (E)-methyl 3-(furan-2-yl)acrylate in acidic methanol? In the organic chemistry textbook by Clayden et al. they mention that the reaction of (E)-methyl 3-(furan-2-yl)acrylate in acidic methanol yields dimethyl 4-oxoheptanedioate without providing the reaction mechanism and leaving it to the reader to figure it out:

I've tried for over an hour now but didn't get to this product. I assumed that since the reactant already has the right number of carbon atoms I won't have some kind of dimerization to consider, only reactions involving one molecule of (E)-methyl 3-(furan-2-yl)acrylate. I think, the first step should be a protonation at the furan 3- or 5-position with a subsequent attack of methanol on the carbonyl group that was formed followed by an acetal cleavage. The problem is that I don't know how to get the second ester group and how to get rid of all 3 $\ce{C=C}$ double bonds? Unfortunately, I don't have access to ChemDraw on this computer so I can't provide some example mechanisms I came up with.

Unfortunately I cannot agree with the currently accepted answer. This 1,5-hydride shift that was invoked seems to have a very strained transition state (the cyclic, planar system means that the "1" and "5" atoms are quite far apart) and with all due respect, I am quite skeptical as to whether it would really work. Furthermore furan does not enjoy a particularly large aromatic stabilisation and "rearomatisation" as a driving force is not too convincing. Furan, in all its reactions, has a very pronounced tendency to lose its aromaticity. For example even nitration with the mild reagent acetyl nitrate ($\ce{AcONO2}$) leads to a 2,5-addition product and you actually need to add a second reagent (pyridine) to eliminate $\ce{AcOH}$ and rearomatise. For all intents and purposes furan reactivity is much closer to that of an enol rather than that of benzene.

Instead of a 1,5-hydride shift, proton transfers should suffice, especially since there are carbonyl groups all over the molecule (i.e. the proton transfer is simply a tautomerism and doesn't go through any particularly unstable intermediates).

One way of thinking about this question would be to try and do it in reverse. 1,4-Dicarbonyl compounds are known to cyclise to form furans in acidic solution, so this should simply be the microscopic reverse of that reaction. The key cyclisation step involves the attack of an enol oxygen onto a protonated carbonyl group. So, in this reaction, the fragmentation step should be the reverse: the ejection of an enol oxygen from a (hemi)acetal.

I would propose this mechanism instead.

So far, this is the same as in the currently accepted answer. However the usage of the furan lone pair to push electrons saves time in drawing all the resonance forms (this is what was meant by K_P's comment also).

Protonation of the furan oxygen (not drawn) brings us to the start of the next step:

As described previously this fragmentation is the reverse of the cyclisation reaction used to generate furans. Following tautomerisation of two of the enols, most of the molecule is already in place and the remaining question is how to generate the second ester functionality. The issue is that the carbon that should become the ester has an extra hydrogen on it. Instead of having a hydride shift it makes much more sense to move it around by tautomerisation!

The ketene acetal can be protonated and attacked by methanol at the carbonyl carbon, but that would be unproductive as there is no way of getting rid of one of the methyl groups on oxygen (unless we invoke a $\mathrm{S_N2}$ reaction on the methyl group with methanol acting as the nucleophile). So, it probably needs to wait until water is added during the reaction workup, at which point it is hydrolysed to give the desired product.


What does the "3,4" mean in 3,4-methylenedioxymethamphetamine (MDMA)? I'm curious to know, what exactly does the 3,4 part of 3,4-methylenedioxymethamphetamine represent?

Numbers in organic molecules tell you where so-called functional groups (important configurations of elements) are attached to a central structure.  The central structure in MDMA is a benzene ring $\ce{(C6H6)}$, with the main functional group attached at a carbon which makes methamphetamine.  This carbon is labeled 1 as the "most important" functional group.  To get to MDMA from methamphetamine, you add a methelynedioxyl group, and this attaches at carbons 3 and 4.  So that's where those numbers come from!

EDIT: Below is a molecule of MDMA:

The hexagon shape near the center is the benzene ring.  The group to the right, attached to the benzene ring at about 2 o'clock is the main functional group that turns benzene into methamphetamine.  That position on the ring is labeled 1.  Now all other functional groups are numbered sequentially in the direction that makes the next functional group nearest to 1.  So that means we go counterclockwise, and the group on the far left with the 2 O's (oxygens) is attached at positions 3 and 4.


Water contamination of Polyol F0350 I just got asked this question on an interview:
I transported Polyol F0350 in 3 tanks. Upon arrival one tank was determined to have an unacceptable water content. It may have been contaminated with seawater, industrial water or distilled water. What tests do I perform to determine the water source and what type of mitigation do I recommend to be most cost productive - off blending the product or offering it for sale at a reduced price?

Tricky question, but it probably has a trivially simple answer if you have experience with this problem, such as the 'look' of the mixture in the tanker. Not quantitative of course but possibly accurate enough, thus it could be worth asking others with experience for their opinion.

Assuming that water is intimate with the organic, i.e. dissolved or as an emulsion, some water needs to be separated out then there is the Distillation method which uses an immiscible organic with a boiling point higher than water, say xylene. The water is quantitatively carried over with solvent and can then be analysed.

Conductivity would be one way but not conclusive, as both industrial water and sea water may conduct or industrial water may not conduct and so be similar to distilled water. However, the distilled water may also conduct as its not de-ionised water. A simple test unlikely to be definitive definitive, but if its sea water, then a simple flame test should show sodium, as may industrial water, but distilled water far less so. Industrial water may contain soluble organics as well as various inorganic ions so may be coloured, acidic or alkaline and have a distinct smell. The sea water and distilled water have known pH.

Ultimately you will want to do a proper analysis, say GCMS.
Finally, after separation, water could still be present in small amounts in the organic then there is the Turbidity method or the Karl-Fisher titration to determine the amount of water left in the organic layer. However, I suspect you are asked to to a test at the back of the tanker or at least with very simple apparatus.

Mitigation is decided by company policy and this probably depends on what the cost is to correct the problem.

I hope that someone with real experience of this can help, that is when they have stopped laughing at my amateurish ideas.


How does the state of iodine affect its HOMO–LUMO gap? My teacher told me that the HOMO–LUMO gap in solid iodine is smaller than that in gaseous iodine.

When I asked about the reason, he told me that it is due to higher interactions in solid iodine as compared to gaseous iodine, which I did not understand. How can the physical state of a compound affect the HOMO–LUMO gap, even though the chemical formula is the same?

I will broadly mention several mechanisms that may influence color of a solid vs. gas, not just the HOMO-LUMO transition.
Without going into the literature and/or performing (extensive) calculations, I cannot confirm or reject your teacher explanation.

In the solid state, you may find intermolecular MO interactions that will affect orbital energies. Depending on the strength, you may even have to abandon the concept of molecular orbitals in favor of crystal orbitals. Quantum dots and their dependence of color on size are somewhat of an example.
Molecules may alter their shape in solid state compared to gaseous state. This is less of an issue in $\ce{I2}$.
In the solid state, intermolecular excitations are more likely (i.e. going beyond HOMO-LUMO transitions). This is especially true if you do not have a pure substance, think doped solids.
Note that supermicroscopic spatial distribution of a substance will have influence on the color as well: finely distributed silver or mercury (as in calomel) may appear as black rather than shiny.


What types of ultraviolet light will ignite hydrogen? I saw a reaction between hydrogen and chlorine ignite in the presence of UV light. I was wondering if the same thing would happen with something like near-ultraviolet light (around 400 nm). What kinds of UV light (if any) would start a reaction of stoichiometric amounts of hydrogen and oxygen gas in something like a balloon?

Hydrogen/chlorine needs only visible light to ignite. You can use a normal security light to do this because chlorine can be dissociated by visible light, $\lt 490$ nm which is of a bluish/greenish colour. The reaction is a radical chain reaction initiated by chlorine atoms.

A security light is suitable, they are cheap and bright and can produce a lot of radicals and thus an explosion. If you put stoichiometric amounts into a large plastic lemonade bottle with a rubber bung in the top it makes a good demo. But make sure to use a transparent screen between you and any audience as the bottle can sometimes shatter. The cork usually hits the ceiling and the bang is loud :)

If you want to do the same with hydrogen/oxygen, which is also a radical chain reaction, you will definitely need UV light as the dissociation energy of both is now far larger, oxygen $\lt 244$ nm, hydrogen $\lt 276$ nm. It probably won't work so well in a balloon as the rubber will absorb a lot of UV. (Also, UV light of this wavelength is dangerous to the eyes and skin).


Why cannot HCN react with ethene? I expected ethene ($\ce{C2H4}$) to undergo electrophilic substitution due to $\ce{HCN}$ but it doesn't happen, as I am told by my textbook.

