How does the proticity of an acid affect the enthalpy change of its reaction with a base? For an experiment I've carried out a series of reactions between monoprotic bases (ammonia and potassium hydroxide) and several acids. Two of these acids were monoprotic (hydrochlori acid and ethanoic acid), another was  diprotic (sulfuric acid) and the other was triprotic (phosphoric acid). My results showed that the triprotic acid released more energy per mole than both the diprotic and monoprotic acids, and that the diprotic acid released more energy per mole than the monoprotic acids. This gave rise to my question: how does the proticity of an acid affect the enthalpy change of its reaction with a base?

To be clear I compared my results between weak acids, strong acids so the strength of the acid has already been accounted for.

You have to remember that enthalpy change is defined at kilojoules PER MOLE. The concentration won't affect enthalpy change because the same amount of heat is given off per mole of zinc , however the reaction will go faster with a higher concentration.


Difference between an absorbance, emission, and excitation spectrum for uv-visible spectroscopy From the list below, circle the wavelength that would be the best setting for the UV-visible spectrophotometer used to determine the FITC concentration in the sample.
\begin{array} {}
\mathrm{wavelengths:} & \pu{400 nm}& \pu{450 nm}& \pu{500 nm}&\pu{550 nm} &\pu{600 nm}
\end{array}

The exam paper answer is $\pu{500 nm}$, because we need the wavelength of "maximum absorbance." But this graph shows two curves, and the absorbance curve is omitted.
But what is the difference between an emission, absorbance, and an excitation spectrum? What are they measuring? How do you interpret/read these graphs - those with multiple curves.

When you have an excitation spectrum and an emission spectrum, it is for fluorescence spectroscopy. To obtain these spectra, you hold either the emission wavelength or the excitation wavelength and scan the other to gather the information. The excitation spectrum shows at what wavelengths the solution uses to produce its fluorescence. The emission spectrum shows what wavelengths are given off from the solution.

For absorbency spectrum, the solution has all the wavelengths shining on the solution, and you monitor the wavelengths of interest. The absorbency shows which wavelengths are absorbed by the solution.

The excitation spectrum has the same peaks as the absorbency spectrum; although the absorbency spectrum may have more peaks.


What, if any, are the differences between hemiwicking and wicking? I'm encountering these two terms, but their uses seem to be different, and there's little documentation out there defining the terms.

Here is an example definition I've found of Wicking:

  Wicking is the spontaneous flow of a liquid in a porous substrate, driven by capillary forces.$^1$

Here is an example definition I've found of hemiwicking (notice little difference between the one above):

  Hemiwicking refers to the spreading of a liquid on a rough hydrophilic surface driven by capillarity.$^2$

Below is a typical depiction showing Wenzel wetting (a), Cassie-Baxter wetting (B), and hemiwicking(c).$^3$ My assumption would be that Wenzel wetting would actually be an example of wicking?

http://journals.sagepub.com/doi/abs/10.1177/004051759606601008
http://fluids.snu.ac.kr/Publication/hemiwicking.pdf
http://jss.ecsdl.org/content/3/1/N3095/F3.expansion.html

I'm not especially familiar with the detailed differences between the two much beyond the definitions you've posted here, but it seems to me that:

Wicking occurs in the bulk volume of a porous substrate, not just on the surface; whereas
Hemiwicking occurs strictly on the surface of an otherwise impermeable substrate

As you note, in both cases the fluid movement is driven by capillarity, however.

  My assumption would be that Wenzel wetting would actually be an example of wicking?

I disagree. All three panels of the figure you've posted look to me as though they are trying to illustrate strictly surface phenomena, and thus none of them are forms of wicking.  The main difference between (a) and (c) is that in (a) the structured surface is wetted only immediately adjacent to the liquid drop, whereas in (c) the surface is wetted in an area beyond the extent of the drop. Thus, the Wenzel wetting is neither wicking nor hemiwicking.


Why does glycogen phosphorylase only act on the non-reducing end of the glycogen chain? I read that glycogen phosphorylase only acts on the non-reducing end of the glycogen chain. This enzyme requires an inorganic phosphate molecule and PLP as a cofactor. The mechanism of the enzyme is as follows:

I attempted to workout the mechanism for the reducing end of the molecule and concluded that perhaps it results in a ring opening reaction.

Is this correct? If not what is the reason?

Glycogen phosphorylase does indeed phosphorylates the non-reducing end of a glycogen chain. It is unclear from the crystal structures how come this is an exo type enzyme. The prevailing catalytic mechanism is almost as you illustrated, involving the protonation of the $\pu{\alpha}$-1,4 link leading to the formation of a oxocarbenium ion and release of the long chain; nucleophilic attack by an inorganic phosphate to the anomeric carbon $\ce{C1}$ of the intermediate, leading to the final product. [1] Overall, the $\pu{\alpha}$ anomeric configuration is maintained from substrate to product, instead of a product with $\pu{\beta}$ configuration as shown on your drawing.
I think there are some unanswered questions about this mechanism:
a) How come there is little $\pu{\beta}$-D-glucose that is produced as a side-product? This is quite unexpected for such an unstable intermediate as an oxocarbenium ion.
b) How does the above mechanism account for the loss of 500-fold activity upon mutation of Glu672? [2,3] This is a huge decrease! I'm wondering whether this residue may act as a nucleophile forming a covalent intermediate with the residual saccharide?
Regarding the second question from the OP: I wouldn't think there is ring opening at the chain anomeric center as there is no evidence that this reducing end of the chain is presented in glycogen phosphorylase catalytic site. It is the non-reducing end that is presented in the catalytic site and the $\pu{\alpha}$-1,4 link that is presented at the catalytic center.
Sources

Duke, E.M., Hadfield, A., Martin, J.L., Clifton, I.J., Hajdu, J., Johnson, L.N., Reid, G.P., Trentham, D.R., Bruce, I. and Fleet, G.W. (1991) "Towards time-resolved diffraction studies with glycogen phosphorylase" Ciba Found. Symp. 161:75-86. DOI: 10.1098/rsta.1992.0064

Schinzel, R. and Palm, D. (1990) "Escherichia coli maltodextrin phosphorylase: contribution of active site residues Glutamate-637 and Tyrosine-538 to the phosphorolytic cleavage of $\pu{\alpha}$-glucans", Biochem. 29:9956-9962. DOI: 10.1021/bi00494a028

Schinzel, R. and Drueckes, P. (1991) "The phosphate recognition sites of Escherichia coli maltodextrin phosphorylase", FEBS Lett. 286:125-128. DOI: 10.1016/0014-5793(91)80956-4


Justify the value of the product of the uncertainty in position and momentum The Heisenberg uncertainty principle states that:
$$\Delta x\Delta p_x\ge\frac12\hbar$$
But more generally for operators $A$ and $B$:
$$\Delta A\Delta B=\frac12|\langle[A,B]\rangle|$$
However, $[x,p_x]=i\hbar$. With this information, I don't see how to get from the second equation to the first.

The commutator $[\hat{A}, \hat{B}]$ of two operators $\hat{A}$ and $\hat{B}$ is also an operator, so when using the Robertson uncertainty relation we have to take the expectation value of the commutator first and then take the absolute value of it.

But for $\hat{x}$ and $\hat{p}$ the commutator $[\hat{x}, \hat{p}]$ is simply a constant $\mathrm{i} \hbar$, so that taking the expectation value is trivial: $\langle [\hat{x}, \hat{p}] \rangle = \mathrm{i} \hbar$. Then we take the absolute value of this result which is $\hbar$ and divide it by two to get $\frac{1}{2} \hbar$ as expected.


How is the yellow colour of nitric acid removed? Nitric acid obtained in laboratory is slightly yellow due to the dissolution of nitrogen dioxide which is produced due to thermal decomposition of a portion of $\ce{HNO3}$.

$$ \ce {4HNO3  -> 2H2O + 4NO2 + O2}$$

Its says here that if $\ce{CO2}$ is bubbled through the acid, it turns colourless- because it drives out $\ce{NO2}$ from warm acid which is further oxidized to nitric acid.

What does it mean by drives out ? How can carbon dioxide drive out nitrogen dioxide that's dissolved in the acid? And also, the oxidation of 'what' is being talked about here?

I think you can explain this using the equilibrium:

$\ce{NO_2(g) <=> NO_2(dissolved)}$

When you pass $CO_2$ through the warm acid, the partial pressure of $CO_2$ ($P_{CO_2}$) above the gas increases (Since partial pressure generally represents the amount of gas above the solution).

Since

total pressure $P = P_{NO_2} + P_{CO_2}$ (Dalton's Law)
And $P$ remains constant at atmospheric pressure (assuming that the system is not closed).

$P_{NO_2}$ will decrease.

According to Le Chatelier's Principle, the equilibrium proceeds backward to compensate the decrease in $P_{NO_2}$ and thus more $NO_2$ gas is released. The gas flows quickly along with the bubbled gas and this prevents the forward reaction.

This seems like quite a roundabout method but it works. I'm sure there could be a much simpler explanation though.

The oxidation mentioned is most probably the disproportion reaction of $NO_2$ when it comes in contact with water:

$\ce{3NO_2 + H_2O -> 2HNO_3 + NO}$


Finding the density of nitrogen at 0 degrees Celsius and 1 atmosphere The density of nitrogen at $\pu{0 ^\circ C}$ and $\pu{1 atm}$ is most nearly equal to which of the following quantities?
(a) $\pu{0.001 g/L}$; (b) $\pu{0.01 g/L}$; (c) $\pu{0.1 g/L}$;
(d) $\pu{1 g/L}$; (e) $\pu{10 g/L}$.

The answer is (d), $\pu{1 g L-1}$.
I think this is how you do it, but please correct me if I'm missing something. I didn't end up using the standard molar volume, so this might be incorrect.
$$pV = nRT$$
We know the pressure is $\pu{1 atm}$, the temperature is about $\pu{273 K}$ and $R = 0.08$.
Solving for $\frac {n}{V}$, we find $\frac {n}{V} = \frac {1}{22}$.  Now, this is in moles per litre, so by multiplying by the molar mass of nitrogen (approximately $\pu{28 g mol-1}$), we find that we have about $\pu{1 g}$ per litre as the density.

Using the ideal gas law
\begin{align}
  pV &= nRT, &
  \frac nV &= \frac{p}{RT},
\end{align}
where
\begin{align}
  T &= \pu{273 K},\\
  p &= \pu{1 atm},\\
  R &= \pu{0.0821 L atm mol-1 K-1},\\
  M(\ce{N2}) &= \pu{28.0 g mol-1},
\end{align}
we get
\begin{align}
  \frac{n}{V} &=
    \frac{\pu{1 atm}}{\pu{0.0821 L atm//mol K} \times\pu{273 K}},\\
  \frac{n}{V} &= \pu{0.0446 mol//L},
\end{align}
and finally
$$
\rho(\ce{N2})
  = \pu{0.0446  mol//L} \times \pu{28  g//mol}
  = \pu{1.2  g//L}.
$$
Therefore answer (d) is correct.


Why do we still need to know about the Rankine temperature scale? I am learning process principles in chemical engineering and I was taught various temperature units like Fahrenheit, Celsius, Kelvin, and Rankine. I see the first three being used quite commonly, but I rarely see the fourth being used. Where is the Rankine scale used then?

