index of hydrogen deficiency I don't really get the concept of index of hydrogen deficiency (IHD). Can someone please use an example to explain what $\mathrm{H_{reference}}$ and $\mathrm{H_{molecule}}$ are? I know the formula:

$$\mathrm{IHD} = \frac{\mathrm{H_{reference}} - \mathrm{H_{molecule}}}{2}$$

You seem to be using a very simplified version of the index of hydrogen deficiency. A more complete expression is:

$$IHD = \frac{2C + 2 - H - X + N}{2}$$

Where $C$, $H$, $X$, and $N$ represent the number of carbon, hydrogen, halogen, and nitrogen atoms, respectively. The logic of the IHD is that any simple acyclic alkane has a formula of $\ce{C_{n}H_{2n + 2}}$, meaning that given $n$ carbons, a simple acyclic alkane will have $2n + 2$ hydrogens. Every increment in the IHD represents either a degree of unsaturation due to $\pi$-bonds, or the presence of a ring. In your formula, $H_{reference}$ should be twice the number of carbons in your molecule, plus two. $H_{molecule}$, on the other hand, is the actual number of hydrogens in the molecule.

For example, if your molecule has the formula $\ce{C6H8}$, $H_{reference} = 2(6) + 2 = 14$ and $H_{molecule} = 8$. The IHD is then $\frac{14 - 8}{2} = 3$, suggesting that the total number of $\pi$-bonds plus rings in the molecule is three. Note that this isn't sufficient to determine the actual structure of the molecule, but it's helpful when combined with spectroscopy data (that is, it could have one ring and two $\pi$-bonds, or just three $\pi$-bonds, or some other combination).


What is the heaviest possible water using oxygen and hydrogen isotopes, and what might it's physical properties be? After a cursory internet search, it appears to me that Hydrogen-3 (a part of Tritiated Water) is the heaviest stable* hydrogen isotope. Oxygen-21 is the heaviest isotope of Oxygen with a half-life of seconds, and Oxygen-16, -17, and -18 are the heaviest stable isotopes (again, this is based on an internet search and may not be accurate).
So could (3H2)(21O) conceivably be a thing? If not, what is the heaviest water possible, and what would some of its theoretical physical properties be? Would the mass overcome surface tension? Would there be stronger or weaker hydrogen bonds? And so on.

The heaviest water obtained with stable isotopes would be $\ce{D_2^{ 18}O}$.  This compound is not well-charactetized itself, but Ref. [1] indicates that the melting point is only about 0.28°C above that of ordinary water.  The implication is that overall physical properties of $\ce{D_2^{ 18}O}$ are close to those of heavy water with oxygen-16($\ce{D_2^{ 16}O}$); apart from density the isotopic effect of $\ce{^{18}O}$ versus $\ce{^{16}O}$ is much less than that of deuterium versus protium/"hydrogen".
References

F. Steckel and S. Shapiro, "Physical Properties of Heavy Oxygen Water.  Part 1.  Density and Thermal Expansion", Trans. Faraday Soc. 1963, 59, 331-343.


Significant figures when standard deviation is 0 For analytical chem lab, our group did a titration in which we got the same exact value every single time. As a result, our standard deviation was 0 and our instructor says to report answer using real rule of sig figs (our last significant figure should be the first digit in the uncertainty measurement) I am not sure what to do as my STDev doesn't have any significant digits?

edit: Molarity is the measurement in question!

Various sources of error are inherent in measurements. The so-called Type A evaluation, i.e. the statistical analysis of series of observations (e.g. calculation of the standard deviation and the standard error of the mean), can be used to evaluate the uncertainty arising from random effects.

However, this evaluation does not include the uncertainty arising from systematic effects. For example, if the concentration of the titrant (the solution of a known concentration that is added during the titration) has a small unknown error and the same titrant is used for all titrations, each of the individual results includes the same systematic error. Or, if the burette has a small unknown error and the same burette is used for all titrations, again each of the individual results includes the same systematic error. You can neither eliminate such errors nor reduce the caused uncertainty by making additional trials, and you cannot evaluate the caused uncertainty by using statistical analysis (i.e. by calculating the standard deviation and the standard error of the mean) of the results. The uncertainty arising from such systematic effects has to be evaluated by other means. The so-called Type B evaluation is usually based on scientific judgment using all of the relevant information available on the possible variability of the results. Finally, you can combine all calculated or assumed uncertainties by using the rules for error propagation.

For example, the uncertainty arising from the unknown error of the burette may be estimated using the manufacturer’s specifications as follows:
Usually, you estimate (interpolate) the measurement between the graduation divisions of the burette. You might be able to estimate the measurement to one tenth of the smallest division. However, volumetric instruments, such as graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes, – like any other measuring instrument – have a limited accuracy. The dimensions, graduating divisions, and accuracy limits of volumetric instruments are standardized. For example, the requirements of burettes are specified in the international standard ISO 385. Two classes of accuracy are specified: Class A or AS for the higher grade and Class B for the lower grade. The nominal capacity, graduation divisions, and accuracy limits are shown in the following tables.

$$
\textbf{Burettes, Classes A and AS} \\
\begin{array}{lll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} \\
\hline
\hphantom{00}1 & 0.01 & 0.006 \\
\hphantom{00}2 & 0.01 & 0.01 \\
\hphantom{00}5 & 0.01 & 0.01 \\
\hphantom{00}5 & 0.02 & 0.01 \\
\hphantom{0}10 & 0.02 & 0.02 \\
\hphantom{0}10 & 0.05 & 0.03 \\
\hphantom{0}25 & 0.05 & 0.03 \\
\hphantom{0}25 & 0.10 & 0.05 \\
\hphantom{0}50 & 0.10 & 0.05 \\
100 & 0.20 & 0.10 \\
\hline
\end{array}$$

$$
\textbf{Burettes, Class B}\\
\begin{array}{lll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} \\
\hline
\hphantom{00}1 & 0.01 & 0.01 \\
\hphantom{00}2 & 0.01 & 0.02 \\
\hphantom{00}5 & 0.01 & 0.02 \\
\hphantom{00}5 & 0.02 & 0.02 \\
\hphantom{0}10 & 0.02 & 0.05 \\
\hphantom{0}10 & 0.05 & 0.05 \\
\hphantom{0}25 & 0.05 & 0.05 \\
\hphantom{0}25 & 0.10 & 0.10 \\
\hphantom{0}50 & 0.10 & 0.10 \\
100 & 0.20 & 0.20 \\
\hline
\end{array}$$

For example, a burette of Class AS with a nominal capacity of 50 ml has graduation divisions of 0.10 ml. By interpolation, you might be able to estimate the measurement to one tenth of the division, i.e. to 0.01 ml. However, the accuracy limits of the burette are ±0.05 ml (at any point on the scale).


weird patterns in solutions When you mix a solid into a solvent you can see weird patterns right after the mixing.
A common example is mixing sugar in water.

What are those wave-like patterns and why do they happen?

Those rippling transparent patterns are caused by the variations in refractive index, which in turn are caused by the variations in concentration of the solute. As soon as the dissolution and mixing is over, and the solution is uniform again, the patterns are gone. I don't know the English word for it, or whether there is one; "Schlieren" seems to be much more general.

It's a physical phenomenon indeed, but chemists might be more used to seeing it.

You may observe a remotely similar pattern around a heated object in the air (see "heat haze"). In that case, it is caused by the variations of temperature and not concentration.


Calculate the atomic radius of iridium atom I was delaying with this question as seen below.

  Calculate the radius of iridium atom, given that Ir has an FFC crystal
  structure, a density of $\pu{22.4 g/cm^3}$ and atomic weight of $\pu{192.2 g/mol}$.

Let me show you how my teacher solved this question.

Givendata:

$$\text {Weight of crystal} = \pu{192.2 g/cm^3}$$
$$\text {Density} = \pu{22.4 g/cm^3}$$
$$\text {Volume of crystal} = V = \frac {192.2}{22.4}$$
$$V = \pu{8.58 cm^3/mol}$$

In FFC, there are 4 atoms in one unit cell.

$$\text {Volume of one cell} = 8.58 \times \frac {4}{6.023 \times 10^{23}}$$
$$ = 5.7 \times 10^{-23}$$
$$V = a^3$$

$$\therefore a = 2.83r$$
and
$$V = (2.83r)^3$$
$$5.7 \times 10^{-23} = 2.83r^3$$
$$r = 1.34 \times 10^{-8}$$

Where did that $a = 2.83r$ come from? That's what I didn't get.

Iridium has a face-center cubic (FCC) structure.

You assume that each iridium sphere has a radius $r$. But the edge of a unit cell has the dimension $a$. The volume of the unit cell, which is a cube, is $a^3$. The relationship between $a%$ and $r$ is that $a = 2.83r$.

FOLLOWING THE TEACHER"S(?) WORK...(or how you copied it?)

The following step is ok...
$$V = (2.83r)^3\tag{1}$$

But the second step here is wrong. The 2.83 should be cubed too.
$$5.7 \times 10^{-23} = 2.83r^3\tag{2}$$

(3) is totally wrong if we accept (2).
$$r = 1.34 \times 10^{-8}\tag{3}$$

From (2)
$$r = \sqrt[3]{\dfrac{5.7\times 10^{-23}}{2.83}} = 2.72 \times 10^{-8}\tag{4}$$

Starting over...

$$V = (2.83r)^3\tag{5}$$
$$5.7 \times 10^{-23} = (2.83r)^3\tag{6}$$
$$r = \dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = 1.36 \times 10^{-8}\tag{7}$$

After calculating the the volume of the unit cell, I'd solved for $a$ by taking the cube root, then divided by 2.83 to get $r$.

You have to be careful with significant figures too. It is not $5.7\times10^{-23}$ but rather $5.70\times10^{-23}$. Using a calculator I'd carried 5 significant figures in all the calculations then rounded to 3 at the end. In my day I used a slide rule so everything was to 3 significant figures.


Possible Products in the Addition To a Double Bond in a Ring I don't understand why both these products should be different. I know hydrogenation is syn addition and both the hydrogens add from the same side, but why should the plane in which both the hydrogen atoms are present even matter?

Also, bond rotation is possible around sigma bond which makes both these products identical.

Please explain why the two products are different and where am I wrong with this logic?

why should the plane in which both the hydrogen atoms are present even
  matter?

In this case it doesn't.  The author of the example is laying the groundwork for the future.

The area above and below the double bond is often referred to as a "face". In your molecule, the two faces are not equivalent. If you build two models of your compound and in both color the top of the double bond red and the bottom blue, you'll find there is no way to superimpose one model onto the other where the red face is superimposed on the blue face.  These two faces are therefore not equivalent, they are said to be enantiotopic (mirror images) in this molecule.

In this case, addition of (an achiral reagent such as) hydrogen to the two faces produces the same molecule.  In more complex molecules, or when using chiral reagents this will not be the case.