I couldn't figure out the reason, as I found out that $\ce{HCN}$ does react with $\ce{CH3CH2Br}$ by nucleophilic substitution.  Upon ionisation, the proton of $\ce{HCN}$ is an electrophile,while the cyanide group is a nucleophile.  Correct?

The "classical" way

Direct addition of $\ce{H-X}$ to alkenes proceeds via the addition of a proton to the $\ce{C=C}$ double bond in the initial step, i.e via a carbocation. This works nicely for strong mineral acids, but not for the weak acid $\ce{HCN}$ with the rather weak nucleophile $\ce{CN-}$.
Synchronous addition?
Q: Can't $\ce{H-CN}$ add synchronously add to $\ce{C=C}$?
A: NO! Such a reaction would involve the interaction of the HOMO (highest occupied molecular orbital) of one reaction partner with the LUMO (lowest unoccupied molecular orbital) of the other. The LUMO of an alkene has a nodal plane orthogonal to the $\ce{C-C}$ bond, while there is - obviously - no nodal plane but a $\sigma$ bond between hydrogen and carbon in $\ce{HCN}$.
The synchronous addition of $\ce{H-CN}$ to $\ce{C=C}$ is symmetry-forbidden!
Catalysis?
Yes! Since the first catalytic hydrocyanation using $\ce{Co2(CO)8}$ (see Philipp's excellent comment above), other catalysts have been developed, namely $\ce{Ni(0)}$ based species using a trypticene-derived chelating phosphine ligand (see image).


Is the Ocean acidification inconsistent with the Bjerrum plot? $\ce{CO2}$ dissolves in water and goes through chemical reactions and exists at different concentrations of,

dissolved $\ce{CO2}$,
carbonic acid,
bicarbonate, and
carbonate.

The Bjerrum plot says that a high concentration of carbonate results in an increased pH (i.e., basic). But many diagrams in the Internet about ocean acidification show that as carbonate is produced, it becomes more acidic (although they also shows as carbonate is produced $\ce{H+}$ is also produced).

Source
I don't know how to reconcile this kind of figure and Bjerrum plot as I feel they are not consistent.

source

I think I see the confusion. You are right that directly adding carbonate, perhaps in the form of $\ce{Na2CO3}$, would increase the pH. However, ocean acidification, as shown in your figure, is from atmospheric $\ce{CO2}$ dissolving in the ocean. The relevant equilibrium: $$\ce{CO2_{(aq)} + H2O <=> H2CO3_{(aq)} <=> HCO3^{-}_{(aq)} +H+_{(aq)}<=>CO_{3(aq)}^{2-} +H+_{(aq)}}$$

The Bjerrum plot is simply a graphical representation of this equilibrium. As you add $\ce{CO2}$, you push the equilibrium towards the right, forming $\ce{H+}$, lowering the pH. Conversely, when you add $\ce{CO3^{2-}}$, the equilibrium move towards the left, consuming some $\ce{H+}$ and raising the pH. Although adding $\ce{CO2}$ leads to an increase in the absolute concentration of carbonate, the ratio of concentrations goes the other way.

You can use this two ways:

If you have a change in pH, you can move along the x-axis to predict how the concentrations of the carbonate species change.
If you change the ratios of the species, e.g. by adding $\ce{CO2}$, you can predict how the pH will change.


Trends of electron affinity Trends for electron affinity going across the group increases negatively.
Does this mean Noble gases' electron negativity is more negative than halogens?
I'm confuse here.
Noble gases should have a positive electron affinity as adding an electron would make it unstable and thus, will not form.
But for halogens, their attraction for that 1 electron to fill up the $\mathrm{p}$ subshell is very strong.
But according to the theory that electron affinity increase negatively across the period, does this theory has an exception of noble gases?
Next, what about $\ce{Be}$ and $\ce{Mg}$? Their electron affinities are not negative.

Not exceptions, but see, theories say that going $L→R$ electron affinity increases up to halogens.
For the noble gases, $\Delta_\mathrm{eg}H$ is positive (energy required). While electron affinity is $0$. Which implies no tendency to lose electrons.
For $\ce{Be}$, the small size and nuclear charge are responsible, which is related to $\ce{Mg}$ due to a diagonal relationship.


Molecular Hamiltonian of Helium hydride ion Currently, I am going through Quantum Simulation of Helium Hydride Cation in a Solid-State Spin Register. I am not a chemist but rather a computer scientist hence having trouble following the paper.

I am using this easy to access note on Hartree-Fock  approximation.

So, I am not sure how the authors created the molecular Hamiltonian in Eq. 2 of the supplementary material.

$$
H = \begin{pmatrix}
-2.85404 & 0.130671 & 0.0 \\
0.130671 & -0.760916 & -0.323568\\
0.0 & -0.323568 & -1.91238
\end{pmatrix}
$$

From the note, I understand that the Born-Oppenheimer approximation (Eq. 3) is the starting point.

$$
H = -\sum^{N}_{i = 1} \frac{1}{2} \nabla^2_i - \sum^{N}_{i = 1} \sum^{M}_{A = 1} \frac{Z_A}{r_i A} + \sum^{N}_{i = 1} \sum^{N}_{j > 1} \frac{1}{r_{ij}}
$$

What should be my following steps to reach at the molecular Hamiltonian.

I am extremely sorry if I am asking for a really long answer.

UPDATE 1:
The molecular Hamiltonian is:

\begin{align}
\hat{H} &= \underbrace{-\frac{\hbar^2}{2 M_{He}} \nabla^2_{He} - \frac{\hbar^2}{2 M_H}}_{\text{Nuclear kinetic energy}} \underbrace{- \frac{\hbar^2}{2 m_e} \nabla_1^2 - \frac{\hbar^2}{2 m_e} \nabla^2_2}_{e^-\text{'s kinetic energy}}
\nonumber\\
& \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{He}}|}
-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 1\text{\textsuperscript{st} } e^- \text{ and nuclei}}
\nonumber\\
& \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{He}}|}
-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 2\text{\textsuperscript{nd} } e^- \text{ and nuclei}}
\nonumber\\
& \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{r_2}|}
 }_{e^-\text{ - }  e^- \text{ repulsion}}
 \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{R_{He}} - \overrightarrow{R_H}|}
 }_{\text{Nuclear - Nuclear repulsion}}
\end{align}

After Born–Oppenheimer approximation:

\begin{align}
\hat{H} &=  \underbrace{- \frac{\hbar^2}{2 m_e} \nabla_1^2 - \frac{\hbar^2}{2 m_e} \nabla^2_2}_{e^-\text{'s kinetic energy}}
\nonumber\\
& \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{He}}|}
-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 1\text{\textsuperscript{st} } e^- \text{ and nuclei}}
\nonumber\\
& \underbrace{-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{He}}|}
-\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_2} - \overrightarrow{R_{H}}|}}_{\text{Attractive potential energy between } 2\text{\textsuperscript{nd} } e^- \text{ and nuclei}}
\nonumber\\
& \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{r_1} - \overrightarrow{r_2}|}
 }_{e^-\text{ - }  e^- \text{ repulsion}}
 \underbrace{+\frac{1}{4 \pi \epsilon_0} \frac{2 e^2}{|\overrightarrow{R_{He}} - \overrightarrow{R_H}|}
 }_{\text{Now a constant since nuclei positions are fixed}}
\end{align}

Your provided note is about the HF Hamiltonian in the one-electron basis set (atomic orbitals). Since a minimal STO-3G basis set is used here, the corresponding matrix representation of this Hamiltonian is a $2\times 2$ matrix.

The $3\times 3$ Hamiltonian from the SI is the FCI Hamiltonian in the many-electron basis set. These basis functions are Configuration State Functions, arising from linear combinations of (excited) Slater determinants (electron configurations), which in turn are built from the HF molecular orbitals.

Configuration Interaction calculations (like FCI) are typically done on top of a HF calculation, to correct for missing electron correlation effects. Your linked note is only about HF. For details about CI, I can recommend the book by Szabo and Ostlund which is also cited in your note.


Reasons for low melting point of gallium J.D. Lee writes in Concise Inorganic Chemistry:

Gallium has an unusual structure. Each atom has one closest neighbor at a distance of 2.43 Å.
This remarkable structure tends towards discrete diatomic molecules rather than a metallic structure. This accounts for the incredibly low melting point of gallium at 30 °C.

That's not all, this surprised me as well:

Note that the boiling point for Ga is in line with the others, whereas its melting point is not. The very low melting point is due to the unusual crystal structure, but the structure no longer exists in the liquid.

Why does gallium exist as diatomic units, that too in a solid crystal, instead of forming the usual metallic crystal structure? And why not in liquid state (where it seems to be more free to make diatomic molecules)?
I might guess that $\ce{Ga2}$ might form stable MOs, but why do the other members of group 13 abstain from such behavior?
It is quite weird to see an element in the middle of a group staging a drama of exceptions like this, it's usually the 2nd period elements which do that.