Rankine is commonly used in the aerospace industry in the United States.

Rankine is to Fahrenheit what Kelvin is for Celsius. So when people in the United States were creating programs and using equations that needed an absolute temperature, they used Rankine before Celsius became dominate for scientific calculations. The reason people still sometimes use it in the aerospace industry is that there are a lot of programs that were developed using Rankine, so in order to be compatible with those old programs, it's often simpler to just use Rankine in the new programs too.

Thus, as an intern at NASA, I was writing simulations that used Rankine in 2009.


Which kind of plastic material would be suitable for a clutch cover on a motorcycle transition? I'm going to have an online machine shop build a part for me using a CAD drawing that I specify.

Typically, these parts are made of aluminium, but this manufacturer also mills parts out of these plastic materials:

Polypropylene (PP)
Polystyrene (PS)
High impact polystyrene (HIPS)
Acrylonitrile butadiene styrene (ABS)
Polyethylene terephthalate (PET)
Polyester (PES)
Polyamides (PA) (Nylons)
Poly(vinyl chloride) (PVC)
Polyurethanes (PU)
Polycarbonate (PC)
Polyethylene (PE)

Polypropylene, I know, is the plastic used in water bottles, so I know that is not what I want.

This fabricated part will be used for the clutch cover on a motorcycle transmission and will have a raised section that needs to be tapped to allow attaching a gear shifter to, like in the aluminium version below.

I'd like something rigid and strong. Plexiglass comes to mind, but there could be a better alternative that I am unaware of, as long as I am not getting into something exotic that would cost more than aluminium.

Could suggest a good plastic for me, which would need to be oil resistant and withstand typical engine transmission temperatures.

Most plastics consist of linear molecules with little inter-molecular interactions, so most plastics melts too easy. The only ones you mentioned having a chance are polyuretanes and polyamides. They have (CO)(NH) fragments, that forms strong intermolecular hydrogen bonds. So, they are ones of the strongest polymeric materials

And I will recommend aluminum anyway as it is much harder to screw things with aluminum.


Vibrational state dependency of the reaction cross section I'm reading p.1238 of Mcquarrie's Physical Chemistry: A Molecular Approach textbook.
In the last line of the page, the author explains that the value of the reaction cross section in the constant total energy depends strongly on the vibrational state of the reactant. Of course, it means that not only translational kinetic energy but also internal energy(such as vibrational state) contributes to the value of the reaction cross section values.

However, the book puts a limit on the 'constant total energy'. If both translational kinetic energy and vibrational energy contribute to the reaction cross section value, does the explanation in the book imply that vibrational energy contributes more to the reaction cross section value than translational kinetic energy?
If that's true, is there a way to explain the difference in contribution qualitatively? Or is it just a matter of quantitative differences?

Vibrations are very important for reactions. In the particular example of $\ce{H2+(g) + He}$

vibrational levels 0-3 have an energy less than E0 (minimum energy for reaction), so additional transitional energy is needed to induce reaction. That’s why we see a threshold.
vibrational levels > 4 already have more than E0 energy so they react even with no additional energy

The greater the vibrational energy level of $\ce{H2+}$ the easier it will be for the substitution reaction, and this makes sense considering the transition state diagram

You may also be interested in the "harpoon" reaction which demonstrates how vibrations are important 1.
This reaction is known as a harpoon reaction because an electron is first transferred from M in $\ce{M...XR}$ or $\ce{[M...XR]^*}$ to $\ce{RX}$ and thereby forms the ionic transition state $\ce{M+...RX-}$. The negatively charged $\ce{RX-}$ is unstable and can dissociate, forming the products $\ce{M+X-}$ and $\ce{R}$.
These type of reactions are, in general, valuable for studying how rovibrational states affect the reaction probabilities because they can be triggered by light, and are small enough to model with extremely accurate theories.
In particular, the reaction probability has notable "resonances" spaced precisely 10.9 $cm^{-1}$ apart. These resonances are most likely due to low frequency bending mode of the T-shaped complex.

References:

Skowronek, S., Jiménez, J. B. & González Ureña, A. Resonances in the Ba...FCH3+hv→BaF+CH3 reaction probability. J. Chem. Phys. 111, 460–463 (1999).


Why does ligand substitution occur when OH- is a better ligand than NH3? $$\ce{[Cr(H2O)3(OH)3](s) + 6NH3 -> [Cr(NH3)6]^3+ + 3H2O + 3OH-}$$

When excess $\ce{NH3}$ is added to triaquatrihydroxychromium(III), why does ligand substitution occur when $\ce{OH-}$ is a better ligand than $\ce{NH3}$?

Wouldn't the 3 $\ce{H2O}$ molecules be deprotonated to $\ce{OH-}$ and therefore form $\ce{[Cr(OH)6]^3-}$?

This also seems to occur when excess $\ce{NH3}$ is added to tetraaquadihydroxocobalt(II):

$$\ce{Co(H2O)4(OH)2(s) + 6NH3 -> [Co(NH3)6]^2+ + 4H2O + 2OH-}$$

This is an equilibrium reaction. Using LeChatelier's principle, you can figure you that if you increase the concentration of NH3 sufficiently, you shift the equilibrium so far to the right that you get the hexamine as the main product. At neutral pH in water, the concentration of hydroxide is quite low (100 nanomolar). You would expect this reaction to be pH-dependent, and of course depend on the concentration of ammonia in solution.


Why does a mixture get warm in an exothermic reaction? The production of ethanol by fermentation is exothermic. My textbook says that the yeast stops working if the mixture gets too warm. Why would the mixture get warm if the reaction is exothermic?

In thermodynamics, we like to divide any situation into two parts, the system and the surroundings. In general the system is the part of the universe that is being studied while the surroundings is everything else that interacts with the system. Between the boundary of the system and the surroundings, energy and heat can be transferred (in an open system).

Now in fermentation, the yeast and glucose molecules are said to be the system while the surround water is the surroundings. Since the reaction is exothermic, this means that energy is moving from the system to the surroundings. Therefore the water should gain more heat energy which will increase its temperature.

However I doubt that the heat generated from the reaction is enough for the enzymes in the yeast to stop working. When I did this experiment we used a Bunsen burner to heat up the solution to approximately 35 degrees. So really what is going to cause the solution to heat up to above 45 degrees (this approximately where yeast begin to stop working) is that you keep the Bunsen burner on and keep heating it.


Proton NMR spectrum of aminocaproic acid I have completely unpacked the NMR spectrum for the drug aminocaproic acid (structure below) and have found a peak that I myself can not explain. The signals for the drug are simple to understand what corresponds to what but the reason behind one of the peaks having six peaks is unknown to me. I had made the assumption that the furthermost up-field signal would have 5 peaks and be a multiplet, due to it having four nonequivalent neighbors and N+1 rule. If there is a reason behind this can some one please explain.

The N+1 rule, in its ideal form, assumes that for multiplets arising from $\mathrm{^3J}$ couplings between neighbouring $\ce{CH_2}$ groups,

shifts for the spins within a $\ce{CH_2}$ group are identical
all couplings between spins in two different groups are identical

While condition 1 usually holds for substituted alkanes of the form $\ce{A-(CH_2)_n-B}$, condition 2  does not. This leads to the title "Is the n+1 rule ever obeyed?" in a section of a textbook [1], like a cry of despair. The text elaborates:

  At other times, when there is a large difference between $\mathrm{^3J_{AB}}$ and
  $\mathrm{^3J_{BC}}$, distinct peaks, more than five in number, can be seen.
  Deviations of this type are most common in a chain of the type
  $\ce{X-(CH_2)_3-Y}$, where X and Y are widely different in character.

So, the exception is actually closer to the rule, and that is because true magnetic (unlike chemical) equivalence is relatively rare.

[1] Pavia, Lampman and Kriz, Introduction to Spectroscopy, 2nd Ed., Saunders Publishing


What are the products of the reaction of methoxypropane with HBr? The products of the reaction of methoxypropane ($\ce{CH3CH2CH2OCH3}$) with $\ce{HBr}$ are?

In my opinion, the products should be: bromopropane ($\ce{CH3CH2CH2Br}$) + methanol ($\ce{CH3OH}$). The first step is the cleavage of the ether. This can happen via the formation of a propyl carbonation or a methyl carbocation. Since the propyl carbocation is more stable, bromine will react with the same to form bromopropane (the other product thus formed is methanol)

But the products given in the solution were propanol ($\ce{CH3CH2CH2OH}$) + bromomethane ($\ce{CH3Br}$). So, what is wrong in my reasoning? And what is the correct reasoning?

Source: NCERT (India) 12th

The first step is protonation of the oxygen of the ether. This cationic species is then subject to nucleophilic attack by Br-. There are two potential sites of attack i.e. the carbons bonded to the protonated oxygen. Bromine is quite a bulky nucleophile, and of the two sites, the methyl group carbon is less hindered so this is attacked in preference, leading to bromomethane and propan-1-ol as your products.


Density of LPG in g/L What is the density of LPG (in vapour state) in g/L?
Please note that I'm asking about the density of vapour LPG when it comes out of the cylinder. NOT the liquefied density!

LPG – propane – gas density is 1.55 times heavier than air. It is  $1.898$ g/L vs  $ 1.225$ g/L for air (both 15°C at sea level).

For more details, please visit this website.


Relationship between freezing point depression and hydrogen bonds Freezing point of a solvent is decreased due to the added solute particles in it. e.g. ions. Therefore electrolytes contribute to lowering freezing point (say for physiological conditions).

If a particular solution has a non-electrolyte solute, which is held together with hydrogen bonds (forming dimers), I want to know how do H-bonds affect the freezing point?

Pure water itself forms hydrogen bonds. Yet at 0'C it freezes theoretically. How to approximate effect of the hydrogen bonds in an acidic solution to its freezing point depression theoretical value? e.g. do H-bonds in CH3COOH halve the freezing point depression value?

The addition of one molecule (or salt) of solute produces $i$ effective solute particles. This defines the van't Hoff factor $i$, which encodes the strength of the solute with regards to colligative properties.

As simple (idealized) examples in aqueous solution, the van't Hoff factor of $\ce{NaCl}$ is 2, as $\ce{NaCl}$ dissociates into the two solute particles $\ce{Na+}$ and $\ce{Cl-}$ in solution, whereas the van't Hoff factor of glucose is 1, as glucose doesn't dissociate in solution.

These examples are idealized because they do not account for intermolecular or interionic interactions, and are therefore most valid in dilute solutions. In concentrated solutions, intermolecular and interionic effects are much more important: the presence of hydrogen bonding, for example, or ion pairs in solution. Attractive interactions "clump" the solute particles together and reduce the van't Hoff factor.

Returning to your example, if we assume that acetic acid does not dissociate in solution, and we're under conditions in which acetic acid dimer predominates, then indeed we will have a van't Hoff factor $i = 0.5$. It's important to note, however, that the hydrogen-bonding interaction in acetic acid is particularly strong, and in general intermolecular and interionic interactions will not change the van't Hoff factor so drastically.