Consider the case where there is a methyl group attached to the methylene group adjacent to the $\ce{C-Br}$ carbon in your molecule.  Now the faces are different (diastereotopic) and addition of hydrogen to the top and bottom faces will produce two different products (methyl and bromine on the same side or opposite sides of the molecule).

The concept of faces is important in chemistry.  Always consider if the faces are equivalent, enantiotopic or diastereotopic. Depending on what you find, the products from attack at the two faces may be the same, enantiomers or diastereomers.  The concept of faces is introduced in your problem.


Partial and total concentration in solutions The total pressure of gas mixture equals the sum of partial pressures of its components.

Can this rule be true for liquid concentration?
I.e. can we say: the total concentration of mixture is equal to the sum of partial concentrations of its components?

It's true and false.

For example:
1 Solution with 1mol/L of NaCl and 1mol/L of KBr.
The total concentration of salts is 2mol/L.

But the ions concentration are:

Na+ 1mol/L
K+ 1mol/L
Cl- 1mol/L
Br- 1mol/L

Is it clear?


Why is the ratio between Silicon and Oxygen 1:3 in single chain silicates? I thought that the ratio of silicon to oxygen in a silica tetrahedron was 1:4, so if a single chain is just many of these linked together, why does the ratio become 1:3?

Silicon has 2 complete oxygen atoms ($2\times1=2$) and 2 equally shared oxygen atoms ($2\times1/2=1$), total 3.


Nucleophilicity in Polar Aprotic Solvent Why is it that fluoride is more nucleophilic than iodide in an aprotic solvent? The Clayden textbook suggests that lower-row elements of the periodic table are more nucleophilic since their valence electrons are in higher energy levels and are therefore more reactive. However according to the slides from my lecture class, the order of nucleophilicity for halides in an aprotic solvent goes from F- > Cl- > Br- > I- which is opposite from what is stated in the Clayden textbook. Can someone please explain this?

Thank you

It's because of the anion stability. Fluoride is less stable than iodide, therefore more likely to donate electrons.
In protic solvents, the fluoride was stabilised by the partially positive hydrogen end. Therefore, it is less nucleophilic then iodide.


Why does this ligand form 6-coordinate complexes with Transition elements and 8-coordinate complexes with Inner transition elements? What change takes place in the denticity of this ligand if it is an inner-transition metal ion vs a common transition metal?
Primarily, why does the ligand behaves differently with the two situations above.

I decided to do some more research for this question as I had faced it before. Turns out the proper name of this ligand is diethylenetriaminepentaacetate ($\ce{DTPA^5-}$). It's an expanded version of EDTA and is primarily used in MRI scanning where it forms complexes with the Gadolinium ion. It forms complexes with a coordination number of six and seven with transition metals simply because these complexes are more stable. Here's Fe(III)-DTPA as an example1:

Experimentally Fe(III)-DTPA has been observed to adopt seven- and six-coordinate
crystal structures. This agrees with our result where the 8-coordinate Fe(III)-DTPA is $\pu{23 kJ mol^-1}$ higher in energy relative to the most stable 7-coordinate structure

With inner transition metals, it forms complexes with a denticity of 8, with Pr-DTPA being an example2:

References:

Sillanpää, Atte J., et al. “Density Functional Complexation Study of Metal Ions with (Amino) Polycarboxylic Acid Ligands.” Phys. Chem. Chem. Phys., vol. 5, no. 16, 2003, pp. 3382–93. doi:10.1039/B303234P.
Fusaro, Luca, et al. “Insight into the Dynamics of Lanthanide-DTPA Complexes As Revealed by Oxygen-17 NMR.” Inorganic Chemistry, vol. 51, no. 15, 2012, pp. 8455–61. doi:10.1021/ic3010085.


Does mercury fulminate really look like crystal methamphetamine? In the AMC series Breaking Bad , Walter (aka Heisenberg) smuggles in a bag of mercury fulminate under the guise of crystal methamphetamine and threatens to blow the room up if he doesn't get his money.

Does fulminated mercury really have the translucent white/blue crystal appearance as was shown in this scene?

Is the compound that sensitive that throwing it against the wall would cause detonation?

Mercury fulminate $\ce{[Hg(CNO)2]}$ can have a white, gray or greenish-brown color depending upon purity and dryness (it is often kept moistened to prevent accidental decomposition).

  Does fulminated mercury really have the translucent white/blue crystal
  appearance as was shown in this scene?

I've never seen a picture where it appears translucent or where it has a bluish color.  The pictures I've seen always show it as an opaque powder.  I suspect that it could be crystallized, but that's probably never done as it would further increase its sensitivity to detonation (e.g. if a crystal cracked, or rubbed against another crystal, the friction, triboelectrification, etc. could cause the material to explode).  Here are some pictures.

  Is the compound that sensitive that throwing it against the wall would
  cause detonation?

Here is a link to an examination of the Breaking Bad "detonation by throwing" experiment (with video).  The conclusion is that it can detonate when thrown, but 1) a larger amount of material than what Walt had, and 2) a much harder throw (greater force, greater shock) would have been required to cause detonation.


Why are sigma bonds stronger than pi bonds? Why are sigma bonds stronger than pi bonds?

The reason behind this is the orientation of the overlapped orbitals. Sigma bonds result from head-on(co-axial) overlapping while pi bonds are outcome of lateral(para-axial) overlapping. Here is a pictorial representation of ethene(sp2 hybridized C atoms) :

The greater the extent of overlapping, the higher the probability of finding the valence  electrons in between the nuclei and hence the bond will be stronger & shorter.

In MOT, this can be explained using Overlap Integral. This is how Atkins depicts it :

In simple terms, after forming a sigma-bond (a pre-requisite for pi-bonds), the two atoms get locked along the inter-nuclear axis. As a result, the orbitals available for pi-bonding can only partially overlap, thus forming a weaker bond.


Why is methyl parathion less toxic than ethyl parathion? According to the PubChem pages on ethyl parathion and methyl parathion (pages 11 and 12 of the toxicity section, respectively), methyl parathion is significantly less toxic than ethyl parathion. The difference in LD50 values depends on the route of administration and the animal to which the pesticides were administered, but it is significant.
For instance, parathion has an intraperitoneal LD50 in mice of 3 mg/kg, the LD50 for methyl parathion administered in the same way is 5.4 mg/kg, making it almost two times less toxic than parathion. Methyl parathion differs from regular parathion in that, in methyl parathion, the ethyl groups are replaced with methyl ones.
How does this structural difference contribute to a difference in toxicity? Why is methyl parathion significantly less toxic than its ethyl analog?

Both methyl parathion (O,O-Dimethyl O-(4-nitrophenyl) phosphorothioate) and ethyl parathion (O,O-Diethyl O-(4-nitrophenyl) phosphorothioate) are organothiophosphates with identical structure except for dimethoxy group in methyl analog has been replaced by diethoxy group in ethyl analog. Both analogs are acetylcholinesterase (AChE) inhibitors, although phosphorothioates ($\ce{P=S}$) are less potent inhibitors compared to corresponding phosphate ($\ce{P=O}$) analogs. The active site of AChE is depicted in following Figure (Ref.1) where acetylcholine (ACh) get hydrolyzed:

The hydrolysis mechanism invoves the $\ce{OH}$ of serine residue and imidazole moiety of histidine residue in the active site as shown. The major inhibition mechanism of all organophosphate esters is involved phosphorylation of the $\ce{OH}$ group of serine residue in the active side. The aging (breakage of the $\ce{O-R}$ ester linkages leaving $\ce{P-O-}$ groups) is followed this phosphorylation making the inhibition irreversible.
Although aging can be visualized by the process of hydrolyzing ester group by water molecule attacking phosphorus atom first, replacing $\ce{R-O-}$ group followed by proton exchange, the literature has shown preferably $\ce{O-R}$ breakage. If that's the case, ethyl breakage is facilitated over methyl breakage just by stability of the resultant products ($\ce{CH3CH2^+}$ vs $\ce{CH3^+}$). Hence, we can conclude that ethyl parathion is more potent than methyl parathion.
An early study of inhibitory activity of various batches of ethyl parathion has revealed that a dithio analog,  O,S-Diethyl O-(4-nitrophenyl) phosphorothioate is much potent inhibitor compared to ethyl parathion (Ref.2). It is possible that this dithio analog is a common contaminate during ethyl parathion sysnthesis. Thus, this factor may contributed to the higher activity of ethyl parathion compared to that of methyl parathion.
It is also worth noting that recent studies on versatility of imidazoles or in reactions with organophosphates (OPs, Ref.3). According to this study, OPs from the $\ce{P=O}$ family, imidazole attacks the phosphorus atom exclusively, making $\ce{P-N}$ bond. With the less reactive $\ce{P=S}$ family, an unusual N-alkylation predominates. Surprisingly, imidazole reacts with methyl parathion exclusively at the aliphatic carbon, whilst for ethyl parathion, imidazole reacts by two pathways: at both the phosphorus and aliphatic carbon, with predominance for the latter:

Since immidazole moiety is present in active site of the AChE, one can suggest the different reactivity of ethyl parathion over methyl parathion with imidazole may play a role in their different activity in inhibition.
References:

Hay Dvir, Israel Silman, Michal Harel, Terrone L. Rosenberry, Joel L. Sussman, “Acetylcholinesterase: from 3D structure to function,” Chemico-biological Interactions 2010, 187(1-3), 10-22 (DOI: 10.1016/j.cbi.2010.01.042).
W. M. Diggle, J. C. Gage, “Cholinesterase inhibition in vitro by OO-diethyl O-p-nitrophenyl thiophosphate (Parathion, E 605),” Biochemical Journal 1951, 49(4), 491-494 (https://doi.org/10.1042/bj0490491).
Valmir B. Silva, Elisa S. Orth, “Are Imidazoles Versatile or Promiscuous in Reactions with Organophosphates? Insights from the Case of Parathion,” J. Braz. Chem. Soc. 2019, 30(10), 2114-2124 (https://doi.org/10.21577/0103-5053.20190084 ).


Determining Fluorescence Quantum Yield of C60 in Toluene For my experiment, I am trying to determine the fluorescence quantum yield of C60-fullerene in toluene (at room temperature).  In order to do that, I need to have a standard sample with a valid and known fluorescence quantum yield. I have been looking for a standard sample with a known quantum yield in toluene, however, I am not sure of what to use.

Would it be better if I used a different solvent in determining the quantum yield and if so what solvent and standard should I use.

Toluene instead of benzene can be a poor choice when H-atom abstraction by the excited state is possible. Here, you're safe.

Try to get a copy of Fluorescence spectra and quantum yields of buckminsterfullerene (C60) in room-temperature solutions. No excitation wavelength dependence.

The authors apparently used 9-cyanoanthracene in toluene as a standard to determine $\Phi_f$ of C60.

You might also want to have a look for review articles by Dirk Guldi from the time around 2000; I remember him giving a number of talks on C60 around that time.


Why do unboiled eggs last longer than boiled eggs? Just like the title says, I watched this video https://youtu.be/CHMY4G9gTPA and then became intrigued.