For your interest, I quote a paragraph from a book "Handbook of Nanophysics: Nanoparticles and Quantum Dots" by Klaus D. Sattler (Ed.). Section 12-8. I don't think people know the reason as to why Ga likes to exist as a dimer (as you know $\ce{S}$ is $\ce{S8}$ and so on).

  The cohesive energy of the elemental solids is defined as the
  difference between the energy per atom of the solid and the energy of
  an individual atom in isolation. However gallium is rather unique
  among metals in that in the solid state it has a very unusual
  structure—an orthorhombic 8-atom unit cell composed of dimers which
  have a bond distance 15% shorter than the next nearest neighbor
  distance. As such, it has been called a “molecular metal,” which
  really contains two distinct kinds of bonding: covalent bonds within
  the dimers which result in a pseudogap at the Fermi level, and
  metallic bonding formed by overlap of the electronic wavefunctions
  along the buckled planes perpendicular to the molecular bond of the
  dimer. As a consequence, conduction in gallium is highly anisotropic,
  with a much greater conductivity along the buckled planes than
  perpendicular to this direction (Gong et al., 1991; Heine, 1968).

  If we estimate the “metallic” cohesive energy of the solid, by
  calculating the cohesive energy relative to the dimer, we immediately
  see that gallium moves much closer to the other elements, as shown in
  Figure 12.6. In fact, in this case we are still overestimating the
  cohesive energy relative to the melting temperature, since the dimers
  in the bulk will be distorted from the free dimer geometry. This
  argument certainly explains the low melting point of gallium relative
  to other metals, and due to the anisotropy of the lattice most likely
  relates to the surface superheating of gallium that was observed very
  early on (Volmer and Schmidt, 1937).


Internal pressure of ideal gas $$\left(\frac{\partial U}{\partial V}\right)_{\!T}$$

I know that this partial derivative is equal to zero for an ideal gas, but how do I determine that? Do I need to use the fundamental thermodynamic relations?

In fact, you need to use two of the fundamental relations.

For an ideal gas, the internal energy $U$ can be considered to be a function of $V$ and $T$. So, the total differential of $U$ would read

$$\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{1}$$

The first partial derivative is the internal pressure, which we are trying to determine.

The other thing that we know is that (assuming only expansion work and no net reaction)

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \tag{2}$$

The idea here is that we want to change the term involving $\mathrm{d}S$ into a combination of $\mathrm{d}T$ and $\mathrm{d}V$, so that we can compare coefficients with equation $(1)$. So, we just express $S$ as a function of $T$ and $V$:

$$\mathrm{d}S = \left(\frac{\partial S}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{3}$$

Substituting $(3)$ into $(2)$

$$\begin{align}
\mathrm{d}U &= T\left[\left(\frac{\partial S}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{\!V}\,\mathrm{d}T\right] - p\,\mathrm{d}V \tag{4} \\
&= \left[T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p\right]\mathrm{d}V + \left[T\left(\frac{\partial S}{\partial T}\right)_{\!V}\right]\mathrm{d}T \tag{5}
\end{align}$$

Comparing the coefficient of the $\mathrm{d}V$ term with $(1)$ we find that1

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p \tag{6}$$

Equation $(6)$ is entirely general and applies to all substances, not just ideal gases. Now, we turn to the Helmholtz free energy:

$$\mathrm{d}A = -S\,\mathrm{d}T - p\,\mathrm{d}V \tag{7}$$

From this we obtain the Maxwell relation

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} = \left(\frac{\partial p}{\partial T}\right)_{\!V} \tag{8}$$

For an ideal gas $p = nRT/V$, so

$$\left(\frac{\partial p}{\partial T}\right)_{\!V} = \frac{nR}{V} \tag{9}$$

Finally, substituting $(8)$ and $(9)$ into $(6)$ gives

$$\begin{align}
\left(\frac{\partial U}{\partial V}\right)_{\!T} &= T\left(\frac{nR}{V}\right) - p \tag{10} \\
&= \frac{nRT}{V} - p \tag{11} \\
&= p - p \tag{12} \\
&= 0 \tag{13}
\end{align}$$

as desired.

Notes

1 Some texts may write that you can obtain $(6)$ from $(2)$ simply by "dividing" by $\mathrm{d}V$ and imposing conditions of constant temperature:

$$\begin{align}
\mathrm{d}U &= T\,\mathrm{d}S - p\,\mathrm{d}V & & \tag{2} \\
\frac{\mathrm{d}U}{\mathrm{d}V} &= T\frac{\mathrm{d}S}{\mathrm{d}V} - p\frac{\mathrm{d}V}{\mathrm{d}V} & &\text{(dividing by }\mathrm{d}V\text{)} \tag{14} \\
\left(\frac{\partial U}{\partial V}\right)_{\!T} &= T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p\left(\frac{\partial V}{\partial V}\right)_{\!T} & &\text{(constant }T\text{)} \tag{15} \\
&= T\left(\frac{\partial S}{\partial V}\right)_{\!T} - p & &\text{(derivative of }V\text{ with respect to itself is }1\text{)} \tag{6}
\end{align}$$

This gives the correct result, and in general this "shortcut" works, but it is mathematically not rigorous.


Why does the solubility of some salts decrease with temperature? It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy. Thus the molecules move randomly and having greater distance from each other which is responsible for the large voids between them and it gives more space for the solute molecules between them to come in. However, there a few salts  like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ?

A saturated solution is at equilibrium (rate of dissolution is equal to rate of crystallization) with some equilibrium constant $K_1$. If you change the temperature of the system at equilibrium, you will observe a different equilibrium constant $K_2$. Whether the equilibrium constant increase or decreases is described by the Van't Hoff equation:
$$\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

So it depends on the sign of the enthalpy of reaction, in this case the enthalpy of dissolution.

  It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy.

The statement about solubility is not always true, and the explanation leaves out something. A lot of things change when you increase the temperature. The solvent gains kinetic energy, the solute gains kinetic energy, and the solid gains kinetic energy. How this influences the solubility depends on the specific system, and is hard to predict.

  However, there a few salts like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ?

Now that we introduced the Van't Hoff equation, we know it must have to do with the enthalpy of dissolution. Usually, you expect the interactions in the solid that need to be broken to be stronger than the gains from solvating the solute. In these cases, it is apparently the opposite.


Can submicron mould particles go through building structures, non permeable foil paper? I read that submicron mould particles and vapours related to mould can go through walls, floors and ceilings?  Is this correct?

You are asking two different questions

  I read that submicron mould particles can go through walls, floors and
  ceilings. How so aren't these structures meant to be air tight?

Not true. Mould particles can not seep through concrete walls or thick wood, even though they are porous structures. The interstitial pore size is too small and tortuous to let any mold particle pass through it randomly.

  What about the lowest thickness of foil paper which is meant to be non
  permeable, can submicron mould particles go through that?

No, an aluminum foil is impervious to solids no matter what their size is. Yes, probably electrons and X-rays will pass through but nothing particulate.


Baking soda & magnesium chloride reactions Woke to this interesting reaction.
Would love any explanation of this occurrence. If it is perfectly common then I apologize for my ignorance. The mag was trace in the clear glass suspended over the white glass jar that had a strong mixture of baking soda and water. They are not touching and are as I found them. Look carefully and you can see the "gel puddle" that formed on my dish drainer. There are slight bubbles in the gel.
Or is it ectoplasm from the ghost?  (Lol)

It is difficult to reconstruct what actually happened based on a single picture.

Nevertheless, the essential effects seem to be the hygroscopy and deliquescence of magnesium chloride. Magnesium chloride has a strong affinity for moisture. When exposed to the atmosphere, the salt will absorb relatively large amounts of water from the air. Magnesium chloride can form several crystalline hydrates: $\ce{MgCl2.2H2O}$, $\ce{MgCl2.4H2O}$, $\ce{MgCl2.6H2O}$, $\ce{MgCl2.8H2O}$, and $\ce{MgCl2.12H2O}$. Furthermore, the salt absorbs moisture from the atmosphere until it dissolves in the absorbed water and forms a concentrated liquid solution.

Apparently, the concentrated liquid solution of magnesium chloride ran out of the glass vessel and formed a puddle on the dish drainer.


Most endothermic room-temperature reaction In terms of energy per gram of reactants, the evaporation of water at -2.2 kJ/g, is much higher than any chemical endothermic reaction that I have seen demonstrated. At extreme temperatures there is no real limit, but for reactions that occur at a reasonable rate at room-temperature it seems water is hard to beat. And it isn't even a chemical reaction.

I have reason to think water is hard to beat: the entropy of gases is much higher than liquids. The strong hydrogen bond network of water makes the liquid phase lower entropy than non-polar liquids because it structures the liquid. Finally, water is a very light molecule so you get many moles of gas per kg of liquid. Despite this huge entropy gain, the phase-transition proceeds sluggishly because it takes so much dry air to shift the equilibrium forward, if it took much more energy it wouldn't happen at a meaningful rate.