It remains to consider how to correct for these interactions. Intuitively, it seems to me that the activity coefficient should be the desired correction factor, considering that the activity coefficient is a measure of how ideal (interaction-less) a solution is, but I do not have a more quantitative argument along these lines.


Why does the ionization energy decrease anytime the atom size increases? I would like to have a better understanding of the relationship that exists between the Ionization Energy commonly known as IE and the size of an atom.

I think that the Ionization Energy can be defined as the smallest energy that is required to remove an electron from a gas phase atom.

Why does the Ionization Energy decreases when the atom size increases?

The ionization energy is the lowest amount of energy required to remove one electron(the most loosely attached electron) from each atom in one mole of an element in gaseous state to form gaseous (positive) ion.

There are certain factors that have an impact on the ionization energy value.

To understand the answer to your question, you must understand the effect of all three

Atomic radius (Inversely proportional to I.E. Value)
Shielding effect of electrons (Inversely Proportional to I.E. Value)
Nuclear Charge (Proportional to I.E. Value)

The Atomic radius is the distance between the to-be-removed electron and the nucleus. The greater the distance, lesser the force of attraction. Hence as this distance increases the electrons are easier to remove; lesser force is required to remove it from that attractive field of the nucleus. (This answers your question) Even the smallest change in this distance will cause a change in ionization energy for example The first ionization energy of Magnesium and Aluminum. The ionization energy should increase as the nuclear charge increases across the period but the I.E. of Aluminum is lesser than that of magnesium because there is a change in orbital which increases atomic radius, decreasing I.E.

Now going for a bit more information.
The Shielding effect of the other electrons decreases the I.E. because it doesn't let the outer most electron feel the full force of attraction.

IF both the first factors are CONSTANT (atomic radius and shielding effect of electron) only then the last factor comes in play.
As the atomic charge/Nuclear charge increases, there is a general increase in ionization energy because of the increased force of attraction by the extra proton. for example, again across the period, the general trend is increase in ionization energy because there is no (significant) increase in atomic radius or shielding effect so the I.E. increases proportional to the Nuclear charge.
(the decrease in sulfur is because of the pair of electron in the p orbital repelling each other. Both electrons apply a force on each other. The looser of the two, is removed when energy is applied.)


the number of significant figures An electronic balance was used to determine that a sample has a mass of 9.00060 g.  If the balance's precision is ± 1 mg, what is the correct number of significant figures for this measurement?
actually the answer is 4 but I cant figure out from where has it come from? Isn't it 5?

The significant digits you use are the values of which you are confident, plus one more value which may be uncertain.

In this case, the balance is accurate to $\pm0.001$ mg, so you're confident about the first 2 decimal places, but the 3rd place is uncertain.

The measurement given is $9.00060$, but the decimal part is smaller than our uncertainty, so we can't count on it.  The most accurate we could record is $9.001$, because the first decimal we're uncertain about is the 3rd place, which leaves us 4 significant figures.

If the measurement had been $19.00060$ instead, then the same process would result in 5 significant figures.


How can I determine if there will be a reorganization in organic reaction? How can I determine if there will be a reorganization of the substrate to stabilize the $\ce{C+}$ in a organic reaction?

Consider an S$_N$1 or E1 organic reaction, when there is a formation of a $\ce{C+}$ during the process. Sometimes, the molecule will reorganize (like one hydrogen will "jump" to another carbon, or a $\ce{CH3}$ will do it) in order to stabilize the $\ce{C+}$. But, sometimes, this doesn't occur. Is there any way of predicting if it will happen or not? Thanks.

Predicting? Not really. You can use sophisticated quantum mechanical calculations to explore the energetics of the reaction channels, and apply let's say trainsition state theory to get dynamical informations. This method is cumbersome, time-consuming and still far from being exact - but this is as close to predicting reaction pathways as it gets.

Now, in organic chemistry, you usually go with your "gut". I will try to present a way which organic chemist can think. When you have seen a lot of mechanisms, then you can guess whether reorganization happens or not. It also helps if you enumerate the circumstances that can affect the reaction. Let's see some of these:

Steric factors

In order to have a rearrangement, you obviously need one part of the molecule (the "jumping" part) to have access in space to the receiving part of the molecule. This requirement  can be satisfied if we let only a hydrogen atom jump, or we let the groups jump only a single bond. Now there are a lot of exceptions of this rule: for example in Benzilic acid rearrangement you can have phenyl groups mitigating. The idea that the groups have to be close to each other is also not always true, because you can have "jumps" between different molecules too. However, if the target part of the molecule is extremly hindered, then no rearrangement can be expected.

Electronic effects

Now, if we look at the electronic requirements of a rearrangement, then we have two different things to look for. One type of reactions can be called "orbital-controlled", because the mechanism depends mostly on the frontier orbital shapes and overlap of the reactants - in this case the rearrangements. A good example of these types of reactions are the sigmatropic processes, like Cope rearrangement.

If we assume that the orbitals are alright - which is not a huge assumption if we consider hydrogen transfer - then we have to look at the free energy profile of the reaction. This means we have to check how much energy we can "win" by rearranging the reactants. In the case you mentioned with the carbocations, this is equal to checking how stable is the carbocation. If it is instable (high energy), then we can "win" a lot by rearranging the molecule to the lower energy - by mitigating a hydrogen atom let's say. The stability of the carbocation then depends on the order of the cation and the electronic effects of different substituents. The general idea is that if you do not have your charge well distributed over many atoms, then you have a high energy - for example, CH3+ cation is extremly instable. If you have a lot of groups to "help" carrying the charge, either by inductive effects, or by resonance, then the carbocation stabilizes, making the rearrangement less favoured.

It is important to note that the things I have mentioned are mostly about the energetics of the reaction. Organic chemists seldom have a way to say anything about the kinetics of a particular reaciton. There are tricks like Hammett's relations and tranisiton state theory, but most of the time, they simply are not applicable. In our case with the hydrogen transfer, we could also look for quantum effects, which is completely out of organic chemistry's predictive reach.


Gibbs free energy where T, V, and P change simultaneously So $\Delta G = \Delta H - T\Delta S - S\Delta T$.

Is there a way to express $\Delta G$ for a process where T, V, and P change simultaneously only in terms of the initial and final T, V, P? Assume this process is for an ideal gas.

Since $G$ is a state function I can calculate it separately for an isothermal process and an isochoric process, but in that case $\Delta G$ for the isochoric process has dependence on the entropy of one of the states.

Your original equation is incorrect.  It should read $\Delta G=\Delta H -\Delta (TS)$, and $\Delta (TS)\neq T\Delta S+S\Delta T$ unless the changes are infinitesimal.

That said, for an ideal gas, irrespective of path, $$\Delta S=S_f-S_i=C_p\ln{(T_f/T_i)}-R\ln{(P_f/P_i)}=C_v\ln{(T_f/T_i)}+R\ln{(V_f/V_i)}$$and$$\Delta H=C_p(T_f-T_i)$$So, $$\Delta G=G_f-G_i=C_p\Delta T-\frac{(T_f+T_i)}{2}\Delta S-\frac{(S_f+S_i)}{2}\Delta T$$Clearly the 3rd term on the rhs depends on either $S_i$, $S_f$, or both, so $\Delta G$ depends on these.  It then all comes down to how you assign your reference state values for G, H, and S, such that $G_{ref}=H_{ref}-T_{ref}S_{ref}$.


How to find chemical similarity between two compounds using Tanimoto method I came across a paper named "Chemical Similarity Searching" by Peter Willett
John M. Barnard and Geoffrey M. Downs, 1998. At one place, this paper talks of finding chemical structure similarity between two chemical compounds using Tanimoto method. Now I want to understand how it is done but could not derive anything just by reading the paper. I wanted to know if someone can explain how the distance is found taking two compounds into consideration. I wanted a step by step explanation. I have failed to derive how to find similarity from the paper. Please help.

Apparently, it is more generally known as the "Jaccard Index":

  In "A Computer Program for Classifying Plants", published in October 1960,[5] a method of classification based on a similarity ratio, and a derived distance function, is given. It seems that this is the most authoritative source for the meaning of the terms "Tanimoto similarity" and "Tanimoto Distance". The similarity ratio is equivalent to Jaccard similarity, but the distance function is not the same as Jaccard distance.

In terms of chemistry, this calculation is carried out on a fingerprint of the molecule. That wikipedia page has a nice reference to this paper, although it is not open access. There are no doubt many resources out there about fingerprints/Tanimoto, even open source implementations...

This also looks interesting: "Why is Tanimoto index an appropriate choice for fingerprint-based similarity calculations?".


Evaporation of water What are the conditions for the evaporation of water?

I know that for the boiling of water, the vapor pressure of water must equal the partial pressure of water vapor.

But what about for evaporation? Is it the same condition above in addition to the requirement that the relative humidity cannot be 100%?

Can we have boiling when the relative humidity is 100%?

Evaporation is a surface phenomenon, boiling is a bulk phenomenon.  Evaporation will always be occurring for any liquid with an exposed surface.  Water molecules in a liquid sample have a distribution of kinetic energies.  It takes less work for a surface water molecule to break free from its neighbors, then for a water molecule in the bulk to escape, since the former has fewer neighbors (no molecules on top).  Those molecules with enough kinetic energy to break away from their neighbors will escape.

For water to boil its vapor pressure must equal the atmospheric pressure.

The only requirement for evaporation to occur is an exposed liquid surface.  I don't think relative humidity affects the boiling point, but I suspect it would affect the rate at which a liquid will evaporate from an open container.  Evaporation is always occurring, but to notice a change in the amount of material in an open system, more material must leave the surface then returns to the surface and the rate of return will be affected by the composition (water content) of the air above the open sample.


Role of acetone in the Finkelstein reaction What is the role  of acetone in the Finkelstein  reaction? It has something to do with Le Chatelier's principle but I am not able to understand why it will not proceed without acetone.

Finkelstein reaction is a halogen exchange reaction between haloalkane and a salt of a different halogenide. $\ce{R-CH2-Cl + KI -> R-CH2-I + KCl}$. The driving force in this reaction is precipitation of $\ce{KCl}$. It is important to find a solvent that would dissolve $\ce{KI}$ better than $\ce{KCl}$. Then solution will always have $[\ce{I-}] > [\ce{Cl-}]$ (more iodide anions than chloride anions). As a result this equilibrium will be driven to $\ce{R-CH2-I + KCl}$. Acetone is exactly this type of a reagent.

For this case Le Chatelier's principle means that reaction will be driven to the direction in which one of the reagents gets outside of the solution. Since $\ce{KCl}$ keeps precipitating the reaction will be driven to the desired product.

This reaction is important because there are many ways to make $\ce{RCH2Cl}$, but making $\ce{R-CH2-I}$ is difficult. Finkelstein reaction allows to convert available $\ce{R-CH2-Cl}$ to $\ce{R-CH2-I}$.


Would equal masses of gold and copper result in the same volume change if added to water? Would equal masses of gold and copper result in the same volume change if added to water?
  Density of gold: $19.3~\mathrm{g/mL}$
  Density of copper: $8.96~\mathrm{g/mL}$

I was thinking the answer would be no since the density is different between gold and copper but was unsure if this was a trick question since it maybe has to do with the mass of the substance that is placed into the water.

"Added to water" is ambiguous.