An eggshell is a complex structure.  From Wikipedia:

Boiling the egg removes the waxy cuticle from the outside of the egg, dissolves a small but not insignificant amount of calcium carbonate from the shell, damages the protein matrix that holds calcium carbonate crystals in place in the shell, and can disrupt or destroy the two shell membranes.  All of these contribute to allowing bacteria a means to enter the egg and spoil it.

Note: Though calcium carbonate is actually less soluble in hot water vs cold, the rate of dissolution is substantially greater.  Unless the egg is boiled in a saturated solution of calcium carbonate, it will lose material to the solution when boiled, and any pores in the shell become enlarged as a result.


Why is it hard to reduce titanium dioxide? Titanium is expensive, although it is not rare and titanium dioxide is a ubiquitous component, it's part of such mundane articles as toothpaste, paper or plastics.

The reason for that is that the processes that extract titanium from titanium dioxide (which occurs naturally and just needs some inexpensive purification) are quite costly. The established industrial method for this is the Kroll process. It works by first chlorinating the titanium to $TiCl_4$ and then moving the chlorine over to some $Mg$, yielding $MgCl_2$ and pure $Ti$. Then follows a costly remelting process under vacuum to further purify the metal.

Recycling titanium scrap is even harder, and removing the other components from the alloys is not entirely possible, so recycling only yields titanium based alloys, not the pure stuff (which is not necessarily a problem since it's very rarely used in its pure form).

What specific chemical properties of titanium make it so hard to purify? For example, the chlorine process at the beginning of the Kroll process takes place with coke in a fluidized bed reactor, very similar to the reduction of $Fe$ in a blast furnace, but if the carbon takes care of the reduction, why is that not sufficient? Why is the chlorine route needed?

Research into titanium production dates back to the late 1880s, and the Kroll process, patented in 1940, made it feasible for mass production, but remains costly and has not been improved upon since, despite much effort. How would a chemist look for suitable reactants during such research?

Reasons why Titanium is hard to extract and purify:

The +4 oxidation state of $Ti$ is the most stable (due to its noble gas configuration) and so reducing $TiO_2$ to $Ti$ is quite difficult.
$Ti$ has chemical properties which are similar to its neighbouring elements (Vanadium, Zirconium, Chromium) and the presence of compounds containing these elements in its ores makes purification even more difficult. For example: Vanadium and Zirconium also form volatile halides ($ZrCl_4$)and oxyhalides ($VOCl_3$) which have to be separated from $TiCl_4$ by fractional distillation in Kroll's Process.

I have tried to explain why the chlorine route is taken using only thermodynamics (I'm not absolutely sure about this but these reasons are what I could come up with).

Consider these reactions which could take place in the fluidized bed reactor:

$\ce{TiO_2(s) + 2Cl_2(g) + 2C(s) -> TiCl_4(g) + 2CO(g)}$
$\ce{Ti(s) + C(s) -> TiC(s)}$

Comparing the Gibbs' Energies of these reactions can tell us which one is spontaneous since if the change in Gibbs' Energy is negative, the reaction will be spontaneous.
And the equation for change in Gibbs' Energy is:

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

At normal temperatures, reaction 2 would occur more readily than reaction 1 since $TiC$ is extremely stable (reflected by its very high melting and boiling points) while $TiCl_4$ is unstable (it hydrolyses immediately on contact with air or oxygen to form $TiO_2$ and $HCl$). And so it is almost impossible to reduce Ti using C at low temperatures.

At higher temperatures (Kroll's Process is carried out at around $1000^oC$), $T\Delta{S}>>\Delta{H}$ so we can neglect $\Delta{H}$.

For the first reaction, since all the products are gaseous (while some of the reactants are solid), $\Delta{S}$ is positive. Therefore $\Delta{G}$ will be negative and the reaction is spontaneous and will occur easily.

For the second reaction, since the number of molecules is reduced, $\Delta{S}$ is negative. Therefore $\Delta{G}$ will be positive and the reaction becomes unfavourable.

Notes:

For the states of the reactants and products, I compared the melting and boiling points of the reactants and products with $1000^oC$ which is the temperature at which Kroll's process is carried out.
Actually, from what I read, there are many possible products for the first reaction. depending on the reaction conditions, a mixture containing $CO$, $CO_2$ and $COCl_3$ can be formed. But all these products are still gaseous so it doesn't affect the explanation.


What is the mechanism by which reactants in endothermic reactions absorb energy from their environments? My understanding of exothermic reactions is that, in the aggregate, the sum of the bond energies of the products is lower than that of the reactants, and definitely lower than that of the sum of any excited state intermediates, and that the resulting difference in energy is accounted for by a decrease in potential energy for the bonding electrons involved, resulting in a release of photons of equivalent energy.

Seeing as how exothermic reactions generally raise the temperature of the environment they are in, it seems logical to me that those photons are rapidly absorbed (inelastically) by atoms/molecules immediately neighboring the reaction products and thus their kinetic energies are raised, although I honestly don't truly understand the mechanism for that, either.

Similarly, for reactions that are isothermal but which are chemiluminescent, photons of a wavelength not easily absorbed by neighboring atoms/molecules simply leave the environment.

However, does that mean that the inverse is true? Is IR radiation from the environment the direct mediator of energy transfer that raises bonding electrons in reactants to anti-bonding MOs so that new MOs can form among different configurations of the original reactant atoms, thus allowing electrons to occupy the bonding MOs of the products?

I ask because the basic collision theory proposed by Trautz and Lewis doesn't seem to account for quantum mechanical models of energy, but I admit my own understanding of the topic is about as limited as my question would demonstrate.

In your first paragraph you state at the end that photons are released. In all reactions, except the rather few chemiluminescent ones, this is not the case.

The energy difference is made up of vibrational and rotational energy of the products, which by collisions with solvent (or inert buffer gas) removes this energy.  This energy transfer occurs as the product is initially 'hot' with respect to the surroundings and thus energy flows out of this into the surroundings, heating it a little, until equilibrium is reached. In femtosecond laser experiments it is possible to observe the flow of energy when in solution and it is very rapid, not much longer than a few picoseconds. Ir photons could be emitted but the radiative lifetime of levels is much, much longer than collision lifetime so that i.r. emission yield will be tiny.

In the low pressure gas phase (where collisions with buffer gas are absent) infra-red can be observed (called infra-red chemiluminescence as it comes from excited vibrational levels) in reactions such as $\ce{H + H2 \rightarrow  HF(v'J') + H} $ where $\ce{v'J'}$ are excited vibrational and rotational levels. Many experiments have been reported by Polanyi and co workers for which Polanyi received the Chemistry Nobel prize. This reference https://www.nobelprize.org/nobel_prizes/chemistry/laureates/1986/polanyi-lecture.pdf gives a clear account of this important work

In the reverse case it is the few molecules that by the Boltzmann distribution have enough energy to surmount the activation barrier that are important not promotion by ir photons.


What is the mechanism of halogen displacement? We teach (ad infinitum....) that a more reactive halogen will displace a lesser halogen from a compound.  And yes, the halogen is a halide ion in the compound.  Now I think about it, it's usually in solutions, so the halide is nominally solvated and surrounded by water (as hydronium ions?).

But how?

Does the halide lose the electron to give a radical, or does the halogen dissociate to radicals, or to ions, or is there an intermediate Cl2- ion, for example?  I realise this is several questions, but the major question is unaffected by my speculation.  All I have found is the bald statement about displacement and reactivity and the table we always draw.

I'm guessing your talking about the halogen displacement reactions:

$$\ce{2NaBr + Cl2 -> Br2 + 2NaCl}$$

which is a redox reaction:

$\ce{2 Br- -> Br2 + 2 e-}$, bromine loses electrons (is oxidised)
$\ce{Cl2 + 2e- -> 2 Cl-}$, chlorine gains electrons (is reduced).

For more detail look in to the redox potentials, this explains what will reduce what.


How much oxygen is in a cup of water? I know that water is 2 parts hydrogen and 1 part oxygen. Hydrogen and oxygen are quite light gasses, they don't weigh very much compared to water. Water is very heavy and it made me wonder exactly how much oxygen I'd need to make a cup of water.

Please answer in terms of cubic meters of air, assuming the air is 21 % oxygen, and that my small cup contains only 100 ml. I suppose the exact question is: How many cubic meters of air contains the same amount of oxygen as found in 100 ml of water?

Since the molecular weight of water is $18\ \mathrm{g/mol}$, $100\ \mathrm{ml}$ (density = $1\ \mathrm{g/ml}$, so $100\ \mathrm{ml} \overset{\wedge}{=} 100\ \mathrm{g}$ of water) of water contains
$$\frac{100}{18} = 5.56$$
moles of water, or 5.56 moles of atomic oxygen ($\ce{O}$), or 5.56/2 = 2.78 moles of molecular oxygen ($\ce{O2}$).

At standard temperature and pressure, 1 mole of a gas occupies 22.4 liters.  Therefore, 2.78 moles of molecular oxygen will occupy 62.22 liters.

If air is 21% molecular oxygen, then we would need 62.22/0.21, or 296.3 liters of air to provide that much molecular oxygen.

Since 1 cubic meter equals 1000 liters, this volume converts to 0.3 cubic meters of air.

Therefore, 0.3 cubic meters of air contains the same amount of oxygen as is present in 100 ml of water.


Schottky Defect in AB2 type crystals Are there any examples of Schottky defect in ${\ce{AB_2}} $ type crystals.
Does Schottky defect require that the cation and anion be of equal size?
According to me it is not necessary that the two ions are of equal size as even ${\ce{NaCl}} $ shows Schottky defect but ${\ce{Na+}} $ has radius is 95 pm and ${\ce{Cl-}}$ has radius of 181 pm which is a bit less than double of that of ${\ce{Na+}} $

Schottky defect occurs when equal number of positive and negative ions are missing from the lattice thereby maintaining net electrical neautrality and leading to a decreased density in the lattice (since there are now less number of ions) .
It is observed to be occurring in cations and anions of similar sizes.

NOW CONSIDER THAT AN ION IS AT THE LATTICE POSITION AND A DEFECT IS TO BE OCCURRED.

There are 2 ways stoichiometric conditions can be satisfied. - either by loosing this ion as well as another of opposite charge in equal ratio(SCHOTTKY) ,
OR not losing this ion at all , but just shifting it's position from lattice position to interstitial one.(FRENKEL)----
If we think logically ,

Since it's a stoichiometric defect it should maintain neautrality as well as loose cations and anions in a certain ratio.

If the sizes were very different , there are chances the smaller ion would occupy an interstitial space causing it to create a hole and still maintain neutrality and ratio without losing any ions. (Which is what happens in Frenkel defect) . Since however ions are of similar sizes , one can't occupy an interstitial space .. and therefore losing equal no. Of anions and cations is the only way to go.
You can think of these as 2 cases of defect... Yes it's not necessary the ions are of similar sizes for a defect to occur BUT IF THEY ARE...They are most likely to cause the Schottky defect and if not , then Frenkel Defect.
As for your NaCl doubt , I would suggest try finding the critical radius ratio for one ion to occupy the interstitial space . This way you can verify whether it can perform Frenkel defect or not ..Or why it's performing Schottky.
Hope this helps.