Can any chemical reaction approach this at room temperature? If so, can any non-gas-evolving reaction come close?

Don't know if these are quite in the spirit of what you are looking for since, with the exception of the first reaction, the entropy change alone is not sufficient to drive them  (i.e., at room temp., their equilibria lie far to the left) (and, for ozone formation, the entropy change is actually unfavorable).  Furthermore, ozone formation and photosynthesis aren't single reactions, but multi-step processes.  But they all do take place at room temperature (or below):

Sublimation of ice (yes, snow can evaporate!):  2.8 kJ/g

Formation of ozone (O$_3$) from O$_2$ (photochemical, occurs in stratosphere): 3.0 kJ/g

Photosynthesis: 7.5 kJ/g

Electrolysis of water: 15.9 kJ/g

The values I've listed are all enthalpy changes (at constant pressure, the heat flow is equal to the change in enthalpy; at constant volume, the heat flow equals the change in energy).  In addition, by convention, the sign of the enthlapy for endothermic reactions is positive, since heat flows into the system, so the enthalpy is increased (not negative, as you've used).


Does pyrite retain the magnetic properties of its element iron? You are walking in the mountains and pick up what looks like a nugget of gold. You're aware that fools gold (iron pyrite, $\ce{FeS2}$) exists. Create an experiment to test if this is pure gold or iron pyrite.

The simplest thing I could think of is to do a magnetic test on the assumption that if it was fools gold it would be attracted to a magnet due to the presence of iron. A quick google search reveals this to be true but I have a sneaky suspicion that I got lucky in assuming, because there is iron in $\ce{FeS2}$ the compound will be magnetic.

I'd like to know if chemical compounds retain the properties of their individual elements or if the compound gains completely new properties some of which may or may not be properties of the elements.

Pyrite is not magnetic. Some related minerals are, but only weakly, not as strong as iron, so the test will likely fail anyway. Magnetism is not an inherent property of an element, but that of a compound. Most compounds of iron are not magnetic (though some are); on the other hand, there are organic magnets made out of totally non-magnetic elements.

Think of something based on chemical properties instead. Pyrite is a salt of a metal and a weak acid. Gold is a metal, of the kind that would not react with acids. See where this gets you.

As for the other properties of the elements, some would be inherited by their compounds and some (arguably, most of them) would not.


What is continuum in line spectra? I read that continuum in line spectra is noticed beyond a certain limit. What does it mean?

The four lines in the visible spectrum (designated by α through δ) were the first observed by Balmer. Notice how the lines crowd together as they approach the ionization limit in the near-ultraviolet part of the spectrum. Once the electron has left the atom, it is in an unbound state and its energy is no longer quantized.
When such electrons return to the atom, they possess random amounts of kinetic energies over and above the binding energy. This reveals itself as the radiation at the short-wavelength end of the spectrum known as the continuum radiation.

Source: Chem1


Why does hand sanitizer turn into a liquid after coming out of the bottle? Hand sanitizer is a gel in its bottle, but when I pour it on my palm, it turns into a free flowing liquid. Why does that happen?

It doesn't turn to a liquid when I pour it on my floor or my table. It doesn't seem to be related to heat, as I tried it also with a hot surface.

A little research and perhaps a small experiment might help you figure something like this out on your own. First, have you tried wikipedia? A small mention of research can go a long way. Describing what product you are interested in never hurts either. Is the polyacrylic acid, a dispersant in this substance? That material can absorb water effectively, so here is what I would do:

Take a very small amount of water and smear it on the same surface that it does not "turn into a free flowing liquid" and see if it does with the addition of water. If it works, you might conclude that your alcohol-based rubs use a suspension material that when contacted with a little water, such as from your palms, will cause the gel to have just enough water absorbed inside to decrease its viscosity.

Perhaps you will like to share the results of your experiment.

EDIT: OP seems to be observing a gel demonstrating Thixotropy, this I have determined from a quick google search


Can Oxybuprocaine hydrochloride react with lidocaine hydrochloride? To anaesthetise the mucous membrane 2 separate drugs were given topically:

Oxybuprocaine hydrochloride and Lidocaine hydrochloride. The reason to use both of them seem to be beginning of action and its duration (the former starts faster but shorter, the latter starts later but longer acting).

If instead having them in separate bottles they will be mixed in one, would it cause any interactions between them?

oxybuprocaine:

Looking at the functionality, oxybuprocaine looks more likely to react with itself rather than lidocaine. As it's a salt and also no doubt from numerous medical trials, this doesn't occur,  the anticipation would be that the two compounds would be fine.  The $pK_a$'s of the tertiary amines would also be expected to be similar, so one shouldn't neutralise the other.


liquid crystal and light The liquid crystal can be control using polarizes light. This means we can control the arrangement of the liquid crystal molecules, then when the light passes through it can be manipulated.
But why exactly is this possible? What is the relationship between liquid crystal material and light pass through it? How do they interact and affect each other?

Liquid crystals are controlled by electric fields and their structure manipulates the polarisation of light
Liquid Crystals are usually made from long rod-shaped molecules which behave in some ways like liquids (they flow) but in other ways like crystals (they form structures with a lot of regularity). The name is appropriate.
The key features that matter are that the structures they form are easy to manipulate with electrical fields and that the rod-like nature of the structure interacts with light in roughly the same way as polaroid: it selectively alters the polarisation of the light passing through.
In LCD devices the liquid crystals are placed between two polaroid filters and in front of a backlight so the the default orientation of the filters plus the liquid crystal layer lets most light through. When a small electric field is applied across the liquid crystal it can be arranged so that it changes the orientation of the molecules in the layer and twists the light so it no longer passes through the polaroid layers. This can be arranged to happen one pixel at a time with suitable electronics. So an electrical field can drive a display.
That's the basic principle. Real displays have evolved to have some very complex arrangements of these effects to improve speed and the twistiness of the LC layer to make faster responding and higher contrast pixels.


Which formula to use to find the pH of a solution? I just had my first exam yesterday, and I'm always tripped with this question that combines neutralization with buffer solution. The task is to find the pH. I did try to understand the general concept behind it. From my understanding, if $\ce{CH3COOH}$ reacts with $\ce{NaOH}$ and the acid has a higher amount of substance ($\pu{0,005 mol}$ vs $\pu{0,004 mol}$), it means that $\pu{0,005 mol}$ of acid is used to neutralize $\pu{0,004 mol}~\ce{NaOH}$, and $\pu{0,001 mol}$ of $\ce{CH3COOH}$ will remain together with $\pu{0,004}~\ce{CH3COONA}$.

To find the pH, one has to use the Henderson-Hasselbach equation and divide the amount of substance with the volume, with $\mathrm{pH} = \mathrm{p}K_\mathrm{a} - (\log c(\ce{CH3COOH})/c(\ce{CH3COONa})$. And then if $\ce{NaOH}$ has a higher amount of substance ($\pu{0,0051 mol}$ vs $\pu{0,005 mol}$ acid), it would create $\pu{0,0001}~\ce{NaOH}$ and $\pu{0,005}~\ce{CH3COONa}$.

So to find the pH, you need to find the pOH of $\ce{NaOH}$ first. So far so good. But in the solution to the problems posted, there's suddenly an acid popping up in the solution and you have to use Henderson/Hasselbach equation. For example $\ce{NH3}$ reacts with $\ce{HCl}$ and creates $\ce{NH4Cl}$, even if you have a bigger amount of substance for $\ce{NH3}$, you still get an acid on the result... I'm so confused. And then there's an example with the coefficient ($\ce{Na2CO3 + 2HCl -> 2NaCL + H2CO3}$). If $\ce{Na2CO3}$ has a higher amount of substance (let's say $\pu{0,224 mol}$) and $\ce{HCl}$ only $\pu{0,0816 mol}$, does it mean that it would result in $0,224 \times 2 - 0,0816 \pu{mol}~\ce{NaCl}$ and $\pu{0,0816 mol}~\ce{H2CO3}$? I feel so stupid now that I can't understand this simple concept and keep making mistakes (and wasted like 8 points during the exam).

From reading your question it is obvious that you are confused. When I studied chemistry in college in the 1970's I don't remember the Henderson-Hasselbach equation being a famous equation. It seems like it is now being used as some magic can opener.

You need to go back to the chemistry and figure out the chemistry first, then do the math.

Ok, let's start by noting that the pKa of acetic acid is 4.76 from Wikipedia. There are two significant figures in the mantissa so a mathematical solution good to 1% will be adequate. Also we are dealing with solutions on the order of $10^{-3}$ molar. This allows us to make two very important assumptions which significantly simply the mathematics.

First we assume that we can ignore the autoionization of water.
Second we assume that since acetic acid is a weak acid the solution ends up being essentially 0.004 molar sodium acetate and 0.001 molar acetic acid.