If the metal objects are shaped and placed such that they float (eg: as a metal boat does, or a light hollow sphere), then they will each displace the same volume of water (the volume of water equivalent to their equal common masses).  So long as they float, they will displace the same amount of water no matter their shape.

If the masses objects are shaped and placed such that they sink (eg: consider a solid sphere or cube), then they will in general displace different volumes.

Bonus.  The second part is slightly more complex than it seems when it comes to calculating how much water they would displace.  If they have no dry internal voids, their respective displaced volumes are simply the mass of each divided by its density.  Others have given the simple calculation.

If they were shaped such that they had dry voids yet were small enough to sink (think of a heavy hollow sphere with overall density greater than water so it sinks, or an overturned metal boat with small air pockets), then the displaced water will be somewhere between the above two calculations (fully floating calc and no-voids calc).

Note that it would be possible to construct the two objects such that they sank and yet displaced the same amount of water, eg: two hollow spheres with the same outer diameter but different internal diameters, both heavier than the same volume of water.  But that would be a very trick question indeed!


Why does calcium carbonate decompose in sulphuric acid? Here's the reaction:

$\ce{H2SO4(aq) +  CaCO3(s)  ->   CaSO4(s)  + H2O(l)  +   CO2(g)}$

I don't understand why calcium carbonate decomposes in sulphuric acid. It's not soluble in water. I could understand the hydrogen ions oxidizing the calcium, yet the reaction isn't a redox reaction; the oxidation states remain the same!

I don't understand why calcium carbonate decomposes... It's not soluble in water.

It's more soluble in water than you might think.  If it weren't soluble, we wouldn't have awesome things such as stalagmites in caves.

Calcium sulfate is more soluble in water than the carbonate, but that isn't saying much.  It's solubility is only about $\pu{2 g/L}$. So writing the product as $\ce{CaSO4(s)}$ is only correct if you are talking about dilute solutions.  Anything too concentrated would decompose to form precipitated calcium sulfate.

  I could understand the hydrogen ions oxidizing the calcium...

Now that would be a very curious reaction.  The calcium in calcium carbonate is already in the +2 oxidation state. Calcium has only two valence electrons and further oxidation would require disrupting an intact octet.  It would be very unfavorable.  $\ce{Ca^{2+} -> Ca^{3+} + e-}$ would be far too unfavorable a reaction for protons to carry out. $\ce{Ca^{3+}}$ would be a much stronger oxidant than aqueous protons.

  Why does calcium carbonate decompose in sulphuric acid?

The decomposition can still happen even with insoluble, solid-phase calcium carbonate because reactions can happen at surfaces. An acidic proton could be attacked by a solid-phase carbonate ion at the crystal surface, forming bicarbonate.  The bicarbonate could attack another acidic proton, forming a surface-bound $\ce{H2CO3}$ molecule, which could spontaneously transform into $\ce{CO2}$ and water, which would be very loosely bound to the surface, allowing them to escape and exposing fresh not-yet-reacted calcium carbonate molecules to the aqueous surface, and the process begins anew...


Why is only one proton transferred in the reaction between sulfuric acid and sodium chloride $$\ce{NaCl + H2SO4 -> NaHSO4 + HCl}$$

The above reaction happens, but why doesn't the following happen? Or does it but only when one of the reactants is in excess?

$$\ce{2NaCl + H2SO4 -> Na2SO4 + 2HCl}$$

The second reaction seems plausible to me. How can I predict whether a similar type of reaction will be like the first equation or the second equation?

Sodium bisulphate (pKa ~ 2 - https://en.wikipedia.org/wiki/Sulfuric_acid) is a much weaker acid that hydrochloric acid (pKa = ~ -7), which means that chloride is not a strong enough base to deprotonate sodium bisulphate.

However all acid pbased reactions are at equilibrium.  Roughly one part in 10^9 of the species involve sulphate anion, which for all practical purposes is negligible.

Table of pKa charts: http://www.lauher.com/che142/acidbase.pdf


Why do chromyl bromide and chromyl iodide not exist? It is known that chromyl chloride exists (popularly known as the product of a test for presence of chloride ion) but chromyl bromide and chromyl iodide are unknown.

Why is it so? Why don't they exist? I'd guess it has something to do with the large size of bromine and iodine compared to chlorine, but I'm not sure.

What are the possible reasons for their non existence?

Chromyl Bromide does exist.  See Halogen Chemistry at page 246:

  Deep violet needles of chromyl bromide are obtained by treating chromium(VI) oxide with $\ce{CF3COBr}$ and HBr in the presence of phosphorus pentoxide. A less pure sample is made by treating chromyl chloride with a large excess of hydrogen bromide. The compound, which is hygroscopic, is purified by vacuum distillation at —135° (Zellner, 1949; Krauss and Starke, 1962).

The book continues:

  Chromyl iodide has not been prepared and may not exist because of the powerful oxidizing property of the chromium (VI) oxidation state.

See similarly The Chemistry Of Chromyl Compounds Chem. Rev., 1958, 58 (1), pp 1–61:

  Chromyl bromide has likewise been isolated within the past few years (273,
  274). The modern work corrects earlier reports of the compound (77).

  While there has been one report of the formation of chromyl iodide as a deep
  red liquid from a mixture of dichromate, iodide, and sulfuric acid (87), a majority of observers (77, 96, 128, 196, 255) deny this observation and the existence of the compound. There is little likelihood that the iodide can exist, in view of the large free-energy change attendant on the reaction:

  $\ce{2Cr^{+6} + I- -> I^{+5} + 2Cr^{+3}}$


What is the difference between ΔG and ΔrG? Consider the reaction

$$\ce{A -> B}$$

The reaction Gibbs free energy, $\Delta_\mathrm{r} G$ is given by the following equation
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT \ln Q$$

Now what is the difference between $\Delta G$ and $\Delta_\mathrm{r} G$ ?

Which one of the above represents $G_{\text{products}} - G_{\text{reactants}}$ ?

I read in my book $\Delta_\mathrm{r} G$ is the slope of $G$ plotted against the extent of reaction at different instants. And at equilibrium, it is zero. So can we say that it has value at a particular instant while $\Delta G$ is for a process ($\ce{A}$ to $\ce{B}$ in my question) i.e. is for a time interval?

Does $\Delta G$ for the above reaction represent change in Gibbs Energy when 1 mole of $\ce{A}$ reacts completely to form $1$ mole of $\ce{B}$ ?

Not for the faint-hearted: There is an excellent, but very mathsy, article here: J. Chem. Educ. 2014, 91, 386 describing the difference.

The Gibbs free energy change, $\Delta G$

You are quite right in saying that $\Delta G$ represents a change in a system over a time interval. The notation $\Delta G$ itself implies that it is a difference in $G$ between two things: an initial state and a final state. Let's use a real example in order to make things clearer. Consider the thermal decomposition of ammonium nitrate:

$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$

If someone were to ask you "what is $\Delta G$ for this reaction?", you should really, technically, be telling them: it is not well-defined. That is because of two things. Firstly, the reaction conditions, e.g. temperature and pressure, are not specified.

But more importantly, a balanced equation does not tell you exactly how much ammonium nitrate is reacting. In your beaker, you could have $1~\mathrm{mol}$ of $\ce{NH4NO3}$ reacting according to the above equation, but you could equally well have $2~\mathrm{mol}$, or you could have $5000~\mathrm{mol}$ (I must say that's one huge beaker though), or you could have $0.001~\mathrm{mmol}$.

The stoichiometric coefficients in the equation do not indicate the amounts of compounds reacting.

So, a more proper question would read like this:

  $\pu{2 mol}$ of ammonium nitrate fully decompose according to the balanced equation $\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. Calculate $\Delta G$ for this process.

Great. Because the decomposition is complete, we now know that our final state is $2~\mathrm{mol}~\ce{N2O(g)} + 4~\mathrm{mol}~\ce{H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$, and our initial state is $2~\mathrm{mol}~\ce{NH4NO3(s)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. So,

$$\begin{align}
\Delta G &= [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot G_\mathrm{m}(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{NH4NO3(s)})]
\end{align}$$

We can't quite find the absolute molar Gibbs free energies, so the best we can do is to use Gibbs free energies of formation.

$$\begin{array}{c|c}
\text{Compound} & \Delta_\mathrm{f}G^\circ / \mathrm{kJ~mol^{-1}}\text{ (at }298~\mathrm{K}\text{)} \\ \hline
\ce{NH4NO3(s)} & -183.87 \\
\ce{N2O(g)} & +104.20 \\
\ce{H2O(g)} & -228.57
\end{array}$$
$$\scriptsize \text{(data from Atkins & de Paula, }\textit{Physical Chemistry}\text{ 10th ed., pp 975-7)}$$

(Note that the use of standard formation Gibbs free energies is only because of the conditions specified in the question, which conveniently corresponds to the standard state. If we specify different conditions, we can still find $\Delta G$, but we would have to use different data.) So:

$$\begin{align}
\Delta G &= [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{NH4NO3(s)})] \\
&= [(2~\mathrm{mol})(+104.20~\mathrm{kJ~mol^{-1}})] + [(4~\mathrm{mol})(-228.57~\mathrm{kJ~mol^{-1}})] - [(2~\mathrm{mol})(-183.87~\mathrm{kJ~mol^{-1}})] \\
&= -1073.62~\mathrm{kJ}
\end{align}$$

Note that we have units of kJ. Since $\Delta G$ is the difference between the Gibbs free energy of one state and another, $\Delta G$ has to have the same units as $G$, which is units of energy.

Now, this does not necessarily mean that $\Delta G = G_\text{products} - G_\text{reactants}$. For example, if I changed my question to be:

  In the Haber process, 100 moles of $\ce{N2}$ and 300 moles of $\ce{H2}$ are reacted at $800~\mathrm{K}$ and $200~\mathrm{bar}$ according to the equation $\ce{N2 + 3H2 -> 2NH3}$. Only 10% of the starting materials are converted under these conditions. Calculate $\Delta G$ for the process. (These numbers are made up.)

then, your final state would not be the pure products. Your final state is not 200 moles of $\ce{NH3}$. Your final state is $90~\mathrm{mol}~\ce{N2}$, $270~\mathrm{mol}~\ce{H2}$, and $20~\mathrm{mol}~\ce{NH3}$.

In general, one could write, for a chemical reaction,

$$\Delta G = \sum_i (\Delta n_i) G_{\mathrm{m},i}$$

where $\Delta n_i$ is the change in the amount of compound $i$ (in moles), and $G_{\mathrm{m},i}$ is the molar Gibbs free energy of the pure compound $i$, under the $T$ and $p$ conditions specified. Going back to our ammonium nitrate example, we would have $\Delta n_{\ce{NH4NO3}} = -2~\mathrm{mol}$, $\Delta n_{\ce{H2O}} = +2~\mathrm{mol}$, and $\Delta n_{\ce{H2O}} = +4~\mathrm{mol}$.

The Gibbs free energy change of reaction, $\Delta_\mathrm{r} G$

As you have correctly stated, $\Delta_\mathrm{r}G$ is the slope of a graph of $G_\mathrm{syst}$ against the extent of reaction, commonly denoted $\xi$. This is the easiest way of interpreting $\Delta_\mathrm{r}G$. This question contains a slightly fuller derivation and explanation of what $\Delta_\mathrm{r}G$ means.