EDIT:- I can't find any examples of ${\ce{AB2}} $ type as such , but you can again try finding critical ratio of ions in ${\ce{AB2}} $ lattices to decide what type of defect they are likely to occur.


Quantifying Ascorbic Acid in Soft Drink I have access to a GC/MS, FID, and a spectrophotometer. My first thought was to derivatize with trimethylsilylating agent, then run it by GC. Unfortunately the water is problematic. Heating it is likely to destroy the ascorbic. I don't have access to a manifold freeze-drier. I do have access to a rotovap, but I don't think that works well for removing water. I do have access to a freezer, but no way to vacuum the sample down to 0.006 atmospheres. Titration is out of the question because I don't have a clue what else is in the sample.

Titration is in the question because ascorbic acid is likely the principal, if not only, reducing agent.

  Using iodine

For example, [this site] describes a titration using the iodine-starch redox indicator (the presence of iodine and starch together creates a blue/black color).

The active oxidant is $\ce{I3-}$, formed from $\ce{I2}$ and $\ce{I-}$:

$$\ce{I2 + I- <=> I3-}$$

The redox reaction then is:

$$\ce{C6H8O6 + I3- ->C6H6O6 + 3I- + 2H+}$$

Add a small amount of a saturated solution of soluble starch to your soft drink sample and then titrate against a known solution that is equimolar in $\ce{NaI}$ or \$\ce{KI}$ and $\ce{I2}$. When the dark color persists in your soft drink sample, you have your end point.

This site provides a full procedure (PDF).

  Using $\ce{KMnO4}$

Potassium permanganate is its own indicator, since $\ce{KMnO4}$ solutions are purple and the reduced version is colorless (technically an insoluble brown solid).

If your soft drink is dark-colored, you may want to dilute it so that you can see the color change.

The redox equation is:

$$\ce{2MnO4- +6H+ 5C6H8O6-> 2Mn^{2+} + 8H2O}$$

  Spectrophotometrically

Ascorbic acid absorbs UV light (but then so does other things in your soft drink, like the preservative sodium benzoate). However, you could use ascorbic acid's reducing power to turn on an absorbance in the visible. Here is one example. This paper may be behind a paywall, but the abstract contains the essential idea. In this paper, the authors describe a system using copper(II) and phenanthroline(phen). Ascorbic acid reduces copper(II) to (copper(I), which creates a coordination complex with phenanthroline that has $\lambda_{\text{max}}=460 \ \text{nm}$. The authors claim a detection limit of 40 $\mu\text{g/L}$, which is pretty good (40 parts per million).

$$\ce{2Cu^{2+} + C6H8O6 -> 2Cu+ + C6H6O6 +2H+}$$
$$\ce{Cu+ + 3phen ->Cu(phen)3+}$$

  Another colorimetric assay

The Folin-Ciocalteu assay is another nice colorimetric assay. Reducing agents react with the phosphomolybdate and phosphotungstate compounds in the reagent mixture and produce a blue color. This method needs to be calibrated against some ascorbic acid standards because the redox chemistry is complex and not fully understood. The absorbance max of the reduced FC reagent is in the 670ish range, if I remember correctly.


Hydrogen evolution during thermal spray Firstly, let me state that I'm no chemist. My specialty is in thermally applied coatings such as electric-arc metalizing.

During the course of my career I've always been presented with vague warnings to use "proper" ventilation to prevent hydrogen buildup within a contained area when spraying Al alloys.

Could someone please explain the process that creates hydrogen gas when arcing between two Al electrodes?

Also, what would the "proper" ventilation be? Would dry dust collection/ventilation suffice or would wet collection methods be required? Would this same reaction occur when using other alloys commonly used in thermal deposition processes?

The culprit in the undesired production of hydrogen gas from electric-arc metalizing is water vapor, as it is the only significant source of hydrogen atoms in the atmosphere. The concentration of water vapor is then of course a major factor in how much hydrogen gas can be produced. In very humid conditions the air can contain as much as a few percent water vapor. The chemical reaction for the conversion of water vapor to hydrogen gas under the conditions of this process is as follows:

$$\ce{H2O <=> H2 + 1/2O2}$$

A few percent water vapor may not seem like enough to produce explosive levels of hydrogen gas, but hydrogen gas becomes explosive in air at levels as low as $\pu{4\%}$.

A recirculating dust filtration system, dry or wet, is not an appropriate system for removing hydrogen gas. The only reliable way to prevent the possibility of hydrogen buildup in a closed area is with an appropriate fume hood (or similar approved device) that vents the gases out of the building.

I can't give an exhaustive list of the metals/alloys that could result in hydrogen production via this process, although base metals like copper, lead, nickel and zinc have the potential to do so.


Why do Organolithium or Grignard reagents act as nucleophiles and not as bases with aldehydes and ketones I've read entire Chapter 14: Organometallic Compounds of  Francis Carey's "Organic Chemistry" but I still didn't get an answer to my question.

Quote from the book:

  Because of their basicity organolithium compounds and Grignard reagents cannot be prepared or used in the presence of any material that bears a $\ce{-OH}$ hydroxyl group. Nor are these reagents compatible with $\ce{-NH}$ or $\ce{-SH}$ groups, which can also convert an organolithium or organomagnesium compound to a hydrocarbon by proton transfer.

Fine. No arguments with that.

  Terminal alkynes also give alkanes with Organolithium and Organomagnesium compounds.

Must be true if he says so. Now coming to the point,

  Ammonia and Amines have $\mathrm{p}K_\mathrm{a}$ around 36.
  Terminal alkynes have $\mathrm{p}K_\mathrm{a}$ of 25.
  Aldehyde $\mathrm{p}K_\mathrm{a} = 17$, Ketone $\mathrm{p}K_\mathrm{a} = 19$ and an Ester $\mathrm{p}K_\mathrm{a} = 25$.
  Water & Alcohols have $\mathrm{p}K_\mathrm{a}$ of 15–18.

If organolithium compounds and Grignard reagents act like bases with alcohols, terminal alkynes, and amines/ammonia, why do they act like nucleophlies and give addition reaction with carbonyl compounds? Why don't they act as bases and give alkanes instead? What am I missing here?

In fact, both reagents you noted here have quite complex structure and are not nucleophiles at all: both are electrophiles, because metallic atom here has too little neighbors to draw electrons from.

Let me explain, using Grignard reagent $\ce{EtMgCl}$ . In reality it has complex structure with $\ce{Mg}$ atom coordinated to alkyl fragment and two diethyl ether molecules. When reacts, it coordinates first with most nucleophilic atom. In case of alcohol it is oxygen of $\ce{OH}$ fragment, and of course it  reacts with closest easily migrating atom: hydrogen, producing alkane. In case of ketone it coordinates with $\ce{CO}$ group, and alkyl group attacks closest electrophile : carbon from $\ce{CO}$ group, whose electrophility was just boosted.

If right compound, effectively bounding to metal atom, is added and no other way of reactions is readily available, both reagents can work as bases. For example, organolithium compounds can replace hydrogen atoms in alkynes, alkenes and arenes as you noted earlier.


Is enthalpy of combustion the 'heat released' or 'enthalpy change'? In a constant volume calorimeter, $\pu{3.5 g}$ of a gas with molar weight $\pu{28 g}$ was burnt in excess $\ce{O2}$ at $\pu{298 K}.$ Temperature of calorimeter increased from $298$ to $\pu{298.45 K}$ due to combustion. If heat capacity of calorimeter is $\pu{2.5 J/K},$ then magnitude enthalpy of combustion of gas in $\pu{kJ/mol}$ is?

In this question, according to me $ΔU = 2.5 × 0.45 = 1.125,$ i.e. $1.125/(3.5/28) = \pu{9 kJ/mol}$, and $ΔH = ΔU + nRΔT = 1.59.$

But according to my teacher, we have to find the heat evolved (which in isobaric process equals ΔU). So answer is $9$.

This is a question from IIT-JEE so I think there is no error in question or answer.

Isn't enthalpy of combustion the change in enthalpy?

The enthalpy of combustion (aka heat of combustion) is defined as the enthalpy change in going from pure reactants (each at 1 bar pressure) to pure products (again, each at 1 bar pressure) at 298 K.

In this problem, it is first of all necessary to assume ideal gas behavior to proceed.  For ideal gases, the change in enthalpy in going from pure reactants each at 1 bar pressure to the starting gas mixture in the calorimeter, there is no change in enthalpy at constant temperature.  Then, application of the first law of thermodynamics to the mixture before and after the reaction leads to the conclusion that $\Delta U=Q$, since the reaction volume is constant.  Since the temperature change between the beginning and end of the combustion in the calorimeter is essentially zero, this means the the calorimeter has measured the internal energy change of the reaction.  As Buck Thorn has pointed out, the change in enthalpy for the gases in the calorimeter is given for an ideal gas by:
$$\Delta H=\Delta U+\Delta (PV)=\Delta U+RT\Delta n$$where $\Delta n$ is the stoichiometric change in the number of moles of gas.  He also correctly pointed out that there is nothing said in the problem statement about the stoichometry of the reaction or the change in the number of moles of gas.  So, as he indicated, the problem statement is flawed.  However, if we arbitrarily assume that the change in the number of moles of gas is zero, we would have to conclude that the change in enthalpy for the process in the calorimeter is equal to the change in internal energy.  Then, since, again, the products form an ideal gas mixture, there is no change in enthalpy in going from the final gas mixture in the calorimeter to the pure products at 1 bar pressure.  So, based on these assumptions, the enthalpy of combustion is equal to the internal energy change in the calorimeter (also Q) per mole of species being burned.


Why does neighbouring group participation happen? Can someone please explain why neighbouring group participation happens?

Does it stabilise the carbocation? If so then is NGP dominant over an intermolecular SN2 reaction (especially in aprotic medium)?

There may be some confusion over what neighbouring group participation (NGP, also sometimes referred to as anchimeric assistance) is. Carey puts it quite succinctly:

  When a molecule that can react by nucleophilic substitution also contains a substituent group that can act as a nucleophile, it is often observed that the rate and stereochemistry of the nucleophilic substitution are strongly affected. The involvement of nearby nucleophilic substituents in a substitution process is called neighboring- group participation.

  Source: Advanced Organic Chemistry (Part A). Francis Carey.

Rather than a discrete carbocation forming, which is then stabilised, NGP usually (though not exclusively) occurs via two consecutive SN2 processes, with overall retention of stereochemistry.

The reason why the NGP happens, rather than an external nucleophile just doing the substitution is generally due to the fact that intramolecular reactions happen at a faster rate than intermolecular ones, since this removes the necessity of encounter between nucleophile and electrophile (the two are essentially tethered together). The initial internal attack forms a ring (usually 3 or 5 membered for reasons that should be familiar to you) which may then be opened by the external nucleophile.

