Transition Metal Compounds So I have a few dumb questions. I'm still in high school and am doing something on magnetic properties of transition metals. I'm facing a few roadblocks.

1) What're compounds like $\ce{CoBr2, CoCl2, AgI, AgBr, AgCl, AgF, MnBr, Co(CN)2}$, supposed to be? Are they the same as the coordinate complexes in something like $\ce{[Fe(H2O)6]}$?

2) If they're not, how do they bond with halogens like $\ce{Cl}$ or ligands like $\ce{OH/CN}$? Is it similar to ligand interaction and d-orbital splitting?

3) If they're neither complex ions nor have split d-orbitals, what determines their magnetic properties?

I'm having a bit of trouble finding the answers. I'm very new to this whole transition metal thing. On some websites compounds like ferrous bromide is considered a and on another it's b. I'd appreciate any help!

What're compounds like $\ce{CoBr2,CoCl2,AgI,AgBr,AgCl,AgF,MnBr,Co(CN)2}$, supposed to be?

That depends.

In a lot of cases, especially if we are talking about large anions with low charge density, such as nitrate and perchlorate, and moderately small cations, like $\ce{Co(II)}$, obtaining an anhydrous salt is a long, tiring quest, if possible at all. For example, anhydrous metal nitrates for small cations in general are obtained in anhydrous solvents only.

In case we are talking about medium anion with moderate charge density, like $\ce{Cl-}$ and the cation binds with its neighborhood strongly, like $\ce{Cr^{3+}}$, a lot of structures with different hydration degree may be obtained, such as
$\ce{[Cr(H2O)3Cl3], [Cr(H2O)4Cl2]Cl, [Cr(H2O)5Cl]Cl2, [Cr(H2O)6]Cl3}$ in addition to anhydrous form $\ce{CrCl3}$ (and again, obtaining the anhydrous salt from water solutions is virtually impossible)

Sometimes a salt is unsolvable in water. This typically means that it is anhydrous, like silver halides and many sulfides, or at least low-hydrated, like $\ce{Cu(CH3COO)2\cdot H2O}$

In some cases a salt may have even more weird structure. For example, $\ce{MoCl2}$ actually contains $\ce{[Mo6Cl8]^{4+}}$ unit with $\ce{Mo}$ atoms forming an octanedron and $\ce{Cl}$ atoms occupying positions on top of each of the octahedron's faces. This unit is rather stable and the remaining chlorine my be exchanged with the unit intact.

Similarly, a series of oxoacetates is known, built around $\ce{M3O}$ unit. And so on....

Unfortunately, most transition metals are rather unique.  There is a strong vertical similarity between second and third rows of d-elements, but even it is not perfect.

  If they're not, how do they bond with halogens like Cl or ligands like OH/CN? Is it similar to ligand interaction and d-orbital splitting?

Typically yes.

  If they're neither complex ions nor have split d-orbitals, what determines their magnetic properties?

d-orbital splitting happens for any atom with said orbtials in neighborhood with symmetry below spherical, i.e. in all chemically viable cases.


Determining the Average Bond Enthalpy for the C-F bond Given the following data, how would you work out the average bond enthalpy for $\ce{C-F}$ bond. I've tried setting up the chemical equations and applying Hess's Law, but that's not getting me anywhere.

$\Delta H_\mathrm f^\circ(\ce{CF4(g)})=-680~\mathrm{kJ~mol^{-1}}$

Bond enthalpy, $\ce{F2(g)}=+158~\mathrm{kJ~mol^{-1}}$

$$\ce{C(s) -> C(g)}\quad \Delta H=+715~\mathrm{kJ~mol^{-1}}$$

EDIT: These are the equations I used:

$$\begin{align}
\ce{C(s) + 2F2(g) &-> CF4(g)}\\[6pt]
\ce{F2(g) &-> 2F-(g)}\\[6pt]
\ce{C(s) &-> C(g)}
\end{align}$$

Your approach to use Hess's Law is reasonable!

\[\Delta H_r = -680 - (715 + 2\cdot158) = -1711\ \mathrm{kJ\cdot mol^{-1}}\]

That's the enthalpy for $\ce{CF4}$ - a molecule with four $\ce{C-F}$ bonds.

The average $\ce{C-F}$ bond enthalpy is smaller:

\[\frac{1711}{4}\ \mathrm{kJ\cdot mol^{-1}} \approx  427\ \mathrm{kJ\cdot mol^{-1}}\]


How to calculate the equivalent weight of KMnO4? Let us consider the following reaction

$$\ce{2 KMnO4 + 16 HCl → 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2}$$

Now, in order to calculate the equivalent mass of $\ce{KMnO4}$, first I need to calculate it's $n$-factor which turns out to be $5$ because the oxidation state of $\ce{Mn}$ in $\ce{KMnO4}$ is $+7$ whereas in $\ce{MnCl2}$ it is $+2$. And the $n$-factor of $\ce{HCl}$ is $1$. So, the equivalent weight of $\ce{KMnO4}$ is

$$\frac{\text{molecular weight}}{5}$$

So, ratio of moles of $\ce{KMnO4}$ reacting with $\ce{HCl}$ should be $1:5$.

But from the balanced reaction we can see that this ratio is $1:8$.

Where am I wrong? Is my understanding and hence calculation of $n$-factor and equivalent weight is wrong?

(Please pardon this childish fashion of writing the question. If there is a problem anywhere, please do comment before flagging or downvoting.)

Refer to the definition of equivalent (IUPAC Gold Book):

equivalent entity
Entity corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions.

In other words, $\pu{1 equiv}$ is the amount of substance reacting with $\pu{1 mol}$ of hydrogen atom.
If hydrogen acts as a reducing or oxidizing agent, then either way $\pu{1 mol}$ of hydrogen atoms liberates or accepts $\pu{1 mol}$ of electrons:
$$
\begin{align}
\ce{0.5 H2 &→ H+ + e-}\\
\ce{0.5 H2 + e- &→ H-}
\end{align}
$$
That's why an equivalent of a redox agent is its amount which liberates or accepts $\pu{1 mol}$ of electrons upon being oxidized or reduced, respectively, assuming electrons don't exist in solution on its own for a significant period of time.
Note that $\ce{KMnO4}$ participates in redox reaction
$$\ce{2K\overset{+7}{Mn}O4 + 16 H\overset{-1}{Cl} = 2 KCl + 2 \overset{+2}{Mn}Cl2 + 8 H2O + 5 \overset{0}{Cl}_2}$$
and the half-reaction for the reduction of manganese is
$$\ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- → \overset{+2}{Mn}^2+ + 4H2O}$$
Also note that $n$-factor of permangante here is not the number of protons $\ce{H+}$, which are also used up in water formation; rather it's the number of transferred electrons, e.g. $n = 5$ and
$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{5} = \pu{31.61 g mol-1}$$
Obviously, the equivalent mass of permanganate isn't a constant and depends on the $\mathrm{pH}$ of the reaction.
For example, in neutral medium half-reaction appears as
$$\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- → \overset{+4}{Mn}O2 + 4 OH-}$$
and there is no explicitly shown protons to count at all!
However, the $n$-factor is $3$, and the equivalent mass of permanganate would be a different value:
$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{3} = \pu{52.68 g mol-1}$$


Can ingredients in Tiger Balm cause any harmful effects on PVC figures? My roommate often uses Tiger Balm and sometimes browses through my collection of vintage Star Wars figures. Can the residue on his fingers cause harm to the PVC figures or cause the paint to crack or chip?

According to this Wikipedia entry on Tiger Balm, there is camphor (among other chemicals) in the formulation of this product. This chart of chemical resistance recommendations for PVC indicates that it is not recommended that PVC should be exposed to camphor. Therefore, it is probably best to reduce or eliminate contact of Tiger Balm with the figures.

As for the paint cracking or chipping as a result of exposure to Tiger Balm, that might depend in part on the composition of the paint, and that cannot be addressed unless you can provide that information.


How much time for a water molecule in pure water to reach the top of a glass of water? I have a curiosity to know how long it will take for a water molecule to move a certain distance in pure water. I can put it simply as how much time for a water molecule to travel the whole length of a typical glass of water if no collective motion is present (therefore very still water with no convective currents).

To simplify things, this is the model:

Suppose I have a completely still and ideally insulated glass of water where the temperature in each part of the fluid is exactly the same (therefore no convective motion).      In addition the top of the glass container is closed and no water air interface is present to avoid evaporation that if present changes the dynamic.    In addition the glass of water is big enough so that the dynamic of the water in the core is not influenced by the the limited range of motion of the water near the borders of the glass.
Since it is a liquid, the water molecules move around each others until the original water molecule is at distance r from where it started(A). In absence of gravity the movement is equally probable in all directions, while with gravity I suppose it may be more likely a layer layout of the water molecules making horizontal movements more likely (see elipse shape of the trace).
The questions are (no need to reply to both):
1) How much time to be at distance r in both situations (gravity or not)? (it will depend on temperature, so let's say at room temperature)
2) How much time roughly will it take for a water molecule to reach reach almost the top of a typical glass of water starting from the bottom?    (in absence of convective motion and evaporation)

I am posting here the answer since it has been quickly found thanks to the useful comments. To give an estimate the best way is to look at the self-diffusion coefficient.
This is defined as the diffusion coefficient when the chemical potential gradient equals zero, or more simply, it is the diffusion coefficient in a medium of the same molecules.
According to Ref.1, the self-diffusion coefficient of water at $\pu{25^\circ C}$ is $D = \pu{2.3 \times 10^{−9} m^2 s^{−1}}$.  The diffusion time can be calculated as: $$t \approx \frac{x^2}{2D}   $$
As pointed out by @Karsten Theis, you can use the calculator and insert the water diffusion coefficient D.
For example, if the distance x is $\pu{8cm}$ like a typical glass of water and we are at $\pu{25 ^\circ C}$, then $t \approx \pu{16 days}$!
Instead if the temperature is $\pu{90 ^\circ C}$ then it's around $\pu{5 days}$.
I have drawn a graph that shows how much time to reach the top of the glass depending on temperature:

References:

Manfred Holz, Stefan R. Heila, Antonio Sacco, "Temperature-dependent self-diffusion coefficients of water and six selected molecular liquids for calibration in accurate $\ce{^1H}$ NMRPFG measurements," Phys. Chem. Chem. Phys. 2000, 2(20), 4740-4742 (https://doi.org/10.1039/B005319H).


Cell overpotential diagrams Is there anyone who knows a book where I can find a detailed explanation of cell overpotential diagrams, like the ones shown here?

Thanks in advance!

Not sure what kind of explanation you are looking for, but

Electrode Dynamics, A. C. Fisher, Oxford Chemistry Press, 1996, ISBN: 019855690X

is a great concise explanation of overpotentials among other things.


Why some cloths become rough after drying? I line dry my cloths and every now, they become rough after drying. This mostly happens with socks. They are regular cotton socks and once I start to wear them they become soft again. Any idea why this is happening?

I believe it is down to water relaxing the fibres in the material. Upon drying in one position the fibres become temporarily fixed in that position. I have also read that the detergents used in washing can cause clothes to dry stiffer. A little experiment such as soaking your clothes in hot water should kill bacteria and then see if they dry stiff. This would give an idea as to whether or not its down to the detergent or just the water.