However, $\Delta_\mathrm{r}G$ does indeed, somewhat, refer to the instantaneous difference between the "molar Gibbs free energies" of the products and reactants. This is different from $\Delta G$ in three main ways.

First, $\Delta G$ is the difference between the Gibbs energy of the entire system at two points in time. Here, we are interpreting $\Delta_\mathrm{r}G$ as the difference between the "molar Gibbs free energies" of two components of the system: reactants and products. In other words, the system generally contains both reactants and products, and $\Delta_\mathrm{r}G$ may be thought of as the difference between the Gibbs energies of the product part of the system, and the reactant part of the system, even though you cannot separate them in the laboratory.

Secondly, it is the difference between the chemical potentials, not the molar Gibbs free energies. The molar Gibbs free energy is simply defined by $G_i/n_i$; the chemical potential is a partial derivative and is defined by $\mu_i = (\partial G/\partial n_i)_{T,p,n_j}$. In the case where there are no other species present (i.e. species $i$ is pure), then the chemical potential is identical to the molar Gibbs free energy.

Lastly, it is weighted by the stoichiometric coefficients $\nu_i$ instead of the change in the amount $\Delta n_i$. The stoichiometric coefficient is a dimensionless quantity, which is negative for reactants and positive for products. So, in the ammonium nitrate decomposition as written at the very top, we have

$$\nu_{\ce{NH4NO3}} = -1; \qquad \nu_{\ce{N2O}} = +1; \qquad \nu_{\ce{H2O}} = +2$$

and our expression for $\Delta_\mathrm{r} G$ is

$$\begin{align}
\Delta_\mathrm{r} G &= \sum_i \nu_i \mu_i \\
&= \nu_{\ce{N2O}}\mu_{\ce{N2O}} + \nu_{\ce{H2O}}\mu_{\ce{H2O}} - \nu_{\ce{NH4NO3}}\mu_{\ce{NH4NO3}} \\
&= \mu_{\ce{N2O}} + 2\mu_{\ce{H2O}} - \mu_{\ce{NH4NO3}}
\end{align}$$

Now, note the units again. The chemical potential is a partial derivative of the Gibbs free energy (units $\mathrm{kJ}$) with respect to the amount of $i$ (units $\mathrm{mol}$), and so it must have units $\mathrm{kJ~mol^{-1}}$. And from our above expression, $\Delta_\mathrm{r}G$ must also have units of $\mathbf{kJ~mol^{-1}}$, since the stoichiometric coefficients are dimensionless.

Does it matter if you start with $1~\mathrm{mol}$ or $2~\mathrm{mol}$ of ammonium nitrate? The answer is now, no. The amount of starting material does not affect $\mu_i$, nor does it affect $\nu_i$. Therefore, $\Delta_\mathrm{r}G$ is independent of the amount of starting material.

Why is $\Delta_\mathrm{r} G$ an instantaneous difference between the chemical potentials? Well, that is because the chemical potentials of the reactants and products, $\mu_i$, are changing continuously as the reaction occurs. Therefore, if we want to calculate $\Delta_\mathrm{r} G$, we have to take a "snapshot" of the reaction vessel: otherwise it makes absolutely no sense to speak of $\mu_i$ because we wouldn't know which value of $\mu_i$ to use.

Compare this with $\Delta G$ above: the quantities of $G_{\mathrm{m},i}$ are constants that do not vary depending on the extent of reaction. Therefore, we do not need to specify a particular extent of reaction to calculate $\Delta G$.

A diagram to sum up

Note that, to adequately define what $\Delta G$ is, you need a starting point and an ending point. I showed two possibilities for $\Delta G$; there are infinitely many more.
Let's say the reaction goes to completion. If you double your starting material and double your product, the difference between $G_\text{products}$ and $G_\text{reactants}$ will also be doubled. So, it's important to specify!
Likewise, to adequately define $\Delta_\mathrm{r} G$, you have to define the single specific point at which you intend to calculate $\Delta_\mathrm{r} G$.
If you double your starting material and double your product, the curve is stretched by a factor of 2 along the y-axis, but it is also stretched by a factor of 2 along the x-axis because $\xi$ is also doubled. (For a mathematical explanation see the definition of $\xi$ given in the linked question earlier.) So, while the difference $\Delta G$ is doubled, the gradient $\Delta_\mathrm{r} G$ remains unchanged.
The units should also be clear from this. Since $\Delta G$ is a difference between two values of $G$, it has to have units of $\mathrm{kJ}$. On the other hand, $\Delta_\mathrm{r}G$ is a gradient and therefore has units of $\mathrm{kJ/mol}$.

A caveat

Unfortunately, the notation $\Delta G$ is often loosely used and treated as being synonymous with $\Delta_\mathrm{r}G$. You will therefore see people give $\Delta G$ units of $\mathrm{kJ~mol^{-1}}$. For more information refer to Levine, Physical Chemistry 6th ed., p 343. I would personally recommend making a distinction between the two.


Does UV water purification cause cancer? I found out that a relative is using a machine that performs UV water purification at home (for drinking).

There seems to be a distinct difference from the water that is not purified by it.

Is it possible that this use of UV radiation may lead to cancer in humans?

Is it possible that this use of UV radiation may lead to cancer in humans?

No!

The lamps used for water treatment are usually low pressure mercury vapour lamps. These lamps emit in the UV-C range with a maximum at $\lambda = 254~\mathrm{nm}$.

No mercury can escape from an intact lamp.
In principle, UV-C radiation may cause severe sun burns and harm your eyes. However, the lamps are housed and the consumer is protected from the direct exposure to radiation.
There is no change to the water due to UV irradiation.
UV-C radiation is efficient in killing microorganisms in water.


Cinnamaldehyde -> trans-3-phenylprop-2-enal When was the name cinnamaldehyde officially changed to trans-3-phenylprop-2-enal?

The rules for systematic nomenclature in organic chemistry have changed several times. They were first issued in Geneva in 1892, followed by the Liége Rules in 1930. The IUPAC rules were issued in 1957 (Sections A and B), 1969 (Sections A, B, and C), and 1979 (Sections A, B, C, D, E, F, and H). The rules were revised in the 1993 Guide to IUPAC Nomenclature of Organic Compounds and amended by official publications from 1993 through 2002. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) was published in December 2013.
According to the current IUPAC recommendations, the preferred IUPAC name (PIN) of cinnamaldehyde is (2E)-3-phenylprop-2-enal.

However, many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. Other names (that are not preferred IUPAC names but that are generated in accordance with IUPAC recommendations) may be used in general nomenclature. In addition, various traditional names are retained for use in IUPAC nomenclature, though the number of retained names has been reduced with each succeeding edition of the IUPAC recommendations.
According to the following subsections of the current IUPAC recommendations, the traditional name cinnamaldehyde may still be used in general nomenclature.

P-65.1.1.2 Retained names only for general nomenclature
(…)
P-65.1.1.2.2 The following names are retained for general nomenclature with
functionalization but no substitution is allowed. (…)
(…)
cinnamic acid
(…)

P-66.6.1.2 Names of aldehydes derived from retained names of carboxylic acids
Names of aldehydes derived from retained names are formed by changing the ‘ic acid’ or ‘oic acid’ ending of the retained names of carboxylic acids to ‘aldehyde’. (…)
(…)
P-66.6.1.2.3 Aldehydes derived from retained acid names given in P-65.1.1.2 are only used in general nomenclature; no substitution is allowed. Preferred IUPAC names are systematic names (…).


What actually is the Wavefunction? I am aware that the square of the Wavefunction gives the probability density of finding an electron at a particular point in space. I have also heard that it's a complex number but since it's a function I am unsure as to how that could be the case (perhaps someone could please clear that up for me as a sub-question). Moreover, I have seen it described as the amplitude of something but I literally have no idea what is meant by that. However, the crux of the question is what is the Wavefunction itself (i.e not what the square of it is). I have read in the book "Why Chemical Reactions Happen" that it is essentially synonymous with the word/concept: orbital. I have also seen in various places w(x,y,x) or w(r,theta,thi) [where "w" represents the sign for a wavefunction] so, from what I can gather it is a function in three dimensions that represents the shape of a particular orbital where the function of (x,y,z) or (r,theta,thi) would, of course, be different for the s,p,d and f orbitals. However, in the book "Why Chemical Reactions Happen" it goes onto say that wavefunctions of different atoms interfere with each other to form molecular orbitals and the atomic orbitals somehow have a phase?! For my idea of a wavefunction to be correct, the idea of interference of these wavefunctions must be an idea that was thought of to try and rationalize the process of the formation of molecular bonding and anti-bonding orbitals from atomic ones to make it easier to understand and conceptualize. If I am right about the wavefunction being a purely mathematical representation of the space where an electron is likely to reside then the molecular wavefunctions are just as curious in terms of their shape and suchlike as the shapes of the atomic orbitals they came from (i.e nobody really knows why they are that shape but when atoms or molecules are arranged in a particular way, orbitals "just are" that shape). Also the idea of phase seemed to bring my argument down but then I thought that perhaps phase is simply defined as being opposite relative to another part of an orbital if it is the other side of an angular node (I suspect this is wrong but could someone please shed some light on this?)

Going back to when I said - "the molecular wavefunctions are just as curious in terms of their shape and suchlike as the shapes of the atomic orbitals they came from" - I thought this because I found it bizarre that empty orbitals could "interfere" with an orbital from another atom. For example one p orbital is empty yet all three interact with the four hydrogen s orbitals. Unless, the electrons can jump between the the degenerate p orbitals at such a frequency that all can be considered occupied I find it incomprehensible that just a portion of space can interfere with an orbital. As a result, I came to the conclusion that I quoted at the start of the paragraph. Basically meaning that when atoms come together to form molecules the electrons find the lowest possible energy spaces to reside; or in other words, a molecuar orbital / molecular wavefunction forms which is equally as curious as the atomic wavefunctions with their strange shapes.

Any help with this concept would be greatly appreciated - I am off to Oxford university this October and Molecular Orbital Theory is a small part of the suggested reading list before arrival into the first year. Also, on that note, please refrain from heavily mathematical answers because I simply won't understand as I have only done A-level Maths and A-level Chemistry. Thanks.

Let me see if I can get at some of your questions. As mentioned above, it's much easier when you ask individual specific questions.

One problem with books on introductory quantum mechanics is that, put simply, the language of quantum mechanics is math. Specifically, most people use the Schrödinger equation which involves second derivatives and differential equations.

When I teach quantum chemistry to undergraduates, there are typically problems due to the mathematical nature.

There are multiple "interpretations" of quantum mechanics. That gets into philosophy. Wikipedia has some nicely-written descriptions.

As you mention, the most widely used interpretation (the Copenhagen interpretation) of the wave function centers on the square of the wave function, or rather $|\psi^*\psi|$ as a probability density. As you mention, the wave function could be imaginary or complex, so this notation indicates a mathematical way of getting a real number for a probability density.

You ask about the shape and why atomic orbitals have particular shapes.