A classical example is of a sulfide displacing a halide, with subsequent opening of the reactive 3-membered intermediate:

  Source: Organic Chemistry. Clayden/Warren.


Molar absorption coefficient - why are the molar absorptivities defined to be the same? In spectroscopy, the isosbestic point is said to be where the molar absorptivities of two species is the same.

$$A= \epsilon(X+Y) \times c(X+Y) \times l$$

reduces to $$A= \epsilon_t \times c \times l$$

the absorbances and the length are the same at this point, but the concentrations are not;

why are the molar absorptivities defined to be the same?

Is c taken to be the relative contribution, rather than the exact concentration?

Actually that's not the correct definition of the isosbestic point. From IUPAC goldbook:

Isosbestic point - Wavelength, wavenumber or frequency at which the total absorbance of a sample does not change during a chemical reaction or a physical change of the sample.
A simple example occurs when one molecular entity is converted into another that has the same molar absorption coefficient at a given wavelength. As long as the sum of the concentrations of the two molecular entities in the solution is held constant there will be no change in absorbance at this wavelength as the ratio of the concentrations of the two entities is varied.
Contrary to a widely accepted idea, the existence of an isosbestic point does not prove that the reaction is a quantitative conversion of one species into a unique other species or that an equilibrium exists between only two species. The observation of isosbestic points only indicates that the stoichiometry of the reaction remains unchanged during the chemical reaction or the physical change of the sample, and that no secondary reactions occur during the considered time range, For the reaction A + B → c C + d D + e E, with c, d, and e the percentages of the products C, D, and E, an isosbestic point will be observed at every wavelength where the condition ɛ A + ɛ B = c ɛ C + d ɛ D + e ɛ E, provided that the values of the percentages c, d, and e remain constant during the chemical reaction or the physical change. The use of the obsolete term isoabsorption point is not recommended.


Regioselectivity in coupling reactions of α-naphthol I have seen in the following reaction of azo coupling where coupling takes place at position 4 of $\alpha$-naphthol.

Now the electron density accumulation is similar at both position $4$ as well as $5$.

However, there are more resonance structures that can be drawn for position 5. It should also be sterically easier to attack position 5.
The only reason I can think of for position 4 over 5 is that $\ce{=OH+}$ has a -I effect, which may somewhat stabilize the carbanion. But at the same time due to large distance this may be very weak and may not efficiently stabilize the negative charge. Then why is it in the para and not the peri position that coupling preferentially occurs?

The product at the para-position is more stable as it allows the compound to exist in the quinoid form, rather than only in the benzoid form:

This is important as in your case, if the substituent $\ce{X}$ is $\ce{H}$, then depending on the solvent, as much as 89% of it may exist in the quinoid form ($\bf{2}$).
References:

Satoshi Kishimoto, Shinya Kitahara, Osamu Manabe, Hachiro Hiyama, “Studies on coupling reactions of 1-naphthol. 6. Tautomerism and dissociation of 4-arylazo-1-naphthols in various solvents,” J. Org. Chem. 1978, 43(20), 3882–3886 (https://doi.org/10.1021/jo00414a020).


Is hydrogen bonding separate from permanent dipole interactions? Is hydrogen bonding separate from permanent dipole interactions? What I mean here is that a molecule such as ethanol has hydrogen bonding, but does it also have permanent dipoles other than the ones on the oxygen and hydrogen atoms that cause permanent dipole attraction?

Consider the three hydroxyphenols catechol (ortho-hydroxyphenol), resorcinol (meta-), and hydroquinone (para-).  All three engage in hydrogen bonding with their hydroxyl groups.  But they differ in polarity.  Hydroquinone, with its hydroxyl groups symmetrically placed on opposite corners of the benzene ring, has a limited (average) dipole moment between two conformations, one polar and one nonpolar, whereas the other two isomers have an additional dipole from their hydroxyl groups being nonsymmettically located and have larger overall dipole moments.


How to read a given chemical formula? I'm trying to read this chemical formula but I'm a bit confused. I'm hoping that someone could help break this down for me to explain what the various parts mean. Someone purchased epigallocatechin in a purchase request and I found that it's synonymous with the following.

(−)-cis-2-(3,4,5-Trihydroxyphenyl)-3,4-dihydro-1(2H)-benzopyran-3,5,7-triol

I believe that the minus up front refers to the way that the polarity changes as light passes through the molecule. What does the 3,4,5 refer to and the 2?

Klaus's answer has all the right explanations but I always find these things hard to follow without pictures, so here are some to clarify.

Note that in the pictures below we don't explicitly draw carbons or the hydrogens attached to them (we just mentally add enough hydrogens to make the number of bonds to each carbon equal 4).

First what is benzopyran?

The two numbers on the first structure tell us where the oxygen appears in the pyran ring and where the saturated carbon (with an extra hydrogen) appears. Both could be in different places. The second picture showns the numbering convention for each atom in the ring so we can keep track of where anything else is attached. The third picture shows what happens if we notionally add extra hydrogens on carbons 3 and 4 to give a fully saturated ring.

Now we add an OH (hydroxy) in certain positions:

The picture on the right shows how we number the parts of trihydroxy phenyl when it is attached to something else (it will be stuck on the 2-carbon on the benzopyran in the final compound).

Finally we have the epigallocatechin:

The "cis" indicates that the two hydrogens (mentally fill them in in positions 2 and 3) are on the same side of the benzopyran (the opposite side to the two wedge-like bonds on the OH and phenyl groups).

I hope that all makes some sort of sense. Note that none of the above bears any relation to how the molecules are made: this is all notional deconstruction for the purpose of explaining how the molecule is named.


What do '(6S)' and '5' mean in (6S)-5-formyltetrahydrofolate? Tetrahydrofolate:

I guess (6S) means some sixth S-center and 5 some other kind of numbering.

I've redrawn a truncated version of the title molecule below with the "5-formyl" group in place.  I've also inserted small numbers so you can see how the base system is numbered.  So "5-formyl" means that a formyl group ($\ce{-CHO}$) is attached to the "5" position.  The 6-position has 4 different substituents attached to it so it is a chiral carbon atom.  The R-S nomenclature system is used to describe the 2 mirror image possibilities at chiral centers.  The enantiomer you've drawn is the "6S" stereoisomer.


Are precipitations exothermic and/or endothermic? Should be an easy one. I'm fumbling a concept. I've read precipitations are exothermic. Is this accurate? Why would there be no endothermic precipitation reactions?

No, absolutely not.
Precipitation reactions can be either endothermic and exothermic.

Table 1. Thermodynamic data of precipitation for some salts
\begin{array}{cccccc}
\hline
\text{Salt} & \Delta G_\text{ppt}^\circ & \Delta H_\mathrm{ppt}^\circ & -T\Delta S_\mathrm{ppt}^\circ(\pu{25 °C}) \\
\hline
\ce{Be(OH)2} & -121 & -31 & -90 \\
\ce{Mg(OH)2} & -63 & -3 & -61 \\
\ce{Ca(OH)2} & -28 & 16 & -44 \\
\ce{Li2CO3} & -17 & 18 & -34 \\
\ce{MgCO3} & -45 & 28 & -74 \\
\ce{CaCO3} & -48 & 10 & -57 \\
\ce{SrCO3} & -52 & 3 & -56 \\
\ce{BaCO3} & -47 & -4 & -43 \\
\ce{FePO4} & -102 & 78 & -180 \\
\hline
\end{array}

It is more than clear that all precipitations are not exothermic.

Also definitely noticable is the fact that $\Delta S$ is a positive quantity.

But that is unexpected, as an increase in the disorder of the system upon precipitation may seem counterintuitive to many.

What is often overlooked is that several other factors are at work.

Paving the way

Ions can be of two types:

Electrostatic structure breakers
Electrostatic structure makers

This classification is based on how ions interact with water.

Electrostatic structure breakers

Sometimes referred to as chaotropic ions, these disrupt the hydrogen bonding interactions between water molecules. Examples include $\ce{NH4+}$, $\ce{Cs+}$, $\ce{Br-}$, $\ce{I-}$.

It is tempting to assume that large cations with higher possible coordination numbers would attach and order water molecules more effectively than small cations, but this is not experimentally observed.

Smaller the radius and/or greater the charge (the more acidic or basic the ion), the greater its structure-making properties in aqueous solution. It is hence clear that the above mentioned ions are structure breakers rather than makers, evidently having very low values of $Z/r^2$.

It is easy to deduce that upon their introduction into aqua, they will correspond to an increase in entropy.

Electrostatic structure makers

AKA kosmotropic ions, these increase the ordering of water. $\ce{Fe^2+}$, $\ce{Al^3+}$, $\ce{Mg^2+}$, are members of this club.

These ions have large charges and relatively smaller radii, resulting in a higher $Z/r^2$ value.

Via hydrogen bonding these ions create relatively larger shells of hydration than their low $Z/r^2$ counterparts.

The differences in hydration for the two types of ions is visible in the following table.

Table 2. Hydration data for some cations

Uh..What does this have to do with the question?

Everything.

The reason that most acidic cations and basic anions combine to give precipitates is due to the disorder that results from the release of large number of water molecules from the well structured hydration spheres.

The second law favours this.

For an example, let us consider the precipitation of $\ce{MgCO3}$

$$\ce{Mg(H2O)36^2+ + CO3(H2O)28^2- <=> MgCO3(s) + 64 H2O}$$

The number of water molecules are derived from Table 2.
The release of $64$ molecules of water per molecule of magnesium carbonate precipitated is very indicative of the positive entropy of the process.

Now for the enthalpy.
Rather than the Gibbs–Helmholtz, let's return to the good old Hess law.

According to Hess' law,

$$\Delta H_\mathrm{ppt}(\ce{M_yX_m)}) = -y\Delta H_\mathrm{hyd}(\ce{M^m+})- m\Delta H_\mathrm{hyd}(\ce{X^y-}) + U(\ce{M_yX_m})$$

The electrostatic structure makers (type 1) would have higher negative

$$y\Delta H_\mathrm{hyd}(\ce{M^m+})+ m\Delta H_\mathrm{hyd}(\ce{X^y-})$$

leading to an overall positive heat of precipitation. e.g. $\ce{FePO4}$, $\Delta H_\mathrm{ppt} = \pu{+78 kJ/mol}$

Vice-versa for type 1: the structure breakers.

I hope I managed to enlighten you on this matter.

P. S. This was a great question. Pardon the short conclusions, I am currently travelling. Members of the community may feel free to edit and make a more dramatic exit.

References:

Hanzlik, R. , Inorganic Aspects of Biological and Organic Chemistry 1976, II, 26-30

https://www.wou.edu/las/physci/ch412/ppt2.htm

https://www.wou.edu/las/physci/ch412/pptrxn.html


Are entries ever removed from the Merck Index? I was re-sorting books the other day, and thought about how many years it has been since I bought my Merck Index. I'm sure there are new entries in the more recent additions, but I do not have such a need to stay that up-to-date.

At first thought, it seemed to me that a book of this nature only adds entries, and does not remove any entries. But is this the case? Are compounds de-listed because they are bad science? Other reasons?