Because we have excess acetic acid we know that the solution will be acidic. If we wanted a more exact solution we'd ignore the second assumption and use the $K_\mathrm{a}$ expression

$K_\mathrm{a} = \dfrac{\ce{[H^+][Ac^-]}}{\ce{[HAc]}}$

to solve the problem more exactly. Letting $\ce{[H^+]} = x$ then $\ce{[Ac^-]} = 0.004 + x$ and $\ce{[HAc^-]} = 0.001 - x$. This requires solving a quadratic equation which is doable, but mathematically messier.

Now back to using our two assumptions we can go to the Henderson–Hasselbalch equation and solve the problem with a simple formula.

$\mathrm{pH} = \mathrm{pKa} + \log{\dfrac{\ce{[Ac^-]}}{\ce{[HAc]}}} = 4.76 +  \log{\dfrac{0.004}{0.001}} = 5.36$

This means that the $\ce{[H^+]} = 4.4 \times 10^{-6}$.

So our assumption that the final solution ends up being essentially 0.004 molar sodium acetate and 0.001 molar acetic acid is reasonable.
The assumption that the autocatalysis of water can be ignored is a bit dicier.


Dissociation equilibria of ammonium bifluoride in water Ammonium bifluoride or ammonium hydrogen fluoride is a salt of a weak base and a weak acid.
The weak acid is because the second equilibria of $\ce{HF}$ written as:
$$\ce{HF + F- <=> HF2-}$$
and it has constant of $\pu{3.963 M}$.
But does the bifluoride anion further dissociate?
$\ce{HF}$ in water has many species: $\ce{H+},$ $\ce{F-},$ $\ce{HF2-},$ $\ce{H2F2}$ and $\ce{(HF)_nF-}.$ The $\ce{NH4HF2}$ has a more acidic $\mathrm{pH}$ than a solution of $\ce{NH4F}$.

You need to write out all the possible equilibria just like you do in the case of a multiprotic acid like Phosphoric acid. You will also need to have a pKa for every equation. The bifluoride ion will obviously dissociate in water


Which is more acidic? Which among these hydrocarbons is more acidic?

Here, I thought about hyperconjugation. For this, I counted the alpha H i.e. the number of hydrogens attached to the carbon attached to a sp2 hybridized carbon. What I get is, (C) has most no. of those Hydrogens. So, my assumption is that  it should be more acidic. But that's not the case. Where am I going wrong?

Acidity is directly proportional to the stability of conjugate base. Aromatic compounds are very stable.

In essence only one compound after deprotonating is aromatic - which is the conjugate base of (A).


What explains the case hardening color effect? Year 11 chemistry student here.
I'm curious about how the case hardening effect works on metals.
I know that when the electrons gain an energy 'level' or lose an energy 'level' they emit light at differing frequencies.
So why does the outside of metal, after being heated, stay those pretty rainbow colors after they cool down?

Case hardening (carburizing or induction or flame) does not color steel. You are seeing “temper” colors which are unrelated to case hardening. It is caused by different thickness of very thin layers of iron oxide. These colors were developed during tempering heat-treatment after case hardening. Starting with yellow, then blue and purple in a continuous transition. Faint yellow starts about 300 °F, longer times give more color development of color.
The same colors also develop on stainless steels and titanium at different temperatures and times. A stainless motorcycle exhaust pipe will show the range of colors as temperature changes along its length. If a titanium weld shows yellow color it has been exposed to air at elevated temperature (and is likely defective because of absorbed oxygen and nitrogen).


How to separate two diastereomeric amines? I am developing a method to synthesize amines using organometallic catalysts and my reaction yield these two diastereomeric amines ($\ce{R2N}$ is, in this case, a morpholine group).

With both being formed in approximate equal amounts, making it almost impossible to identify their structures by NMR without separating them. All the NMR spectra look like a mess since some signals get duplicated and some not. I have a hint that they are actually these two amines because they show the same molecular ion in the MS spectrum and the methodology used for synthesis offers little space for doubt, since it always forms the predicted structure.

I would like to know, if there is a practical way of separating those two compounds without using something super fancy like chiral column chromatography decorated with gold and antibodies kind of stuff. I have tried column chromatography but couldn't find a solvent system that could separate them. I also tried recrystallization (though I don't know exactly why I tried it), it also didn't work.

When you mentioned column chromatography I had to assume you meant flash chromatography or biotage, which by implication is normal phase. When people mention HPLC they are normally defaulting to reverse phase. Now, if you can get reverse phase TLC plates (C18 are available) , you could make some advances.  The issue though is then getting the silica and you also mention large quantities.

Large lab scale to me is say, 50g plus. For this, I would advocate crystallisation,  but to guarantee some differentiation,  I would make salts with organic acids. Nitrobenzenesulfonate salts are renowned for giving well defined crystals.

If you needed something to better differentiate again the two amines, I would try a camphorsulfonate salt (pick an enantiomer).  These tend not to be too expensive.  The salts then have different solubilities and the right solvent becomes a screening exercise. Otherwise, tartaric acid is very cheap and easy to salt break after crystallisation.


How to rationalise the increasing bond length order in the carbonate ion, carbon monoxide, and carbon dioxide? I am unable to rationalise the order of increasing bond length in $\ce{CO3^2-}$, $\ce{CO}$ and $\ce{CO2}$.
Having gone through the factors affecting bond length in two different books, my approach to the problem was:

$\ce{CO}$ has bond order 3 while carbonate and $\ce{CO2}$ have bond order 4 so
$\ce{CO}$ should have longest bond length.

$\ce{CO2}$ is sp hybridized while $\ce{CO3^2-}$ is sp2 hybridized.
Since carbonate has lesser s character and therefore it should have a greater
bond length than $\ce{CO2}$

Thus,
  $$\ce{CO2  <  CO3^2- < CO}$$

However, I am unsure about my method. What would be the best approach to determine the relative bond length order?

Actually, in regard to bond length, you should be looking at the bond order of a single bond within the molecule. The bond order is 2, not 4, for $\ce{CO2}$, and the bond order for the carbonate ion is somewhere between 1 and 2 due to resonance. Thus, the carbonate ion has the longest bond length, followed by carbon dioxide, and finally carbon monoxide.


How is the Hessian evaluated? I was reading about the various ways to optimise the geometry of the system, and found that none of the sources mention how the Hessian matrix is obtained.

How does one obtain the Hessian matrix? I’m thinking it’s either some analytical formula or just numerical differentiation.

The documentation for the Gaussian software is actually pretty complete, even if qualitative. On the subject of optimizations, I recommend you look up the freq and opt keywords.

The manual explains the procedure followed by the Berny optimization as involving construction of an initial analytical estimate of the Hessian using a simplified force field, with subsequent optimization steps improving on that estimate using computed gradients. A number of options are provided for improving the initial estimate of the force constants where the defaults derived from the estimated Hessian are not good enough.

For frequency calculations, accuracy requires a good estimate of the full Hessian, better than estimates that might suffice for geometry optimizations. Gaussian allows for analytical computation of force constants for some of the methods (e.g. RHF) and uses analytical gradients to compute a numerical Hessian with some other methods, or if requested. In optimization cases where convergence is an issue, Gaussian facilitates convergence to a real minimum by employing the Opt=CalcAll keyword, which results in analytical (where possible) computation of the full Hessian (at some cost though).


How to get specific concentration of carbonic acid in solution? I wanted to know if there is a way of getting specific quantities of carbonic acid? Carbonic acid is water + $\ce{CO2}$ but not all the $\ce{CO2}$ dissolves in water. So if I wanted $5\:\mathrm{g}$ of carbonic acid solution, how much $\ce{CO2}$ should I add to water? Is there a way to know how much has been added?