Interestingly, a recent study suggests that fabric softeners reduce the wettability of materials. So if you use fabric softener on your towels, over time they will become less effective at drying (being towels).


Choice of solvent during thin layer chromatography I understand most of the theory behind TLC and that different volatile solvents have different polarities and cause the 'spotted' substances to move at different rates, but what is the point of creating mixtures and then using them in different ratios? How could the results change if I changed the ratios?

For example, during our lab, we synthesized acetylglycine and made three spots on our TLC plates, one spot was pure glycine, one spot was glycine + acetylglycine, and the third spot was just acetylglycine. the chamber was equilibriated with Dichloromethane & Hexane in an 8:2 ratio.

Glycine didn't move, and the other two spots did move up.
How would the results have changed if I used only dichloromethane? only hexane?

What if I used a 7:3 ratio? 5:5 (1:1) ratio?
I have a feeling the answer is really obvious and i'm overthinking this but it's really not hitting me right now.
Thank you for reading!

Chromatography is really an empirical science. The fluid mechanical side and the adsorption phenomenon is somewhat well understood but choosing mobile phases is essentially based on trials, sadly to date. If someone has given you a ratio, say 8:2, it means someone must have tried it and they found that this mobile phase works the best. The main criterion for choosing the mobile phase is that all the mixture components must be soluble in it. Thus if glycine did not move, this means that the mobile phase has no affinity for this solute (it is perhaps insoluble in it) and it is a poor choice. One then starts to tune the polarity of the solvent mixture. Think about a petroleum sample, which literally has thousands and thousands of different molecules. Now think yourself: will they all dissolve in 100% hexane? You do need an additional solvent.

There is no theoretical restriction in chromatography to use multiple solvents or a single solvent. For example in gas chromatography, you always use a pure gas. The reason for using a mixture of solvents in liquid chromatography originates from the concept of "general elution problem of chromatography" which basically says that there is no single adsorbent or a single mobile phase which can separate everything. Not only separation scientists use a single mixture composition of mobile phase (isocratic), they can also continuously change the mobile phase composition as a function of time. This is called gradient elution. You cannot do that with thin layer chromatography without using fancy stuff.

People are trying to implement machine learning, so that one can enter the molecular structures contained in the mixture, available stationary phases, and with a lot of statistics and tons of experimental data, the computer can tell you to try a certain mobile phase composition.


Esters of catechol - Fries rearrangement Esters of catechol undergo the Fries rearrangement

But one product is formed more than the other.

I believe it is due to the electron withdrawing nature of the $\ce{OCOCH3}$ group which prefers the attack on meta position.I was able to write a reaction mechanism for the product which is minor in this case. But I couldn't write one for the major product. I wonder how it is formed?

Does the minor product itself rearrange to give the major product or are both of  the products formed directly in this reaction?

This Wikipedia article provides a nice overview of the Fries rearrangement. They note that the mechanism of the rearrangement is not fully understood, but the following mechanism is generally accepted.

The key step in the reaction is thought to involve the interaction of a lewis acid with a phenyl ester to generate an acylium ion ($\ce{[R-C=O]^+}$).  Once the acylium ion is generated the reaction proceeds like any other electrophilic aromatic substituion. The phenolic oxygen is a strong ortho-para director, so both ortho- and para-products are produced by the same mechanism. The $\ce{OCOR}$ group does not play a role in "directing" the substitution pattern.

The ratio of ortho/para isomers in most aromatic electrophilic substitution reactions is very temperature dependent. This is because of kinetic vs. thermodynamic control. Typically, the para product is the kinetic product (first formed), while the ortho product is the thermodynamically more stable product and will form if the reaction is run at higher temperatures and allowed to equilibrate. For example, when phenol is mono-brominated at low temperature, the para-bromo product is formed in >95% yield (reference). So your specific reaction is likely being run under kinetically controlled conditions to produce the para product as the major product. As the Wikipedia link points out, the para product can equilibrate at higher temperatures to produce more of the ortho product.

Finally, just a word to address @bon 's comment.  Indeed, Fries rearrangement of catechol diesters does produce mono-acyl catechols (reference, p A-172) - apparently one of the acyl groups is lost in the process.


Phase transition for an ideal fluid Does a phase transition occur also for an ideal fluid (from ideal gas to ideal uncompressible liquid)?
Or ideal fluids can be modelled only as ideal gas or uncompressible liquid at all, without the possibility for a phase change?

I hope the question is well posed, thanks

Ideal gas and ideal fluid are not the same. Ideal fluid has molecular interaction that is even throughout different entities, and has no viscocity, and no compression.

Yes, ideal fluid is modeled so that it will go through phase change, but you would have to consider the gas from ideal fluid as a pseudo- real gas, since it has molecular interaction.

An ideal gas has no volume, has no interaction, will not go through any sort of phase change.

Hope this helps.


What is the safest way to hook up car battery jump cables and why? I know that when both connecting and disconnecting battery jumper cables between two car batteries there is a prescribed safe way to do it. I can never remember the correct way because I can’t understand the reasoning behind the prescribed sequence. Is there an easy explanation / (chemical?) principle that can make this procedure (the sequence of both connecting & disconnecting the four posts) more logical and therefore easier to remember?

My confusion is that at some point you must make a connection that causes sparks (which risks battery explosion) because it completes the circuit. Completing the circuit can’t be avoided, so why does it matter which post you are connecting to when you complete the circuit?

Many recommended procedures such as this one call for connecting (+) to (+), then (-) on the "good" battery to a metal component in the "dead" car engine.  The negative terminal of the battery is grounded to the metal components, so we may think of the last connection as (-) on the dead car.  If a spark were to form when the circuit is established on the last connection, it would occur at the metal component and not at the terminal where it could interact with electrolytically generated hydrogen or oxygen gas.


Fajans' Rules Versus Lattice Enthalpy In analyzing the solubility among $\ce{CaF2, CaCl2, CaBr2}$, I found that $\ce{CaF2}$ is insoluble due to high ionization enthalpy.

But comparing among $\ce{AgF, AgCl, AgBr }$, it was said  that $\ce{AgF}$ is more soluble because of Fajans' rules (consequence of polarization). $\ce{AgCl, AgBr}$ have high polarizing effect and so they have high covalent characteristics.

Now question is why can't I use the Fajans' rules at $\ce{CaF2}$? Where can I use Fajan's Rule and Where should I use Lattice enthalpy rule?

This problem is also present at comparing the melting point of $\ce{NaCl, KCl}$. Fajans' rules cannot be applied here also.

The thing you have to understand is that the enthalpy of dissolution is actually the sum of lattice enthalpy and hydration enthalpy. So being a thermodynamic concept, we will have to turn to Gibbs free energy i.e. ΔG= ΔH-TΔS for completely understanding the solubility trends. For the process to be spontaneous, ΔG(dissolution)<0 . Now, entropy will generally increase while dissolving, so we can safely assume ΔS>0 for now. So obviously, the factor of -TΔS will be negative for T>0, so this contributes to the solvation of the said compound.(From now on, ΔH will be referring to enthalpy of dissolution)

Now, it all comes down to the ΔH part : it's magnitude and sign. If ΔH is positive and initially greater than -TΔS, then ΔG will be positive, and as temperature is increased, the -TΔS factor will ultimately overcome the positive ΔH and then, ΔG will become negative. And indeed, this is why you will find that $\ce{PbCl2}$ is insoluble in cold but quite soluble in hot water. Similarly, you can consider the implications of variations in the magnitude and sign of ΔH on solubility (in a polar solvent) yourself.

Now, coming to your specific problem, you see,  you first of all have to consider the configurations of $\ce{Ag+}$ and $\ce{Ca^2+}$

$\ce{ Ag^+ -  [Kr] 4d10}$ and
$\ce{ Ca^2+ - [Ar]}$

So you see, the calcium ion has 8 electrons in it's outermost shell ( 2 in 3s and 6 in 3p) while the silver ion has 18 electrons in it's outermost shell( 4s2 4p6 4d10). Now, due to the ineffective shielding by the d-electrons, effective nuclear charge on the valence shell and hence it's polarising power increases tremendously. Here, the calcium ion possesses an "inert gas configuration" while the silver cation has a kind of  "pseudo noble gas configuration"  (as it is not exactly a noble gas configuration, but still the cation is quite stable). On moving further down the periodic table,you get cations like $\ce{Tl^+, Bi^3+}$ etc. which have neither 8 nor 18 valence electrons (hence they can be called "non-inert gas configurations" ) and owing to the even poorer shielding by f and d electrons, there polarising power is even higher than the previous two configurations. So, the general order of polarising power of cations can be : Non-inert> Pseudo-inert> Inert gas configuration .( Note: All this can also be derived quantitavely by actually calculating effective nuclear charge on valence electrons by Slater's rule)

So, between $\ce{Ag^+}$ and $\ce{Ca^2+}$, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as compared to $\ce{Ca^2+}$ . As to your question about when to apply which concept: You are basically required to compare the sum of the hydration enthalpy and lattice enthalpy. If the lattice is very strong(example: LiF due to compatibility of ions and less interatomic distance) then the hydration enthalpy can be ignored, while for something like $\ce{LiNO3}$ the lattice is weak and hydration enthalpy is highly negative, so lattice enthalpy can be ignored for the final ΔH. You can use any concept as per your convenience, be it Fajan's rule, or extent of hydration or lattice strength; they will genrally complement each other provided you apply them correctly

For alkali metal chlorides like KCl and NaCl, the cations have only got a unit positive charge (which is not much to cause a lot of polarization) and more importantly, these cations belong to the inert gas configuration type, so here lattice energy will be the dominant factor


Rationalizing Strength of Binary Halogen Acids with MO Theory Help me walk through the logic here:

1) HF is a weaker acid than HCl.

2) This is as far as I can understand. How does sigma star affect the acidities of the binary halogen acids?

  If a covalent bond was formed, you would expect a decreasing acidity
  down the periodic table of the halogens, because the energy of the σ∗
  would increase forming a less attractive LUMO (lowest unoccupied
  molecular orbital), which would result in less hydrogen atoms being
  pulled of the halide (i.e. lower acidity, higher pKa).

Here are some questions about the above excerpt:

1) Covalent bond formed between what?

2) Why would the sigma star orbital be populated if the hydrogen were removed and a bond formed in place of it? Or is a bond going to be formed with the hydrogen still present?

Update:

https://usflearn.instructure.com/courses/986898/files/31117039

Some initial comments on the above:

1) Bonding in the $\ce{HX}$ series is a varying mix of ionic and covalent bonding.  $\ce{HF}$ has the largest covalent bonding component and $\ce{HI}$ the smallest.  This trend goes against what electronegativities alone would suggest, but is consistent with what overlap integrals (e.g. overlap between a $\ce{1s}$ orbital and a $\ce{2sp^3}$ orbital is better than overlap between a $\ce{1s}$ orbital and a $\ce{5sp^3}$ orbital) would predict.  Therefor, the latter effect appears to predominate.

2) An MO diagram depicts only the covalent portion of the bonding situation.  Therefor it is unlikely to accurately predict acid strength over the entire $\ce{HX}$ series where the (covalent\ionic) contribution to bonding is varying.