Iit would be great to know other details about the editorial process, as well, if relevant.

Without several print editions side-by-side, I restrain my answer to the electronic version. This public link or here regarding the online version based on the 14th edition may be read like it may be the case.  To quote:

 "The 14th Edition also contains "retired" records from earlier editions of The Merck Index which still contain valuable information but are no longer being updated with new information."

Of course, beside RSC you may contact them (Merck in Rahway, NJ), too.


Why is it that neither silver nor copper react with a strong acid? $\ce{Fe}$, $\ce{Mg}$, $\ce{Ni}$, $\ce{Pb}$, $\ce{Sn}$, and $\ce{Zn}$ all react when they (in solid form) are submerged in an acid solution with the presence of a strong acid like $\ce{HCl}$, but silver and copper do not.

I have examined the electronegativity values and both $\ce{Ag}$'s and $\ce{Cu}$'s electronegativity value is only different from $\ce{Ni}$ by $\mathrm{0.1}$ on the Pauling scale. I don't see a pattern; the only thing I see which is common between copper and silver is that they both may have an oxidation number of +1.

So, why is it that $\ce{Ag}$ and $\ce{Cu}$ are immune to the H cations in the acidic solution?

This will involve some degree of hand-waving. Let me first limit the analysis to the transition metals so that more effective comparisons can be made.

The reaction between a metal and an acid requires the metal to be oxidized, i.e. lose electrons to the acid, creating positively-charged metal ions in solution. The general idea is that as you go along the transition metals in a period from left to right, there is an increase in effective nuclear charge felt by the valence $ns$ electrons (due to the additional $(n-1)d$ subshell electrons poorly shielding the $ns$ electrons from the nucleus) making the valence electrons harder to remove. This trend can be seen to some extent in the increase in ionization energies of the elements, for example, and also in the standard reduction potentials for metal cations of same charge (they become more positive, in other words making oxidation and hence attack by acid more difficult). Thus, metals at the left of the transition metals tend to be more easily attacked by acids than the metals at the right.

There is a rather interesting exception though: zinc strongly breaks the trend, being far easier to oxidize than copper, the element before it, even though zinc has a much higher ionization energy, clearly indicating its electrons are held tighter (cadmium also exhibits this anomaly, to a milder extent). What gives? The thing is that the tendency for a metal to oxidize is the result of several combined factors, one of which is also how strongly bound the solid metal is in the first place. If a metal contains atoms which are strongly bonded to each other, then oxidation tends to be more unfavourable as it would require these bonds to break. When comparing copper and zinc, it is clear that the latter has far weaker metallic bonding in the solid (their melting/boiling points are $1360\ \rm{K}$ / $2835\ \rm{K}$ and $695\ \rm{K}$ / $1180\ \rm{K}$, respectively). Thus, if you include the energy necessary to separate the atoms from the solid before ionizing them, it turns out that the process is easier for zinc than copper.

Comparing different rows of the transition metals is a bit less clear, in part because not all elements have ions of the same charge which can be directly compared. The general tendency is that the metals become tougher to oxidize as you go down the rows. For the transition metals in the sixth and seventh period this is probably a consequence of lanthanide/actinide contraction and relativistic effects, which decrease the energy of the valence $ns$ orbitals.

Now for some other elements. Magnesium is very reactive towards acids because it is both a metal containing relatively weak bonds and because its ionization energy is comparatively low, being an alkaline earth element. For tin, the metallic bonding is not weak, but its ionization energy is not too high, so it will oxidize in acids, though not as strongly. Lead is somewhat less reactive to acids, possibly because its oxidation tends to stop at +2 instead of tin's +4, creating a significantly softer cation which isn't as well solvated by water.


Rotary Evaporator Cleaning I am using a rotary evaporator ('rotovap') in my lab mainly with dichloromethane as a solvent.

Usually between experiments I clean the receiving and boiling flasks with a brush using soap and water and finally rinse with alcohol.
For the condenser I try to do the same, but it is next to impossible to brush the water serpentine.

I want to try and distil some coffee and tea for consumption, but I am afraid of any possible contaminants.

What is an efficient and safe method to clean the glassware?

There's no further need to comment on the coffee maker, but you might give the following cleaning method for the condenser and the rotation tube a try:

Note that you might want to first check whether this is in accordance with local safety regulations!

Take a small or medium-sized (max. 500 mL) 2-neck flask (NS29 socket connects to the rotavap, NS14.5 socket is for your thumb) and fill it with max. 250 mL acetone
Hook it up to the rotavap, support it with your hand
No spinning :D
Start the vacuum pump
Close the NS14.5 socket with your thumb
Watch the acetone starting to boil under reduce pressure
Rapidly remove your thumb from from the socket and let the acetone "spit" all through the rotavap


What will form in the hydrolysis of urea in urine? I just wanna know what is the end product of hydrolysis of urea in urine. I know that the hydrolysis of urea will form ammonia and a subsequent rise in pH but will the ammonia react with the water from urine to form ammonium and hydroxide?

The hydrolysis is the first of two reactions that occurs in aqueous urea. This reaction forms one equivalent of ammonia and one equivalent of carbamic acid. Carbamic acid rapidly decomposes to form gaseous carbon dioxide and another equivalent of ammonia:

The ammonia then simply participates in an equilibrium where it reacts with water to form ammonium hydroxide, or so it is commonly described. There is actually a fair amount of debate on this topic, but it typically written like this: $$\ce{NH3 + H2O <<=> NH4OH}$$


Can I expect 18O-hydroxide when adding slight amount of NaOH-16O to 18O-water? In adding $\ce{^{16}O}$-sodium hydroxide to $\ce{^{18}O}$-water to get a $0.25\pu M$ solution, is the hydroxide in form of $\ce{^{16}O}$-hydroxide or $\ce{^{18}O}$-hydroxide?

is there an oxygen exchange between  $\ce{^{16}O}$-hydroxide and  $\ce{^{18}O}$-water?

Yes, it will but it will be a hydrogen exchange, not an oxygen exchange.
$$\ce{2 H2^{18}O <=> \color{blue}{H3^{18}O+} + ^{18}OH-}$$
$$\ce{\color{blue}{H3^{18}O+} + ^{16}OH- <=> H2^{18}O + H2^{16}O }$$
The oxygen will mix with water and hydroxide to a near equivalent ratio of $\ce{^{18}O:^{16}O}$ in both $\ce{H2O}$ and $\ce{OH-}$ but not perfectly equivalent as isotopes do not have equivalent chemical potentials.


Why alkyl shift happens during hydroboration oxidation? In this oxidation step, where alkyl shift happens for 3 times to kick out the OH- to obtain the boron-ester. But why does this happen? Won't the boron atom become less stable as the empty p orbital is regenerated? Won't this cause the OH- (strong nucleophile) to attack the boron (electrophile) again?
What is the mechanism of trialkylborane oxidation with hydroperoxide? (I have read this but sorry I don't get it. before this reaction, isn't the boron atom also bonded to an oxygen atom? Why would this reaction make it more stable?)

But why does this happen?

Well, if it was a plain old carbon in the middle, this migration wouldn't happen. It would sit there forever. Boron isn't carbon, though: it has fewer protons, is less electronegative, and doesn't like holding onto electron density all that much. That electron density gets pushed onto the adjacent carbons, which make them rather more nucleophilic, and happy to jump next door into the $\sigma_\ce{O-O}^*$ orbital.
You can see, after all, that there's something about that boron that's not quite perfect: it has a formal negative charge. While this shouldn't be taken as a sign of great instability, it does suggest that boron won't mind losing that formal negative charge too much.

Won't the boron atom become less stable as the empty p orbital is regenerated?

The energetics of this migration aren't just about the boron atom, it's about the system as a whole. You are also breaking a weak O–O bond, forming a stronger C–O bond in replacement, and so on. And indeed, even if this one step is energetically unfavourable, it might not even matter as long as the overall reaction is spontaneous.
Is the overall reaction energetically favourable? I'll leave it to you to think about it, based on the types of bonds that are broken and formed over the course of the entire reaction. You already saw some of the considerations in the linked question. (I don't get what you mean by the boron atom being bonded to oxygen before the reaction; it isn't, as it has three bonds to carbon.)

Won't this cause the $\ce{OH-}$ (strong nucleophile) to attack the boron (electrophile) again?

Well, yes, but not much happens if you go down that route. It's far more interesting if $\ce{OOH-}$ attacks the boron again, because that means the second alkyl group can migrate. That is precisely the mechanism that you have learnt.


Measuring pH of pasty solutions/mixtures I'm trying to measure the pH of a very viscous paste mixture and have been told to dilute the sample for more accurate results, because the viscosity of the sample may prevent the sensor to detect the real pH of the sample.

Does this sound right to you? Wouldn't diluting the sample alter the pH?

Thank you

Diluting the sample does alter the pH, but it does so in a predictable way.

When we are measuring pH, we are measuring the concentration of hydrogen ions (or hydornium ions, or whatever ) via the negative logarithm:

$$\mathrm{pH} = -\log_{10}\left([\ce{H+}]\right)$$

Diluting a solution has a predictable effect on concentration, since the total amount (in moles) of $\ce{H+}$ will not change by the addition of more solvent. This relationship is summarized by the formula (where $c$ represents concentration, $v$ represents volume, and $1$ and $2$ represent the solution before and after dilution, respectively:

$$c_1v_1 =c_2v_2$$

Let's take an example. Suppose you have 1.000 mL of a viscous acidic aqueous solution. The viscosity of the solution will impede easy measurement of pH because of the kinetics of mass transport. If you dilute this solution with water to a total volume of 10.00 mL, you now have a less viscous solution with the same number of moles of $\ce{H+}$, because sources of $\ce{H+}$ were neither added nor removed.

If this new diluted solution has $\mathrm{pH = 3.03}$, then $$[\ce{H+}] = 10^{\mathrm{-pH}}=10^{-3.03} = 9.33\times 10^{-4}\ \mathrm{M}$$.

The concentration of the original solution can now be determined:

$$c_1 = \dfrac{c_2 v_2}{v_1} = \dfrac{\mathrm{9.33\times 19^{-4}\ M)(10.00\ mL)}}{\mathrm{1.000\ mL}}=9.33\times 10^{-3}\ \mathrm{M}$$

The pH of the original solution is:

$$\mathrm{pH = -\log_{10}(9.33\times 10^{-3}) = 2.03}$$

Note, that since I chose to dilute by a factor of 10, the pH of the dilute solution is 1 pH unit higher than the original solution.

This approach works for solutions of strong acids and strong bases and will work over much of the pH range, but may get a little awkward around neutral. The approach breaks down for weak acids and bases since changing the concentration changes the position of equilibrium. For weak acids, diluting likely results in additional ionization events. However, even without knowing the acid/base and its equilibrium constant, you could perform multiple dilutions and do some curve-fitting and get back to the original pH.


Which one has the highest boiling point? 10 g of substance below were dissolve in each 1 L water:
$\ce{Na2SO4}$
$\ce{NaCl}$
Glucose
$\ce{MgCl2}$
Which one has the highest boiling point?