One thing that's amazing about quantum mechanics is that we put in some relatively simple equations to describe the motion of an electron in a hydrogen atom, and the solutions are exactly these atomic orbitals (s, p, d, f, etc.).

So I disagree that "no one knows" why atomic orbitals have particular shapes. The answer is that they are the mathematical solution to the Schrödinger equation. That is, we want to know the energies of the system and there are only certain quantized solutions.

And yes, the whole notion of chemistry is that electrons (and hence orbitals) on different atoms interact. Otherwise, there is no bonding.

I'll leave you with this. Without the math, it can be exceptionally hard to understand quantum mechanics. When you do understand the math, I think the results are quite beautiful. The shapes of orbitals aren't arbitrary. This is actually just a reflection of the mathematical nature of quantum mechanics.


Finding formula of reduced scandium oxide using its mass and mass of obtained metal An oxide of scandium, with a mass of $\pu{1.423 g},$ is chemically reduced with $\ce{H2}$ to give $\ce{H2O}$ and $\pu{0.929 g}$ of $\ce{Sc}$ metal. What is the formula of scandium oxide? How much water is formed?

I know that the reaction is of the form:
$$\ce{Sc_xO_y + H2 -> H2O + Sc},$$
where $x$ and $y$ are the units to define scandium oxide molecular formula (not empirical).
I don’t know how to get the constants using the info given, and what the info given tells me. I typically use a linear system of equations to solve for the reactants and products. In this case, we’d have:
$$
\begin{align}
\ce{Sc}&: &\quad  Ax - D &= 0;\\
\ce{O}&: &\quad   Ay - C &= 0;\\
\ce{H}&: &\quad   2B - 2C &= 0.
\end{align}
$$
Usually, we would have an equation where $x$ and $y$ aren’t present and you can solve the system of equations. Here, it prevents you from doing the balancing. Instead, you’re just guessing. Any suggestions when linear algebra doesn’t work?

Since scandium oxide is an ionic compound, the method is exactly the same as obtaining the empirical formula, from the ratio of the amounts of elements $n(\ce{Sc})$ and $n(\ce{O}).$
$$n(\ce{Sc})=\frac{m(\ce{Sc})}{M(\ce{Sc})}=\frac{\pu{0.929 g}}{\pu{44.96 g mol-1}}=\pu{2.07E-2 mol}$$
The mass difference comes from the loss of oxygen atoms:
$$n(\ce{O})=\frac{m(\ce{Sc_xO_y})-m(\ce{Sc})}{M(\ce{O})}=\frac{\pu{1.423 g}-\pu{0.929 g}}{\pu{16.00 g mol-1}}=\pu{3.09E-2 mol}$$
The ratio between the amounts of elements is close to $2:3$, therefore the formula is $\ce{Sc2O3}$. You can check that using oxidation states.


Why is the electronegativity of hydrogen 2.20? Well, I was doing research about the ways we can measure electronegativity and I just realized there must be an arbitrary electronegativity set for an specific element to give us the relative amounts of the other ones. Pauling used hydrogen and suggested 2.1 for its electronegativity which later was altered to 2.20. Now, the question is, why was the electronegativity of hydrogen suggested as 2.1 and why is it 2.20 right now? What are the secrets in those numbers?

Interestingly, the first paper from Pauling on electronegativity "The Nature of the Chemical Bond. IV. The Energy of Single Bonds and The Relative Electronegativity Of Atoms" published in 1932 gives a different scale with H = 0.0 and F = 2.0!

Later, he changed to 2.1 for H and 4.0 for F, e.g. http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-01.html

If we look through his book, on p. 90 we find the following text:

  Following these there are given the values obtained by adding 2.05 to them...

The resulting numbers are then rounded to one decimal, from H = 2.05 to H = 2.1 and F = 4.0.

His notes are online (e.g., here) but there's no obvious explanation I can find yet.

My hunch is that he started with H = 0.0, and realized that the alkali metals would be negative numbers. The shifted scale gave Li ~ 1.0 and F ~ 4.0, which seem like "nice" numbers.

I tried to hunt down the reason for the move to 2.2 as the standard for hydrogen. The first reference seems to be by Huggins (J. Am. Chem. Soc., 1953, 75 (17), pp 4123–4126) but the paper itself doesn't really explain the reasoning.

Wikipedia suggests a paper by Allred, but that cites the Huggins paper for the 2.2 number.

Presumably more could be found with a very careful search through the papers of Pauling and Huggins, but I've never heard any clear explanations and the documents online don't give them either.


Does the delocalization of p-orbitals ensure conductivity? For example, polyacetylene is a conjugated system, i.e. with de-localized p-orbitals, but accroding to Chiang et. al. (1977) polyacetylene is semiconducting, and becomes conductive only when doped with Cl/Br/I/AsF5.

Isn't the conjugation alone enough to allow conductivity?

(source)

UPDATE

One reason they mention in the paper is:

  In the idealized situation of a uniform chain, the resulting
  conduction band would give rise to metallic behavior. However, such a
  system is unstable with respect to bond alternation, which causes the
  formation of an energy gap in the electronic spectrum.

But if this is true then (1) how the halogens improve the situation, and (2) why isn't there a similar phenomenon in all conjugated systems?

Delocalized $\pi$ orbitals do allow for mobile charges. The catch there is "allow." Just because these systems can have mobile charges does not make them conductive.

Conductivity can be defined as:
$\sigma = n e \mu$
where $e$ is the charge on an electron, $\mu$ is the charge mobility, and $n$ is the number of charges.

In the case of $doped$ conjugated polymers like you indicate above, the halogen vapors or other dopants usually oxidize the polymer and create more charges (i.e., $n$) and the conductivity increases. This does happen with pretty much every conjugated system I've seen: oxidize or reduce the neutral system and you'll have more charge carriers = greater conductivity. (NB: most conjugated organics are easier to oxidize than reduce, so you get + charges (holes) as the charge carrier, rather than electrons.)

That's the quick explanation as to why adding halogens increases the conductivity -- you get more charge carriers.

Incidentally, there are other ways to increase the charge carriers. Chemical or electrochemical doping usually makes a conductive system, but you can also use electrical bias (e.g., a field-effect transistor) or light (i.e., a photovoltaic) which will also increase carrier concentrations.

As to why trans-polyacetylene isn't a metal in the first place?

If we use a simple particle-in-a-box model, the difference in energy levels goes to zero as the box becomes very large. This would suggest trans-polyacetylene should be a metal (i.e., no band gap).

Consider also the Hückel model for polyacetylene. We can build up the nearly-infinite system by solving for ethylene, butadiene, etc.

Notice that the energy gap between levels becomes smaller and smaller. We can plot the HOMO-LUMO gap as a function of chain length:

The gap becomes smaller and smaller!

Wikipedia even gives the general formulas:

$$E_k = \alpha + 2\beta \cos \frac{k\pi}{(n+1)}$$
$$\Delta E = -4\beta \sin \frac{\pi}{2(n+1)}$$

So both particle in a box and Hückel predict no band gap when $n$ gets very large. Theory predicts a metal!

Why doesn't the real polyacetylene behave as a metal. For any one-dimensional system, there will be a Peierls distortion which lowers the total energy by distorting the geometry (long-short bond alternation) and increasing the band gap.1

In short, a truly one-dimensional metal is impossible, but doped polymers are certainly very good electrical conductors. Record films of trans-polyacetlyene have conductivity close to conventional metals.2

1 The long-short bond alternation is the Peierls distortion. It means some electrons are localized (shorter double bonds) and thus lower in energy. This creates a band gap (i.e., lower E for valence band).

2 The film isn't a 1D material, but the trans-polyacetylene chains are each 1D.


How do you calculate the heat of a reaction given a table of heat of formation values? So I've been trying to answer this question, and as far as I've learned it's $\Delta H^0_{\mathrm f}$(products)-$\Delta H^0_{\mathrm f}$(reactants), multiplying each by number of moles and such.

Trying to use that on this problem I came up with $\ce{0.008 OH- + 0.008 HCO3- -> 0.008 H2O + 0.008 CO3^2-}$ after (possibly incorrectly) noting that $\ce{HCO3}$ was in excess. However, this does not yield anywhere near the correct answer of C. What am I doing incorrectly?

I can obtain the correct answer using your values. Maybe double check for mistakes and typos?

Here are my quick calculations:
The overall reaction is:

$\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$

The equation used is:
\begin{equation}
\Delta H_\text{reaction}^\circ = \sum {\Delta H_{f}^\circ(\text{Products})} - \sum {\Delta H_{f}^\circ(\text{Reactants})}
\end{equation}

$\Delta H_\text{products}^\circ= (\pu{0.008 mol}\times\pu{-286 kJ/mol} +\pu{0.008 mol}\times\pu{-677 kJ/mol}) = \pu{-7.704 kJ}$

$\Delta H_\text{reactants}^\circ= (\pu{0.008 mol}\times\pu{-692 kJ/mol} +\pu{0.008 mol}\times\pu{-230 kJ/mol}) = \pu{-7.376 kJ}$

$\Delta H_\text{reaction}^\circ = (-7.074 - (-7.376))~\pu{kJ/mol} = \pu{-0.328 kJ} = \pu{-328 J}$


What is exposed in the surface tension of water? (I guess) I understand the covalent bonding of water and the hydrogen bonding of water between two different molecules, but I would like to know which part is the part that that exposes itself to the air in the surface tension of water. There should be something different to the rest of the liquid, otherwise there would be no frontier, right?
Is it the electropositive part of hydrogen or is it the electronegative part of oxygen that faces the air? (or may be none of them).
Surface tension in water owes to the fact that water molecules attract one another, as each molecule forms a bond with the ones in its vicinity. At the surface, though, the outmost layer of molecules, has fewer molecules to cling to, therefore compensates by establishing stronger bonds with its neighbors, this leading to the formation of the surface tension.... more here:

https://www.usgs.gov/special-topic/water-science-school/science/surface-tension-and-water?qt-science_center_objects=0#qt-science_center_objects
Thanks to all for your help (I'm in chemistry '101' for dummies so please try to be as basic as possible.
Video: https://www.youtube.com/watch?v=moITG5Q7zzI
(Hydrogen bonds https://www.khanacademy.org/science/high-school-biology/hs-biology-foundations/hs-water-and-life/v/hydrogen-bonding-in-water)

Surface tension may be described regardless if all / some / none of the water molecules expose one or both hydrogen atoms or the oxygen atom toward air, even if there are studies in this field (e.g., example, example).
Key points however are

the movement (translation, vibration) of the water molecules in the bulk of water droplet in restained, because of intermolecular attractive forces (e.g., hydrogen bonds), as well as intermolecular repulsion (molecules normally do not come closer to each other than the sum of the corresponding van der Waals radii).  Like with humans, there is an optimal intermolecular distance where overall the energy passes a minimum.
In extension of your drawing of the situation in the water droplet, I add the following figure where molecules are reduced to a black sphere; these are either in the inner of the droplet (light blue background), or at the interface (dark blue background) exposed to air.  The black arrows indicate intermolecular attractive forces (responsible for cohesion).

(derived from wikipedia)

For a molecule in the inner of the droplet, surrounded completely by molecules of the same type, the net force on this molecule is zero.