3) The MO diagram should probably have hybridized the $\ce{Cl}$ AO's prior to mixing them with the $\ce{H}$ AO.  That is, I'd guess that the covalent portion of the $\ce{HX}$ bond is between a $\ce{H}$ $\ce{1s}$ orbital and a $\ce{Cl}$ $\ce{3sp^3}$ orbital, not a $\ce{Cl}$ $\ce{3p}$ orbital.  This could be assessed by looking at the photoelectron spectrum of $\ce{HCl}$ and analyzing it (e.g. how many chlorine lone pair signals?) in terms of hybridization.

4) $\ce{\sigma^{\ast}}$ orbitals don't really need to be brought in to the discussion since they are unoccupied throughout the $\ce{HX}$ dissociation process.

5) Increased $\ce{HX}$ acidity correlates inversely with $\ce{HX}$ bond strength. $\ce{HF}$ with the best overlap and strongest bond is the weakest acid; $\ce{HI}$ with the poorest overlap and weakest bond is the strongest acid.

Response to OP update

I'm not sure I understand just what % ionic means in the Table you posted.  When bonding is discussed, usually 3 types of bonding are presented (see this link for example) 1) covalent bonding where electrons are roughly equally shared, 2) polar bonding where we have a covalent bond with unequal electron sharing and 3) ionic bonding where electrons are not shared.  For $\ce{HX}$ you can draw 2 resonance structures, one covalent and one ionic that both contribute to the true picture of the covalent bond in $\ce{HX}$.  Is the % ionic in the Table referring to the polarization in the $\ce{HX}$ covalent bond, or is referring to a truly ionic bond?  How does the increasing covalency suggested by the Table as you move down the $\ce{HX}$ series square with decreased bond strength down the series?  Since bromobenzene and methyl bromide have dipole moments, the equation used to generate the % ionicity in the Table could also generate a percent ionicity for these compounds, but they are certainly not "ionic" compounds.  I suspect the Table is measuring the polarization of a covalent bond and not referring to the ionic character found in a salt.


Why is cyclopentyne unstable? Why has no one ever been able to synthesise cyclopentyne as a stable molecule? I've been thinking about this for quite some time and I can't seem to figure it out. Something to do with angular strain in the molecule?

Stable alkynes are normally linear, that is, they have bond bond angles of 180°. That is necessitated by the fact that alkynes are formed by two π-bonds and one σ-bond, which must be mutually perpendicular. If cyclopentyne were to exist, it would be forced to adopt bond angles of roughly 108°, equal to those of the interior angles of a regular pentagon. This deviation from ideal geometry causes extreme instability by virtue of poor orbital overlap. The rigidity of the alkyne moiety would also exacerbate the inherent conformational inflexibility of small rings, wherein the inability of bonds to freely rotate results in eclipsing between atoms.

The only cyclic alkyne I'm aware of is benzyne. However, it's far from stable, existing only as a fleeting intermediate in specific nucleophilic aromatic substitution and pericyclic cycloaddition reactions.

Edit: After a quick search, I found the cycloalkyne Wikipedia page, which claims that cyclooctyne is the smallest cycloalkyne stable enough to isolate and observe. I don't know whether this information is completely current, however.


Could detergents dissolve cell membranes? Animal cells have a lipid membrane and detergents are great at emulsifying lipids. Yes, there are proteins, cholesterol and other numerous substances in the lipid bilayer, but the lipids make up the majority of it. If we dipped a bacterium or an isolated animal cell in detergent, would their membrane be dissolved?

Yes they can dissolve membranes and are even used to isolate active forms of membrane proteins by solubilization. The product are micelles of a lipid-detergent mix and a protein-lipid-detergent complex. Maybe you find this short abstract interesting.

  A critical step in any in vitro analysis of membrane proteins is the
  solubilization of the membrane to extract the protein of interest in
  an active form to obtain an aqueous solution containing the membrane
  protein complexed with detergents and lipids in a form suitable for
  purification and further analysis. This process is particularly
  delicate as the aim is to maximally disrupt the lipid components of
  the membrane while putting the protein components in an un-natural
  detergent environment without perturbing them. Looked at this way, it
  is remarkable that it ever works. Although the process is difficult
  and hard to master, an increasing number of membrane proteins have
  been successfully solubilized in active forms, allowing some general
  principles to be established that we illustrate in the method
  developed in this chapter.

Duquesne K., Sturgis J.N. (2010) Membrane Protein Solubilization. In: Mus-Veteau I. (eds) Heterologous Expression of Membrane Proteins. Methods in Molecular Biology™ (Methods and Protocols), vol 601. Humana Press


Why do iron(II) ions and chromium have different electronic configurations? What I know:

Empty 3d orbitals are higher in energy than empty 4s orbitals
Aufbau Principle (electrons always go into an empty orbital with the lowest energy)
Partially/half/fully filled 3d orbital are lower in energy than 4s orbitals
$\ce{Cr}$ is an exception: $\ce{[Ar] 3d^5 4s^1}$. It breaks the "rule" because the extra stability from half-filled 3d orbitals is favoured.
From (1) and (2) => After the $\ce{[Ar]}$ core, electrons will fill 4s before occupying 3d
When forming ions, electrons of the highest energy are removed first.
Iron(II) ions exist as $\ce{[Ar] 3d^6}$ in reality.

What I understand:

I understand why Fe is $\ce{[Ar] 3d^6 4s^2}$. [from (5)]
I understand why Cr is $\ce{[Ar] 3d^5 4s^1}$.

From (4), (5) and (3), I believe for $\ce{Cr}$ the 4s is filled before 3d too, i.e. it is $\ce{[Ar] 3d^3 4s^2}$ before the last electron is added. After the last electron is added to form $\ce{[Ar] 3d^4 4s^2}$, a 4s electron immediately moves to the 3d orbital which is of lower energy.

From (6) and (3), for transition metals, 4s electrons are removed first. i.e., I understand why iron(II) ions may be $\ce{[Ar] 3d^6}$

What I don't understand:

I know $\ce{[Ar] 3d^6}$ is correct for iron(II), but I don't understand why?

Why doesn't iron ions adjust itself as Cr does to achieve the extra stability? Is it because, in this case, it needs to excite an electron to an orbital of higher energy, whereas in Cr, the 4s electron just moves down to an orbital of lower energy?

P.S.

I've read What is the electronic configuration of Fe(II) ion? and I am not satisfied with the answers in that thread.
I've read Do ions also form special electron configurations? (just as Copper and Chromium do), and it only tells me "it is possible for ions to have special electronic configurations too!"

For starters, it isn’t nearly as simple as it is usually put in chemistry classes to explain chromium’s ground state electron configuration.

$\ce{Cr(0)}$ and $\ce{Fe(II)}$ have the same number of valence electrons. You might have heard of the concept of isolobality, which, briefly explained, suggests that compounds with similar electronic configurations (isolobal configurations) will react similarly; so you could replace $\ce{F}$ with $\ce{OMe}$ or even $\ce{[Co(CO)4]}$. The concept has its values and its strong simplifications which I will not go into here.

However, even assuming the isolobality concept’s validity, it would not count $\ce{Cr(0)}$ and $\ce{Fe(II)}$ as isolobal to each other. The reasoning is rather simple: One is a neutral metal, the other is a cation with $+2$ charge. For every electron an atom looses, the atom’s orbitals will contract and the atomic (or ionic) radius will decrease. This also changes the energy levels between orbitals and will affect lower d-orbitals differently than higher s-orbitals.

The result is that the $\ce{4s}$ and $\ce{3d}$ orbitals are now so dissimilar in energy, that electrons hopping back or forth is now no longer an option.


Lithium ion battery - measure C rate How does one determine the C rate of an aged cell?

I would like to find the discharge rate just before any cliff edge of capacity loss. Can the C rate of a cell be estimated by the cells rated heat dissipation and measured internal resistance? Where I could find the current that will use up all the cells heat dissipation capacity.

I have a number of recycled laptop battery 18650 cells.

Are there any negative effects of the cells being below room temperature?

To determine the C-rate, you should charge and discharge two cycles very slowly say at 0.1 mA, between say 1.5 V to 4.3 V for your Li-ion (unless it's a very old laptop battery it should be LiCoO2), and measure how long it'll take. The C-rate would be your capacity divided by one hour.

Don't know of a way to measure it using heat or internal resistance.

Batteries generally have an adverse effect when heating up, so unless it is much below freezing, they should be ok.


Analysis of multiple spectra to determine an unknown structure I've been trying to solve the structure of a molecule from a $^{13}\ce{C}$ and proton NMR as well as a MS spectrum, but I've been stumped, I've been staring at it for a while with no improvement.
I've attached the images below. From the MS, I understand the Mr of the molecule to be 122, the $^{13}\ce{C}$-NMR suggests six different environments.
The part that confuses me the most is on the NMR. There is a 3 hydrogen doublet and a 4 hydrogen singlet.  I went on to assume that the structure might be some sort of aromatic and symmetrical, and maybe 2 carbons in the same environment so that the formula might be $\ce{C7H10N2}$, this accounts for the Mr and the fact there is an exchangeable proton peak.
The spectra I've been provided are poor quality, so I can't even work out the coupling constants, or properly view the peaks.
I still can't figure out what it might be. Can you point me in the right direction please. I have a feeling I've been confusing myself over something that's simple.

A very quick solution to this problem before my train stop.

Mass Spec gives MW of 122.

1H spectrum gives 5 aromatic, a 1H quartet at 4.8ppm, a 1H exchangeable proton at 2.4ppm and 3H doublet at 1.5ppm. The 4.8/1.5 splitting indicates a substituted ethyl chain with the C1 group having two substituents (CHRR'-CH3). The chemical shift of this 4.8ppm proton indicates some substantial substitution, such as a heteroatom. We also have a labile 1H, which could come from an alcohol or an amine.

13C spectrum shows monosubstituted benzene ring (4 carbon environments with ipso carbon downfield shifted). Also shows two alkyl carbons corresponding to the ethyl chain we have worked out above.

To summarise data to date:

1 monosubstituted benzene (C6H5)
1 ethyl chain, doubly substituted on C1 (CH-CH3)

That gives us a molecular formula of C8H9 which adds up to 105. That leaves 17...... which includes one labile proton, leaving a mass of 16.

As this is clearly a homework question, I'll leave you to join the pieces.


Why are pi antibonding orbitals lower in energy than sigma antibonding orbitals? I know when a carbonyl group is attacked by a nucleophile, the electrons of the nucleophile enter the antibonding π orbital of the $\ce{C=O}$ bond, which breaks the π bond. My textbook says that it enters the antibonding π orbital instead of the antibonding σ orbital, because the antibonding π orbital is lower in energy. This is my point of confusion: why is it lower in energy?

When two atoms form a bond, if two atomic orbitals mix, then two molecular orbitals are created, one of lower and one of higher energy than that of the atomic orbitals.

Now, just how much the energy changes is dictated by how well the atomic orbitals overlap. Overlapping depends on three things:

Orbital energy
Distance between atoms
Orbital symmetry (they must be "aligned")

Now, assuming that the atomic orbitals are similar in energy and the symmetries are correct, which is the case for C and O.

The sigma bond is a result of two orbitals very close together, directly facing each other frontally. The overlap is great. The pi bond emerges from two orbitals overlapping side-by-side, which isn't as efficient because it happens at a greater distance.