To get a specific amount of $\ce{H2CO3}$, you should monitor and adjust the $\mathrm{pH}$ of the water to the $\mathrm{pH}$ that would correspond to $5\:\mathrm{g}$ of $\ce{H2CO3}$. For this you would need to know the acid dissociation constant, $K_\mathrm{a}(\ce{H2CO3})=1.738 \times 10^{-4}$, and the following equations:
$$\ce{CO2(g) +H2O(l) ->H2CO3(aq)}$$

$$\ce{H2CO3(aq) +H2O(l) -> HCO3- (aq) +H3O+(aq)}$$

So now we can calculate, using:

$$K_\mathrm{a}=\frac{[\ce{HCO3-}]\cdot [\ce{H3O+}]}{[\ce{H2CO3}]\cdot [\ce{H2O}]}$$

$[\ce{H2O}]$ is omitted, as it remains mostly constant ($55.5084\ \mathrm{M}$). The only time it needs to be considered is when working with extremely concentrated solutions. From this we now get:

$$K_\mathrm{a}=\frac{[\ce{HCO3-}]\cdot [\ce{H3O+}]}{[\ce{H2CO3}]}$$

Disregarding any autoprotolysis, the $[\ce{H3O+}]=0$ before any $\ce{CO2}$ is added and any $\ce{H2CO3}$ is formed. Therefore, we know that all $\ce{H3O+}$ comes from the carbonic acid, and that $[\ce{H3O+}]=[\ce{HCO3-}]$. Therefore:

$$[\ce{H2CO3}]=\frac{[\ce{H3O+}]^2}{1.738 \times 10^{-4}}$$

Here, we can greatly simplify this equation, and introduce a parameter that we can actually measure--$\mathrm{pH}$. We know:

$$\mathrm{pH}=-\log[\ce{H3O+}] \Rightarrow [\ce{H3O+}]=10^{-\mathrm{pH}}$$

Substituting, we get:

$$[\ce{H2CO3}]=\frac{(10^{-\mathrm{pH}})^2}{1.738 \times 10^{-4}}$$

This, however, only gives you the amount of $\ce{H2CO3}$ that is not dissociated into $\ce{H+}$ and $\ce{HCO3-}$. If we want to calculate exactly how much $\ce{H2CO3}$ is in solution, even in the dissociated form, we must add to it the amount that is dissociated, which as you may recall is $[\ce{H3O+}]=10^{-\mathrm{pH}}$:

$$\text{Total}\ \ce{[H2CO3]}=\frac{(10^{-\mathrm{pH}})^2}{1.738 \times 10^{-4}}+10^{-\mathrm{pH}}$$

From this you can calculate the mass of the $\ce{H2CO3}$ in solution by multiplying this value by the volume of solution present.


How is Zn not a transition metal? A transition metal can be defined as an element that possesses an incomplete sub-level in one or more of its oxidization states. In the textbook I'm reading, it claims that zinc is not a transition metal because it has a full $d$-sub-level in all its oxidization states.

A quick google reveals that zinc has oxidization states $-2, 0, +1$, which means that zinc(with oxidization number $+1$) has an incomplete d-sub-level and is a  transition metal.

What's going on here? Is my textbook incorrect?

Zinc in the +1 oxidation state is $\text{[Ar]}3d^{10}4s^1$, and even in its highest, most common known oxidation state +2 (which the quoted values above seem to have forgotten) it's still $\text{[Ar]}3d^{10}$. No known zinc species in what we normally consider the realm of chemistry breaks that complete $3d^{10}$ subshell, and we would need a major revamp of our calculations and models if any ever does.  Moreover, the thermophysical properties of zinc also betray a loss of transition-metal character.  Zinc is just not a transition element.


Why does a molecular sieve work for gas separation? I was looking at membrane separation of air to nitrogen / oxygen and saw some details on the molecular sieve which on Wikipedia states you can get much higher purity from an adsorber / sieve, but when I look at these images, it doesn't make any sense.

Molecular Sieve material / Zeolite seems to be supplied in granules that are fairly large, why doesn't the compressed gas just go around all the material instead of going through it?

Without melting it / packing it into a solid column, it doesn't seem like it would be much more efficient than just glass beads since there are so many ways to get around each piece. Are you really supposed to just dump it in and have it work?

http://www.instructables.com/id/Nitrogen-gas-generator/

I have seen finer zeolites but this one does not seem as coarse as you sounded it. Consider it again, likewise in fractional distillation, there will be some molecules with never being an equilibrium with another phase. However, much more of them will go through zeolites and with the passage of each unit distance and time, you will have more of the compound you want to get and less of the redundant one. If your column is enough long and zeolites are sufficiently fine and ordered, it will be a good purifier.

Try to visualize it in more detail, while some molecules try to pass through the cavities between zeolites, other molecules will be forced to collide and travel through directly zeolites.


Does any chirality-dependent inorganic chemical reaction exist? In the particle physics, there are non mirror-symmetric reactions.

In the organic chemistry, particularly in complex proteins, there are reactions producing only a single chirality of molecules.

But what is in the intermediate regime? Does any inorganic chemical reaction exist with an asymmetric chirality behavior?

I suspect, maybe it would be possible with highly asymmetric electron clouds, probably with a significant EM octopole moment.

A crystalline form of silicon dioxide known as α-quartz is optically active and can spontaneously crystallize in one of the two optically active form. However, which one of the two you get is defined purely by chance.

This, however, is different from "asymmetry" in particle physics, see here for details.


Why do spectra of elements never overlap? I understand the concept behind how spectra is formed, through exciting the electron beyond it's ground state, however my question is why do the spectra of the elements never overlap?

From my understanding, the wavelengths of light produced are decided by the distance the electron travels when falling back down.

However, what is the factor that guarantees that all of these spectra will be unique? I think part of my confusion also stems from the fact I do not know which electrons "jump" - is it just the valence ones or also the ones in the middle? I thought that perhaps different energy levels have different distances but this sounded a bit funny to me.

My guess would be that every element is different. Going from the basic shell model of the atom to orbitals and transition probabilities there are huge lists for all different transitions. I think there is a green emission for boron where a $11f$-orbital or so is populated and relaxes to give off a green light. And this is so unlikely and weak that you don't even see this on a regular spectrum. So this may be one hint, there are so many possibilities. On the other side if we agreed on one electron, let's say the $1s$-electron, because this is present in all atoms, just exciting that would require different energies for all atoms because the number of protons increase and therefore the positive charge pulling it closer and strong increases as well. It may happen that one transition of element A and a completely different one  on element B have the exact same wavelenght, but the spectrum has so many possiblities and there are only very few elements in the PSE for comparison that I doubt that there will be two identical ones. And even if this happens you can strech that spectrum to the UV, X-ray region as well. What is more likely to happen is that the characteristic emission of an element, let's say the one you observe strongest may be similar, like for Lithium, Calcium and Strontium. I doubt that I'd be able to distinguish them by their red color, but then I'd just take a look at the rest of the spectrum.

But I once had a related case. I was preparing a Dysprosium compound and used a sodium based flux. After solving the structure we had the idea that traces of sodium might occupy long anionic-channel in the compound. So we analyzed the elements using energy-dispersive x-ray spectroscopy. And in the spectrum I got were the characteristic Dysprosium emission line and the one from Sodium pretty much on the same spot. So we had to do a wavelenght-dispersive test as well and then I was able to confirm the sodium. This was a case where the strong x-ray emission line of Sodium almost identical to the one of Dysprosium. But in regular spectra you have much more intense emissions so I guess the probability will be really small.


Relation between chemical kinetics and chemical equilibrium In my chemistry book, the law of chemical equilibrium is derived from the law of mass action:

For a reversible chemical reaction $$\ce{aA +bB\rightleftharpoons cC + dD}$$ where $a$, $b$, $c$ and $d$ are stoichiometric coefficients in the balanced equation, we have, from the law of mass action, the rate of the forward reaction $$r_\mathrm{f} = k_1 \ce{[A]^a[B]^b}\label{1}\tag{1}$$ where $[\ce{A}]$ represents the concentration of $\ce{A}$ and so on. The rate of the backward reaction is $$r_\mathrm{b} = k_2\ce{[C]^c[D]^d} \label{2}\tag{2}$$
At equilibrium, $r_\mathrm{f} = r_\mathrm{b}$, so, we have $$\frac{k_1}{k_2} = \frac{\ce{[C]^c[D]^d}}{\ce{[A]^a[B]^b}}$$ Replacing $\displaystyle \frac{k_1}{k_2}$ by $K_\mathrm{c}$, we have
$$K_\mathrm{c} = \ce{\frac{[C]^c[D]^d}{[A]^a[B]^b}}\tag{3}\label{3}$$
which is the law of chemical equlibrium. $K_\mathrm{c}$ is supposed to be constant, for any concentration of the reactants, at a particular temperature.

However, in chemical kinetics, they tell us that the rate equation, derived from the law of mass action is not always correct. That is, the exponents of the concentration terms in the rate equations may not be equal to their stoichiometric coefficients. If this is true then, the above two rate equations ($\eqref{1}$ and $\eqref{2}$) may not be correct for a particular reversible reaction.
Then, how can we apply the law of chemical equilibrium on every reaction, without experimentally verifying the rate equations? If the rate equations derived from the law of mass action are wrong, then the value of $K_\mathrm{c}$ obtained by the final equation will not be constant.

The derivation that you cite only holds true if both forward and backward reactions are elementary steps. In this case, the rate of the forward reaction is indeed given by

$$r = k[\ce{A}]^a[\ce{B}]^b$$

and so on, and hence the derivation is logically sound. If they are not elementary steps, then even though the conclusion is correct, the method of arriving there is not, for the reasons you have already pointed out. See also: hBy2Py's answer and Ben Norris's answer where the assumption of an elementary step is explicitly mentioned.