For a molecule at the interface water / air, the attractive forces to molecules of water in the inner of the water droplet exceeds the attractive forces between this water molecule and air molecules.  Thus, there is a net force pointing to the inner of the water droplet; this is symbolized by the red arrows.
Because of this inward directed force, the droplet thus tends to contract, building an inner pressure until compensated by the earlier mentioned intermolecular repulsion.

Overall, the lowest energy of the water droplet would be achieved as a sphere.  Sometimes, this is shown in microgravity (Scott Kelly at 8:28min).  On earth, however, things like gravity, contact with a surface (adhesion), relative air speed (e.g., if a drop falls down the rain gutter) etc. lead to deviations from this ideal.  However no, because the molecules in the inner of the water droplet are the same as the ones at the interface exposed to the air, there is no difference in the intermolecular, pairwise attractive forces between molecules of same type.


Proton leak during ATP synthesis I was told there is a proton leak during ATP production. When going through the mechanisms we went over in class, I was unable to find when this occurs.

During what part of ATP production is there a proton leak and what are the ramifications of this?

This occurs during Oxidative phosphorylation and the "proton leak" process is also called Uncoupling of Oxidative Phosphorylation
The mitochondrial electron-transport chain
Electron transport, causes Complexes I, III, and IV to transport protons across the inner mitochondrial membrane from the matrix, a region of low [$\ce{H+}$] and negative electrical potential, to the intermembrane space (which is in contact with the cytosol), a region of high [$\ce{H+}$] and positive electrical potential.

Three of the four electron-transport complexes, Complexes I, III, and IV, are involved in proton translocation.
Two mechanisms have been put across that would couple the free energy of electron transport with the active transport of protons: the redox loop mechanism and the proton pump mechanism.
Uncoupling of Oxidative Phosphorylation
Electron transport (the oxidation of NADH and FADH2 by $\ce{O2}$) and oxidative phosphorylation (the synthesis of ATP) are normally tightly coupled due to the impermeability of the inner mitochondrial membrane to the passage of protons. Thus the only way for $\ce{H+}$ to re-enter the matrix is through the $\ce{F0}$ portion of the proton-translocating ATP synthase.

In the resting state, when oxidative phosphorylation is minimal, the
proton-motive force across the inner mitochondrial membrane builds up
to the extent that the free energy to pump additional protons is
greater than the electron-transport chain can muster, thereby
inhibiting further electron transport.
However, many compounds,including 2,4-dinitrophenol (DNP) and
carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP), have been
found to “uncouple” these processes. In a pH gradient, they bind
protons on the acidic side of the membrane, diffuse through, and
release them on the alkaline side, thereby dissipating the gradient.

The chemiosmotic hypothesis has provided a rationale for understanding the mechanism by which these uncouplers act.
Uncouplers act by dissipating the proton gradient across the inner mitochondrial membrane created by the electron-transport system.

As uncouplers, they function by carrying protons across the inner
membrane. Their tendency is to acquire protons on the cytosolic
surface of the membrane (where the proton concentration is high) and
carry them to the matrix side, thereby destroying the proton gradient
that couples electron transport and the ATP synthase.
In mitochondria treated with uncouplers, electron transport continues
and protons are driven out through the inner membrane. However, they
leak back in so rapidly via the uncouplers that ATP synthesis does not
occur. Instead, the energy released in electron transport is
dissipated as heat.

The presence in the inner mitochondrial membrane of an agent that renders it permeable to $\ce{H+}$ uncouples oxidative phosphorylation from electron transport by providing a route for the dissipation of the proton-motive force that does not require ATP synthesis. Uncoupling therefore allows electron transport to proceed unchecked even when ATP synthesis is inhibited.

Uncoupling of oxidative phosphorylation:  The proton-transporting ionophores DNP and FCCP uncouple oxidative phosphorylation from electron transport by discharging the electrochemical proton gradient generated by electron transport
References

Voet and Voet 4th ed. Biochemistry: Oxidative Phosphorylation
R.H Garret & C.M Grisham Biochemistry 4th ed : Electron Transport and Oxidative phosphorylation


What would be different about a heavy-water ocean world? The water on Earth is primarily made of $\ce{H2O}$. I'm wondering about the conditions in a hypothetical $\ce{D2O}$ ocean world.

Would it be stable over geologic timescales (~$\pu{0.1–2 Ga}$)? What would be expected to be different, if anything? Would it affect the chemistry of its atmosphere, or appreciably change the incident radiation from its host star?

For one thing, it would be hugely more unlikely to exist.  We are more likely to have worlds where there is a little more deuterium in the hydrogen, not 100%.

Most of the hydrogen in our Universe is protium (hydrogen-1) because the vast majority of unfused nuclear particles in stars are protons.  Neutrons must either fuse with something or they decay to make protons with a half-life of ten minutes.  Deuterium and tritium are formed initially when the protons are fused, but they are just intermediate towards the formation of more stable helium-4 nuclei (alpha particles) and so do not accumulate to large concentrations.

If there were a heavy water world, life would have to adapt to it.  Prolonged consumption of pure heavy water is deadly to life on Earth.


Oxidation States of Actinides It is observed that actinides do not exhibit +2 oxidation state, contrary to that of some of the lanthanide elements. Why?

See also: Why is WF6 stable whereas CrF6 is unknown?

Whenever one wants to compare oxidation states, there are a couple main factors to take into account. To reach a higher oxidation state, one obviously has to pay for it in the form of ionisation energy/energies. However, there is a compensatory effect in that elements in higher oxidation states generally get more out of bonding. For example, the lattice energy of $\ce{FeCl3}$ is larger than that of $\ce{FeCl2}$, or the four covalent bonds in $\ce{XeF4}$ are collectively stronger than the two covalent bonds in $\ce{XeF2}$.

The higher ionisation energies of the actinides tend to be smaller than those of the lanthanides, for the same reasons as explained in the linked question (5f orbitals have one radial node; 4f orbitals do not). This generally favours higher oxidation states and disfavours low oxidation states in actinides.


Why does permanganate oxidation of 2-nitronaphthalene not form phthalic acid? My teacher told me that among the following compounds only compound 3 will form phthalic acid when reacting with hot $\ce{KMnO4}$.

I do understand why the other compound will not form phthalic acid.

Looking at the substituents; NH2- activates the ring, making it more reactive than dibenzene ring system (easily oxidised), and the NO2- substituent is very deactivating - (strongly electron withdrawing), making it less reactive (not easily oxidised).

Although  a powerful oxidising, optimised oxidising agent (KMnO4) is being used, it is only able to oxidise the 2-naphthylamine compound to produce phthalic acid.

and 2-nitronaphthalene is very unlikely or at least takes longer to completion with very little yield.


Are all lattices Bravais lattices? Can all lattices be described as one of the fourteen Bravais lattices?
Is the hexagonal close packed structure also one of the fourteen Bravais lattices?

From Ashcroft and Mermin's Solid State Physics:

  A fundamental concept in the description of any crystalline solid is that of the Bravais lattice, which specifies the periodic array in which the repeated units of the crystal are arranged. The units themselves may be single atoms, groups of atoms, molecules, ions, etc., but the Bravais lattice summarizes only the geometry of the underlying periodic structure, regardless of what the actual units may be.

So, a real crystal structure is defined by the Bravais lattice, and the unit placed at each point on the Bravais lattice.  Now, it turns out that the hexagonal close packed structure (hcp) is not, in fact, a Bravais lattice.  Hcp is based on the simple hexagonal Bravais lattice, but with two atoms as the basis (one could also call it two interpenetrating simple hexagonal Bravais lattices).  The bottom line is the hcp is not itself a Bravais lattice (see further details in Ashcroft and Mermin). I'll skip any discussion of point groups and space groups, since those are just further refinements of what it means to combine the symmetry of a Bravais lattice with the symmetry of the repeating unit.

There is a hierarchy of symmetry - 7 crystal systems, 14 Bravais lattices, 32 crystallographic point groups, and 230 space groups. For hcp, the point is that it can be represented as a simple hexagonal Bravais lattice with a two-atom unit, so having hcp as another "Bravais lattice" would totally change the definition of what a Bravais lattice is - the fundamental symmetry is simple hexagonal.

The general idea is to start with something you can repeat in a specified way to fill space.  For example, starting with a cube, it is pretty easy to see how to stack them up to fill space.  If, instead of being cubes they are stretched along one axis to get a rectangular solid, you can still fill space, but it will look different depending on what direction you are looking.

So, we get the seven crystal systems: Cubic (with simple cubic, body centered, and face centered cubic Bravais lattices); Tetragonal, which is a cube stretched along one side (with simple tetragonal and centered tetragonal Bravais lattices); Orthorhombic, a unit with perpendicular faces but unequal lengths on each side (with 4 Bravais lattices); Monoclinic, which tilts an orthorhombic unit to make one angle non-90 degrees (2 Bravais lattices); Triclinic, which has no two faces perpendicular to each other (1 Bravais lattice); Trigonal, which is obtained by taking a cube and stretching it along a body diagonal (1 Bravais lattice), and Hexagonal, which has one Bravais lattice.

You get those 7 crystal systems and 14 Bravais lattices by being able to stack the appropriate units to fill space.  Now, while this concept is pretty fundamental, there are further symmetries imposed on the crystal by what you stack up.  Lets consider something fairly simple, simple cubic.  If you start with a big pile of white cubes, the resulting object you put together will look the same regardless of which direction you look at it, or how you rotate it around.  BUT - if you start with a pile of white cubes with one face painted red, you just changed things. To conform with the Bravais lattice, you would stack them so that all the red faces pointed in the same direction.  You would still get a big cube made of the smaller cubes.  But now it would look different depending on what direction you looked at it - you have reduced the symmetry of the crystal.

For simple hexagonal, one starts with a hexagonal right prism (hexagon base, some height perpendicular to the base).  Stacking those one on top of each other results in a simple hexagonal Bravais lattice.  Lets make the hexagonal prisms out of clear acrylic, but place a marble at the center of the prism.  Now when you have stacked them up you would see a simple hexagonal lattice of marbles, or heck, lets call them atoms.  Now, exactly where you put the marble in the clear prism doesn't matter - you would still see the exact same arrangement of the marbles in space.  OK, lets place the first model in the center of the base of the prism, and place a second marble halfway up one of the edges.  When you stack them up, you will now get an arrangement of marbles in space that looks like the hexagonal close packed structure.  But you stacked the prisms following the rules of the simple hexagonal Bravais lattice.  Placing that second marble in the Bravais prism generated a two atom basis.  It clearly is different from simple hexagonal, but the reason is not how you stacked the prisms, it is what the prism contains.  The underlying Bravais lattice does not change, but the symmetry of the resulting crystal is changed by the basis unit used.

Now, use different colored marbles in the two positions, and it changes once again.  Throw three marbles in in different places in the prism and it changes yet again.

So, the Bravais lattice occupies a fundamental level in understanding the crystal structure - it is the rules for stacking things up (and only the rules for stacking things).  The objects you stack up may have their own properties, meaning that two things stacked using the same Bravais lattice may in the end look very different.  This combination of the lattice and the unit are what create the 230 space groups.