Here is a pic to exemplify:

Thus, not only is the bonding combination of a sigma bond lower in energy, but also the antibonding combination is higher in energy.


Synthesizing 2-methyl-2-butanol I have recently been trying to come up with a method to synthesize 2-methyl-2-butanol or tert-amyl alcohol. I wish to avoid complex/expensive steps since as a home chemist/student, I don't have access to much money or equipment.

What I have done so far: I noticed that methyl ethyl ketone has pretty much the correct structure, minus a methyl group.

I then realized this meant I could use a methylating agent to get the correct molecule. What I needed was to break the $\ce{C=O}$ bond from the carbonyl group and add a methyl group.

Long story short, I found that either Grignard reagents or something like methyl lithium would do it since it would donate an electron pair to the carbon, and form a bond with the methyl. I would then have the 2-methyl-2-butanol as a conjugate base with $\ce{Li}$ ions present. I could then react them out with $\ce{HCl}$ to precipitate out the $\ce{Li}$ salt and add a hydrogen to the negatively charged oxygen.

My problem: I don't know if this would work at all. I didn't find any documentation. Also, methyl lithium is very expensive and difficult to make so I can't use it.

Does anyone have any other route suggestions? Would it be possible to make methyl sodium and if so, would it work the same? Is it possible to make Grignard reagents easily at home?

I'm a hobby chemist, so please be detailed with how you predicted the organic reaction.

Also, I don't necessarily care about what chemicals I use for the reaction, so if anyone can suggest a different starting chemical, I'd be glad to hear it. I was even thinking of using an alkene group of some kind and methylating the $\ce{C=C}$ bond, but I don't know for sure, as I'm pretty new to organic chem.

Isopropyl lithium and ethyl bromide would be my go-to if I were to try to prepare this stuff (which I will call TAA). This requires a solid fume hood, and the resulting mixture of salt, TAA, and isopropyl lithium solvent (usually n-hexane) should be easy to separate. The hexane can be stripped from the TAA via fractional distillation, with more distillation passes resulting in a more pure product. TAA boils at $\pu{102 ^\circ C}$ (Sigma Aldrich), and n-hexane boils at around $\pu{70 ^\circ C}$ - this $\pu{30 ^\circ C}$ gap in BP passes the rule-of-thumb "can I separate with distillation?" question.

All that said, even pricey Sigma Aldrich can get you 99% for $115 a liter.

I am aware that TAA is an experimental recreational drug, and a possible replacement for ethanol. I advise caution in sourcing this material for that purpose, as the impurities present are likely to be more toxic than TAA itself. I believe the industrial synth is from 2-methyl-2-butene (2m2b) with the addition of water (this would be cheapest); TAA prepared this way can be purified further by refluxing with an open condenser, enabling any unreacted alkene to evaporate (Boiling point = $\pu{39 ^\circ C}$). Acetone and ethyl Grignard would also form TAA, but with much more hassle and hazard, so the remnant ethyl chloride and acetone from such a synthesis are not likely impurities in any cheap sample.


What determines humidity limit / dew point of the air? - Why can air only hold a certain amount of water? From Wikipedia:

  The dew point is the temperature at which the water vapor in a sample of air at constant barometric pressure condenses into liquid water at the same rate at which it evaporates. At temperatures below the dew point, water will leave the air.

I am wondering what is is about the air which determines how much $\ce{H2O}$ vapour can be stored in a set volume of air at a set temperature without the water condensing back to liquid water.
Also, why is it that at higher temperatures the air can hold a greater amount of water vapour?

One way to understand this is if you think of air as bunch of atoms and molecules moving around in random fashion. The speed at which they move is measured by the temperature of air. (This is true even if we are not talking about air, but we don't need to bother about that here.) So when we say higher temperature we mean that the speed at which these particles move is higher. Furthermore the faster they move the less likely they are to stick together.

Now water molecules like to stick to each other and when too many of them are together they "condense". But if you were to increase the temperature of the air (and hence the water molecules), they end up moving around faster and less likely to stick together. This way you can add more water molecules for the same amount of air.

The other thing to think about is pressure. In some sense it describes how crowded the place is. The higher the pressure, the more crowded it is. So when you increase the pressure, you increase the chances that the water molecules will stick together and hence "condense".

Another way to think about this is through Le Çhatelier's principle, which states that when a system in equilibrium (especially chemical equilibrium) is disturbed it tries to adjust in a way which minimizes the disturbance.

So now think of a box with some water and air. If you leave it long enough the system will reach an equilibrium, wherein no more water will evaporate or condense. But say now you increase the temperature of the air, the system will try to readjust so as to reduce the change in temperature. To do this it will cause the water to evaporate (as evaporation requires/consumes heat) and thereby reducing the change in temperature.
Similarly when you increase the pressure which is similar to decreasing the volume of the box, the air tries to account for this by condensing out some of the water (since liquid water occupies less volume than gaseous water), thereby reducing the effective decrease in volume seen by the air.

You may wonder why other particles in air do not "condense." They do, but only at much lower temperatures and/or higher pressures.


Numerical on change in degree of dissociation wrt pressure The degree of dissociation of $\ce{PCl5}$ is 0.4 at 2 atm pressure and T kelvin. At what pressure will degree of dissociation of $\ce{PCl5}$ be 0.6?

Taking the degree of dissociation as 0.4, I got the partial pressure of $\ce{PCl5}$ to be $\frac{2}{2.33}$. I then thought of taking the degree of dissociation as 0.6 and multiplying the mole fraction of $\ce{PCl5}$ with the total unknown pressure and equating it to the partial pressure of $\ce{PCl5}$. But I then realized that wouldn't the partial pressure change with change in total pressure?
I'm not able to figure this question out. Any help would be greatly appreciated!

The dissociation of $\ce{PCl5}$ is given by the equation:

$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$

with $\displaystyle K_\text{p} = \frac{p_{\ce{Cl2}}p_{\ce{PCl3}}}{p_{\ce{PCl5}}}$. Putting this in terms of total pressure, $\displaystyle K_\text{p} = \frac{x_{\ce{Cl2}}x_{\ce{PCl3}}}{x_{\ce{PCl5}}}p_\text{T}$.

In terms of the degree of dissociation, $\alpha$, assuming an initial amount $y$ of moles of $\ce{PCl5}$, at equilibrium, we have $n_{\ce{PCl5}} = y - \alpha y\,\text{mol}$, $n_{\ce{PCl3}} = \alpha y\,\text{mol}$, and $n_{\ce{Cl2}} =  \alpha y\,\text{mol}$. Thus, their molar fractions are $x_{\ce{PCl5}} = \frac{1 - \alpha}{1 + \alpha}$, $x_{\ce{PCl3}} = \frac{\alpha}{1 + \alpha}$, and $x_{\ce{Cl2}} =  \frac{\alpha}{1 + \alpha}$, so:

$$K_\text{p} = \frac{\left(\frac{\alpha}{1 + \alpha}\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right)}p_\text{T} = \frac{\alpha^2}{1-\alpha^2}p_\text{T}$$

Since at $p_\text{T} = \pu{2 atm}$, $\alpha = 0.4$, we get $K_\text{p} = \pu{0.381 atm}$. Hence, for $\alpha = 0.6$, since $K_\text{p}$ doesn't change (i.e. assuming it's an isothermic system), we get $p_\text{T} = \pu{0.677 atm}$. This is in agreement with Le Chatelier's principle: reducing the total pressure will push the equilibrium towards the side with more gaseous moles, so the dissociation constant in this case will be higher.


Confusion about boiling point and vapor pressure It is said that a liquid boils when its vapor pressure equals to the external pressure of its environment (e.g. atmospheric pressure). That’s because the bubbles of water vapor that form will be crushed if the vapor pressure is below the atmospheric pressure. However, it’s the fluid’s pressure that actually does the crushing, isn’t it? And the fluid pressure varies with depth, with only the fluid at the top of the liquid being equal to atmospheric pressure. Shouldn’t we say that the liquid boils when its vapor pressure is equal to the greatest fluid pressure in the liquid?

Shouldn’t we say that the liquid boils when its vapor pressure is equal to the greatest fluid pressure in the liquid?

No.  If you heat up water at the ocean's surface to 100 °C, it will boil, even though the pressure at the bottom of the ocean is far higher.

As you note, pressure is not guaranteed to be homogeneous in space (i.e. the same everywhere), especially for deep bodies of water.  But neither is nearly any other fluid property.  Temperature can vary with position in poorly-mixed waters.  So can whether the water is boiling or not.

A given small-ish region of water with nearly constant temperature and pressure will boil if and only if the (P, T) condition of the water is at or above the water's vapor pressure at the same temperature.


Coordination numbers of ionic and intermetallic compunds If $\ce{NaCl}$ crystallizes in an fcc lattice, its coordination number should be 12. Then why do we say that $\ce{NaCl}$ has a coordination number of 6?
Also, if $\ce{CsCl}$ crystallizes in a simple cubic lattice, why is it's coordination number 8 and not 6, as for other simple cubic lattices?

NaCl

NaCl is not monoatomic. When discussing coordination number of cations in ionic compounds, the number refers to the anions the cations is surrounded by (or bonded to, or touching).

The blue Na+ cations are bonded to six other Cl- anions: top, bottom, left, right, forward, backwards (in this image). They are not bonded and do not touch the other Na+ cations in the diagonals.

CsCl

CsCl is not a "simple" simple cubic. Each of the different ions, Cs+ and Cl- by themselves are simple cubic, but they are interpenetrating to form something that looks like a body centered cubic:

A mono atomic simple cubic will have coordination number of six, but this is an ionic compound. Thus, each cation or anion that sits in the center of the cube is bonded to the eight anions or cations (respectively) that sit in the vertices of the cube.


Thermoplastic & Thermoset adhesives: I have a question that is bothering me since few weeks. I'm not a chemist, nor I have any experience with polymers theory, but:

Epoxy - is defined on Wikipedia as a thermoset polymer as a process of polimerization occurs that can not be reversed.

PVA - Is defined (again on Wikipedia) as a thermoplastic polymer.

So why the cured epoxy has a glass transition temperature, and therefore when heated above that temperature it softens and its shape can be modified. But when PVA glue hardens (by loss of humidity) it can not be softened by heat?

First to clear up some terms. Glass transition just means that the intermolecular bonds are weak enough for the material to continue crystallizing—it is not the melting temperature.  This can be accompanied by softening of the material but that does not mean the polymer will flow.  Thermoplastics can flow at elevated temperature but thermosets (which includes cross-linked thermoplastics) do not by definition.

Epoxy has a glass transition due to a loss of crystallinity but is not usually shapeable unless formulated to or poorly prepared.  PVA is a thermoplastic that melts around $\pu{200^\circ C}$  which is also the temperature that it starts to pyrolyze and char. With careful temperature control you can melt and form PVA and there are 3D filaments dissolvable in water based on PVA.  Without careful control, you decompose the plastic before melting it.


Really how rare are hydrogen atoms? Hasok Chang, a chemist, but perhaps more a philosopher recently published the book Is Water H2O? . Chang used the question and an extensive web of arguments ultimately as a platform to support a pluralistic approach to scientific endeavor. I don't know that I buy it all, but in a latter chapter he made some statements regarding 'fundamental' components of chemical reactions, and one in particular that interested me and brings me to the point of my question.