Glucose has London dispersion force.
$\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}$ have ionic forces.
Ionic forces is stronger than London dispersion force, so $\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}$ have higher b.p. than glucose.
Now, how to decide which one has highest b.p. among $\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}?$

Boiling point elevations proportional to the molar concentration of dissolved particles.  So that's what you have to figure out, being careful to (1) properly convert mass to moles and (2) recognizing that some of the solutes can form ions so you can get more than one particle per "molecule".

Let's look at glucose.  If you work out the molecular mass it's about 180 grams per mole.  Divide that into 10 grams and you get about 0.056 moles of glucose molecules in your one liter of solution.

Now try sodium sulfate.  For this one you have a "molecular" mass of 142 grams per mole, so you seem to get about 0.070 mole of particles.  Except ... Sodium sulfate forms ions, each "molecule" gives one sulfate ion and two sodium ions.  Thus three particles overall, and you have to multiply your 0.070 figure by that factor of 3.  Thus the sodium sulfate solution will have 0.21 mole of ions in one liter.  The sodium sulfate solution will therefore give a higher boiling point than the glucose solution.

Work it out now with the other two choices and pick the one with the most particles in solution including ion formation.  Be sure to use reasonably accurate atomic masses because this set of choices could throw you for a loop.


Oxide content description From time to time, I come by to see what people do with my mhchem package. Here, I came across a notation that I did not know yet: $\ce{K2O-nSiO2-xH2O}$ and $\ce{K2O - 3.2SiO2 - 2.7H2O}$

What is this? Is this really a bond? Is this a short form for writing $\ce{K2O-Si_nO_{2n}-H_{2x}O_x}$ and $\ce{K2O - Si_{3.2}O_{6.4} - H_{5.4}O_{2.7}}$? Is this an established notation that mhchem should support? (It doesn't yet with the numbers.) Do you have authoritative References?

The notation, upon replacing potassium with sodium, can refer to a water glass composition. The unique notation, in my opinion, is warranted based on the following comment from a paper1:

Despite of years of investigations, there are still questions about
the molecular species configuration of these solutions that need to be
answered to gain clarity over the best conditions for the various
applications
When adding sodium hydroxide ($\ce{NaOH}$) to a water glass, the ratio (molar
or weight) $\ce{SiO2/Na2O}$ is decreased. This ratio is called water glass
modulus (n) and determines various physical and chemical properties
such as the pH and the viscosity of the solution.
[...]
the viscosity is extremely dependent on the modulus and increases as
the solution becomes either more siliceous or more alkaline, i.e. at
both higher and lower $\ce{SiO2/Na2O}$ molar ratios (n). Since the viscosity
of such disperse systems like this is given by the silicate
conformation i.e. extended chain conformation and the degree of
polymerization
[...]
the degree of polymerization of the predominant silicate species in
the alkali-solution [...]

So we have apparently a weakly specified structure characterized by extended chain conformation and degrees of polymerization with its rather unique notation (seems right to me).
Reference

Helén Jansson, Diana Bernin, and Kerstin Ramser , "Silicate species of water glass and insights for alkali-activated green cement", AIP Advances 5, 067167 (2015), DOI: 10.1063/1.4923371


Principle of potentiometric titration In potentiometric titration between ascorbic acid and iodine, when we plot the graph of EMF with volume of iodine consumed, we get a sudden decrease in EMF at equivalance point--why? It is also negative--why?

Using Nernst equation at equivalence point we get
$$E_{cell}=E_{STP}~~~(voltage ~at ~STP)$$

What is its physical meaning?

In potentiometric analysis, there are two electrodes, an indicator electrode (platinum here) where reduction takes place and reference electrode (calomel electrode) where
oxidation takes place.

So here the reference is not $E_{STP}$ cell but the $E_{cathode}$ that is the calomel electrode, so there is drop to $-241mV$ after the equivalance point. Hence,
 $E_{cell} = E_{cathode}$


Carbonation of Grignard reagent with a carbonyl group present in it Here's the question in my book (see picture)

Option A is a standard reaction which gives a carboxylic acid.

Option B is definitely wrong because an acid base reaction will take place and CH4 will be formed after the first step. CO2 will not react with the methane gas.

So without knowing about option C, I can safely tick option A. But what happens in option C?
Here's the solution provided in the book.

Why cannot a carbonyl group and Grignard reagent be present in same substance? Also, what will be the reaction?

You should look at two things:

1)  Will the reaction stay on course?  If you add water to the Grignard reagent first, without the carbon dioxide initially, what do you get?  Is it something that will get you your carboxylic acid when you add the carbon dioxide?

2)  Can you make the Grignard reagent to begin with?  Here again you are asking the reaction to stay on course; if the Grignard function can react with another functional group in the molecule, it will do so either within the molecule or as an intermolecular reaction between two of the molecules.
Then you no longer have your Grignard reagent.

Only one of the given possibilities works against both of these questions, so there is your answer.


Reactivity of fluorine compound I have tried to make sense of the article on fluorine on Wikipedia but I am still very perplexed.

I would like to know generally how reactive the fluorine compound β-(4-Fluorophenyl)-γ-aminobutyric acid; β-(4-Fluorophenyl)-GABA or "Fluorophenibut" would be as it passes through your body. Is fluorine a generally reactive element irrespective of what it is bound to or is it generally stable since it is bound to a carbon atom on an indole ring?

Fluorine the element is very reactive, but that doesn't mean its compounds are

Fluorine the gas is a very reactive element because it loves to take electrons from other compounds. But this is irrelevant to whether compounds containing fluorine are reactive. Consider the more common chlorine. Chlorine gas is less reactive than fluorine, but is still very reactive. It is a dangerous and corrosive compound that has been (and probably is still) used as a military weapon. But you encounter very unreactive compounds of chlorine every day. Common salt is a chlorine-containing ionic compound that makes up a significant portion of the electrolytes in your body; PVC is a fairly inert plastic made from a polymer containing chlorine carbon bonds.

The very reactivity of fluorine makes some of its compounds very unreactive. F-C bonds, for example, are very strong and often very unreactive even compared to other carbon-halogen bonds. PTFE (a long carbon chain with all the hydrogens replaced by fluorine) is a very inert polymer that is safe enough to use at the high temperatures achieved in frying pans, for example.

Some fluorine compounds are, however, very biologically active but not because their fluorines are "reactive". Fluoroacetic acid and its salts, for example, are dangerous (and naturally occurring) poisons. But their mechanism of action relies on their stability not their reactivity: they are similar enough to acetic acid that they get involved in the standard citric acid cycle but end up blocking one of the key enzymes to further reactions because the product is too stable.

Whether or not fluorophenibut is "reactive" in the body in the sense you mean is unlikely. It is a neurodepressant similar to gabapentin and pregabalin, though little is known about its safety.

Complicated molecules can't be judged as dangerous, safe or reactive just because of the elements they contain. You need to know the detailed and specific ways they interact with bodily processes. Prozac, for example, is a widely used anti-depressant that contains fluorine. Just because it does doesn't make it reactive or dangerous.


Nickel plating with nickel acetate, nickel chloride, sulfur, saccharin and coumarin I am a hobbyist trying to learn/apply a nickel electroplating process that I can safely do at my shop.  My substrate is a machinable non-stainless steel with unknown carbon content, and I'm looking to do a close variation of the "Watts Bright nickel" formula.  I have a 99.9% pure nickel anode, nickel acetate, nickel chloride, boric acid, sulfur powder, saccharin and coumadin powders.

The formula calls for nickel sulfate; is it possible to simply add the powdered Sulfur to the solution, or do I have to use a different process to synthesize the nickel sulfate? If so,  can I do so with the materials that I have?
Also, the Watts formula does not include the saccharin or coumarin, but I know that these chemical additives will increase the odds of a smooth mirror like finish.

Also, is there a recommended quantity range, or at least a starting point, for these two chemicals for this application? Any insights here from an experienced chemist/electroplater would be greatly appreciated.

The formula calls for nickel sulfate; is it possible to simply add the powdered sulfur to the solution, or do I have to use a different process to synthesize the nickel sulfate?

This will not work. Sulfur ($\ce{S}$) and sulfate ($\ce{SO4^{2-}}$) are two pairs of shoes. The oxidation state is different and there is no way that sulfur powder will turn into sulfate during the electroplating.

  If so, can I do so with the materials that I have?

It might be possible to heat another nickel salt in sulfuric acid, cool down and wait for the nickel sulfate to precipitate, but I wouldn't do it!

Given that the nickel salts are highly toxic, might cause cancer on inhalation and are harmful for aquatic organisms, I wouldn't mess around with crystallisation and the remaining solution for which you would have to find a safe way for disposal.

It might be much easier to aquire nickel sulfate and set up the bath for electroplating according to the given specifications.


Question on Hydrogen Balloon I tried to extract hydrogen for my kid's birthday, by mixing caustic soda with water and aluminum foil (mixed in a glass bottle). A lot of heat and (hopefully)hydrogen was released.
However, within 2-3 hours, the balloons started losing altitude and began falling down. I supposed it was caused by losing the gas slowly.
But when I tried to burn the ballon the next day it did not create the pop sound for the hydrogen test.
Please excuse me if this is a naive question. I'm curious since this is the first time I'm dealing with hydrogen and balloons. I attempted this for almost 4-5 time. And every time the balloon stopped flying within 2-3 hours. All other balloons filled with air did not lose gas.
I want to know what happens to the gas. Does it go out? Is it some sort of diffusion?
Thanks.

Hydrogen, having very small and fast moving molecules,  diffuses faster and through smaller holes than air. Your balloons seem to be permeable for hydrogen. Try a different brand, if it helps. But as others have mentioned, rubber is not the best material to keep hydrogen. Better is a metallised plastic.
Pure hydrogen just burns in a flame.  If you have ever  watched the film footage (1937)  of the catastrophic accident of the airship Hindenburg , it was just a huge scale of burning in a flame.
What makes an explosive "bang"is a hydrogen mixture with air, even better with oxygen. Because  hydrogen does not explode, until is contains >6% v/v of oxygen.
Any binary chemical explosion of gases requires a self-sustaining chain reaction. That does not happen if a mixture contains too little of one of mutually reacting gases.


An explosion in a custard factory I have a vague memory of a question on the BBC panel game QI which started with a long chemical reaction equation on the screen, asking what it might be.

The answer was definitely "an explosion in a custard factory", due in part to high quantities of fine powders in the air, but also some of the chemical constituents of custard powder.

Does anyone know what that equation would be, or how I might re-find it?

How would one go about getting back from that equation to custard?

but also some of the chemical constituents of custard powder

Don't blame it on the custard. As long as there's

dust of a combustible material, preferably with a small particle size
sufficient oxygen in the athmosphere
an igniter, such as a spark or an open flame

there's a good chance for a devastating dust explosion.

Your combustible material can be something seemingly unspectecular, coal dust, saw dust, flour, or custard powder.