In general the equilibrium constant does not need to be derived in this manner; a more conventional method of arriving at it is by using (pressure here is taken to equal an arbitrary but specified standard pressure $p^\circ$)

$$\mu_i = \mu_i^\circ + RT\ln a_i$$

where $\mu_i$ is the chemical potential of species $i$, $\mu_i^\circ$ is the standard chemical potential of species $i$, and $a_i$ is the chemical activity of species $i$. We now have, for a reaction

$$\ce{a A + b B -> c C + d D},$$

the following equation:

$$\begin{align}
\Delta_\mathrm r G^\circ &= c\mu_\ce{C}^\circ + d\mu_\ce{D}^\circ - a\mu_\ce{A}^\circ - b\mu_\ce{B}^\circ \\
&= c(\mu_\ce{C} - RT\ln a_\ce{C}) + d(\mu_\ce{D} - RT\ln a_\ce{D}) - a\mu_\ce{A} - RT\ln a_\ce{A} - b(\mu_\ce{B} - RT\ln a_\ce{B}) \\
&= c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B} - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right] \\
&= \Delta_\mathrm r G - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right]
\end{align}$$

However, at equilibrium, $\Delta_\mathrm r G = 0$ and hence

$$\Delta_\mathrm r G^\circ = - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right]$$

The term in square brackets can then be identified as the equilibrium constant, $K$. As you can see there is no need for any recourse to kinetic rate laws.

Why is $K$ a constant? Well, the standard chemical potentials $\mu_i^\circ$ are thermodynamic constants, just like standard enthalpies of formation, standard entropies, etc. So $\Delta_\mathrm r G^\circ$ - a sum of constants - is itself a constant, and it follows that $K = \exp(-\Delta_\mathrm r G^\circ/RT)$ must be a constant.


Functional Groups Identification Identify functional groups of the following organic molecule:

A. ketone, alkene, carboxylic acid, ester
B. alkyne, ester, carboxylic acid, aldehyde
C. carboxylic acid, alkene, ketone, ester
D. ester, aldehyde, carboxylic acid, alkene

I am not sure what the answer it. I know it can't be B because there is no alkyne triple bond. However, shouldn't an alkene only exist in a straight chain? I know there is a COOH carboxylic acid, R-O-R' ether and a CHO aldehyde, and a double bond O for a ketone.
I was wondering if someone could tell the right answer.

The only right choice is D, because there is an ester, even though this ester is intramolecular, being made of the bridge -CO-O- in the lower part of the molecule. The aldehyde group is -CHO in the R.H.S. The carboxylic acid is in the upper part of the middle. And there are two alkene groups in the cycle at L.H.S.


Why are multiple oxidation states useful for a catalyst? Transition elements are good catalysts because they have multiple oxidation states?Why multiple oxidation states matter?

The answer has a lot to do with "what is a catalyst in the first place?"
A catalyst is a compound that can affect the rate of a chemical reaction
by providing an alternative and lower energy profile or pathway. That is,
it is a substance that makes it so that A -> B at a lower energy cost. But
it only has an effect on the rate of the reaction. That is, it only changes
the cost of the activation energy. It is not related to the thermodynamics
of the process and hence, the final product distribution. That is, if the
above process gives 90% B via a non-catalyzed pathway in 4 days, then a
catalyst will help you get 90% B in say 4 hours.

Okay, so a catalyst affects the transition state and activation path. How
does it do this? Typically, by complexing one of the reagents. Complexation
by transition metals affords access to a wide variety of oxidation states
for the metal. This has the property of providing electrons or withdrawing
electrons from the transition state of the reaction. That is, if the
transition state is electron rich, then the transition metal might hold
some of that electron density and those prevent too much from building up
on the reagent. This would then facilitate the reaction. Or the transition
metal might undergo formal oxidation/reduction to achieve electron transfer
to a substrate, thereby allowing a reaction to occur. This is "complexation
and electron storage" taken to the extreme but is a common mechanism in
organometallic chemistry. Indeed, a variety of catalytic pathways rely on
a two electron transfer between the metal and the substrate (e.g.
hydroformylation). It is the ability of the transition metal to be in a
variety of oxidation states, to undergo facile transitions between these
oxidation states, to coordinate to a substrate, and to be a good
source/sink for electrons that makes transition metals such good catalysts.

Of course, this is for transition metal complexes. Most industrial used
catalysts are the transition metal in a bed, as a metal or bound structure.
The above considerations are important but also the physical properties of
absorption/adsorption and the electron band structure of the material.


Protons and osmotic potential In a hypothetical enclosed membrane only permeable to water (including hydronium), where inside the pH is lower than the outside, would water move inside?

No, because osmosis is not an electrical phenomenon, but rather a chemical one. The concentration of protons would diffuse across the membrane if the fluid, I'm assuming which is water, will be allowed to interact with the other side of the membrane. However, water will not seek to increase the pH of an acidic solution within a membrane the same way it will to dilute concentrations of ions. The resulting balance of pH across the membrane will be as a result of the water to water interaction across the membrane and not because of the pH being imbalanced.


Absolute limits on solubility of immiscible solutes? I know that for a particular, immiscible solute, there is a limit on how much of it can be dissolved in a solvent at a given temperature. However, I was wondering if there is only so much "total solute" that a given solvent can dissolve. As a simple example, suppose you saturate water with salt, then start dissolving sugar into it. Does the fact that it is saturated with salt limit how much sugar it can dissolve? My mental image is one of each water molecule being able to suport only so many solvation structures, regardless of the chemicals participating in the structure.

If you speak about solubility, you are out of domain of simple approximations and the only correct answer is the experiment. For the example given, see http://dx.doi.org/10.1111/j.1365-2621.1992.tb01192.x

The problem is the difficulty to quantify the effect of temperature, which is very important here.


Calculate ozone layer thickness at a given temperature and pressure The ozone concentration in the upper atmosphere averages $\pu{3.0 \times 10^13 molecules per cm3}$ in a region between $15$ and $\pu{35 km}$ high, at an average pressure of $\pu{0.001 atm}$ and temperature of $\pu{–25 °C}$. How thick would the ozone layer be, if these same molecules were at a pressure of $\pu{1.0 atm}$ and a temperature of $\pu{20 °C}$?

I understand that I need to use the ideal gas law, but I really have no idea how to approach the question, or even how to work with $\pu{molecules per cm3}$. I see that we have been given the values for temperature and pressure, but am unsure how to go about finding volume, and then the thickness.

I am also confused with how to deal with the two separate situations. Would you use $p = \pu{0.001 atm}$ or $p = \pu{1 atm}$. And would you use $T = \pu{248 K}$ or $T = \pu{293 K}$? And where do you find the value for the amount of substance?

Final conditions ($\pu{1.0 atm}$) give a hint at what altitude the thickness should be estimated – at the Earth's surface. Let's denote the unknown thickness as $h_x$, radius of the Earth as $r_\mathrm{E} = \pu{6371 km}$; also take $h_1 = \pu{15 km}$ and $h_2 = \pu{35 km}$ and put these dimensions on a rough 2D sketch:

Fig. 1. Schematic representation of two ozone layers at given altitudes.

To find $h_x$, we can use the volume of the new hypothetical ozone layer $V_x$ as the difference in volumes of two spheres with corresponding radii:

$$V_x = \frac{4 \pi (r_\mathrm{E} + h_x)^3}{3} - \frac{4 \pi r_\mathrm{E}^3}{3} \tag{1}$$

and express $h_x$ from (1) as follows:

$$h_x = \left(\frac{3V_x}{4 \pi} +r_\mathrm{E}^3\right)^{\frac{1}{3}} - r_\mathrm{E} \tag{2}$$

From ideal gas law for a relatively thin gas layer without temperature and pressure gradients:

$$V_x = \frac{nRT_x}{p_x} \tag{3}$$

where $n$ is the amount of gas in ozone layer, $T_x = \pu{20 ^\circ C}$ and $p_x = \pu{1 atm}$ (new conditions). Let's also denote ozone concentration as $N = \pu{3.0e13 cm-3}$, then total amount of ozone is to be found via initial volume $V_0$ and Avogadro constant $N_\mathrm{A}$ as

$$n = V_0 \times \frac{N}{N_\mathrm{A}} \tag{4}$$

Initial volume can also be found from Fig. 1 using the difference in volumes of the corresponding spheres with radii $r_\mathrm{E} + h_1$ and $r_\mathrm{E} + h_2$:

$$V_0 =  \frac{4 \pi (r_\mathrm{E} + h_2)^3}{3} - \frac{4 \pi (r_\mathrm{E} + h_1)^3}{3} = \frac{4 \pi}{3}((r_\mathrm{E} + h_2)^3 - (r_\mathrm{E} + h_1)^3) \tag{5}$$

Now all we have to do is carefully plug in the values with the corresponding units carefully and find all the unknown variables. From (5):

$$V_0 = \frac{4 \pi}{3}((\pu{6371 km} + \pu{35 km})^3 - (\pu{6371 km} + \pu{15 km})^3) = \pu{1.03e10 km3} = \pu{1.03e25 cm3} \tag{6}$$

Now let's find the amount of ozone using $V_0$ and equation (4):

$$n = \pu{3.0e13 cm-3} \times \frac{\pu{1.03e25 cm3}}{\pu{6.023e23 mol-1}} = \pu{5.12e14 mol} \tag{7}$$

and, subsequently, the unknown volume $V_x$ using (3):

$$V_x = \frac{\pu{5.12e14 mol} \cdot \pu{8.314 J mol-1 K-1} \cdot \pu{293 K}}{\pu{101325 Pa}} = \pu{1.23e13 m3} = \pu{1.23e4 km3} \tag{8}$$

Finally, plugging all the values in (2), we can find the thickness:

$$h_x = \left(\frac{3 \cdot \pu{1.23e4 km3}}{4 \pi} + (\pu{6371 km})^3\right)^{\frac{1}{3}} - \pu{6371 km} = \pu{2.4e-5 km} = \pu{24 mm} \tag{9}$$

The result is somewhat close to the fact posted by NASA:

  Around the world, the ozone layer averages about 3 millimeters (1/8 inch) thick, approximately the same as two pennies stacked one on top of the other.