Hydrogen gas as I understand it naturally occurs as a hydrogen molecule - two hydrogen atoms stuck together as a diatomic molecule. If ionized, the electron is stripped from the atoms and you have hydrogen ions or rather protons. But what about single hydrogen atoms? Are these non-existent or just a rare thing? Is there some way I can arrange to have a container of single hydrogen atoms and prevent their grouping into hydrogen molecules without ionizing them?

For what answers that follow - do these apply in general to the stable nature of all diatomic gases?

While usually short-lived, isolated hydrogen atoms (also called hydrogen radicals) are well studied and chemically important species. Bond formation in simple species is always an exothermic reaction (usually highly so) with zero activation energy, which means bond formation is thermodynamically very favourable and has no kinetic barrier stopping it from happening. Simply put, the isolated atoms really want to react with each other, and they'll attract each other at a distance, so if at some moment you create a container with pure monoatomic hydrogen, the hydrogen atoms will react as soon as possible to regenerate dihydrogen molecules (hydrogen gas).

The best you can do is expose the dihydrogen molecules to energetic conditions, so that a fair amount of bonds are being constantly broken at a rate comparable to which they're formed. A simple way to do this is to strongly heat the hydrogen gas; enough thermal energy will cause the bonds to break, though the dihydrogen bond is unusually strong ($\rm{436\ kJ\ mol^{-1}}$) so even at several thousand degrees kelvin there will be only a small portion of free hydrogen atoms in the gas (see Mithoron's comment). A more interesting option is to shine a strong ultraviolet light into the gas, as the dihydrogen molecules can separate into their constituent atoms upon absorption of photons with energy around $\rm{4.52\ eV}$, which corresponds to a photon wavelength of $\rm{274\ nm}$.

Any substance can have its bonds broken through the above two methods (among others), so the same applies to all other diatomic molecules. The stronger the bond between the atoms, the more uncommon is it to find the isolated atoms. For substances with very weak bonds, such as molecular dibromine or diiodine, mere exposure to visible light is enough to trigger formation of a fair amount of isolated atoms, something which is exploited in several reactions.

Keeping in mind that isolated atoms are very reactive, they are rather abundant in space. The average density of matter in the Universe is very, very low (about 1 atom per 5 cubic metres of space!), so isolated atoms can go a long time before finding anything to react with. This makes them extremely important for understanding astrochemistry.


Why do you need excess LAH to reduce esters? I am little confused on this question. On one hand, I remember my teacher saying that $\ce{LiAlH4}$ is like $\ce{BH3}$ of hydroboration oxidation of alkene, where you one mole of $\ce{BH3}$ can oxidize 3 mole of alkene molecule. but on the other hand my textbook keeps on mentioning excess LAH when reducing ester? Can't one mole of LAH reduce four mole of ester molecules as well? Thanks.

Lithium aluminium hydride or LAH is the more everyday form of $\ce{LiAlH4}$. When originally produced, most of it is indeed $\ce{LiAlH4}$ but that quickly partially hydrolyses giving the grey powder you may or may not know from the lab. As such, $\ce{LiAlH4}$ is practically never entered into reactions in pure form. Since it is impure, excess reagent is added to make sure the reduction job is performed as desired.

Also, remember to consider the reaction meachnism:

$$\ce{R-COOR' ->[LAH][- HOR'] R-CHO ->[LAH] R-CH2OH}\tag{1}$$

Note that the second step, the reduction of an aldehyde, is more rapid than the first as aldehydes are more reactive towards nucleophiles than esters. Thus, if you attempted to bargain, hoping to arrive at the aldehyde selectively, you would end up with under half of your reactant reduced to the alcohol and half of it still as an ester. Another reason to always add excess just to be sure.


What is change in internal energy of a system in which combustion occurs at constant temperature? We got a question in a test, in which we were asked which system has zero change in internal energy and it had an option which was combustion of methane at constant temperature. I imagined this to be a situation in which the combustion is carried out in a system, where the change in internal energy of system due to heat released, is cancelled by the work done by the system. The reason why I thought of this is because of the formula,
∆U=nCv∆T,
But this was wrong. When I asked my sir how this was wrong, he was not able to say why the ans using above stated formula is wrong. He tried giving me a logical explanation in which he said that a small part of the system releases heat which is used by the remaining part for it's combustion, thus the total temperature remains constant. but I am unable to understand why we choose to only study a part of the system while we have the entire system. It makes no sense to me. Could someone please explain how my sir's explanation is valid here and why the formula is wrong to use here?

The change in internal energy of a reaction mixture is a function not only of the temperature change but also of the changes in the amounts of reactants and products in the mixture.  This is because energy is consumed and released when we break- and make chemical bonds.  The equation you wrote for the internal energy change applies only to a pure species or to a mixture of constant chemical composition.  The occurrence of chemical reactions causes a change in the internal energy even at constant temperature.


Why are polysaccharides non-reducing sugars? My book says that polysaccharides are non-reducing sugars, and they form of condensation of >6 molecules of monosaccharides. The condensation involves the carbonyl groups of the sugars, leaving only one free carbonyl group at the end of a big molecule.

My question is: since there is one free carbonyl group why aren't polysaccharides reducing sugars?!

For polysaccharides made with only glucose (starch, cellulose, glycogen, etc), only 1 unit can be reduced from hundreds, thousands or tens of thousands of units. With the same mass of dextrose and starch, the amount reduced for starch is 1/100, 1/1000 or 1/10000 the amount for dextrose. This is usually below the detection limit for most ways of detecting the reaction.


When does ratiometric measurement account for signal variations? I've been reading about pH and ion sensors that rely on a ratiometric approach, which supposedly circumvents many problems of intensity-based methods such as dye bleaching, dye leaching, and fluctuations in source intensity. I can think of situations though when this might not be the case.

For example, fura-2 is excited at two independent wavelengths, and the two emissions are detected sequentially. If the excitation source intensity at either wavelength were to fluctuate, the ratio of the emissions would be off from the calibration. Another example would be a FRET sensor in which the acceptor dye leaches at a faster rate than the donor dye; the emission ratio of acceptor to donor would no longer match the calibration.

So when does ratiometric measurement account for signal variations like dye bleaching, dye leaching, and fluctuations in source intensity? Is it only when there is one dye and one excitation source? Am I missing any other requisites? It seems like if there's only a single dye that bleaches/leaches or a single excitation source that fluctuates, the change in dye fluorescence would remain proportional such that the emission ratio remains the same.

Thank you!

National Instruments$^{TM}$ gives a good explanation here, in a paper titled "Measuring Bridge-Based Sensors with the Ratiometric Approach, Publish Date: Feb 11, 2016".  Here are a couple excerpts relevant to your question:

  1) "The fact that the mV/V ratio stays constant for changes in the excitation voltage increases measurement stability. Small changes in the environment due to temperature differences typically affect excitation voltages. Because the excitation voltage is sensed and fed back to the ADC, any changes play a smaller role in the accuracy and stability of the measurement."

  2) "By using the excitation as the source for the ADC reference, the output of the ADC provides measurements that are in units of V/V. This reduces the amount of scaling required to return the appropriate units."

So, the short of it is that noise, particularly that arising from power fluctuations, in either the source or detector in the measurement systems as you described is largely cancelled out when ratioed to a reference system.

This approach is most effective when you are rapidly switching, i.e. via a chopper mirror, between the sample measurement source and detector and that of your reference or blank measurement.  When the signal from the source is then ratioed to that of the reference system, then any noise common to the two is effectively cancelled out, increasing your signal to noise ratio.  This method addresses your examples of "sequential" measurements. In the system describe above, you are effectively taking your reference measurement simultaneously or in parallel rather than sequentially.


What would be the reaction equation between ammonium nitrate and a powdered metal? The SDS for ammonium nitrate says it will react with powdered metals spontaneously. What would be the reaction equation, say with iron or copper?

This mixture is used in various pyrotechnics, and it is intentionally allowed to combust with heat and shock, but how does the reaction proceed spontaneously as described in various SDSs in the absence of any ignition source? Is this due to the mix attracting small amounts of water, allowing the nitrate to attack a metal instead?

Would the theoretical products always just be various oxides of the metal involved?

I'm betting they mean that you can inadvertently make an explosive (technically a blasting agent) or highly flammable mixture by mixing powdered metal and ammonium nitrate.  I used to be a hard rock miner many centuries ago and was well read on explosives.  Aluminum powder and ammonium nitrate is a classic solid rocket fuel and industrial blasting agent/military explosive, for instance ammonal in bombs.  Ammonium nitrate/aluminum powder mixes (magnesium works, too) have better explosive properties than the usual mining ANFO (ammonium nitrate/fuel oil) mixtures, but the latter are preferred usually since ANFO is less expensive.

While blasting agents technically need a booster explosive to detonate, they will burn ferociously.  I'm guessing when a writer of a SDS sheet talks about powdered metal and AN, they are thinking about the potential of creating an extremely flammable mixture.

I looked at some MSDS for ammonium nitrate, and I see powdered metal listed along with reducing agents as incompatible mixtures.  Could you post a link to the MSDS you are reading?  I'm interpreting 'incompatible' as referring to explosive potential and not to a slow spontaneous reaction.

EDIT:  Well.  The usual metal powders that react with AN used in mining and explosives are aluminum and magnesium, so that is all I considered.  However, a bit of research shows that powdered zinc does react with ammonium nitrate, as in this video (note that it didn't actually ignite until set afire, but it definitely reacted).


The dissolution of ammonium chloride The dissolution of ammonium chloride is used to cool a container of water placed in the solution. It's an endothermic process. What absorbs the heat and what loses it?

When discussing this question, one needs to think about all the processes involved (or at least those that account for most of what happens). In general, any process will occur spontaneously, if $\Delta G$ is negative. To determine $\Delta G$, equation (1) is used.

$$\Delta G = \Delta H - T \Delta S\tag{1}$$

There is an enthalpic term related to the heat a reaction produces or requires and an entropic term that is temperature-dependent. We cannot say a lot about the entropic term a priori, but we can about the enthalpic term — which is luckily what the question is about. For that, we first need to write a reaction equation (2):

$$\ce{NH4Cl (s) + n H2O <=>> NH4+ (aq) + Cl- (aq) + m H2O}\tag{2}$$

Dissolving of the ammonium chloride can also be thought as two separate processes, see equation $(2')$.

$$\ce{NH4Cl (s) -> [NH4+ (g)] + [Cl- (g)] -> NH4+ (aq) + Cl- (aq)}\tag{2'}$$

Meaning that first we break up the ammonium chloride crystal structure and second we dissolve the bare ions. If we want to write that in single enthalpy terms, we can do that as shown in equation (3).

$$\Delta H_\mathrm{tot} = -\Delta H_\mathrm{lattice}(\ce{NH4Cl}) + \Delta H_\mathrm{solv}(\ce{NH4+}) + \Delta H_\mathrm{solv}(\ce{Cl-})\tag{3}$$

All of these values can be looked up. I have given those as calculated by Jenkins and Morris in table 1.[1]

$$\textbf{Table 1:}\text{ Values of enthalpies used in this answer as}\\\text{quoted from Jenkins and Morris (reference 1).}\\\begin{array}{cccc}\hline
\text{compound} & \Delta H_\mathrm{lattice} [\mathrm{kJ/mol}] & \Delta H_\mathrm{solv} [\mathrm{kJ/mol}] & \Delta H_\mathrm{tot} [\mathrm{kJ/mol}] \\\hline
\ce{NH4Cl} & -709.1 & - 694.7 & 14.4 \\ \hline\end{array}$$

So we need energy (a lot of it!) to break up the crystal lattice of $\ce{NH4Cl}$. We then re-gain energy by creating the hydrated, i.e. dissolved, ions in solution. If energy is required, it is typically (excluding photochemical reactions — not the case here) heat energy which is simply drawn from the surroundings. If energy is released, it is typically (same caveat) released as heat into the surroundings.