UPDATE

The incident in the BBC game most likely took place at a factory in Banbury, UK in 1981. Their traditional custard powder, known as Bird's Custard, is egg-free and mostly consists of cornflour.

  how you might use it to tell that the reaction was an explosion in a custard factory

In risk assessment for dust explosions of organic materials, product analysis rarely plays a role; the "equation" isn't interesting and it is usually assumed that all carbon ends up as $\ce{CO2}$.

In order to establish workplace safety, it is more important to determine, how "strong" a possible dust explosion might get.

A measure for the "strongness" of an explosion is the rate by which the pressure (rapidly) increases. This is expressed in the $\mathrm{K_{St}}$ value (in $\mathrm{bar\cdot m \cdot s^{-1}}$), and experimentally determined in standardized tests.

Based on the $\mathrm{K_{St}}$ value, materials of certain properties (particle size, rest humidity) are assigned to the explosion classes (ST 0 to ST 3), and suitable workplace safety measures are taken.


How to calculate entropy change for a gas reaction using $S^\circ$? I'm calculating molar changes in thermodynamic properties due to reactions between gasses (assumed to be ideal gases).  I can calculate $\Delta H$ easily enough, because it's just $\sum_i \nu_i\Delta_f H^\circ_i$, with $\nu_i$ the stoichiometric coefficients. $\Delta G$ (at standard pressure) can be calculated from $\sum_i \nu_i\mu_i$, with $\mu_i = \Delta_f G^\circ_i + RT\log[i]$. Since $\Delta G = \Delta H - T\Delta S$, I can calculate $\Delta S$ as $\frac{1}{T}(\Delta H - \Delta G)$.

So far so good. But I'm inexperienced in working with tabulated quantities (I'm a more of a theoretical physicist than a chemist) and keep making mistakes with signs and units. So I figured that as a sanity check I would calculate $\Delta S$ directly, using tabulated values of $S^\circ$. But then I realised I don't know how to do this, and it seems that none of the textbooks on my desk explain it either.

Based mostly on intuition, it seems like it should be $\Delta S = \sum_i\nu_i(S^\circ_i + R\log[i])$, with the $R\log[i]$ term having something to do with the entropy of mixing. For the example reaction I chose, this gave something with the approximately correct magnitude but the wrong sign (-15.6 instead of 14.2, which is what I get by calculating it from the Gibbs energy and the enthalpy).

So my question is, what is the correct way to calculate the entropy change due to an ideal gas reaction if, for some reason, you only have access to the concentrations and the standard entropies of the reactants?

Just to make sure that I get your question correctly; the process in question is a chemical reaction that takes place at a temperature and pressure close enough to the reference temperature and pressure (those used for the tabulated values) that any changes to the tabulated values are neglible? (I assume this, as this is what you have done in the enthalpy calculation)

In that case, it is probably easier that you would expect - you can use (almost) the same approach as you used to calculate $\Delta H$ as both entropy and enthalpy are state functions. That is:

$\Delta_{rx} S = \sum_i \nu_i S^\circ_i$

where $S^\circ_i$ is the tabulated value for each of the components. This is described on Wikipedia.

If the temperature deviates sufficiently from the reference temperature used then you should adjust the entropy change accordingly

$\Delta S_{T_1 \rightarrow T_2} = \int_{T_1}^{T_2}\frac{C_V}{T}dT=C_V\ln(T_2/T_1)$

where the last equality holds if $C_V$ can be assumed to stay constant in the interval $T_1 \rightarrow T_2$. A similar adjustment can be made if the pressure, volume etc. changes during the course of the reaction. Again, as the entropy is a state function, it is enough to adjust the total change in entropy for the reaction.

Hope this helps! Please let me know if I am not being sufficiently clear, and I'll try to elaborate things a bit.


What kind of materials are plastic syringes made from? I'm trying to figure out what kind of materials my disposable plastic syringe is made from. I have a Medefil MIS-1130 syringe and when I visited the company website I couldn't find any information on what the disposable syringes were made from (although they did tell me a lot on what's NOT in them).

I'm trying to find a regulation or something that could help me identify what kind of plastic and rubber are used to make the shill, plunger and seal that make up the disposable syringe. Can anyone on here please give me some guidance?

The syringes are advertised as “ISO compliant” on the website of the manufacturer. This is suspiciously vague and does not necessarily mean that the product complies with all requirements of all relevant ISO standards. (Also note that, concerning compliance and quality, the website mentions that the manufacturing facility is “ISO certified”, which is not permissible. However, the manufacturer actually has been certified to comply with ISO 9001:2008 and ISO 13485:2003 – that’s more like it.)

Ideally, such prefilled syringes are made in accordance with the applicable standards of the ISO 11040 Prefilled syringes series and other documents that are referenced in these standards (e.g. ISO 7886-1:1993 Sterile hypodermic syringes for single use – Part 1: Syringes for manual use).

However, the normative minimum requirements for materials are very general and not very restrictive. For example for the material of the barrel according to ISO 11040-6:2012 Prefilled syringes – Part 6: Plastic barrels for injectables:

  For the manufacturing of the barrels, suitable polymers shall be selected and used, based upon the intended application and processing techniques, which include, for example, the method of sterilization or the method of decontamination. The material shall exhibit the appropriate performance properties, e.g. oxygen and water permeability.

or the material of the plunger stopper according to ISO 11040-5:2012 Prefilled syringes –
Part 5: Plunger stoppers for injectables:

  The elastomeric material used shall meet the requirements specified in Clause 7.
  Plunger stoppers shall be made from the elastomeric formulation originally tested and approved by the end user. The manufacturer of the plunger stopper shall ensure the conformance of each delivery with the type sample and compliance with previously agreed functional and compendial requirements.

  The elastomeric material shall withstand two sterilization cycles when autoclaving in saturated steam at (121 ± 2) °C for 30 min without impairment of its function under conditions of normal use. If other sterilization methods are used, e.g. irradiation, the suitability of the material shall be evaluated.

The mentioned Clause 7 includes minimum requirements on hardness

  The hardness agreed between the manufacturer and the user shall not differ from the nominal value by more than ±5 Shore A when tested in accordance with ISO 7619‑1 on a special test specimen. Alternatively, the hardness can be tested on the plunger stoppers in accordance with ISO 48. If tested according to ISO 48, the microhardness shall not differ by more than ±5 IRHD from the type sample.

resistance to ageing

  The maximum time between the date of manufacture of the plunger stoppers and their pharmaceutical use should be agreed upon between the manufacturer and the user.

  The plunger stoppers shall maintain their performance characteristics throughout the entire shelf-life of the medicinal product, which is tested as part of the stability test by the user.

chemical requirements

  The aqueous extractables of the plunger stopper material shall not exceed the limits specified for elastomeric parts defined in ISO 8871‑1.

and biological requirements

  The requirements in ISO 8871‑4 shall apply.

In addition to the normative minimum requirements, ISO 11040-6:2012 also includes informative guidance on materials:

  Materials used in the construction of syringe barrels for use in pre-filled syringes should be suitable for the process to be used, the sterilization method to be applied and for the specific requirements as primary packaging for injectable drug products. Materials used in the construction of pre-filled syringes should not cause the syringes to be detrimentally affected, physically or chemically, by the normal use of injectable preparations and should be safe for the intended use.

  Certain grades of polypropylene (homo-polymer and/or co-polymer polypropylene), cyclo-olefin copolymer and cyclo-olefin polymer have been used for the barrels of pre-filled syringes in applications with a variety of injectable products (drugs and biologics).

  The organization responsible for the marketing approval of the injectable drug product is required to demonstrate the appropriateness and compatibility of the selected material and the suitability of the prefillable syringe as a container closure system. This includes, among others, the verification of drug/container interactions (leachable, extractable and sorption assessments) in reference to regulatory guidance (e.g. FDA and EMA), pharmacopoeias and other guidelines (e.g. ICH).

  To develop the necessary data as outlined above, a collaborative approach is recommended between the pharmaceutical company that holds the marketing authorization of the injectable product and the supplier of the packaging system, with reference to ISO 15378.

Furthermore, ISO 11040-6:2012 includes a list of polymer materials codification:

Cyclo-olefin copolymer (COC)
Cyclo-olefin polymer (COP)
Polypropylene (PP)
Other polymer (OTH)

Remarkably, the manufacturer mentions the materials glass, COP, and COC (without further context) for prefilled syringes on another page, whereas the package insert mentions “The luer lock syringes are made using polypropylene.”

For reliable information, you should directly inquire from the manufacturer of the prefilled syringes. Apparently, this company does not make the empty syringes but only fills syringes that are obtained from another supplier.


Activated Complex Products/Reactants We know that an activation complex is the temporary unstable compound formed when activation energy is reached. But what dictates whether the products are indeed going to be products or if they become reactants?

Does the activation complex form when orientation is both correct and incorrect, and only forms products when the orientation is correct? Or is the decomposition of the complex purely a chance matter?

Transition state theory or activated complex theory is a very advanced topic and deals with chemical reaction dynamics. It goes beyond the scope of my knowledge, however, I will attempt my best to explain with what I know.

As stated, the activated complex is a temporary unstable product. As two particles with sufficient kinetic energy collide; along the reaction path, they distort the bonds and rearrange into new atoms. The location of the transition state for the activated complex is at the highest point; or peak, of an potential energy diagram.

As stated in my general chem textbook Principles of Modern Chemistry by Oxtoby et. al, p. 685:

  The activated complex is assumed to exist as if it were in equilibrium
  with the reactants and the theory focuses on calculating the rate at which the
  activated complex passes through the transition state to form products.

Statistical thermodynamics is required to calculate all motions of particles, including both rotational and vibrational energy.

In the transition state, change from products is mediated when bonds between particles or molecules become out of phase (or antisymmetrically stretched) by vibrational energies. The frequency of the molecules becoming out of phase with one another decreases as the reaction proceeds forward and the new product is transformed. The energy transferred from vibrational energy to achieve the activated complex is turned into kinetic energy.

The Eyring formula makes it possible to calculate reaction rates using transition state theory:

$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}K^†$$

The kappa term is referred to as the transmission coefficent and refers to the probability that the particles will create the transition state. $K_\mathrm{B}T/h$ measures the rate at which the activated complex dissociates to form into it's products.

This equation is unwieldy to us; as you stated previously, it is extremely hard to determine equilibrium constants for the activated complex since the instability achieved at the height of the potential energy for the reaction exists for only a brief period of time.

Instead, we can relate the Eyring formula to thermodynamics and substitute $\exp (-\Delta G^†/RT)$ for $K^†$.

$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta G^†/RT)$$
$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta H^†/RT)\exp(\Delta S^†/R) $$

The equation that Arrhenius derived is:

$$k_r = A\exp (-E_a/RT)$$

Activation energy is expressed as activation enthalpy and activation entropy. The pre-exponential factor $A$ is substituted with $K_\mathrm{B}T/h$. This equation and the use of enthalpy and entropy as driving forces is used extensively in enzyme catalyzed reactions.