Is there a method to recover camphene from sublimation at ambient conditions? Camphene solidifies and sublimates at ambient room temperature below 45°C. I'm aware that a glass funnel is often used to separate mixtures through sublimation of the volatile component (in this case camphene) from another material.

Is it possible to recover sublimated camphene using this or another method?

MIT has a brilliant technique for sublimating and condensing:

https://www.youtube.com/watch?v=dBNELFi5XiY

The sample is placed in a culture dish sitting on a hotplate. The lid is placed on top.  Finally a beaker with ice is placed on top of the dish to condense the sublimate. After a while crystals form on the top of the lid. These pure crystals are periodically scraped with a policeman and collected.


How to prove it mathematically that either rhombic or the monoclinic sulfur is denser? For the transition, rhombic sulfur → monoclinic sulfur, the value of $ΔS$ is positive. The transition temperature increases with increase in pressure. Which is denser, the rhombic or the monoclinic form? Prove your answer mathematically.

I tried by using Clapeyron equation but still failed. How should I approach and solve it? This is so tricky.

$$\frac{\mathrm dP}{\mathrm dT} = \frac{\Delta_\mathrm{trs}S}{\Delta_\mathrm{trs}V}$$

$\Delta_\mathrm{trs}S$ is positive. $\displaystyle\frac{\mathrm dP}{\mathrm dT}$ is said to be positive as well. Thus, $\Delta_\mathrm{trs}V$ has to be positive, i.e. monoclinic sulfur has a greater molar volume, and thus rhombic sulfur is denser.


Plot Binary Eutectic from Temperature of Melting and Enthalpy I'm a little confused about plotting a binary eutectic diagram (temperature vs mole fraction), and finding the eutectic point.  I understand the eutectic point is the lowest temperature that a mixture of the two substances will melt, and that this is somehow related to minimizing the Gibbs free energy of the system. However, I can't figure out how to actually solve for the eutectic point or how to graph it based on the melting temperature and enthalpy of fusion of two compounds.

I found a formula $$\frac{X_A}{X_B} = \frac{T_B-T_0}{T_A-T_0}$$ where X is the mole fraction of each component, but I was getting unrealistic answers for certain mole fractions, e.g. for $$X_A=.9$$ I was getting a value greater than the melting point of A.  And for $$X_A=X_B$$ this equation can't be solved.  Also, this formula doesn't take into account the enthalpy of fusion, so doesn't seem to relate to the free energy.

The volume & entropy of the system is unknown, so I wasn't sure how to set up an equation to solve for the Gibbs free energy of either component, but I was thinking that if I had an equation, I would be able to  solve a system of equations for the lowest free energy and plot it (somehow?).

Can anyone explain the relationship between the making a eutectic diagram and the free energy of components in the system?

Edit: Thank you for the answer! That was really helpful conceptually in relating the free energy of a mixture to that of the pure substances, but I actually have a problem from the class that says:

  Calculate and graph the binary eutectic melting diagram for the system diopside - anorthite using the data given below, and assuming for both diopside and anorthite that the activity in the liquid is equal to the mole fraction.

  \begin{array}{lll}
\text{Diopside}\, (\ce{CaMgSi2O6}): & \Delta H_\mathrm{fusion} = \pu{20,000 cal/mol} & T_\mathrm{fusion} = \pu{1390^\circ C} \\
\text{Anorthite}\, (\ce{CaAl2Si2O6}): & \Delta H_\mathrm{fusion} = \pu{29,000 cal/mol} & T_\mathrm{fusion} = \pu{1553^\circ C}
\end{array}

So it must be possible to find based on these parameters right?  Is the fact that the activity is equal to the mole fraction make the difference in solving it?

The reason that you can't figure out how to solve for the eutectic based on the melting temperature and enthalpy of fusion of the two constituents is because, well, you can't. Those bits of information, while useful in constructing the phase diagram (well, at least the melting temperatures), are not what determines the eutectic. You will have a eutectic if and only if there are interactions between the constituents in the liquid phase that reduce the Gibbs free energy below the ideal value.

Lets make this more concrete. Take elements A and B, each with a solid and liquid phase, no intermediate phases. Taking just one phase (solid or liquid) for the moment, what happens to the Gibbs free energy when we mix them together? The first bit you get is the weighted combination of the pure phase free energies, that is:
$G_{mixture} = x_{A}G_{A} + x_{B}G_{B}$.  But is that enough? Well, no, we forgot the entropy of mixing term. So, for an ideal mixture,

$G_{ideal} = x_{A}G_{A} + x_{B}G_{B} + RT(x_{A}\ln x_{A} + x_{B} \ln x_{B})$

What does this look like? Well, taking 'fake' elements, A that melts at 950K and B that melts at 1000K, just using ideal mixtures for solid and liquid, one gets a pretty boring diagram that looks like:

with the slight separation in solidus and liquidus from the slight asymmetry in the melting points. Hmmmm... - no eutectic (since the ideal mixing impacts both liquid and solid in the same way, that is to be expected).

OK, now lets move to a regular solution model. The difference is the addition of an enthalpy term, that describes whether an A atom would rather see A or B around it (and vice versa). In the simplest way, this can be represented by a heat of mixing term using one parameter, $\Omega$. $\Omega$ is positive if the atoms don't want to mix (so, it can cause a miscibility gap), or negative if they actually prefer to mix. So, for a regular solution one gets:

$G_{regular} = x_{A}G_{A} + x_{B}G_{B} + \Omega x_A x_B + RT(x_{A}\ln x_{A} + x_{B} \ln x_{B})$

Setting $\Omega$ to -1kJ/mol, the previous diagram turns in to:

and you see that a eutectic has now formed at about $x_B = 0.25$.

Now, back to the parameters you thought were important:
The melting temperature is kind of important, in so far as an asymmetry between them shifts the eutectic of the regular solution towards the lower melting element.

The enthalpy of fusion is not a factor at all in determining where phase boundaries are.

The interaction term(s) are highly relevant, since they determine where and how deep the eutectic is.


Keto Enol Equilibrium for Claisen Condensation Products Do Claisen Condensation products prefer the di-keto form, or a keto/enol form due to increased conjugation and hydrogen bonding?

The Claisen reaction yields a product with an acidified proton in $\alpha$-position of carbonyl and ester, right at the same carbon atom of R depicted in blue.

In presence of a base, yes, $\beta$-dicarbonyl compounds undergo deprotonation at their methylene unit to yield a keto-enolate, stabilized by mesomerism with the secound carbonyl group in conjugation. Compare the $pK_a$-values of $\beta{}$-dicarbonyl compounds with ketones, for example (one reference), to estimate if the equilibrium is on the left, or on the right.
.

I think cations of the right size and charge equally favour the formation of the keto-enolate.  If you synthesize $\beta$-dicarbonyl compounds, the risk to loose product due to chelation  is lowered if initial work-up is concluded by drying the washed organic phase with sodium sulphate ($\ce{Na2SO4}$), rather than by magnesium sulphate ($\ce{MgSO4}$).


What prevents everyday objects from reacting? This may be a stupid question but I would like to know what prevents chemical reactions from happening among common objects in everyday life? The opposite would be, what requirements must be met for chemical reactions?

For example, as I am typing this what prevents no reaction between my fingers and the keyboard, or everyday objects such as chairs, tables, etc. In other words, why do we not constantly see chemical processes happening all around us constantly? Why don't my feet chemically react with the ground, my pencil reacts with the air, my fridge with the floor, etc...

There must be electron transfer such as conductivity, and charging but why no chemical reactions? Even as I am in AP Chemistry and understand Thermodynamics, Hess' Law, and so on, I am too embarrassed to ask my teacher with such a silly question.

As I see it, two factors: thermodynamics and kinetics.

Thermodynamics: Many of the reactions that "don't happen" in everyday life are not favored thermodynamically, we call them non-spontaneous. This means that the products would be higher in energy than the reactants. Such reactions can certainly occur if you supply external energy (charging your cell phone battery, for instance, is a non-spontaneous process). But, without applied energy, non-spontaneous processes generally will not occur. Reactions that we observe in nature are generally spontaneous, favored by thermodynamics (with lower-energy products). Some thermodynamically spontaneous  processes are not observed to occur, even when we might naively expected them to. Example: Sugars decomposing to $CO_2$ and water is a very energetically favorable reaction, in fact the energy released from this process fuels nearly all known life. However, we do not observe our table sugar decomposing in its container because without outside help like enzymes and applied heat, the process is terribly slow kinetically.

Kinetics: For us to observe a reaction happening, it must proceed at rate that is fast relative to our human scale of time. Perhaps your fingers fusing to your keyboard is thermodynamically favored/spontaneous, but if it takes hundreds of years to occur then for our human purposes we may regard it as "no reaction". Indeed, the decomposition of most types of the biomolecules constituting our bodies is favored thermodynamically, but the kinetics are so slow (440 years for table sugar)°. There are a few ways to increase the rate at which a reaction occurs. To occur, a reaction requires a certain amount of activation energy. Think of this as hill you have to climb before you can sled down to the products (or up, potentially). The reaction does not occur if the starting material molecules do not possess enough energy to climb the activation energy barrier. You may help them overcome the barrier by supplying heat; thus, heating your table sugar will allow it to decompose into $CO_2$ and water before your eyes (the water will be released as steam of course). The other main way to increase the kinetics of the reaction is to lower the hill, instead of adding the energy to climb it. That is, use a catalyst like an enzyme to lower the activation energy barrier of the reaction. Without such external factors like applied heat and use of a catalyst, many reactions that are thermodynamically favored will be disfavored by kinetics, and not proceed at an observables rate.

So the reason the universe isn't a useless, constantly rearranging chemical soup boils down to thermodynamics and kinetics. Even if a process is thermodynamically spontaneous, it may be kinetically slow. I should point out that in a lot of places, such as the centers of stars, there is indeed a copious abundance of uncontrollable chemical and nuclear reactions. And planets like Jupiter pretty much are a useless chemical soup (no offense to Jupiterians). But in our favorite places on Earth, at room temperature and atmospheric pressure, the carbon in our bodies isn't going to undergo nuclear fission and become lead (thankfully).

By the way, this is in not a silly question. In fact, if a chemistry student doesn't wonder this, one may question if they are actually learning the subject. I applaud you for your curiosity.


Are phenylboranes susceptible to ipso or meta substitution? Are phenylboranes susceptible to ipso or meta electrophilic aromatic substitution, i.e. what's the regioselectivity?

A SciFinder search turns up very little on this type of transformation.

Qui and coworkers run halogenations of phenyl boronate esters using N-halosuccinimide using gold(III) catalysis. The regioselectivity was essentially 1:1:1 for ortho:meta:para. When the reaction was carried out with other substituents as well, the other substituent's directing group preference dominated the regioselectivity.
Gold(III)-Catalyzed Halogenation of Aromatic Boronates with N-Halosuccinimides

There is one other reaction in SciFinder that is relevant, however I don't have access to the full text (and it's in Russian, so I wouldn't be able to read it anyway). The substrate is a benzene ring substituted with a borane cluster. The reported regioselectivity is 80:20 for meta:para.

Electrophilic substitution in the benzene ring of 9-phenyl-o- and -m-carborane. Zakharkin, L. I. et al Zhurnal Organicheskoi Khimii, 23(8), 1691-5; 1987.