Thus, the answer to your question is:

The heat is absorbed by the solid ammonium chloride. It is used to break up the salt crystal according to equation $(2'.1)$. Part of it is re-released (unnoticed) by forming solvent–ion hydrogen bonds (the solvation mechanism of $\ce{NH4Cl}$ — equation $(2'.2)$).
This heat is drained from the surroundings, which in this case is primarily the water in which you wanted to dissolve the ammonium chloride. (Of course, the solution will then further exchange heat with whatever is around it, meaning that either the air around a bottle/flask or an outer flask as in your described experimental setup will lose heat at the benefit of the $\ce{NH4Cl}$-solution.)

Reference:

[1]: H. D. B. Jenkins, D. F. C. Morris, Mol. Phys. 1976, 32, 231. DOI: 10.1080/00268977600101741.


Why is the vanadium(3+) ion paramagnetic? I know that the electron configuration of vanadium is $[\ce{Ar}]\mathrm{4s^2 3d^3}$.

None of the electrons in the 3d subshell are paired. Once it loses these three electrons, shouldn't the remainder of the electrons be paired? How can $\ce{V^{3+}}$ be paramagnetic if it loses all its unpaired electrons?

In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the $\mathrm{d}$-block (a.k.a. the transition metals) obey one special rule:

In general, electrons are removed from the valence-shell $\mathrm{s}$-orbitals before they are removed from valence $\mathrm{d}$-orbitals when transition metals are ionized.

(I took this formulation from these online lecture notes, but you will find equivalent statements in your textbooks.)
So, what that does mean is that if you remove electrons from vanadium(0), you will remove the $\mathrm{4s}$ electrons before you remove the $\mathrm{3d}$-electrons. So, you have the following electronic configurations:

$\ce{V}$ is $\ce{[Ar]} \mathrm{4s^2 3d^3}$
$\ce{V^2+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^3}$
$\ce{V^3+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^2}$
$\ce{V^4+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^1}$
$\ce{V^5+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^0}$

And thus, $\ce{V^3+}$ is paramagnetic, because it has two unpaired $\mathrm{3d}$-electrons. In fact, all the ions above are paramagnetic, except $\ce{V^5+}$.


How do you explain very high vanadyl phosphate NMR chemical shifts? Generally, phosphates have chemical shifts in the range -20 to 10 ppm. For vanadyl phosphates containing $\ce{V^{4+}}$, the isotropic chemical shift is about 1500 ppm. Why?

$\ce{V^{4+}}$ (and $\ce{V^{3+}}$) is paramagnetic, and therefore you would expect a very large change in chemical shift. Paramagnetic shifts arise from the interaction of the observed nucleus with unpaired electron spin of the paramagnetic metal ion.
The observed isotropic shift comes from two interactions: the contact shift and pseudo-contact shift.

The contact shift comes from the direct interaction between the observed nucleus and the unpaired electron - a through bond effect. This gives very large shifts - many hundreds of ppm.

The indirect pseudocontact shifts essentially come from the interaction of the unpaired electron density with the electron density of the orbitals of the observed nucleus, and are effectively compensating for the nonzero magnetic field of the paramagnetic source - a through space effect. These are much smaller shifts, often within the usual observed chemical shift range.

Whilst the measurement of contact shift information is infrequently used for structural insight, pseudocontacts using paramagnetic lanthanides are used extensively for 3D solution structures of proteins.


What is the pKa of the second ionizable group? One compound contains an amino group with $\mathrm{p}K_\mathrm{a}$ of $8.8$, and one other ionizable group with a $\mathrm{p}K_\mathrm{a}$ of $5$ to $7$. To $100\ \mathrm{mL}$ of a $0.2\ \mathrm{M}$ solution of this compound at $\mathrm{pH}\ 8.2$ was added $40\ \mathrm{mL}$ of a solution of $0.2\ \mathrm{M}\ \ce{HCl}$ and the $\mathrm{pH}$ changed to $6.2$. What is the $\mathrm{p}K_\mathrm{a}$ of the second ionizable group?

I am a little confused because I am not sure what to do when the $\mathrm{pH}$ does not equal the $\mathrm{p}K_\mathrm{a}$ of the amino group. I'm thinking since the known $\mathrm{p}K_\mathrm{a}$ of $8.8$ is highest at that point there should be no free $\ce{H+}$.

Since you're given that the maximum $\mathrm{p}K_\mathrm{a}$ of the second group is $7$, you can assume that the amino group will fully protonate before the second group starts accepting electrons.  (Perhaps this is what you meant by "at that point there should be no free $\ce{H+}$"?)
Therefore, from Henderson-Hasselbalch$^\ddagger$, you can calculate the fraction of the amino group that is protonated based on the starting $\mathrm{pH}$ and the amino group $\mathrm{p}K_\mathrm{a}$. (Hint: Yes, the amino group should be mostly, but not entirely, protonated.)
It's straightforward to then calculate:

The total number of moles of acid added
The number of moles of not-yet-protonated amino groups before adding the acid

which is equal to

The number of moles of acid consumed to protonate the rest of the amino groups

and, therefore, you can find

The number of moles of acid left over to protonate a portion of the second ionizable group in the molecules.

Having calculated #4, you can then calculate what fraction of the second ionizable group is protonated by the leftover acid.  This feeds back into Henderson-Hasselbalch, from which you can calculate the second $\mathrm{p}K_\mathrm{a}$ value since you are told the value of the final $\mathrm{pH}$.
$^\ddagger\,\mathrm{pH} = \mathrm p K_\mathrm a +\log\left(\frac{\ce{[A-]}}{[\ce{HA}]}\right) = \mathrm pK_\mathrm a + \log \left(\frac{[\mathrm{A_{total}} ] - [\ce{HA}]}{\ce{[HA]}}\right) = \mathrm p K_\mathrm a + \log\left(\frac{1}{f_\mathrm{protonated}}-1\right)$


In cleaving ethers with both SN1 and SN2 mechanisms possible, which occurs first? What are the products when tert-butyl ethyl ether is cleaved with concentrated $\ce{HI}$?

My answer is iodoethane and tert-butanol, but the correct answer is ethanol and 2-iodo-2-methylpropane.

I know that the ethyl side goes by an SN2 mechanism because it is a primary R group, and the t-butyl group would be SN1 because it is tertiary. However, why would the SN2 reaction not happen first?

One "side" of a molecule doesn't react by one mechanism and the other "side" by another mechanism. Generally speaking, a molecule reacts by a single mechanism.  Now, the mechanism can change depending upon the other reactants that are present as well as the reaction conditions.  In the case of t-butyl ethyl ether, if the reaction were carried out with $\ce{KI}$ under non-acidic conditions an $\mathrm{S_N2}$ reaction would occur and your answer (iodoethane and tert-butanol) would be correct.  On the other hand, if the reaction were carried out under acidic conditions then an $\mathrm{S_N1}$ reaction would occur.  As you pointed out a tertiary carbocation is more stable than a primary carbocation so the expected products in this case would be ethanol and 2-iodo-2-methylpropane.  Apparently, the reaction conditions in your reaction must have been such that the reaction occurred by the $\mathrm{S_N1}$ mechanism.


Best reagent for reducing a ketone in the presence of an alkene The following reduction is best accomplished with:

  (a) Zn/Hg, conc. HCl
  (b) N2H4, C2H5ONa
  (c) H2, Pd
  (d) All of the above

My thoughts: By looking at the question it is evident that only the C=O group should be affected. H2/Pd is used for hydrogenation which might affect the C=C double bond as well. Option (a) gives the reagents which are used for the Clemmensen reduction which is used to convert carbonyl groups to alkanes. So, I think (a) is the correct answer. Is that right?

The Clemmensen reduction is really only effective at reducing ketones to alkenes when one group is aromatic. Additionally, the Clemmensen reduction uses concentrated HCl, which might cause problems. Another functional group in your molecule, the tertiary alcohol, will be removed by HCl via an elimination reaction.

Choice (b) is the Wolff–Kishner reaction, which also converts ketones into alkenes. It requires strong base (not acid) and works on almost all ketones.

Choice (c) will react with the alkene also, and so cannot be correct.


What kind of 'product' is the 'product' in 'ionic product of water'? My language has different words for product (=the result of a process) and product (=the sum of multiplied quantities), so I need to know the right meaning to choose the right word.

Which of the meanings above is present in the English term 'ionic product of water' ($K_\mathrm{w}$)?

This definition from http://www.chemguide.co.uk/physical/acidbaseeqia/kw.html seems to suggest it is a sum of multiplied quantities:

  The units of $K_\mathrm{w}$: $K_\mathrm{w}$ is found by multiplying two concentration terms together. Each of these has the units of $\mathrm{mol~dm^{-3}}$. Multiplying $\mathrm{mol~dm^{-3}} \times \mathrm{mol~dm^{-3}}$ gives you the units above.

However, the use of the verb produce in this sentence from https://en.wikipedia.org/wiki/Self-ionization_of_water suggests something that is produced as a result of a process:

  According to the theories of Svante Arrhenius, this must be due to the presence of ions. The ions are produced by the self-ionization reaction $\ce{H2O + H2O <=> H3O+ + OH-}$

Alternatively, could both meanings apply depending on your perspective?

Thanks in advance for any clarification.

Per Zhe's comment, in the specific context of the term ionic product of water, the word product is used in the sense of the result of the mathematical operation of multiplication:

  The ionic product of water equals the concentration of $\ce{H+}$ times the concentration of $\ce{OH-}$:
  $$K_\mathrm w = \ce{[H+][OH-]}$$

(Strictly, per MaxReid's comment, the ionic product water is of course the product of the activities of $\ce{H+}$ and $\ce{OH-}$, but concentrations are commonly used in routine/non-rigorous calculations as they provide good accuracy for dilute solutions.)

This sense of the word product applies any time one is referencing the mathematical description of the equilibrium among the ionic forms of a given species. This usage also includes the solubility product (again with the above caveat about activities):

  The solubility product of silver chloride equals the concentration of $\ce{Ag+}$ times the concentration of $\ce{Cl-}$:
  $$K_\mathrm{sp} = \ce{[Ag+][Cl-]}$$

As an aside, note that the main exception to this terminology is when the expression describing the distribution of species requires division by one or more quantities. In this case, the term to use is quotient, for which I expect little ambiguity exists:

  The reaction quotient of the reaction
  $$\ce{A + B->C + D}$$
  is
  $$ Q = {\ce{[C][D]}\over \ce{[A][B]}}$$

However, to be complete: chemistry does also use the word product in the other sense you describe, as in something being produced by a reaction.  Due to this, the word product in the phrase the ionic product(s) of X could also bear this meaning.  For example:

  The ionic products of the aqueous decomposition of dinitrogen tetroxide at neutral $\mathrm{pH}$ are nitrite and nitrate ion:
  $$\ce{N2O4(aq) + 2H2O -> NO2-(aq) + NO3-(aq) + 2H3O+}$$

In general, I think the difference boils down to whether one is talking about the ionic distribution of a specific species of interest, as opposed to the transformation that occurs in a particular reaction of interest.