What is the difference between the roles of a surfactant and a stabilizer in the process of nanocrystal growth? The crystals that I am synthesizing use two different stabilizers CTAB and SDS, but they say one is used as a surfactant and since they are both stabilizers what is the need for both?

CTAB is a cationic surfactant, SDS is an anion surfactant.

These surfactants act to selectively stabilize ionic faces on the crystal as it grows, causing those faces to grow more rapidly. This leads to a change in geometry or crystal shape. CTAB causes one morphology being cationic, while SDS will cause another, its head ion having the opposite charge. Further, a mixture of the two will cause an intermediary morphology.

For instance, CTAB will cause an aqueous solution of thioacetamide and Lead(II) acetate to form PbS nanocrystals in a octahedron shape, while SDS will cause them to form a dendritic fern like shape, while a CTAB/SDS mixture (5:1) will cause a stellated octahedron or star shape to form.


Do alkyl halides with an alpha nitrogen atom undergo SN1 reaction? Today, in my organic chemistry class, I learnt about the nucleophilic substitution reaction. The notes have stated that:

  Alkyl halides with $\alpha$ oxygen atom can also undergo SN1 reaction because of the heteroatom stabilised cation (oxonium ion)

An example was then given of a bromide leaving from a bromoether to give a resonance-stabilised carbocation.

I was wondering if alkyl halides with an $\alpha$ nitrogen atom can undergo SN1  reaction as well. When asked, the teacher responded by saying that with $\alpha$ nitrogen atoms, the carbocation would experience greater stabilisation and result in a faster rate of SN1 reaction for such alkyl halides.

The reactivity of haloalkyl ethers are well documented (e.g., they were originally used to protect $\ce{-OH}$ groups during total syntheses). For instance, chloromethyl methyl ether had been used to protect alcohol and phenol $\ce{-OH}$s as $\ce{-OCH2OCH3}$ or commonly known as $\ce{-OMOM}$. They should be sealed and stored in freezer to protect from moisture because they are susceptible to decompose by reacting with water, showing their reactivity. They are also well known carcinogens. An early review on $\alpha$-haloalkyl ethers can be found in Chemical Reviews (1955)([Ref. 1]).

A recent example in Ref. 2 shows in situ preparation of $\alpha$-halo ethers to protect $\ce{R-OH}$ groups, in order to prevent researchers from exposure to carcinogenic $\alpha$-halo ethers (see Figure below):

For example for carcinogenic activity of a $\alpha$-halo ether: Human health assessment information on bis(chloromethyl)ether can be found on the IRIS website (https://www.epa.gov/iris). IRIS stands for Integrated Risk Information System, a part of EPA’s mission to protect human health and the environment.

Based on the high reactivity of  $\alpha$-halo ethers, one can expect $\alpha$-halo amine would have been even higher reactivity than them. To my knowledge, they do not exist, mainly because, they'd undergo elimination quickly to give relevant  Schiff bases (see below):
$$\ce{-> [R-CH(Cl)-NHR1] -> R-CH=NR1 + HCl} \ \mathrm {or}$$
 $$\ce{-> [R-CH(Cl)-NR1R2] -> R-CH=N+R1R2Cl-}$$

[Ref. 1]: The alpha-Haloalkyl Ethers: Lawrence Summers, Chem. Rev., 1955, 55(2), 301-353 (DOI: 10.1021/cr50002a003).

[Ref. 2]: Simple, Rapid Procedure for the Synthesis of Chloromethyl Methyl Ether and Other Chloro Alkyl Ethers: M. A. Berliner, and K. Belecki, J. Org. Chem., 2005, 70(23), 9618-9621 (DOI: 10.1021/jo051344g).


Why are the three most simple alcohols completely miscible? I want to know why the three most simple alcohols (methanol, ethanol and propanol) are completely miscible in water?

I want to know what happens at molecular level.

I think hydrogen bonding plays an important role.

This happens because the intermolecular forces between different low molecular weight alcohols are very similar; as you say, they are dominated by hydrogen bonding. You can consider mixing on a molecular level to involve exchange of molecules from the two phases at their interface; if the energy changes involved are very small, mixing will occur as it is entropically favoured (mixed phase is more disordered) and the average thermal energy is far higher than the barrier to diffusion.


Identifying if reaction is reduction or oxidation. Is there an error in this question? I am looking at this reaction:

I am trying to determine if it is reduction or oxidation.

In the initial molecule I calculate the overall oxidation numbers as:

ox(Cl)*6 + ox(H)*6 + ox(C)*6 = 0
-6 + 6 +ox(c)*6 = 0
hence, ox(c) = 0

In the subsequent molecule, I calculated the oxidation numbers as:

ox(Cl)*4 + ox(H)*6 + ox(C)*6 = 0
-4 + 6 +ox(c)*6 = 0
hence, ox(c) = 6/2 = 1/3

However, my textbook says the oxidation number of ALL carbon in the second molecule is 2. I am very confused by this, since if all the carbons have an oxidation number of 2 then the molecule would be charged overall? Then I am unsure if my oxidation numbers can be fractional. Is my textbook wrong, and I wrong or are we both wrong?

Overall, what I am really trying to get at is if the molecule is being oxidized or reduced, and how many electrons are being transferred. Is there a better way of doing this?

Thank you kindly

Before the reaction, every carbon is attached to to other carbon atoms, one hydrogen atom and one chlorine atom. So oxidation state of each carbon = 0+0+-1(due to C-H bond) +1(due to C-Cl bond)= 0; average oxidation state of carbon is also zero.
In the product only two carbon atoms suffer a change in bondings; now each of these two carbon atoms is attached to one carbon through double bond, another carbon through single bond and a hydrogen atom; so O.S. of this carbon = 0+0+-1 = -1; so to conclude four carbon atoms remain in zero oxidation states, where as the doubly bonded carbon atoms are in -1 oxidation state. so average oxidation state of carbon = -2/6 = -1/3

P.S.:Fractional oxidation state results only when we take average, and that too in some cases


Which element is being reduced in this reaction? Which element is reduced in this reaction:
  $$\ce{2KMnO_4 +3NaSO_3 +H_2O ->2MnO_2 +3NaSO_4 +2KOH}$$

I am trying to figure out which elements lose their electrons, but how am I supposed to know when elements are neutral and when they are ionized? I have tried the following, but I don't believe it to be correct

$$\ce{2K^+Mn^+O_4^{-2} +3Na^+SO_3^{-2} +H_2O ->2Mn^{+2}O_2^{-2} +3Na^{+}SO_4{-2} +2K^+OH^-}$$

My reasoning being that:

Oxygen atoms are always doubly anionized
Sulfate and sulfite are also doubly ionized
Sodium and potassium are always singly cationized

Based on these rules I tried to make the bonds neutral, but my assumptions must be incorrect. Any help would be greatly appreciated

Your reasoning is not entirely right. In those compounds, O is always (-II), K and Na always (+I). However, it's not true for S, which can have different oxidation states. So does Mn.

So, in KMnO4, Mn must be (+VII). In MnO2, it is (+IV). Thus, Mn is reduced.

For sulfur: in NaSO3, it must be be (+V). In NaSO4, it must be (+VII). Thus, sulfur is oxidized.

However (big however, in fact): the sulfate anion is not SO4–, but SO42–. Similarly, sulfite is SO32–, not SO3–. So, your equation should be instead:

$$\ce{2KMnO_4 +3 Na_2SO_3 +H_2O -> 2MnO_2 +3Na_2SO_4 +2KOH}$$

which means that sulfur is oxidized from (+IV) to (+VI).