As I noted above, though: In the specific case you are asking about, product is used in the sense of the result obtained from multiplication.


Is redox potential a valid term? I mean doesn't every compound has its own different reduction and oxidation potential ? Or its just one potential that can be compared to understand which one will get reduced and which one will get oxidised ?

I mean doesn't every compound has its own different reduction and oxidation potential ? Or its just one potential that can be compared to understand which one will get reduced and which one will get oxidised ?

A compound has neither. You need to specify the half reaction. For example, take an aqueous solution of $\ce{Fe^2+}$ and hydrogen peroxide, among others. What are possible half reactions for $\ce{Fe^2+}$?

$$\ce{Fe^2+ -> Fe^3+ + e-}\tag{1}$$
$$\ce{Fe -> Fe^2+ + 2e-}\tag{2}$$

and maybe some more exotic ones. For hydrogen peroxide, you would have e.g.

$$\ce{H2O2 -> O2 + 2e- + 2H+}\tag{3}$$
$$\ce{2H2O -> H2O2 + 2e- + 2H+}\tag{4}$$

Notice that these are all written in the direction of reduction, and are associated with a reduction potential. To see which reactions are feasible, you would compare reduction potentials and come up with a redox reaction combining a half reaction as written with a half reaction (the oxidation half reaction) in the reverse direction. For example, hydrogen peroxide could undergo disproportionation:

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

This is a slow reaction, but the enzyme catalase (which happens to have an iron(III) bound to its active site) increases the rate substantially, as you can see from the oxygen evolving when you put some hydrogen peroxide on a freshly cut slice of potato.

  Is redox potential a valid term?

Yes, under certain circumstances it makes sense to talk of redox potential as a property of an aqueous solution, similar to discussing its pH. For example, if a biological cell contains glutathione at fairly high concentration in the reduced and oxidized form (abbreviated as GSH and GSSG), this will determine the redox state of other substances in the cell. If a strong oxidation agent is introduced, it will react with GSH. If a strong reduction agent is introduced, it will react with GSSG. Like mainining a pH with a buffer, the redox potential of the cell is maintained through the ratio of GSH and GSSG (and we could call it a redox buffer).

The analogy between pH buffer and redox buffer only goes so far. The biggest difference is in the role of water. While water acts as an acid and base at any pH (i.e. whenever acids and bases are added), it does not undergo reduction and oxidation for many redox reactions in aqueous solution. Any pair of acid and base will react (because water provides the "reservoir" of hydrogen and hydroxide ions). Many pairs of possible oxidizing and reducing agents will not react, even when thermodynamically favored because there is no good mechanism for the electrons to be transferred (a question of kinetics).


Do maxima in Gibbs energy also correspond to equilibrium positions? ΔrG=(∂G/∂ξ ) at constant p and constant T, where ξ is the extent of reaction.

By Second Law implication, ΔrG= 0= (∂G/∂ξ ) at equilibrium. If we consider a graph of G against ξ, this would mean  we have a turning point (where first derivative=0) when ξ is that extent of reaction when equilibrium attained:

  (∂G/∂ξ )= 0 when ξ= extent of reaction until equilibrium
  attained

So, ΔrG=0 when equilibrium attained.

However, first derivative=0 does not mean minimum point only, it can mean maximum point too! Let's say you do work until you reach Gibbs energy maxima. My question: Does a Gibbs energy maxima correspond to equilibrium state or not?

(a) If yes, doesn't this violate the Second Law which implies that Gibbs energy should be minimized whenever possible? No matter which direction you move, your Gibbs energy will always decrease. Just simply move from the maximum point and you will at least be better than your current state:

(b) If no, then it will mean that we have to move from the maximum point, but in which direction? In case the graph is a symmetrical parabola, where same dξ results in same dG, what will you observe in real life? Shift to reactants or products?

G vs. $\xi$ does not have a maximum

If all reactants and products are pure liquids and solids, G vs. $\xi$ is linear. If some of the species are in mixtures, the entropy of mixing is responsible for the "sagging" shape of the curve. If the curve has an extreme value, it will be a minimum.

Analogy to mechanics

In mechanics, this situation would be called metastability. Giving a nudge in either direction will allow the system to reach lower potential energy.


Kinetics of interstellar chemistry: Applying steady state to formation of H₂ I am trying to solve problem 2 from the 2009 International Chemistry Olympiad (IChO), which is as follows:

  If two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\ce{H2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\ce{H2}$ rapidly desorbs. This question examines two kinetic models for $\ce{H2}$ formation on the surface of a dust particle.

  In both models, the rate constant for adsorption of $\ce{H}$ atoms onto the surface of dust particles is $k_\mathrm a = \pu{1.4*10^-5 cm3 s-1}$. The typical number density of $\ce{H}$ atoms (number of H atoms per unit volume) in interstellar space is $[\ce{H}] = \pu{10 cm-3}$. (You may treat numbers of surface adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations).

  Desorption of $\ce{H}$ atoms is first order with respect to the number of adsorbed atoms. The rate constant for the desorption step is $k_\mathrm d = \pu{1.9 * 10^-3 s-1}$. Assuming that only adsorption and desorption take place, calculate the steady-state number, $N$, of $\ce{H}$ atoms on the surface of a dust particle.

First off, what do they mean by steady-state number here? I tried searching online but I could not find anything. The solution they give has the equation:

$$k_\mathrm d N − k_\mathrm a[\ce{H}] = 0$$

from which they solve for $N$. I know this has something to do with steady state, but in this case why must the rate of absorption be equal to the rate of desorption? Is it because you are considering a single $\ce{H}$ atom which is considered to be an intermediate? Also, why is it $k_\mathrm dN$?

Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?

In the above, how did they come up with those rates? Why does absorption have $\ce{[H]}x_m$ but desorption have $\ce{m}x_m$? Finally, for the reaction rate, where did they get $m(m-1)$? I think the $\frac{1}{2}$ comes from the fact that the rate of production of $\ce{H2}$ is half the rate of absorption/desorption, is that right?

From these, they calculate the rate of change of $x_0$, $x_1$, and $x_2$ as follows:

  $$\begin{align}
\frac{\mathrm dx_0}{\mathrm dt} &= -k_\mathrm a[\ce{H}]x_0 + k_\mathrm d x_1 + k_\mathrm r x_2 \\
\frac{\mathrm dx_1}{\mathrm dt} &= k_\mathrm a[\ce{H}]x_0 - (k_\mathrm a [\ce{H}] + k_\mathrm d) x_1 + 2k_\mathrm dx_2 \\
\frac{\mathrm dx_2}{\mathrm dt} &= k_\mathrm a[\ce{H}]x_1 - (2k_\mathrm d + k_\mathrm r)x_2
\end{align}$$

I sort of understand how they got the value for rate of change of $x_0$ as it is the rate of production minus the rate of consumption, and desorption from state 1 and reaction produce state 0, while absorption consumes it. However, why does the $k_\mathrm a$ term come with $[\ce{H}]$ while the other terms do not? What happened to the $m x_m$ that was mentioned above?

Further, how did they get the rate of change for $x_1$ and $x_2$?

OK, let's step through your questions one at a time.

  First off, what do they mean by steady-state number here?

Intermediates formed in reactions usually exist in low concentration.  The steady-state assumption assumes that the concentration of this (low-concentration) intermediate will not change appreciably over the course of the reaction once steady state has been established.  The rate of its formation is assumed to be equal to the rate of its consumption.  Making the steady state assumption allows us to recast the previous statement as a mathematical equation as we'll see below.

  In this case why must the rate of absorption be equal to the rate of desorption

Just like in a chemical equilibrium where it takes a finite time for the equilibrium to establish itself, it takes a finite time before the steady state is established. Once there is a finite, non-zero concentration of the intermediate and we have arrived at the steady state, then the rate of absorption and desorption will be equal and the non-zero concentration of the intermediate(s) will remain constant – a steady state.

Applying the steady-state assumption to the problem at hand allows us to state that for all of the intermediates (dust particles with 1 or 2 hydrogen atoms on them), the rate of hydrogen atom adsorption on an intermediate must be equal to the rate of hydrogen atom desorption from that intermediate. This can be written as

$$k_\mathrm a [\ce{H}] = k_\mathrm d N$$

and then rearranged and solved:

$$\begin{align}
N &= \frac{k_\mathrm a}{k_\mathrm d}[\ce{H}] \\
&= \frac{\pu{1.4 * 10^-5 cm3 s-1}}{\pu{1.9 * 10^-3 s-1}} \cdot \pu{10 cm^-3} \\
&= \pu{7.4 * 10^-2}
\end{align}$$

  Why is it $k_\mathrm d N$?

Because $k_\mathrm d N$ represents the total rate of desorption of hydrogen atoms from dust particles, the right side of our verbal equation. It represents desorption from $x_1$ and $x_2$ - the only two species that can desorb, and it is equal to

$$k_\mathrm d N = k_\mathrm d x_1 + 2k_\mathrm d x_2$$

the factor of "2" arising because there are 2 hydrogen atoms on each dust particle in $x_2$, and both are viable candidates for desorption.

  Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?

It is the same rate, that's what $k_\mathrm a [\ce{H}] = k_\mathrm d N$ means.  Rates are always a positive number, we are multiplying a rate constant times a concentration, both positive numbers.  However, when we write an equation and insert the rate expression into it, we can insert a minus sign before the rate expression if the rate expression represents a pathway that is causing the concentration of a product, reactant or intermediate to decrease.

  How did they come up with those rates?

Adsorption rate: $r_\mathrm a = \sum_m k_\mathrm a [\ce{H}] x_m = k_\mathrm a [\ce{H}]x_0 + k_\mathrm a [\ce{H}]x_1$
This is just the typical kinetic expression – adsorption rate constant multiplied by the concentration of hydrogen atoms available to adsorb times the concentration of the various dust particles that can adsorb a hydrogen atom, which are only $x_0$ and $x_1$.
Desorption rate: $r_\mathrm d = \sum_m k_\mathrm d m x_m = (k_\mathrm d \cdot 0 \cdot x_0) + (k_\mathrm d \cdot 1 \cdot x_1) + (k_\mathrm d \cdot 2 \cdot x_2)$
Here $m$ is a concentration term analogous to $[\ce{H}]$.  It represents the number of hydrogen atoms on a dust particle, and can be 0, 1 or 2.  Note how it "removes" $x_0$ from the desorption equation as it should since a particle with no hydrogen atoms cannot desorb.  Notice too how it adjusts $x_2$ for the fact that there are two adsorbed hydrogen atoms in this case, so desorption is statistically twice as likely to happen.  It differs from $[\ce{H}]$ in its units: $[\ce{H}]$ is in $\pu{atoms/cm3}$ while "m" is in units of atoms/dust particle.
Reaction rate: $r_\mathrm r = \left(\frac{1}{2} \cdot k_\mathrm r \cdot 2 \cdot 1 \cdot x_2\right) \left(\frac{1}{2} \cdot k_\mathrm r \cdot 1 \cdot 0 \cdot x_1\right) = k_\mathrm r x_2$

EDIT 1:

  Also, after your explanation I understood how they got the absorption and desorption rates, but I'm still confused on how they got the reaction rate $\left(\frac{1}{2} \cdot k_\mathrm r \cdot 2 \cdot 1 \cdot x_2\right) \left(\frac{1}{2} \cdot k_\mathrm r \cdot 1 \cdot 0 \cdot x_1\right)$. Could you please elaborate further on where all the numbers came from?

Note we have a factor of $1/2$ because it takes 2 hydrogen atoms to produce 1 hydrogen molecule.  The $2$ and the $1$ come from the concentration terms $m(m-1)$.  In the "Reaction" case we have to combine two hydrogen atoms.  It is a bimolecular reaction and just like in the bimolecular reaction $\ce{A + B -> C}$ where the rate is proportional to $[\ce{A}][\ce{B}]$, here the analogous concentrations are $m(m-1)$.  Initially the concentration is $m=2$, but after that first hydrogen atom is "selected" the remaining concentration from which to pick our second hydrogen atom has been decreased to $m-1$.  What we're saying is that when we pick the first hydrogen we have two atoms to choose from, when we pick the second hydrogen atom we have one atom to choose from. If we had 5 hydrogen atoms on a particle and were still going to make 2 picks the equation for our two picks would become $5(5-1)=20$.  We would expect this case to proceed $(20/2)=10$ times faster than the case with only 2 atoms on the dust particle.

There are 2 ways to form $\ce{H2}$ in our problem. Label the hydrogen atoms A and B, then we can either pick via an AB or BA sequence. $m(m-1)$ is exactly analogous to $[\ce{H}][\ce{H}]$, it's just not a square because there is a significant difference between $2\cdot 2$ and $2\cdot 1$. When dealing with a mole of atoms ($N = 6 \cdot 10^{23}$, or Avogadro's number) and they react with one another in a bimolecular reaction, there is no need to write $(6\cdot 10^{23})(6\cdot 10^{23} - 1)$, but in our case with such small numbers the difference matters.

  Why does absorption have $[\ce{H}]x_m$ but desorption have $mx_m$?

As noted above, $m$ is a concentration term analogous to $[\ce{H}]$, but with different units.

  Finally, for the reaction rate, where did they get $m(m−1)$?

See above.

  why does the $k_\mathrm a$ term have $[\ce{H}]$ while the other terms do not?

The other terms use $m$ which, as explained above, is analogous to $[\ce{H}]$ in that they both reflect concentrations, $[\ce{H}]$ for hydrogen atoms in space and $m$ for hydrogen atoms on dust particles; they use different units as explained above.

  What happened to the $mx_m$ that was mentioned above?

Since the full rate expressions have been written out, they have been replaced with integers, 0, 1, or 2, representing the concentration of hydrogen atoms on a dust particle.

  Further, how did they derive the rate of change for $x_1$ and $x_2$?

The rate of change of $x_1$ is the rate of formation minus the rate of destruction.

The rate of formation is given by $k_\mathrm a[\ce{H}]x_0$ (the rate of a bare dust particle picking up a hydrogen atom) plus $2k_\mathrm d x_2$ (the rate of a dust particle with two hydrogen atoms losing one).

The rate of destruction is given by $k_\mathrm a[\ce{H}]x_1 + k_\mathrm d x_1$ (the rate of a dust particle with one hydrogen atom either picking up one more, or losing it).

For the time being, I'll leave the rate of change of $x_2$ as an exercise for you to test yourself on what you've learned.


What does it mean to take reading up to 0.05 mL? In titration we usually take burette reading upto $0.01~\mathrm{mL}$, but my teacher says that we need to take reading up to $0.05~\mathrm{mL}$. What does that mean? Does it mean if I read $23.56~\mathrm{mL}$ that I should record it as $23.60~\mathrm{mL}$ and if I read $23.44~\mathrm{mL}$ I should record it as $23.40~\mathrm{mL}$?

I believe what your teacher means by this is that the error for your measurement is within $0.05~\mathrm{mL}$ of the estimate, since $0.05~\mathrm{mL}$ is half the value of the smallest measurable volume in a typical $50~\mathrm{mL}$ burette. This is typically how the error of an instrument reading is determined.

The readings you gave would then be: $23.56~\mathrm{mL} \pm 0.05~\mathrm{mL}$ and $23.44~\mathrm{mL} \pm 0.05~\mathrm{mL}$.


Effect of phenyl and vinyl substituents on Acidity of carboxylic Recently I came across this statement in my textbook:

  Direct attachment of groups such as phenyl or vinyl to the carboxylic
  acid, increases the acidity of corresponding carboxylic acid contrary
  to the decrease expected due to resonance effect.

On researching online, I found this statement as an answer:

  Although we expect a decrease in acidity due to the resonance effect
  of phenyl and vinyl groups, but because of the greater
  electronegativity of the the $\ce{sp^2}$ hybridized carbon which is
  directly attached to the carboxylic carbon.

My Thoughts

Please keep in mind that the carbonyl carbon(here carboxylic carbon) is also $\ce{sp^2}$ hybridized. So $\ce{sp^2}$ hybridized carbon is directly attached to the carboxylic carbon. Therefore there is no electronegativity difference between the two carbons and as a result of this only +R effect should work and direct attachment of groups such as phenyl or vinyl to the carboxylic acid should not increase the acidity of corresponding carboxylic acid rather they should decrease its acidity due destabilization (due to electron donating nature vinyl and phenyl group by means of resonance) of conjugate base so formed.

The question

Kindly explain it theoretically by giving plausible reasons as to why the statement in my textbook is true?

Note: avoid quoting pKa values

When the cases of carboxilic acids are considered, we must note that the conjugation of $\ce{COO-}$ group is so strong in itself, that it's conjugation with the rest of the compound is rendered relatively weak. In such cases, the inductive effect is dominant.

Let us take the example of benzoic acid. The carbon to which $\ce{COOH}$ is attached shall be named as the key-carbon $\ce{C}$1. The conjugate base would be $\ce{PhCOO-}$, and the +I(Inductive effect) of $\ce{COO-}$ group will be dominant on $\ce{C}$1. Any group attached to this compound will further increase or decrease the acidic strength of the compound as a whole, depending upon the attached group's contribution towards electron density at $\ce{C}$1.

For example, in case of para-nitrobenzoic acid, electron density at $\ce{C}$1 decreases due to $\ce{NO2-}$'s strong -R (Mesomeric) Effect(partial positive charge is developed on $\ce{C}$1 which is stabilised by $\ce{COO-}$'s +I). Hence it turns out to be a strong acid.

The opposite case is that of para-methoxybenzoic acid. The electron density at $\ce{C}$1 increases due to +R of $\ce{OCH3}$ group and makes the conjugate base unstable, hence acid weaker. This can be cross checked by the $\mathrm{p}K_\mathrm{a}$ values of above mentioned two compounds as well.

Now coming to the compounds mentioned in the question, which are phenyl and vinyl, one must note that both of these would reduce the charge density on $\ce{C}$1 because of the $\mathrm{sp^2}$ hybridised Carbon atom which contributes to the overall electro-negativity of these two groups.

Long story short, the resonance effect between $\ce{COO-}$ and the rest of the compound is suppressed by a relatively stronger Inductive effect which works mutually between $\ce{COO-}$ and the rest of the compound.


Recognising a rate-determining step in E1 If there is a hydride/methyl shift possible in E1, then is the carbocation formation the rate-determining step (RDS), or the shift, or is it considered a cumulative step and happens all at once?

Anindya Prithvi has said everything you needed to hear in two sentences. It is true that rate determining step for E1 mechanism is the carbocation formation. Thus, I present here the detail energy diagram for reaction progress of acid-catalyzed dehydration:

The first step is the protonation of starting compound (SM), which is shown in the right hand top corner in blue box. The next step is the rate determining step, which has the biggest activation energy and progress through the transition state 2. The resultant intermediate is the relatively higher energy 2°-carbocation, which is quickly rearrange to lower energy 3°-carbocation by 1,2-hydride shift through the transition state 3. It is worth noting that this 1,2-hydride shift does not involve net bond-breaking or bond-forming (it is believed to be undergoing concomitant bond breaking-making mechanism involving orbital overlapping). Thus its transition state (transition state 3) would not be higher in energy than that of transition state of RDS. The 3°-carbocation would finally give away the expected product through the transition state 4. It doesn't matter if all other steps happen at once or separately. All those steps have to wait to happen until the formation of the 2°-carbocation. That's why it is the rate determining.


Reaction of potassium with water In general, when we put potassium into water, it will produce potassium hydroxide and hydrogen:

$\ce{2K + 2H2O ⟶2KOH + H2}$

However isn't the following reaction also possible?

$\ce{2K + H2O ⟶K2O + H2}$

Why isn't the oxide produced?

The reactions are ongoing this way:

Relatively free electrons of  potassium reduce water:

$$\ce{2 e- + 2 H2O -> H2 + 2 OH-}\tag{1}$$

That leaves metal positively  charged.

Liquid ammonia, if exposed to alkali metal, reacts with electrons much slower than water, forming a dark blue solution of solvated electrons. As electrons progressively kick out protons from ammonia, forming hydrogen, the solution finally turns to a colourless solution of NaNH2.

But back to water.

The potassium ions get hydrated, decreasing the charge.....

$$\ce{K(s)^{n+} -> K(s)^{(n-m)+} + m K+} \tag{2}$$

forming $\ce{KOH}$ solution in form of mixture of hydrated ions $\ce{K+ + OH-}$

But heavy potassium ions cannot keep pace with light and fast electrons and the drop of melted metal progressively gains positive charge and finally ends up by - as authors call it -  Coulombic explosion.

The hydrogen gets eventually ignited by microsparcs due the charge instability  even before the explosion. As "Terminator T1000-like" spikes of liquid metal eventually pierce isolating vapour+hydrogen layer, getting into contact with ignitable hydrogen-air mixture.

It was recently theoretically predicted by quantum chemistry simulation for several dozens of alkali atoms by Czech chemist Pavel Jungwirth And col. Chemistryworld-Alkali metal explosion explained

They have experimentally verified it by high speed 10000 f/s camera, using sodium/potassium alloy forming an eutectic with low melting point.

I knew that from the popular science radio broadcast interview, finding backwards some reference for it.

See also their article in Nature ( which I forgot about and found later):

Coulomb explosion during the early stages of the reaction of alkali metals with water

  Abstract
  Alkali metals can react explosively with water and it is textbook knowledge that this vigorous behaviour results from heat release, steam formation and ignition of the hydrogen gas that is produced. Here we suggest that the initial process enabling the alkali metal explosion in water is, however, of a completely different nature. High-speed camera imaging of liquid drops of a sodium/potassium alloy in water reveals submillisecond formation of metal spikes that protrude from the surface of the drop. Molecular dynamics simulations demonstrate that on immersion in water there is an almost immediate release of electrons from the metal surface. The system thus quickly reaches the Rayleigh instability limit, which leads to a ‘coulomb explosion’ of the alkali metal drop. Consequently, a new metal surface in contact with water is formed, which explains why the reaction does not become self-quenched by its products, but can rather lead to explosive behaviour.